I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

[muriel_volestrangler]

In memory of Martin Gardner, who died a few days ago - this puzzle was posed at the "Gathering for Gardner" convention for mathematical puzzles earlier this year. The write-up of it in New Scientist is here, including an explanation.

Gary Foshee, a collector and designer of puzzles from Issaquah near Seattle walked to the lectern to present his talk. It consisted of the following three sentences: "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

The event was the Gathering for Gardner earlier this year, a convention held every two years in Atlanta, Georgia, uniting mathematicians, magicians and puzzle enthusiasts. The audience was silent as they pondered the question.

"The first thing you think is 'What has Tuesday got to do with it?'" said Foshee, deadpan. "Well, it has everything to do with it." And then he stepped down from the stage.

This still blows my mind a little, though I have now managed to convince myself that the New Scientist explanation is perfectly correct. I hope there's an easy way of guarding against the logic I used to come to my first, incorrect, answer, but I can't see how to specify how to take the care. There's a lesson for the use of statistics and probability here, but I can't put it into words.

And before anyone comes up with 'real world' objections, this does not depend on slightly different chances of girls or boys being born or surviving, or of twins who are very likely to be born on the same day of the week, or anything like that. The maths are done with a girl or boy being equally likely, and any birth equally likely on all different days of the week.

(Mods, if you feel this isn't right for the Science forum, I apologise. Perhaps the Lounge, if that's the case?)

[katmondoo]

1. Both my kids were born on a Saturday on th 9th of the month

Daughter April 9th Son December 9th I think it means nothing just that was the way it was.

[bananas]

252. I posted a poll on this question

Here's the poll: http://www.democraticunderground.com/discuss/duboard.php?az=view_all&address=228x67455

Here are the current results:

Poll result (42 votes)
1/4 (2 votes, 5%)
1/3 (1 votes, 2%)
1/2 (22 votes, 52%)
13/27 (15 votes, 36%)
Not sure (1 votes, 2%)
Other (1 votes, 2%)

[Jim__]

263. What your poll most clearly demonstrates is why math problems are not decided by poll.

Tinrobot created a chart demonstrating the answer to this (here). People can claim he is wrong, but then anyone making that claim has to either tell us what's wrong with the sample space he generated; or why the entries are not equiprobable.

[hlthe2b]

2. Well, I would have ignored the Tuesday issue

and then it simply becomes an easy, what is the probability of having a boy child, given you already have a son. The probability of a male for every birth is 50/50 so 1/2. See, if you just ignore the extraneous crap, you luck out and get it right anyway....LOL ;)

[muriel_volestrangler]

7. I'm not sure how to interpret your winking smiley

because, if you read the solution in the article (look for 'Trouble with boys'), you'll see it's not 1 in 2. This is why I find the answer a little mind-blowing, and I have to consider what is truly 'extraneous'.

[hlthe2b]

25. ....

"It seems remarkable that the probability of having two boys changes from 1/3 to 13/27 when the birth day of one boy is stated – yet it does, and it's quite a generous difference at that. In fact, if you repeat the question but specify a trait rarer than 1/7 (the chance of being born on a Tuesday), the closer the probability will approach 1/2."

[Blues Heron]

3. of all the people who have two children and one of them a boy born on tuesday

half have two boys and half have a boy and a girl. So 50/50 probability of either case. Am I right? Somehow I think this might be a trick question, hinging on the meaning of the word "probability"........ :)

[fl_dem]

16. I have 2 sons, the 1st one was born on Tuesday!

things that make ya go hmmmmmmmmmmmm

[frazzled]

4. Yes, the 1 in 3 probability that they are both boys is the easy part ...

Now, what is the probability that one of your two children is born on your own birthday? That happened to me!

[muriel_volestrangler]

5. Have you read the entire article? "1 in 3" is the answer to a different question

and "1 in 2" is not right either, for those who have answered that so far.

[frazzled]

8. Oops, no I didn't, but it still doesn't answer my probability question

I actually have no head for math whatsoever--even though this child to whom I gave birth on my own birthday is completing his doctorate in pure mathematics at a well-known university. Now that makes the probability factors even more complex.

[muriel_volestrangler]

11. For your problem:

"what is the probability that one of your two children is born on your own birthday?"

Assuming every year has 365 days (more complicated if you think about leap days),

the probability that neither of your 2 children is born on your own birthday = (364/365)*(364/365) = 0.99453
the probability that the elder is, but not the younger = (1/365)*(364/365) = 0.00273
the younger, but not the elder = (364/365)*(1/365) = 0.00273
both: (1/365)*(1/365) = 0.0000075

So the chances that exactly one of them is born on your birthday is about 1 in 183.

[frazzled]

26. Um, I really didn't want to know, but while we're at it ...

what's the probability that you possess a sense of humor and the ability to read such into peoples' posts?

[muriel_volestrangler]

28. Heh , sorry, when you repeated it, I thought it meant you were serious (nt)

[dalaigh lllama]

270. My son was born on my birthday, too

I've always thought that was so cool!

[d_r]

6. I'm sorry but

I completely disagree with this write up.

The correct answer is 50$. Well, it isn't exactly 50$ because boys and girls aren't born 50$ even, but it is awfully close and you get my point.

What happens the first time has no impact on the second.

It is the same as flipping a ooin.

If you flip a coin and get heads on a Tuesday, what is the probability that you will get heads on a second flip?

50%,

What happens the first time has no implication on what will happen the second.

They are independent events. Each throw you have a 50-50 shot.

It is a fallacy to think in terms of "sets" or "patterns." Each trial is a brand new shot. Just because you have gotten heads five times in a row, the sixth throw has a 50-50 chance of being heads or tails (unless the coin is rigged of course). Each trial stands alone. This is a fallacy for gamblers, for example, to fall in to. That a number is "due."

Random is random. That's my opinion.

[muriel_volestrangler]

9. I think you are interpreting the question as "the oldest is a boy born on a Tuesday"

when it is actually "one is a boy born on a Tuesday". Try working it through again, making sure to take that into account.

[d_r]

10. I don't think it matters

they are both isolated events. It doesn't matter which comes first. The chance of each child being a boy is 50-50, regardless of whether the other child is a boy or girl. The chance each child is born on a given day is 1/7, regardless of the day the other is born. I don't think it matters the order.

[muriel_volestrangler]

12. Can you understand the argument for "one is a boy. What are the chances both are boys?"

That shows, fairly simply, the difference between asking "the elder is a boy. What are the chances both are boys?" and "one is a boy. What are the chances both are boys?".

From there, you can work up to the case when Tuesday is brought into it.

[d_r]

17. Hi, yes, sorry

but I just don't think it matters.

I mean we could make this more complex by thinking about the biology. I mean it could be that having one boy causes changes in a woman's uterus or hormones that has some impact on the sex of subsequent children, but I know nothing that suggests that. I mean there could be some change that makes an egg more or less likely to "accept" a sperm with an x or a y chromosome, but there is no evidence of such a thing. Also, it was a male making the statement,so we don't know if the children were with the same woman or if the woman had other pregnancies, so I don't think that is part of it honestly.

No, I think this puzzle is only about probability. The assumption that I have is that there is basically a 50-50 chance of having either a boy or a girl.

Those odds are the same every single time. 50-50.

The outcome of one event has no impact on the outcome of the other. That's how "random" works.

Think of this example. Say that you want to play a lottery game, and I'll oversimplify it. Say that each week the lottery is to pick one number between 1 and 10. And for the last five weeks in a row the lottery has picked the number 7. What is the probability that this week another 7 will be picked? I think it is 1 of 10.

One fallacy that gamblers fall in to is thinking "it always comes up 7, so surely this time it will be 7." The opposite fallacy is to say "it has come up 7 so many times, something else is due." According to this line of thinking, it would be so rare to get a 7 six times in a row that it will surely not happen again, the odds must be against it.

But you can't think in terms of sets. You can't think "what is the probability of getting a 7 six times in a row?" Because if it is random, then the probability for each trial that you will get a 7 is one of ten. Each trial stands alone and is not impacted by the others.

Here's the proof.
Ten trials:
7-7-7-7-7-1
7-7-7-7-7-2
7-7-7-7-7-3
7-7-7-7-7-4
7-7-7-7-7-5
7-7-7-7-7-6
7-7-7-7-7-7
7-7-7-7-7-8
7-7-7-7-7-9
7-7-7-7-7-0

For that event, the chance of getting a seven is the same as the chance for getting any other number. It is an independent event.

Now, this is hard to get my head around, but I don't think the order of the lottery picks, or the coin flips, or the baby births matter. I'll call it the "Lost series finale" factor. There is no now here. Each trial is independent of the others, if they are random, so it doesn't matter what result other trials had either before or after. It just makes no difference.

Unless there is some biological change that an earlier pregnancy has that impacts subsequent ones, which is why I put that up there, but again, I'm sure that isn't the point of this puzzle. And the day of the week really does not matter at all if you think about it in terms of each trial being a random event of its own.

[muriel_volestrangler]

21. Have you tried the explanation in #14 or #15 (basically the same) ?

The approach is: work out all the possibilities. Eliminate any that we are told cannot be true. Then the probability is "how many of the remaining possibilities fit the pattern we're looking for?"

So, for the boy-girl combinations, we start with 4 equally likely combinations:
1) GG
2) GB
3) BG
4) BB

but we're told 'one is a boy'. That eliminates 'GG', but still allows 'GB', 'BG' or 'BB'. (Since the final question is 'what are the chances both are boys', we know that 'one is a boy' must mean 'at least one is a boy' here, for the question to be meaningful). So there are 3 equally likely combinations, but only one of them satisfies 'both are boys'. So the chances are 1 in 3.

[d_r]

27. I understand but I think that isn't correct

I'm sorry, but if each event is random, you can't think in terms of sets of two. It doesn't matter what the other one is, each birth has a 50-50 chance. My belief is that you can't think in terms of sets of GG-GB-BG-BB. For each child (whether first or second) it doesn't matter what the other child's sex is. There is a 50-50 shot each time.

Take this statement:

"I flipped a coin twice. One flip was heads. What is the probability the other flip is heads?"

The answer is not 1/3, it is 1/2.

The idea that it would be 1/3 follows this logic:

There are four possible "sets:

H-T
T-H
H-H
T-T
t
Since I told you one was heads, it can't be T-T, so it has to be one of the other three combination, thus 1/3.

But random doesn't work that way. Random doesn't care what you got on another flip before or after. Random is random each time. Other trials have no implication on it. There is no set. Each coin flip is z 50-50 chance to be either heads or tails, no matter what coin flips come before or after, or what flip somebody else makes, or what a flip is another day or anything else. Each flip stands alone, a 50-50 chance to be either heads or tails. If you flipped a coin 1,000,000,0000 times and recorded every single flip, you would have no better guess at what the next flip would be than if it was your first flip. That 1,000,000,0001 flip will still have a 50-50 chance of being either a head or tail. There is no improving the prediction of the outcome. That is what "random" means and how it works. Our minds really, really want to work in sets and patterns, but random doesn't care.

[muriel_volestrangler]

29. Do you agree you are twice as likely to get a head and a tail, in any order, than 2 heads?

And that also you are as likely to get 2 tails as 2 heads? Those are all the possibilities. So when you have been told "it's not 2 tails", then you know that it's twice as likely to have a head and a tail than 2 heads. So the chances of a head and a tail (in any order) are 2 in 3, while the chances of 2 heads are 1 in 3.

But random doesn't work that way. Random doesn't care what you got on another flip before or after. Random is random each time. Other trials have no implication on it. There is no set. Each coin flip is z 50-50 chance to be either heads or tails, no matter what coin flips come before or after, or what flip somebody else makes, or what a flip is another day or anything else. Each flip stands alone, a 50-50 chance to be either heads or tails.

Yes, that's all true. But the question is not "what is the result of one particular flip?" We don't know which flip is "the other flip", because we don't know which flip we have been given the result for. So you can't just make the question a simple "what is the outcome of one flip?"

[d_r]

31. I'm sorry, I don't.

"Do you agree you are twice as likely to get a head and "a tail, in any order, than 2 heads? And that also you are as likely to get 2 tails as 2 heads?"


No, I'm sorry I don't agree with the basic premise. I don't think the order matters because each flip is unique. You can't look at it as a set. What if you throw a head and I throw a tail, is that a part of your set? What if somebody in China throws one, does that have any implication on your flip? The "set" is an artificial concoction.

What our difference is, is the idea of the order. I don't think it matters, because each is an isolated event. I understand your point, I just don't agree with it. If I did, honestly I'd be working with spss on the past powerball numbers trying to right a regression equation or something.

I understand the importance of the semantics of the puzzle, but I'm stuck with random is random.

[muriel_volestrangler]

32. The thing is, you are the one who brings order into it

Because you are saying "we've looked at one of the flips, and eliminated it; now let's work our the chances for the other".

The flips have already been made. We are then given some information about the pair of flips that have happened. We cannot tie that information to any bit of that pair. We can use the chance of each coin flip to work out the combinations possible for the pair; but we can't say "there is a new random flip to be made".

Would this help:

I have picked a number between 0 and 3, inclusive, with all four of the possibilities equally likely. I then tell you the number was not 0. What are the chances the number was 3?

Then write that out in binary, with leading zeros where needed:

I have picked a number between 00 and 11, inclusive, with all four of the possibilities equally likely. I then tell you the number was not 00 (ie it contains at least one digit '1'). What are the chances the number was 11?

Substitute the '0' and '1's with 'H' and 'T', and you have the coin flipping problem.

00 = TT
01 = TH
10 = HT
11 = HH

[d_r]

41. our basic belief is just different

I understand the logic, I do, but I just think that random is random. I don't think you can put order to it. Each new trial is a new random shot - I just don't think you can make it part of a set. I don't think random works that way.

[Thor_MN]

46. Actually, you don't. Random is random, but you are changing the meaning.

"One is a boy, what are the odds the other is a boy?" is NOT the same as "One is a boy, what are the odds I have two boys?". By equating those two sentences in your mind, you are collapsing possibilities.

To illustrate:
One family names their first child Andrew and the second Bill. One down. 2 boys
Second family names first child Angela and the second Bill. Two, all is well. 1 girl 1 boy.
Third family, first child Angela is a boy and Bill is a girl. Three, 1 boy one girl.

See the significance? If you can't, Angela will, when he gets the crap beat out of him every day in grammar school.

[qazplm]

54. four

the family names their first child Bill and the second Andrew.

If all we are doing is changing names to change probabilities...

[Thor_MN]

106. Practically infinte by that "logic"...

Guess it wasn't obvious enough, so I'll spell it out. The actual names are not relevent, they only serve to illustrate that Child 1 and Child 2 are not interchangeable. With two children, knowing that one is a boy, there remain three possibilities. One can not say that Girl, Boy is no different than Boy, Girl.

[qazplm]

177. sure they can

because one of those two can't happen once we know that one of them is a boy.

The wording of the question matters. But hey, enjoy the unnecessary attitude simply because you are so certain you are correct.

[Thor_MN]

196. You are absolutely correct

One of them can't happen, but since you have no clue which one, both are still possible. Is there any doubt in your mind that of all the families with two children, where there is at least one boy, that the number of families with 2 boys is one third of the total?

Just because you feel that you can simplify the question by subsituting "the other" for the original question doesn't make you right. Please do not go to Vegas and bet the mortgage payment on anything where you feel you should have even odds due to your sense of how you should be able to alter the wording of the wager.

[qazplm]

210. I find it funny

that folks go to the "experts" and I don't use the quotes sardonically simply because this isn't a question of, is the Earth round, or something of that level of certainty, but that folks say, hey the experts say it's 2/3rds, which most of them do, but then say, but the word order/choice doesn't matter, which even the experts say in fact DOES matter, which a brief visit to these links and to the wiki page on this issue immediately show.

So folks on your side say, look at the experts, then when we say, yeah but the experts agree with us that phrasing the question a different way leads to a different result, then it's yeah no that isn't true.

I'm not going to repeat myself ad infinitum, but suffice to say, there are SOME wording of the problem whereby the question asked will lead to 2/3rds and OTHER wordings of the problem which will lead to 1/2. You can feel free to find where I've made those arguments, I'm not repeating them again and again in this thread.

The wording is not "at least one boy"...start from there. IF the wording were "at least" one boy, the question is different than if the wording excludes those two words. This has nothing to do with Vegas because the "wording" of the wager is the same every single time. This problem, the wording changes.

Even the original links to the NY Times, the understanding is the originally worded question is problematic and could be correctly interpreted as 1/2, but hey, clearly because YOUR "interpretation" of what variable(s) the question is looking for is different from my "interpretation" of what variables are being looked for, clearly you must be right.

[qazplm]

38. well if they are twins

and for the sake of this hypo born at the same time, then GB and BG collapse into one.

But it seems to me, GB and BG ARE the same thing, in both scenarios, you have one boy and one girl. So it comes across as a rather artificial approach to it. I could as easily have said you have three options B X2, G X2, or BG. Since we know it isn't G X2, that just leaves B X2 and BG, or 1/2, which makes more logical sense.

[rucky]

253. What are the chances that twins would be born on either side of midnight?

Tuesday.

Identical or Fraternal? Which is more likely?

[caraher]

116. Try it for a simpler case

Suppose the statement were, "I have two children. One is a boy born during a PM hour. What is the probability I have two boys?"

For the first child, we can have

AM boy, AM girl, PM boy, PM girl

For the second, we have the same possibilities. Let's adopt a shorthand wherein the four cases outlined above are AB, AG, PB, PG. We then get sixteen possible cases for having 2 children, all equally likely:

AB AB, AB AG, AB PB, AB PG,
AG AB, AG AG, AG PB, AG PG,
PB AB, PB AG, PB PB, PB PG,
PG AB, PG AG, PG PG, PG PG

Of these sixteen possibilities, there are 7 where there is a boy born during a PM hour:

AB PB,
AG PB,
PB AB, PB AG, PB PB, PB PG,
PG PB

Of those seven, there are 3 where both are boys: AB PB, PB PB, PG PB. So the chance is 3/7 that the speaker has two boys.

[Travis_0004]

14. You are not thinking it through.

If you have two children, then it can be BB, BG, GB, GG. Obviously GG is impossible, since one must be a boy, so the odds of both being a boy given the above information has to be 1/3.

If they were to state "My FIRST born child is a boy", what is the odds the second is a boy, then its 50%, but since it says 'a' child is a boy, then you have to figure out all the combinations.

[d_r]

19. I'm sorry, but

I honestly believe that if each trial is random then the results from one trial have no impact on the others. The odds of having a girl or a boy are 50-50, regardless of other births before or after. You can not think in terms of sets, each event is a random event on its own. It is not 1/3 it is 1/2.


Take a quarter and a pen and paper and do some flips - try it, random is random.

[Thor_MN]

108. Each trial is random. Would it help if we said that you haven't made any wrong assertions

other than you understand the question being asked?

You are constrained to think in sets, when that is what the question is about. The odds that the child that isn't the boy mentioned is a boy IS 50%. However, that is not the question. The question is what are the odds that they are both boys, and you haven't been told if the boy mentioned is the elder or younger child. If you don't think that has any consequence, try telling the eldest they can not get thier driver's license until after their younger sibling gets their's.

[rock]

20. And I agree with you

Grice's maxims (concerning conversational implications) tell me you're giving me information to identify the child (and that the other child was not born on Tuesday or is a girl). So the question is I have another child: what's the probability it's a boy?

[muriel_volestrangler]

23. The conversational implication is not *the child*, it's *at least one child*

So the question is not just "I have another child ...".

[rock]

24. If I had asked the parents

"Do you have a boy born on Tuesday?" and they replied, "yes" and then I asked, "do you have another child?" and the answer was "one other", then surprisingly the logic in the given answer works. My question doesn't identify the child (they could both have been born on Tuesday).

If we take it that the 'boy born on Tuesday' is to identify him then there are only two combinations are BB and BG.

[muriel_volestrangler]

30. I'm not sure what you mean by "to identify him" in your 2nd paragraph

Yes, your 1st paragraph sets it out well. The information in the original question is exactly the same - "I have a boy born on Tuesday, and one other child".

I'm not sure what you're saying in your 2nd paragreaph, therefore.

[rock]

33. Well that's what makes this puzzle fun!

It's difficult to make my point. Try this.

Parent: I have two children. One is a boy born on a Tuesday.
Me: What about the other one?
Parent: He was born on Tuesday too.

I'd look at the parent funny. I'd feel like he/she had flaunted Grice's maxims.

[muriel_volestrangler]

34. But if it were "I have two children. Only one is a boy born on a Tuesday", it'd be 12 out of 26

(or 6 out of 13, if you prefer).

Adapting the explanation from the article, we'd have:

When the first child is a BTu and the second is a girl born on any day of the week: there are seven different possibilities.
When the first child is a girl born on any day of the week and the second is a BTu: again, there are seven different possibilities.
When the first child is a BTu and the second is a boy born on any day of the week apart from Tuesday, there are six different possibilities.
Finally, there is the situation in which the first child is a boy born on any day of the week apart from Tuesday, and the second child is a BTu. We find an extra six possibilities here.


Summing up the totals, there are 7 + 7 + 6 + 6 = 26 different equally likely combinations of children with specified gender and birth day that is unique to that child, and 12 of these combinations are two boys. So the answer is 12/26, or 6/13.

[rock]

35. That's the clearest explanation I've heard yet!

I am (tentatively) switching my answer. Thanks!

[muriel_volestrangler]

36. I think I have a visual look at this now, too

Which builds on this comment someone gave to the New Scientist piece:

You are not changing the probability of having a boy, nor are you changing the probability of having two boys once you already have two kids. What you are changing is the group of parents we are talking about. We aren't talking about all parents who have two kids, we are only talking about parents who have two kids and one of which is a boy born on a Tuesday. When the speaker says "I have two children. One is a boy born on a Tuesday.", he is narrowing the population of parents down so that it is more likely he has two boys. That is, he is not just any parent with two kids, he is a parent with two kids but at least one boy born on a Tuesday.

While there aren't as many parents with a boy born on a Tuesday as those parents with any two kids, a greater proportion of parents with a boy born on a Tuesday have two boys compared to parents with any two kids.

Imagine a 14 x 14 grid setting out all the possibilities for 2 children. I've labelled it '1' for a boy born on Monday, '2' for a boy born on Tuesday, ... '7' for a boy born on Sunday, 'a' for a girl born on Monday, 'b a girl on Tuesday etc. '*' denotes 2 boys, '+' 2 girls, and '.' a girl and a boy.

So you start with:

  1234567abcdefg

1 *******.......
2 *******.......
3 *******.......
4 *******.......
5 *******.......
6 *******.......
7 *******.......
a .......+++++++
b .......+++++++
c .......+++++++
d .......+++++++
e .......+++++++
f .......+++++++
g .......+++++++

When you say 'at least one is a boy born on a Tuesday', you cut down the possibilities to those in just the row marked '2', or the column marked '2':

  1234567abcdefg

1  *
2 *******.......
3  *
4  *
5  *
6  *
7  *
a  .
b  .
c  .
d  .
e  .
f  .
g  .

And if you say 'only one is a boy born on a Tuesday', you cut out the one at the intersection of the row and column as well:

  1234567abcdefg

1  *
2 * *****.......
3  *
4  *
5  *
6  *
7  *
a  .
b  .
c  .
d  .
e  .
f  .
g  .

[Blues Heron]

13. hmmmmmm

What are the chances one child is a boy or a girl? 50/50

what are the chances a second child is a boy or a girl? 50/50

what are the chances of having two boys? 1/4 (4 equally likely cases BB GB BG GG)

what are the chances of having a boy and a girl? 1/2 (two of the 4 equally likely cases)

what are the chances of having 2 girls? 1/4 (same as BB)

given that one child is a boy, what are the chances the second is a boy? 1/2 (two cases BB and BG)

[muriel_volestrangler]

15. You're OK for the 1st 5 lines

What are the chances one child is a boy or a girl? 50/50

what are the chances a second child is a boy or a girl? 50/50

what are the chances of having two boys? 1/4 (4 equally likely cases BB GB BG GG)

what are the chances of having a boy and a girl? 1/2 (two of the 4 equally likely cases)

what are the chances of having 2 girls? 1/4 (same as BB)

then:

given that one (meaning either of them, not ncessarily the first) child is a boy, can we eliminate any possibilities? Yes, we can eliminate the GG case.

The remaining cases are still equally likely as each other.

So the equally likely possibilities are: BB GB BG

So the chances of the combination being BB is 1 in 3.

[Blues Heron]

18. I'm getting it ever so slowly!

If I say "of all the people who have two children, one of which is a boy, how many have a boy and a girl, and how many have two boys" easy - half and half

If I say "of all the people who have two children - ignoring those that have two girls- how many have two boys, how many have a boy and a girl - thats 2/3 BG and 1/3 BB

the tuesday thing just makes my head hurt, so I won't go there!! :)

[muriel_volestrangler]

22. Almost there; your first statement isn't quite correct

When you say, in the 2nd statement, "ignoring those that have two girls", that's exactly the same as saying "at least one of which is a boy". You could say: "of all the people who have two children, how many have a boy and a girl, and how many have two of the same sex" - half and half. And then go from there to say "but ignore those with 2 girls ..."

Yeah, the Tuesday thing is what does my head in - and many at the convention too, it seems.

[qazplm]

39. disagree

if the boy is the second one born, then we'd eliminate BG.

if the boy is the first one born, then we'd eliminate GB.

We KNOW there are only two children, and we know one of them is a boy, that means not only is GG eliminated, but ONE of GB or BG is eliminated as well, we just don't know which one.

So, IMO, there are only two possibilities remaining, BB or whichever one of the BG/GB that applies depending on whether the boy is born first or second.

[Thor_MN]

197. If you don't know which one, both are still possibilties.

and thus can not be eliminated. Think of it this way, it is not possible for them to have GB and BG, but it is also not possible for them to have GB and BB, can you eliminate one of those two conditions, based on your reasoning?

Short of gender re-assignment surgery, the sex of the children, elder and younger, is not interchangeable. I think we can eliminate that for the purposes of this discussion.

So it remains that they will either have GB, BG or BB. You can't tell which, since the information has not been provided. They are all mutually exclusive conditions, none of the three can be eliminated.

[hunter]

37. The guy asking the question knows the answer.

It's a poker game.

Maybe he's bluffing. Maybe he's got three kids at home, all girls. Ha, ha, you are all wrong!

[pokerfan]

40. We haven't discussed the car and goat problem lately

That's always good for some laughs.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

[muriel_volestrangler]

43. I thought about bringing it up

since it's another thing that confused me a lot, at first, but with which I feel happy now. But since it's not precisely the same problem, I thought it'd muddy what I was saying about the boy problem. So I'll leave it to you.

Good luck!

[Lithos]

42. 100%

Now if you had asked the question before you procreated - 25%, but at this stage, it is 100%... Got to love the semantics.

L0

[muriel_volestrangler]

44. Well, if that's the way you look at it, the probability could also be 0%

because the man might have a boy and a girl; he knows, but we don't. The point is that from our point of view, we don't have enough information to know for sure, but we can put a probability on what the actual situation is, given what we do know.

If you want, think of it as a Schrodinger's Cat situation; rather than a 50% alive, 50% dead cat, we have a 13/27 all-boy, 14/27 boy-girl pair.

[Lithos]

45. Yep

something like that...

The beauty of the logic here to get to 13/27 is that it forces you to move beyond the BB/BG/GB/GG permutation and think differently.

[hunter]

47. I could think of it as a signal processing exercise.

You have a noisy source, a random stream of ones and zeros, 50% ones, 50% zeros. You slap a "Boy born on Tuesday" filter on the machine and what do you get?

But mostly the kind of logic I enjoy is not this binary Yes/No or True/False logic (which is a human contrivance, just as the original problem is a contrivance) but the "Yes/Don't know, Don't care" logic that's representative of reality.

In the case of Schrödinger's cat, you ask "Is the cat alive?"

Since you are not doing any measure, in fact the possibility of measure is specifically excluded in the puzzle, the answer from the universe is "Don't care."

"Is the cat dead?"

"Don't care."

Even from the cat's perspective there is a clear answer. "Am I alive?" Yes.

Otherwise, Don't care.

This logic is even clearer when you ask a photon if it is a particle or a wave. Without measure, the universe answers "Don't care." With measure, the answer is "Yes"

In reality everything is a transaction. There are no hidden cards.

[qazplm]

48. it's 50%, not 33% in the original hypo

you are right, we don't know whether the second child is a girl or a boy, but we know one of the children is a boy, and we know there are only two children.

The argument is that you have four possibilities

GG
GB
BG
BB

Assumedly representing birth order. Everyone agrees that GG is excluded.
However, the idea that GB and BG are both still possible. By definition one of them is excluded, we just don't know which. If the boy is born first, then GB is excluded, and if the boy is born second, then BG is excluded.

More logical then to collapse it into one possibility, because there is only one possibility, not two.

That leaves BG and BB (ignoring birth order, the same as if they were magically born simultaneously, or removed as identical twins in total by C section).

[muriel_volestrangler]

49. No, it's not 50%; the 1 in 3 is positively, definitely, absolutely the correct answer for ...

... the simple "one of them is a boy; what is the probability that both are?" question.

Have you seen the argument in #32, where I make it like a random number in binary?

The argument is that you have four possibilities

GG
GB
BG
BB

Assumedly representing birth order. Everyone agrees that GG is excluded.

Fine so far.

However, the idea that GB and BG are both still possible. By definition one of them is excluded, we just don't know which.

No. Neither of them are 'excluded'; by 'excluded', we mean that we, as the puzzle solvers, know that a combination is not possible. We know that at least one of the children is a boy; so we can exclude GG. We cannot exclude BG, GB or BB.

If the boy is born first, then GB is excluded, and if the boy is born second, then BG is excluded.

But we don't know (and don't really care) which was born first, so the 'if' here isn't going to get us anywhere.

More logical then to collapse it into one possibility, because there is only one possibility, not two.

What do you mean by "collapse it into one possibility"? It's not a phrase that means anything to me. There are still two possibilities for a boy-girl combination. You can't just say "we'll say it's just one".

Would a diagram help? If you look at #36, I drew a grid for all the combinations of all sexes and days of the week. If you just ignore the differences in the days of the week, you'll still see that, in the first diagram in that post, the boy-girl combinations take up twice the area that boy-boy does.

The only other thing I can suggest is you toss 2 coins at once; if the result is 2 tails, then ignore it, otherwise write down whether it's 2 heads, or 1 head and 1 tail. Repeat it a decent amount of time (50 times?), and see what the proportion of 2 heads:1 head+1 tail is.

Or, if you program, you could write a small progrma to demonstrate it.

[qazplm]

50. I simply disagree

no, we don't know WHICH of GB or BG is excluded but we know one of them is, thus effectively, BG = GB for purposes of this.

There are simply NOT two possibilities. There is only one possibility between BG and GB, we simply don't know which one.

Logically speaking, knowing the birth order should have zero effect on the probability of the other child being a boy or a girl. Neither should not knowing the birth order.

You say we don't care which is born first, ignoring the reality isn't compelling to me. The boy is either born first or second, therefore saying that both BG and GB are available alternatives is simply wrong, only one of them is. The fact that we don't know which one does not mean we should count both as being alternatives. There is a reason why it doesn't make any sense that it is 1/3 because it truly isn't 1/3.

If we want to get really technical, the proposed question doesn't say they aren't born simultaneously. The assumption is that there is a birth order, which is the only relevance of having BG or GB. If you aren't looking at birth order then BG truly does equal GB. It is like saying someone with blue eyes and black hair is different from someone having black hair and blue eyes.

There is nothing that differentiates BG from GB. They are equivalent.

This might be a different scenario, if we said, two children, we don't know what either child is, what are the odds that it is BB, GG or BG. That would be 1/3.

You and IMO the person in that article are falsely making a distinction between BG and GB when there is none. I thought the argument was that the distinction was birth order, but you seem to be disavowing that.

I agree with this comment from the article:

"As someone has pointed out, you can't count the girl/boy as two separate possibilities unless you count the two boys as two separate possibilities. If you give these people names, it is easy to see. You can have Tom then Andy (BB) or Andy then Tom (BB) or Tom then Kate (BG) or Kate then Tom (GB). So, if you tell people you have one boy (say Tom), any of these could be true. There are two ways you can have two boys and two ways you can have a boy and a girl. So before we start introducing Tuesdays into it, the probability of having two boys is two out of four - 0.5."

Said a different way, if I have two coins, and I flip the one in my right hand and get heads (boy), do you really think the odds are only 1 in 3 that if I flip my left hand that I will get another heads (boy)? The odds are clearly 50/50 as I will either flip a heads (boy) or a tails (girl). 50% of the time I will get two heads (BB) and 50% of the time I will get a head and a tails (BG).

In effect, making BG different from GB means switching hands too.

[pokerfan]

53. How about this

Go here: http://www.random.org/coins

Set the number of coins to two and start flipping and record the results.

After 100 trials I got:

HH: 23
TT: 26
mixed: 51

[gcomeau]

55. We're not flipping *two* coins.

One kid is a boy. Period. That's not a coin flip. The only thing we don't know is what the other kid is. If you want to make "heads=boy" and "tails=girl" to use the analogy then one coin is glued to the table heads up before we start.

So set the number of coins you're flipping to 1.

[qazplm]

57. I said the same thing

you just said it much more concisely. :)

[muriel_volestrangler]

58. No; the coins were flipped before we knew anything about the outcomes

Imagine 1,000,000 2 child families. Do you agree there will be approximately 250,000 families with 2 girls, 500,000 families with a girl and a boy, and 250,000 families with 2 boys? (If you don't, then this is hopeless - you have to understand that, or there's no way to talk about probabilities at all)

What the puzzle setter has said is: "I have at least 1 boy, ie I do not have 2 girls". So we know his family is in the 750,000 families of 'a girl and a boy' or '2 boys'. We have no more information about his family; so there is a 250,000/750,000 (ie 1 in 3) chance he has 2 boys.

We cannot liken it to 'gluing a coin to the table'; because that would be equivalent to "I have 1 child now, who is a boy; what are the chances the next child is a boy?" When he tells us at least one of his children is a boy, he has not specified which one. There is more information in saying "my eldest child is a boy" than is saying "one of my 2 children is a boy".

[qazplm]

60. and that's the illogicalness of this

if birth order determines the probability, that isn't logical.

If knowing that his boy was born first makes a difference, then a logical argument can be made that we don't know enough to know the probability.

Put another way, as you said, he hasn't said which one, so there are two possibilities:

A. the boy was born first. You appear to agree that in that scenario, there is a fifty percent chance another boy is born.

B. the boy was born second. You appear to argue that in this scenario, there is only a 33% chance the other child is a boy.

The witholding of information should not change the probabilities.

[muriel_volestrangler]

63. Witholding information always changes probabilities

If I throw a die, and ask you "what are the chances it came up as 6?", you'll say "1 in 6". But if I tell you "it was even", you'll say the chances are 1 in 3. By witholding the information that it was even, I changed the probabilities. These are the probabilities for the person who doesn't know what the die/coin/pair of boys are. They have some information about them.

[qazplm]

65. not really

the probability of what was thrown was always 1/6.

The probability of me guessing correctly has changed to 1/3, but the probability of what was thrown was always and will always be 1/6.

The probability that I will guess what the sex is may change, the probability of the sex will not.

[muriel_volestrangler]

68. Well, if that's your definition, then the probability of 2 boys is 1 in 4, isn't it?

Not 1 in 2. By your definition of probabiliity of "we must ignore any information we have about the children", the odds of them being 2 boys will always be 1 in 4. Even is the father tells us "they're both boys", the odds of them both being boys would remain 1 in 4, according to the definition you've just given.

Of course we're talking about the probability from the point of view of the person asked the question.

[qazplm]

70. yes and no

by way of example.

We start off with two coins under a cup.

I ask you the probabilities of what each of the coins are. In that scenario, with that question, yes the probability doesn't change.

However, I then turn over one of the cups. I don't tell you whether it was the coin I flipped first or second.

You and I see that it is heads. I ask for the probability that the other coin is also heads. That's a DIFFERENT question and it is now focused on one coin flip, not two.

So yes, at the end of the day, the probability of there being two boys IS always 1 in 4. However, knowing that one of them is a boy, you ask a different question seeking IMO a different probability.

[gcomeau]

67. Yes, they were. Not the point. Doesn;t matter.

The point is we DO know the outcome of one of them now when we're doing our probability calculation. Known variables have probabilities of 100%. Always.

If you want to put it your way... I flipped two coins 5 minutes ago. One of them was heads. It's sitting right here on the table in front of us and we both see it's heads. I just told you "look, see that coin I flipped? It's heads. See?". The other one I'm hiding under my hand.

NOW... given what you know, what are the odds the other one is heads? Thus giving us two heads?

Imagine 1,000,000 2 child families. Do you agree there will be approximately 250,000 families with 2 girls, 500,000 families with a girl and a boy, and 250,000 families with 2 boys?

Of course I do. It's just doesn't matter to this question. You're letting irrelevancies confuse you about an extremely simple problem.

What the puzzle setter has said is: "I have at least 1 boy, ie I do not have 2 girls". So we know his family is in the 750,000 families of 'a girl and a boy' or '2 boys'.

You're lost in the weeds. You're supposed to be calculating the odds of the unknown variable, not looking at things we DO know and are thus 100% probabilities and not a factor in the calculation.

The unknown variable is the gender of the second kid. If it's a boy he has two boys. If it's a girl he doesn't have two boys. Now... what are the odds a kid is a boy and not a girl?

When he tells us at least one of his children is a boy, he has not specified which one.

Which has nothing to do with anything! You just keep chasing off after these shiny distractions. He just asked what the odds of him having two boys were... not what the odds of the boy he told us about having an older brother were, or the odds of him having a younger brother were... just the odds of him *having a brother at all*. Since we know he has one sibling he either has a brother, or he has a sister. The odds are 50/50 that sibling is a brother.

[muriel_volestrangler]

72. OK, if you can't follow my argument (and I accept I may not be at all clear), how about the NYT?

(in addition, that is, to the New Scientist writer).

Here is a UCLA professor, Craig Fox, writing in the New York Times, explaining why it's 1 in 3: http://tierneylab.blogs.nytimes.com/2008/04/14/mr-smiths-gambling-problem/

Or here's a presentation by a couple of people from the Australian National University, Canberra: http://www.kent.ac.uk/secl/philosophy/jw/2005/progic/papers/Puza.pdf

[qazplm]

77. Mr. Fox puts the word "correct" in quotes for a reason

I think he recognizes the same problems with this as we do but IMO makes the same mistake of separating out GB and BG, when in my mind, just like we know GG is eliminated we also know that one of BG or GB is eliminated because the boy was born either first or second.

The post by Andrew in the comments section is also interesting:

"Your original question was posed like this:

1. Mr. Smith has two children, at least one of whom is a boy. What is the probability that the other is a boy?

As I responded in comments to the original question, it really does depend on how you know the two parts of the question, particularly the second part. If you learn this by first meeting a boy child (doesn’t matter if he’s elder or younger, or called Fred or Bill) you can assume a 50:50 probability that there are two boys.

However, if you use the formulation proposed by Dr Fox

“Mr. Smith says, ‘I have two children and at least one of them is a boy.’ Given this information, what is the probability that the other child is a boy?”

then you should conclude that the probability of two boys is 1/3, based on the fact that of the four possible variants, BB, BG, GB and GG, only GG is now ruled out.

I'd also say I think this problem is different from the Monty Hall problem, which makes logical sense despite the apparent linkage in that article.
So this is a very subtle question, and exactly how it’s posed really does matter."

[gcomeau]

78. And here's me...

http://www.democraticunderground.com/discuss/duboard.php?az=show_mesg&forum=228&topic_id=67044&mesg_id=67117

...explaining how they screwed it up. They made the exact same mistake the New Scientist article did. "BG" and "GB" are NOT separate combinations in this scenario. They are the exact same thing.

They are inserting conditions into the problem which were not presented in the problem itself. But if you don't want to listen then I'll demonstrate it your way by pretending that not knowing WHICH kid is the boy is actually relevant. Now, this is what the question explicitly states:

1. There are two children. (blank spaces below indicate unknown variables)

____ _____

2. One of them is a boy. (fill in a blank, either the first kid or the second kid is a boy)

Scenario A: ___B___ _____

OR:

Scenario B: _____ ___B___

The answer is going to come out exactly the same doing it this was as doing it the way I already showed you.

Now, what are the possible combinations we're faced with at this point, *given what we know*?

Scenario A:
___B___ ___B___ or ___B___ ___G___

Scenario B:

___B___ ___B__ or __G___ ___B___

Oh look. FOUR possibilities. Half of them involve there being two boys. So what does that make the odds doing it this way? That's right. Still 50/50.

[qazplm]

80. I agree with you but

now you are kinda adopting the same tone others in this thread have which is unnecessary from both sides IMO.

This is a problem/issue that clearly very smart people see different ways.

[gcomeau]

82. Yes, I got a little irritable...

...when it switched from arguing the question to "I have links that agree with me. <link>".

I just finished an exchange in the religion forum where a certain poster refused to argue their position for a dozen posts in a row while just constantly declaring that some website said he was right so therefore he was right. Has me a little touchy on the subject.

[muriel_volestrangler]

84. The point about using links is to get a professional to explain it

in the hope that they'll explain better than me.

[qazplm]

87. I don't think the problem is that you aren't explaining it

sufficiently, simply that we disagree with you (and I suppose by extension, him).

But in reading your links, even he admits that phrasing of the question makes a difference and that the originally framed question may have been flawed.

So as I've said in other places, this may boil down to parsing out phrasings and words, which isn't exactly probability or math.

[muriel_volestrangler]

83. Here's the flaw in that reasoning:

1. There are two children. (blank spaces below indicate unknown variables)

____ _____

2. One of them is a boy. (fill in a blank, either the first kid or the second kid is a boy)

Scenario A: ___B___ _____

OR:

Scenario B: _____ ___B___

fine so far

Now, what are the possible combinations we're faced with at this point, *given what we know*?

Scenario A:
___B___ ___B___ or ___B___ ___G___

Scenario B:

___B___ ___B__ or __G___ ___B___

In scenario A, you cannot assign equal probabilities to "___B___ ___B___" and "___B___ ___G___". If both children were boys, then the choice between the revelation in Scenario A (that the child on the left is a boy) and that in B is equal, and so each scenario is equally likely to be followed. But if the child on the right was a girl, then Scenario A had to be followed, which makes that scenario twice as likely to come about from that combination of children. And similarly, if the child on the left was a girl, then Scenario B had to be followed. So the probabilities are:

Scenario A:
X or 2X
Scenario B:
X or 2X

So the total probability of 2 boys is 2X / 6X , or 1/3.

Would this help, as wording?

"I have 2 children. They are not both girls. What is the chance they are 2 boys?" Notice that, with this wording, we have not looked at a child at all. All we know is something about the pair of them, collectively. However, "they are not both girls" is logically identical to "at least one is a boy". So this is the same question.

[qazplm]

85. I think the real issue

is you are assuming changing the wording doesn't change the question asked.

After reading all of your links, it becomes pretty clear that wording is paramountly important to determine what question asked.

I think the real issue again goes back to wording, one reads it one way and gets one answer/question, one reads it another and gets another, and both end up being right from their point of view.

[muriel_volestrangler]

90. No, the change in wording is not changing the substance of the question

It is being done just to try to persuade you of how the maths works.

You started this sub-thread, in reply #48, by saying you wanted to "collapse it into one possibility"; but you still haven't explained what you mean by that. Nor, for that matter, did you justify your claim that we can 'exclude' one of BG or GB. We know that they can't both be true at the same time, but we could say the same about BB as well. Yet you don't try to 'exclude' BB, whatever you mean by exclude. But the meaning of 'exclude' should be clear; it means "this cannot be the true situation", and thus the only situation we can exclude is GG.

The only way for the answer to become "1 in 2" is to fundamentally change the question, to, for instance, "here is a boy; what is the chance of the other child being a boy?" And that was not the question that was asked.

[qazplm]

91. I dont know how else to explain

the collapse to you then I already have. I believe artificial the distinction between GB and BG.

Put another way, you have three outcomes, two boys, two girls, or a mix.

Does that explain it better?

As for being able to exclude, the other guy has done a better job laying that out then I have, but logically, we know that the boy is born first or second.

If he is born first then it is clear that we either have BB or BG but not GG or GB.

IF he is born second (or flipped second) then it is clear we either have BB or GB but not GG or BG.

Both scenarios are 50% and BB is not excluded in either scenario.

I believe that in fact was the question asked, I find completely artificial the argument that saying I know one child is a boy, what are odds that the other child a boy is any different from saying I know one child is a boy, what are the odds that both children are boys. If the other child is a boy then both children are boys, if the other child is not a boy, then both children aren't boys. I think it's word play.

[muriel_volestrangler]

96. You hgave 3 outcomes, but you have not said anything about their relative likelihood

And we know (from reality, even if you won't accept from logic), that a mix is twice as likely as two boys. And a mix is twice as likely as two girls.

Your 'collapse' seems to be 'calling these things the same'. That's fine, but it doesn't alter the number, or likelihood, of them. If I have one '69 Mustang, and one '70 Mustang, and I say "meh, a Mustang is a Mustang", then I say I have 2 Mustangs. I have to add together the numbers for the 2 earlier classes; and if I am taking 2 permutations, and saying "order doesn't matter, I am just concerned with the one combination", then I have to add the probabilities of the 2 permutations together, to get the probability of the combination.

No, the other DUer is not 'excluding'. I repeat: only 'GG' (or 'TT' for coins) is excluded. Exclusion means something has been ruled out as a possibility.

"I believe that in fact was the question asked, I find completely artificial the argument that saying I know one child is a boy, what are odds that the other child a boy is any different from saying I know one child is a boy, what are the odds that both children are boys. If the other child is a boy then both children are boys, if the other child is not a boy, then both children aren't boys. I think it's word play."

If you try to interpret the question as ending "what are the odds that the other child a boy", then you have assumed that you have already selected one child as 'this child'. And that has not happened in the question.

Have you thought about the illustration in #36? Or the analogy to binary numbers?

[qazplm]

97. I've shown elsewhere

where I believe in fact one child has already been selected as a boy. The differentiation you make is wordplay IMO.

I also linked to an explanation in the other post why I believe this to be so from someone other than me. Again, from my POV we are focused on one binary condition, not two, once you are told that one of the kids is a boy.

[gcomeau]

89. You're not following.

And I think I know why. Let me remind you that you were the one that insisted that the placement of the identified kid as either the first or second kid mattered and had relevance, which is th only reason I'm doing things in this unnecessarily complicated manner. So let's make it more explicit. The identified boy with the Tuesday birthday is going to be indicated by "B(Tuesday)".

So, let me alter that initial table of what we know from the conditions in the question:

Scenario A: _B(Tuesday)_ _____

OR:

Scenario B: _____ _B(Tuesday)_

With there being a 50% possibility of it being either scenario.

Still fine?

[muriel_volestrangler]

95. If you want to bring Tuesday into it, have you tried the visual aid in #36?

No, I haven't insisted that the placement as either 1st or 2nd mattered; in fact, I've said more than once that order does not matter. What I have said is that you cannot just say "we're not interested in the order, therefore we can call the chances of getting a mixed pair must be the same as getting a boys-only pair".

As an example: flip 4 coins, at once. Getting 2 heads and 2 tails is more likely than getting 4 heads. And similarly, when you flip 2 coins, you are more likely to get 1 head and 1 tail than 2 heads.

As for your scenarios, remember that "_B(Tuesday)_ _B(Tuesday)_" fits both scenarios. With that in mind, what is the possibility that you are measuring?

[gcomeau]

98. First...

No, I haven't insisted that the placement as either 1st or 2nd mattered;

Yes you did. You may not have realized you were doing it, but by insisting that "BG" and "GB" are separate and distinct possible combinations, and arguing that it is relevant to the calculation that we haven't been told which child was the boy born on the Tuesday (which you thought was so important you bolded it in post 58), you are saying that order does matter.

Second, please answer my question. Do you or do you not accept that the two scenarios, as listed in the last post, accurately describe the initial set of possible conditions described by the content of the question?

And finally... no, "_B(Tuesday)_ _(B(Tuesday)_" does not fit both scenarios. It does not fit any scenario. "B(Tuesday)" is not indicating any old boy who happens to have been born on a Tuesday. It is a unique identifier of THE boy we have been told was born on a Tuesday and only him. Change it to "B(identified) if you find "B(Tuesday)" confusing.

[muriel_volestrangler]

101. BG and GB are distinct *permutations*

and each permutation has the same likelihood (apart from any that get eliminated altogether, such as 2 girls). A boy-girl pair is one *combination*, but that *combination* has a higher likelihood than the all-boy combination. This is just like throwing 2 dice; the combination of a 1 and a 2 is twice as likely as the combination of a double 1.

I bolded "he has not specified which one" (is a boy) in #58. That's the point - I'm saying he has not specified a difference between the children. You, by showing one of the coins and 'gluing it to the table', as in that post, are specifying a difference between the information about each child - you are applying the information we have to only one child. Thus, you are changing the situation by doing that.

The father has said one of the children is a boy (ignoring Tuesday for now, and concentrating on the 1 in 3 problem). If you want to write this as "B(identified) _____________" or "_____________ B(identified)", you can, and we can say both are equally likely, but remember that is a probability about the father's actions in setting the problem, not the children's sexes; and proceed with care when you get to the situation where both children turn out to be the same.

[hunter]

104. The sample size is one and there is no transaction here.

There is no verb in the sentence.

"Two children. Boy born on Tuesday."

That's what we know. That's all we know.

If we make a query of what we know, "Is the other child a boy or a girl?" then the answer is: "Yes, the child is a boy or a girl." But that's only if our database hasn't been set up to account for humans of indeterminate gender. Accounting for everyone, even children of indeterminate gender, the answer is "Don't know." In any case, the course of inquiry becomes extinct here. There are no odds, there are no probabilities.

If you add a transaction to the problem then the answer you get will be shaped by the nature of the transaction, by the "verb" in the sentence so to speak.

Let's suppose we had one mathematician standing at the podium who said "I drank Holy Water and my cancer went into remission." Logically it's the same problem. Two nouns, no verbs, no transaction, a sample size of one. Well, okay, dear... and?

Now, in a great gathering of ten thousand mathematicians if we asked everyone to stand up who had two children, and we asked all those standing who had two girls to sit down, etc., etc... then we can start talking about probabilities and distributions.

[muriel_volestrangler]

105. I have absolutely no idea what you're trying to say

Your last paragraph does look rather like my earlier post about "if there are 1,000,000 families ...". But the stuff about nouns, verbs, databases and transactions means nothing (I notice that, for instance, you claim there are no verbs in the 'Holy Water' sentence, when there are 2 - 'drank' and 'went').

[hunter]

110. It's a fixed state, "drank" and "went" are not "verbs" in this case.

That's the problem with our language, actually. The demarcation of equivalence and transaction is blurred.

Maybe this makes it clearer--

A: "Holy water drinker"

B: "Cancer in remission"

There's nothing else there, and it's exactly equivalent logically to--

A: "Two children"

B: "Boy born on Tuesday"

Did the holy water cure the cancer? Is the other child a girl? :shrug:

We don't know. There are no odds, no probabilities.

[muriel_volestrangler]

111. OK, you don't believe in the common concepts of 'probability' or 'verb'

There's not much we can do to help you, then. But no, you have not made your position any clearer at all. If you don't believe there is such a thing as probability, then I doubt there will be much you can do to be clear.

[hunter]

115. Gah. I understand the statistics.

It's the flaw in our language that makes this "counter-intuitive" that I'm trying to tease out.

[gcomeau]

114. Read post 113

I am quite certain you are making the error I described there.

[pokerfan]

59. No we're not

So set the number of coins you're flipping to 1.

That would be valid if one of trials had been identified (i.e. the oldest is a boy). Then we have:

BB
BG
GB
GG

But since we did not, GB and BG are separate events.

[gcomeau]

69. Why are you talking about "oldest"?

Was the question "what are the odds the boy born on Tuesday has an older brother?" No it was not.

It was "what are the odds I have two boys"

Whether the identified boy is an older or younger sibling is completely irrelevant.

[pokerfan]

71. And that is exactly where you go off the track

Do you agree that if we sample all two child families that the breakdown is roughly...

25% - BB
25% - GG
50% - BG (or GB)

[qazplm]

74. and for a setup with no initial conditions

you would be correct.

and if we were asking what are the odds the following outcomes would happen, that would also be correct.

But that isn't what is being asked. What is being asked is that, having one binary outcome already determined, what are the odds that the other binary outcome will be B or will be G. The odds of that binary outcome are 50%.

The odds of both binary outcomes collectively are different. But we already know one of the outcomes, which changes the question asked.

If I put the two kids in a giant cup, and you couldn't see under them, and I picked up one of the cups with a crane, and it was a boy, and I asked you, what are the odds that the child under the other cup is a boy or a girl, it would be 50%. It is one or the other. You've collapsed some of the probabilities from a set of two, to one.

[gcomeau]

81. No, it's actually where you go off track.

Do you agree that if we sample all two child families that the breakdown is roughly...

25% - BB
25% - GG
50% - BG (or GB)

Yes. It also makes no difference because you are ignoring the conditions set by the question. As I just finished explaining in another post, we'll do it your way and pretend whether the boy is the oldest or the youngest actually matters... and the answer isn't changing when we properly acount for the conditions set by the question. Now, what do we KNOW? We know there are two kids. And we know that:

Either the youngest of the two kids is a boy:

___B___ ___?__

Or the oldest of the two kids is a boy:

__?___ ___B___

If you want to say that's wrong, say how.

So, what are the possible combinations we're dealing with given the conditions specified in the question?

If the identified kid is the youngest (of which there is a 50% chance) then the older sibling is either a boy or a girl, either being equally likely:

BB (25%)
BG (25%)

If the identified kid is the oldest (of which there is a 50% chance) then the younger sibling is either a boy or a girl, either being equally likely):

BB (25%)
GB (25%)

Oh look, those BB combinations add up to 50% of all possible outcomes. Imagine that.

[Thor_MN]

198. What rails are you on?

BB where the idenitfied child is youngest and BB where the identified boy is oldest ARE THE SAME CONDITION. Namely, having two boys. So the "sum" of the BB combination is 25%, or 1/3 of the possibilities. BB = BB. BG ≠ GB ≠ BB.

[qazplm]

56. but you haven't set any initial conditions

Yes, if we are starting from scratch, then you'd have things correct, but we aren't.

We are asking a specific question, with known conditions, and that is that one of the kids is a boy, what are the odds that the other one is also a boy i.e. that you have two boys.

Similarly, if I already have one heads, what are the odds that when I flip the other coin, it will also be heads. The fallacy here is you aren't flipping two coins, you are only flipping one because you've already flipped one (which has no impact on the odds of your flipping the other one).

There are only two options then, not three.

[pokerfan]

61. Had I identified one of the coins as heads

Say the leftmost one, then your analysis would be correct.

But since I did not, then HT is not the same as TH.

[qazplm]

75. let me try this another way

If I plug in the boy into the "left" hand I come up with 50%.

If I plug in the boy into the "right" hand I come up with 50%.

Those probabilities are the same regardless of which hand I plug the boy into.

The fact that it is left unidentified which hand does not change the probabilities.

Put another way, if I reveal a coin in my left hand to be heads, and ask you what the odds are that the coin in my right hand is heads or tails, does it matter which hand flipped first? I don't think it does.

[muriel_volestrangler]

62. No; the puzzle setter has already flipped both coins

The puzzle setter has already looked at both coins; and then the puzzle setter has decided to say "one of the coins is 'heads'".

Say the coins are a dime and a nickel. The possibilities are:

\nickel|       |
 \     |       |
  \    |       |
   \   | heads | tails
    \  |       |
     \ |       |
 dime \|       |
-------+-------+------
       |       |
 heads |  HH   |  HT
       |       |
-------+-------+------
       |       | 
 tails |  HT   |  TT
       |       |

There are 4 possible outcomes of tossing the 2 coins. The tosser has told us at least one of the coins came up heads. So we know the result was not 2 tails. Yes, you can call 2 of the results "a head and a tail", and for the purposes of answering the question, that's all you need say about them. But that doesn't mean that the chances of those 2 outcomes suddenly decreases, just because you've decided to call them the same thing. 'HT' is twice as common an outcome as 'HH'.

[qazplm]

64. we dont know that

the puzzle setter hasn't given us that information.

He hasn't told us if he flipped the boy first, or if he flipped the boy second.

Again, this is why this may fit probability theory but is simply illogical.

If he flipped the heads first, then there is only a fifty percent chance he flipped heads second or tails second.

If he flipped the heads second, then there is only a fifty percent chance he flipped heads first or tails first.

Him not telling us which doesn't change that reality that in either scenario the other coin is either heads or tails.

If he hasn't flipped either, THEN, there is what you've laid out.

[muriel_volestrangler]

66. He has flipped both already, and he hasn't told us which is a head

If you mean "it's not as I expected without working it out", then fine - it's not what you expected. But 'illogical' means 'not according to the rules of logic'. And the rules of logic are exactly what we're using here.

Have you tried flipping a series of coins? Notice that one DUer has already, and the New Scientist's odds of 1 in 3 were shown to be right.

[qazplm]

73. again

we arent asking about two coins, we are only asking about one.

one coin was already flipped, we already know what it is, the probability is now set to 100% for that coin.

We are now focused on the other coin, and that coin only. That coin only has two logical states, heads or tails.

That's the whole problem here, you are ignoring that we already know one of the coins and treating the probabilities as if we really don't.

I'm also not sure why the tone simply because someone disagrees with you.

[muriel_volestrangler]

76. No; we are talking about 2 coins, which have both already been flipped

Notice that, in #62, I said nothing about whether the dime or the nickel was flipped first or second. It doesn't matter which was flipped first or second.

We are not focusing on one coin only. The question is about the pair of coins (children, etc.). We do not know the state of a coin; we know something about the pair. There is an important difference.

I'm sorry if my tone comes across badly, but I'm trying to get across the important bits of the argument. I know I have not yet done so successfully. But I also know the '1 in 3' answer is right. There is absolutely no doubt about this.

[qazplm]

79. disagree

We absolutely do know the state of one of the "coins", we know it is a boy, or heads (or tails or whatever).

The only thing we don't know is the state of the other binary outcome.

I think there is doubt about whether the one in three answer is right. I think even Mr. Fox lays out how the wording can make a difference, and I think the comments lay out why presentation can change what question you are asking and thus what the answer is.

I also think that is the crux of the problem, disagreement on what question is being asked. You think they are asking about the pair, and I think when you know half the pair, and you are then left figuring out what the other half was, it isn't about the pair anymore, it's about the remaining one, even if you "phrase" it as being about two boys versus the other child.

[muriel_volestrangler]

86. No, we do not know the state of one of the coins

We have partial information about the pair of coins. We know that it is not a pair of tails (or we know the children is not a pair of girls). We do not know the state of a given coin.

Consider a random binary number between 0 and 15 (decimal), inclusive - writable with 4 binary digits. If we are told "at least one of the digits is a '1'" (which is exactly the same as being told "the number is not '0000'"), and then asked "what are the chances the digits are all '1's" (ie what are the chances the number is '1111', ie 15 in decimal), then we'd say: it's a random number between 0 and 15, except we know it's not 0. So it's a random number between 1 and 15, so the chances of it being 15 are 1 in 15".

If it were a 3 digit binary number (0 to 7), and we are told it's not '000', then there's a 1 in seven chance it is seven.

And if it's a 2 digit binary number (0 to 3), and it's not '00', then there's a 1 in three chance it is three. And that is exactly the setup with the 2 coins/children.

[qazplm]

88. I have two kids, one is a boy

I have two coins, one is heads.

We know one of the coins is heads, we are now focused on the other.

We do in fact know the state of one of the coins. We may not know which hand that coin is in, but we know one of them is heads.

And the problem doesn't establish HOW we know one of them is heads (or a boy).

You used the term "at least one of them is..." but the word "at least" isn't used, from the original link:

"So, consider this preliminary question: "I have two children. One of them is a boy. What is the probability I have two boys?"

I have two coins, one of them is heads, what is the probability that the other one is heads is the same things as asking what are the odds that both are heads. Perhaps if it had been phrased as, if "at least" one is heads, what are the odds, I'd be more prone to believe you are correct.

[muriel_volestrangler]

92. The 'at least' is implied in the question

If the question had been "I have two children. Only one of them is a boy. What is the probability I have two boys?", then the probability would be zero, I think we can all agree. So "one of them is a boy" has to be interpreted as "at least one of them is a boy", or it's not really a problem at all.

In your example above, you say "we know one of them is heads". We do not know the state of either coin; we know something about 'them'. In a murder mystery, we might say "one of the people in this room is the murderer". But we don't know which person is the murderer. Knowing that a particular person is the murderer would be extra information.

"I have two coins, one of them is heads, what is the probability that the other one is heads is the same things as asking what are the odds that both are heads."

Yes, but both versions are different from "this coin is heads; what is the probability the other coin is heads".

"And the problem doesn't establish HOW we know one of them is heads (or a boy)"

The father knows the state of both children, and has given us a piece of information about the pair.

[qazplm]

94. we've ended this IMO

I think they've given us information about one of the two, you think it's only the pair.

I think saying I have one of x, what are the odds that the other is x, is no different from saying, I have one of x, what are the odds that I have two x you think there is.

I think wording makes a difference (as does your link), you think it doesn't.

I'm done, but if you want an interesting look, one of the folks in the comments to your links argues we are both wrong and the real answer is 75%.

For the "at least part" I found a comment that explains the distinction:

re #82. To Bill:

Here is my take on the difference between stating, “at least one is a boy” and “one is a boy.”

“At least one is a boy” is a statement about the set of two children. It is not a statement about any one of the two children.

“One is a boy” (after you eliminate the “exactly one is a boy” interpretation) is a statement about one of the two children.

In the former case, I believe the only way to interpret “other child” without extra assumptions is as the child left over after the “at least one” requirement is satisfied.

In the latter case, the “other child” is the one not referred to in the “one is a boy” statement.

A solution to the former cannot at any point identify one of the children as a boy without changing it to the latter, even though we know one of the two must be a boy.

Example:
“At least one is a boy”
“One, Fred, is a boy.” (I add the name in there, because 1) it makes the demonstration clearer without changing the problem and 2) as a nod to one of the other posts)

If you say, “Fred is a boy,” it certainly follows that “At least one is a boy.” A statement about a member of the set allows one to make deductions about the set as a whole.

If you say, “At least one is a boy” it does not follow that “Fred is a boy.” Fred could be a girl (Winifred) and the statement could still be true.

So the two can’t be equivalent.

“Fred is a boy”
the possibilities are
Fred – brother
Fred – sister
and the probability is 50%. Every solution getting 50% is, at some point, identifying a member as a boy, and reducing it to the above problem.

“At least one is a boy”
Wilfred – brother
Wilfred – sister
Winifred – brother
Here, at no point, are we specifying the gender of either member. Using the (to my mind) only meaningful definition of “other child” given above, the answer is 1/3. Two observations: 1) the fact that either Wilfred or his brother could be used to satisfy the “at least one” requirement does not mean you count it twice. 2) The statement “Fred is a boy” amounts to eliminating option 3, and reducing it to the 50% solution.

I suppose the confusion comes down to whether we’re abiding by the unwritten law of solving math problems — that you use only the information given you in the math problem. I accept the “at least one” statement as true because it’s given. If I start asking how it was established to be true, then it’s very easy to change the problem to the 50% one, as demonstrated repeatedly in these posts.

[Thor_MN]

199. The problem comes down to believing that "the other" has any bearing

The question does not mention "the other". It only talks about the odds of the pair. It gives information about one child, making it very easy to think that the question is about "the other". But the only deduction that can be made about the pair, knowing that one, or at least one (really doesn't make any difference, assuming that the children are whole and not sliced up melanges of boy and girl parts) is that two girls are not possible.

[qazplm]

205. absolutely wrong

now you used the word deduction, we can absolutely deduce that one of the two pairs of GB/BG is not possible. Either GB/BG has a distinguishable difference between the two or it does not. If it does not, then they are equivalent, and it's mere word/number play to distinguish them. If it does, say birth order, then we already know that one of those two pairs are not possible. We can logically deduce that only one of those two pairs are possible.

[Thor_MN]

215. BG and GB are both possible, because you don't know who the "A boy" is.

BG, GB, BB. All mutually exclusive possibilities for the children. You have no clue which is correct, so all are possible. You can not remove one of the possibilities because you feel it shouldn't be there. The arguement for removing one of the 1 girl one boy combinations is no more valid than saying that BB is not possible because it is not possible for both BB and GB.

I think what's baking your brain is the fact that there is only one BB combination. I can't really explain that one myself, but there it is, the three possibilities giving rise to the 1 in 3 odds for two boys.

[qazplm]

217. it IS possible for both BB and GB

That is the difference, saying there is a boy does not mean either BB or GB is knocked out. In fact, it tells us nothing about which one is left.

It DOES however mean that one of GB and BG is knocked out. There is a difference.

And my brain is "unbaked."

[Thor_MN]

218. So now you have a meta child that is both boy and girl?

You have repeatedly insisted that it can't be both BG and GB. It can't be both GB and BB either. Do you understand the difference between one set of children and the possibilities for the entire population? In any one pair of children, GB, BG and BB are mutually exclusive. The children are not interchangeable, GB is not the same as BG and not the same as BB.

How does saying there is a boy knock either BG or GB out? Both possibilities fit the given conditions of one boy.

If I tell you I have two children and one is a boy, I could have an elder girl and younger boy, an elder boy and a younger girl, or I could have two boys. Period.

Please explain why one of the two mixed pairs is not possible. Distinguish why you would remove one of the mixed pairs and would not exclude BB for the exact same reason.

To date, the arguements that you have used to remove one of the mixed pairs also removes the BB pair, making the odds, by your reasoning, 0%.

[qazplm]

221. no

the fact that one of the children is a boy, does not mean the other child can't be either a girl or a boy.

It does not logically cancel out either BB or Mixed. What it does do is logically cancel out one of the two of GB/BG, assuming BG is different from GB which again I and others believe is wrong.

You could as easily say mixed as to say GB/BG. It goes back to the idea that this is two separate binary possibilities completely independent of each other.

[Thor_MN]

222. You did not answer the questions. Should I assume "LaLaLa" is the best you have

You are back on the concept of the "the other" again...

Please, we removed the, on average, 25 pairs of bills marked GG per the information in the original question.

Leaving 50 pairs of GB or BG and 25 pairs of BB, again on average. You are argueing that 25 pairs of the GB or BG must be removed. Why? You keep saying that GB and BG are not both possible. There they are, you have to remove 25 of them for some reason. WHY? You say that they they both can't be true, yet there they are, lying on the table.

You see what I have done there? I don't care which of the 50 pairs that contain one bill of each mark MUST be removed to make your 50% chance true. You have to take 25 pairs off the table, for a reason stated in the original question. WHY? Please tell us why they have to be removed, for a reason that does not automatically force the removal of the BB bill pairs as well.

Got a deck of cards? Shuffle them up, boys are black, girls are red. Lay em down in pairs, remove the two red card pairs. Then explain why you have to remove half the mixed pairs, I'd really love to hear why they are not possible when they are sitting right there in front of you.

[qazplm]

255. I've said why repeatdly

that you choose not to understand why or agree with it is not my problem, nor is it the apparent difficulty you appear to have with my not agreeing with you as evidenced by the rather d*ckish posture you appear to want to take in this thread with me and others.

Heck, if I'm so wrong, one would think you'd chalk it up to me being stupid, and you'd not bother with me any further. I know it's for me not worth engaging you in the same circular process where I explain the same thing over and over again and you tell me you don't see what I am talking about over and over again.

[Thor_MN]

257. Hence my comment that LaLaLa is the best you have to offer.

You go on and on repeating that one of the mixed pairs must be removed because it can't be both. I've shown you the fallacy of that argument, since it can not be both BG and BB either. You have avoided answering questions, simply repeating that one of the mixed pairs must be removed. Since you have offered no reasons that do not also exclude the condition of 2 boys, one must conclude that the answer is either 0% chance of 2 boys, or there must be something wrong with that line of reasoning.

[qazplm]

259. no you havent

the fact that you don't understand does not equate to fallacy. I've addressed the difference between why BG/BB is different.

The fact that you don't understand that also does not equate to fallacy.

The fact that you feel you need to demean my position suggest either that you are a guy who can't stand the idea of someone thinking you are wrong, or you aren't as strong in your position as you aggressively pretend.

Either way, I'm done with you. Feel free to respond, again, with lalala or whatever else you'd like. I'm sure someone will find that really convincing.

[Thor_MN]

260. I understand why you feel you must remove one of the mixed pairs, 50% requires it

What you seem to not understand is that every argument you have raised for removing a mixed pair, by extension, also removes BB. You are shifting frames of reference at whim. In any INDIVIDUAL family, that has a boy and 2 children, you can not have both BG and GB. Likewise in any INDIVIDUAL similar family you can not have both a mixed pair and 2 boys. Both of which have squat all to do with the probabilities of all families fitting the initial criteria. In terms of ALL families, that have a boy and 2 children, there exists BG, GB and BB. Three possibilities, held up by normal distributions, and observable in nature.

I'm glad you are done and can get on with your life and vision of probability. Maybe you could go on vacation to Vegas and take your theories with you. Or sit and wait for the Nobel committee to inform you of your prize, as you have created an entirely new and contrary view of genetics that will shine a new light on all work done in the last 144 years since Mendel started playing with his peas.

The only thing I ask of you is to not teach any children, there is enough crap science out there as it is with the obvious disdain for science currently being pushed by conservatives.

[muriel_volestrangler]

265. It's a good point about Mendelian genetics

The boy-girl problem maps directly to a simple dominant-recessive gene choice, where you can tell the difference between a pair of recessive inherited genes (which would be the 1 in 4 that show the recessive characteristic, and would breed true with those of a similar appearance), and an organism with one, or two, inherited genes of the dominant form. The situation has been looked at extensively for over a century, and the probabilities of the offspring from the individuals with the appearance from at least one dominant gene are completely agreed on.

[gcomeau]

51. The New Scientist explanation is wrong.

""I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?""

The question is what the probability of having two boys is given the stated conditions. That's it. Two boys. So then the New Scientist write-up immediately screws it up by saying this:

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG.

Wrong. BG and GB are not seperate combinations, they're the exact same thing. One boy and one girl. Writing one of them down first isn't a relevant distinguishing characteristic that makes them distinct. It doesn't matter which one you write down first you still have one of one, and one of the other.

Given the conditions of the question the available options are that the other kid is either a girl, or they're a boy. That's two possibilities. Not 3.

They also then go on to screw it up even worse by committing the gamblers fallacy and declaring that the odds of already determined events impact the odds of future outcomes. It doesn't matter what day the first boy was born on. It's specified as a given condition by the question so it isn't a variable that comes into play in the probability calculation of the unknown condition. If the question had been "What are the odds of any given family having two boys with one born on a Tuesday" then the odds of one kid being born on a Tuesday would matter, but that wasn't the question.

This is all silly misdirection. No, the boy being born on a Tuesday has nothing to do with it, the odds he has two boys are 50/50. Period. Odds he has another kid? 100%. He already said he did. Odds he ONLY has one other kid? 100%. He already said so. Odds the other kid is a boy? It's that or a girl if we're not bringing alien mutants or hermaphrodites into this. So it's 50% if we're ignoring that in reality that's not exactly the right ratio for boy/girl births. If it's a boy he has two boys, so the odds of him having two boys are 1/2. Done.

Any other answer relies on changing the conditions of the question from what they were stated to be.

[qazplm]

52. absolutely right

The idea that BG is different from GB is ridiculous.

They give stated conditions which means that one of the two kids is a boy, therefore, we are left with figuring out whether the remaining child is a boy or a girl, not figuring out all possible combinations as if we didn't already have the condition present.

[Lithos]

99. Actually no, it is correct

BG means that the boy happened first, followed by a girl. The GB is the case where there was a Girl first followed by a boy. They represent different positions of a permutation tree.

L-

[gcomeau]

100. Which is only relevant...

If you're calculating the odds of something where such permutations matter.

Which we're not.

[Lithos]

102. If "Tuesday" were left off, it would matter significantly

Nowhere in the question was the birth order mentioned... So, in order to determine odds, you have to start at the beginning of the permutation and then discard what no longer is applicable, ie he first had a boy, then another boy; he first had a girl, then a boy; he first had a boy, then a girl; he had two girls. In this case, because you know he had one boy the two girls option is removed as not applicable. This leaves only three possible permutations of which two boys is only one - hence 1/3.

The article describes the above as the normal mental path people take. However, to properly work the "Tuesday" piece in, the logic takes you down a different path.

L-

[gcomeau]

113. Stil wrong.

And it's because you insist on unnecessarily complicating the calculation by pretending it matters that you weren't told the birth order. But fine, I'll do it your way. We'll pretend as if birth order matters. In that case, let's use more precise terminology.

B(y) = Youngest Boy
B(o) = Oldest Boy
G(y) = Youngest Girl
G(o) = Oldest Girl

So, the possible combinations before we know anything at all about this family except that they have two kids are:

B(y) B(o) = 25%
B(y) G(o) = 25%
G(y) B(o) = 25%
G(y) G(o) = 25%

Ok?

Now, he tells us one of them is a boy. You throw out the G(y) G(o) option. Fine. But that's not the only thing we know. We also know that EITHER B(y) G(o) OR G(y) B(o) is ALSO eliminated. We don't know which one, but we DON'T CARE. Because for the purposes of the question they're both NOT TWO BOYS which is all we're trying to determine the probability of. Half of the NOT TWO BOYS options are now out of play. We know that for a fact. Don't know which one, don't care which one, just care that we're down to either BB or ONE of the BG/GB combos... NOT BOTH of the BG/GB combos, which is what people who say the odds at this point are 1 in 3 are wrongly claiming.

Which means the odds of it being BB is 50/50.

The faster way to get here of course is to just recognize what your variable is from the beginning. There's exactly one thing you don't know after he asks the question that actually matters. The sex of the second child. It's either a girl or a boy. Voila, you can jump straight to your 50/50 answer without twisting your mind into a pretzel like this.

[muriel_volestrangler]

117. No.

"Now, he tells us one of them is a boy. You throw out the G(y) G(o) option. Fine."

OK so far.

"But that's not the only thing we know. We also know that EITHER B(y) G(o) OR G(y) B(o) is ALSO eliminated. We don't know which one, but we DON'T CARE."

No. This is where you go wrong. We throw out (aka 'eliminate', 'exclude') G(y) G(o) because we know it's impossible. The remaining 3 options are all possible. We cannot throw any of them out. We know one of them is the actual case. But we have no way of knowing which one. You cannot just decide to throw one out becuase you think it's too similar to another one for your taste.

Can I take you back to dice? Do you agree that if you throw 2 dice, you are twice as likely to get a 2 and a 1 than a double 1? Please, answer whether you accept that.

Staying on dice, if you threw a die, and I asked "what is the chances of a 6", you'd say "1 in 6". You said "Because for the purposes of the question they're both NOT TWO BOYS which is all we're trying to determine the probability of"; if I said "for the purposes of the question, 1, 2, 3, 4 and 5 are all NOT SIX which is all we're trying to determine the probability of", and then I proceded to say that chances of throwing a 6 were therefore 50/50, you'd see I was talking rubbish. Wouldn't you?

[gcomeau]

118. Yes.

Answer me this. We're playing poker.

You are currently holding a 2,3,4,and 6.. and have one card to draw. You're trying to draw that 5 and finish off your straight.

What are the odds you're drawing the 5?

What are the odds if I tell you one of the 5's in the deck is out of play.. EVEN IF I DON'T TELL YOU WHICH ONE? Are they the same? Do you still have 4 possible straights you can draw into just because you don't know which 5 you can no longer draw? Hmmm? Of course they're not. Knowing one of the 5's is gone immediately changes the odds and it doesn't make one single tiny little difference that you don't know which one it was. You have to account for it being gone. There are now only 3 possible straights for you to draw into. (If you disagree with this, I really would like to play a cash game of poker with you.

Now, apply the same principle to this problem. YOU HAVE JUST BEEN TOLD ONE OF THE GIRLS IS OUT OF PLAY. That changes the odds of "drawing" one of the boy-girl combinations. You cannot still draw into both of them, only one. That you don;t know which one is *completely irrelevant* to the calculation of the odds. Get it?

And stop asking if I accept odds I already told you I accepted. Particularly in response to a post where I just posted those odds. The problem isn't any disagreement over what the initial odds are before we are told what one of the kids is. It's how you are incompletely accounting for being told that information.

[muriel_volestrangler]

119. "you have just been told one of the girls is out of play" - what on earth does that mean?

I cannot fathom how you think that has something to do with anything we've discussed so far. There is not some finite number of girls, and neither have we been told that any girl has turned up already, in any form. All we know about girls is that the pair of children is not 'two girls'.

Because there is not a limited number of boys or girls to 'draw', your poker analogy fails completely. However, the analogies with dice are standing up - because each die has a fixed number of possibilities it can land on, and dice do not affect how other dice land.

Actually, you've never said anything about accepting the odds of dice throws. If you mean that you accepted that the odds, when we know nothing about the pair of children, are 25% for 4 different permutations, then that is indeed a start, as I said. But you still haven't faced up to the way you are 'throwing out' a possibility when you should not be.

So now I can take it you've accepted that, when throwing 2 dice, a one-and-two combination is twice as likely as a double one. What we have with the girls and boys is exactly the situation of: "I have thrown 2 dice. The total is either 2 or 3. What are the chances it is 2?" "The total is either 2 or 3" eliminates all possible combinations except one-and-two or a double one. Since the one-and-two combination is twice as likely as a double one, then the chances the total is 2 are 1 in 3.

"The problem isn't any disagreement over what the initial odds are before we are told what one of the kids is. It's how you are incompletely accounting for being told that information."

Your problem is that, having established those initial odds, you then 'throw out' a possibility, with no justification at all. I am not "incompletely accounting for being told that information"; the information is that the pair of children is not '2 girls', and so I eliminate the '2 girls' possiblity.

[gcomeau]

127. That means...

...that when he told you one of the kids was a boy, he was also telling you one of the kids was NOT a girl. And you are ignoring that.

And yes, there ARE a limited number of girls to "draw". You started with two possibles... an older girl sibling and a younger girl sibling. ONE of those is no longer possible once you are told one of the kids is a boy. If the boy that was identified is the younger sibling then that eliminates the possibility of having a younger girl sibling. Same goes if the boy is the older sibling, that eliminates the possibility of an older girl sibling.

[muriel_volestrangler]

134. No, I am not ignoring that "one of the kids was NOT a girl"

That is exactly equivalent to "the pair of children is not '2 girls'", and we use that to eliminate the previous 25% possibility that it is 2 girls.

"If the boy that was identified is the younger sibling then that eliminates the possibility of having a younger girl sibling. Same goes if the boy is the older sibling, that eliminates the possibility of an older girl sibling." And all you are saying here is "if the younger sibling is a boy, then the younger sibling is a boy; if the elder sibling is a boy, then the elder sibling is a boy". It is a tautology.

You said "you started with two possibles... an older girl sibling and a younger girl sibling" - but that is not the complete list of possibilities (there is the possibility that neither child is a girl). Your tautology that follows that does not get us any further - it says "if (A is true), then (A is true); if (B is true), then (B is true)".

Really, the poker analogy is not working. We are not 'drawing' a new outcome; we are revealing an existing situation, bit by bit. A deck of cards will not give a good analogy - because the moment you know what one card is, you get more information about all the other cards you have not yet seen.

[gcomeau]

178. Yes you are.

Really, the poker analogy is not working. We are not 'drawing' a new outcome; we are revealing an existing situation, bit by bit.

Fine, you have been dealt a 5 card poker hand, you have looked at 4 cards and one is laying face down on the table. I have also been dealt a 5 card hand.

The 4 cards you have looked at are as stated before.

Oh look, now we're just "revealing an existing situation".

Now, according to your approach to odds calculation we have these four possible straights:

1. 2,3,4,5(spades), 6
2. 2,3,4,5(hearts), 6
3. 2,3,4,5(clubs), 6
4. 2,3,4,5(diamonds), 6

And you calculate your odds of actually having each one. And you add them together. And those are your total odds of having a straight. Fine so far.

Now, I tell you my hand has a 5 in it. I do not tell you which 5.

According to the approach you have argued we have to use:

You can't eliminate combo 1, because I didn't tell you it was the 5 of spades.
You can't eliminate combo 2, because I didn't tell you it was the 5 of hearts.
You can't eliminate combo 3, because I didn't tell you it was the 5 of clubs.
You can't eliminate combo 4, because I didn't tell you it was the 5 of diamonds.

So the fact that I am holding a 5, or 2 fives, or even THREE FIVES has no impact on the odds you're holding a straight. Those are ALL still in play, so the odds of ALL of them get added together.

I am quite certain you know that's ridiculous.

What you are not understanding here is that you don't have to eliminate a specific combo. But you DO have to take into account that the cumulative odds of you holding one of them is lower because one of those 5's is not possible any more even though you don't know which one it is.

They are still EACH possible, but they are not ALL possible. And you need to understand the distinction.

Returning to the BG/GB situation.

YES, it could still be a BG situation.

YES, it could still be a GB situation.

They are both still EACH possible. But they are not still BOTH possible.

Either the boy that was identified was the first one and that one being a girl is impossible, or the boy that was identified was the second one and that one being the girls is impossible. ONE of those slots being a girl is now impossible even if you don't know which one that is. So the odds of it being one of those two combinations are now cut in half... and we are down to a 50% chance it is ONE of BG/GB, and a 50% chance it is BB.

[muriel_volestrangler]

179. I think you're mixing up 'possible' and 'true'

You say "They are both still EACH possible. But they are not still BOTH possible."

Look at what you wrote again. What do you really mean? 'Both' appears in both sentences. What is the difference in meaning between "they are both still each possible" amd "they are still both possible"? I think that the two sentences are equivalent.

Your explanation after that seem to point to you meaning "They are both still EACH possible. But they cannot BOTH be true." In other words, you are saying that you know that at least one of them will turn out to be false, when all is revealed. But we can say the same thing about BB too. We know that BB and BG cannot both be true; but both are still possible, as is GB. When you say "ONE of those slots being a girl is now impossible even if you don't know which one that is", you're wrong - both are still possible. We don't know which is correct, but either one might be true. So your following "So the odds of it being one of those two combinations are now cut in half" just doesn't apply.

Now, I tell you my hand has a 5 in it. I do not tell you which 5.

According to the approach you have argued we have to use:

You can't eliminate combo 1, because I didn't tell you it was the 5 of spades.
You can't eliminate combo 2, because I didn't tell you it was the 5 of hearts.
You can't eliminate combo 3, because I didn't tell you it was the 5 of clubs.
You can't eliminate combo 4, because I didn't tell you it was the 5 of diamonds.

Correct; one cannot eliminate any 'combo' (I'm not keen on the term 'combo' here, because it may imply 'combination', which has a specific mathematical meaning; and if we were being specific about the suit of the 5 we are contemplating, I think we should have been specific about the suits of the cards we hold in our hand. But if it's meant as 'possible hand', then I'm OK with this).

So the fact that I am holding a 5, or 2 fives, or even THREE FIVES has no impact on the odds you're holding a straight.

No; and that form of thinking is nothing like the reasoning I have used in this thread. We work out the odds based on how many cards remain unseen, and how many 5s there are available. When you know the number of fives that still might be that one card has changed, then the odds change. But in the boys question, we have not changed 'the number of girls that might be a child'. Even when you are dealing with the 'Tuesday' form of the question, a girl born on a Monday is exactly equivalent to one born on Tuesday, and all the other days, for the purposes of the question. What the information does is rule out the possibility that both children are girls.

Notice that, in the 'boys' question, there are 2 children that are not yet known. All we know (in the simple form of the problem, that doesn't mention Tuesday) is that they are not both girls. If you want to construct a poker analogy, then a question involving the discovery of 2 cards is needed, in which we know they cannot both be a particular rank (eg when it is known 3 cards of that rank have already been seen somewhere).

[gcomeau]

182. Ahem...

No; and that form of thinking is nothing like the reasoning I have used in this thread.

It is exactly the reasoning you were using.

We work out the odds based on how many cards remain unseen, and how many 5s there are available.

Yes! Exactly!

But when you start dealing with the kid situation you refuse to properly take into account how many kids are left unseen. That being ONE. You keep insisting on calculating the odds based on random 2 kid combos when the gender of one of the kids is no longer a random variable. Instead you keep doing this:

Notice that, in the 'boys' question, there are 2 children that are not yet known.

No there is not! There are 2 kids total, and we've been told the gender of one, leaving ONE unknown. We don't know whether it's the oldest kid or the youngest kid whose gender is unknown, but we know it's ONE. Not TWO.

[Lithos]

120. You are adding an additional constraint which is simply not there

There is no knowledge that the boy was born either first or second - just that there are two children.

Given two children, there are four possible outcomes, these are:

1) Boy born first, followed by a second boy (BB)
2) Boy born first, followed by a girl (BG)
3) Girl born first, followed by a boy (GB)
4) Girl born first, followed by a girl (GG)

So, before the introduction of any constraints/knowledge of the outcome, you would have the following odds:

25% of two boys, 50% of a boy and a girl, and 25% of two girls.

However, because we *KNOW* that there is at least one boy, this means that there is zero chance for two girls. This means that there are only three available options to chose from -

Alternatively, consider that we wrote down those four options above on four pieces of paper. But, because we know there is at least one boy you have to remove the two girls option from consideration. So instead of 4 options, there are only three - BB, BG and GB. This means that you have a 1 in 3 chance of having two boys.

[gcomeau]

131. No, I'm not.

Nothing I just wrote involved knowledge of whether the boy was born first or second. Try reading it again, and refer to post 129 for the explanation of the mistake you're still making. You're ignoring half the information you've been provided with. Being told one of the kids is a boy doesn't JUST tell you there's a boy in the mix.

[dnbn]

124. You cannot eliminate either B(y)G(o) or G(y)B(o).

Both are not excluded by having at least one boy and therefore they are possible combinations. You may not count them as a positive case of two boys but they must be counted against the probability. Only G(y)G(o) is excluded - and only because the condition of at least one boy excludes that. We are left with 3 combinations that have a total non-conditional probability of 0.75. This is the un-conditional probability of having at least one boy. The non-conditional probability of B(y)B(o) is 0.25. The ratio between them (latter divided by former) is the probability of having two boys GIVEN that there is at least one boy. This a basic rule in probability theory- the conditional probability of x given y is:

p(x|y) = p(x,y)/p(y).

In our case x is the probability of having two boys without any condition (i.e. 0.25) and y is the probability of having at least one boy (=0.75 because it is equal to 1 minus the probability of having two girls). In case you are not familiar with the notation, p(y,x) is the probability of the union of the two events x (two boys) and y (at least one boy). Since the former (x) is a subset of the latter (y) we have by definition p(x,y)=p(x).

[gcomeau]

129. You not only can, you have to.

Both are not excluded by having at least one boy and therefore they are possible combinations.

No, EACH is a possible combination... but not BOTH.

Let's play poker.

You are holding a 2,3,4 and 6 and are drawing one card. You need a 5 to complete the straight. For the sake of simplicity we're playing with a single deck, with 42 cards remaining in it.

The possible straight combinations you can get are:

2,3,4,5(spades),6
2,3,4,5(hearts),6
2,3,4,5(diamonds),6
2,3,4,5(clubs),6

Agreed?

And your odds of getting one of them given what you know is 4/42. Agreed?

Now... I tell you that the other hand that was dealt in the game had a 5 in it. But I do NOT tell you WHICH 5.

So, according to your logic, since I didn't tell you which 5 is out of play and you can't eliminate any specific straight up there as being possible, you can still score a straight 4 ways and your odds haven't changed. I trust you can tell THAT is wrong? Because it does not matter if you don't know WHICH combination just became impossible. It only matters that you know that ONE OF THEM DID. Your odds just dropped to 3/42.

Now let's go back to our Boy/Girl matrix:

We started with these possibles:

B(y) B(o) 25%
B(y) G(o) 25%
G(y) B(o) 25%
G(y) G(o) 25%

When we were told one of the kids was a boy, that did NOT just tell us we could eliminate that last one. It also told us we are now in one of two possible situations:

1: The boy that has been identified is B(y). If that is the case G(y) G(o) AND G(y) B(o) are impossible. Knock them both off, odds of having two boys is 50/50.
2: The boy that has been identified is B(o). If that is the case G(y) G(o) AND B(y) G(o) are impossible. Knock them both off, odds of having two boys is 50/50.

Follow? Now we don;t know which scenario we're in, but that doesn't matter. Whichever one we're on ONE of those BG/GB combos is no longer available.

[muriel_volestrangler]

135. See #83, again

We know that the sex of each child is random. But we don't know that, in your situation (1), the chances of B(y) G(o) and B(y) B(o) are equal. Because we don't know if the parent was forced into revealing the sex of the younger child as a boy (since the elder one is a girl).

What we do know is that the chances that the family already is a boy and a girl are twice that of 2 boys. Both of those combinations fit "one of the children is a boy"; their relative probabilities cannot change.

Do you understand simple programming? If so, I have written a simple JavaScript program that will go through the steps, and show the ratios come out approximately as 2:1. I'll post it if you want, and then you can run it (then you don't have to flip a lot of coins).

[gcomeau]

141. Post #83 doesn't address this.

We know that the sex of each child is random. But we don't know that, in your situation (1), the chances of B(y) G(o) and B(y) B(o) are equal

Yes we do. Now you're denying this:

GG - 25%
BG - 25%
GB - 25%
BB - 25%

You are constantly asking me if I agree that those are the initial odds. Now you just throw them out the window. 25% = 25%.

Because we don't know if the parent was forced into revealing the sex of the younger child as a boy (since the elder one is a girl).

Irrelevant. Nothing I just posted required we know that.

What we do know is that the chances that the family already is a boy and a girl are twice that of 2 boys.

We knew that BEFORE they identified one of the children as a boy, fixing the probability of the gender of one child at 100%. Being given that information CHANGES THE ODDS.

EDIT: Ok, let's try this yet one more way. This time we're going step by step to identify exactly the point where you stop calculating probabilities the way I do. New situation, I tell you this information:

"I have a boy. He has one sibling."

Now my first question for you is, what are the odds he has an older sibling, and what are the odds he has a younger sibling?

Leave gender out of it for now. What are those odds? Assume no twins.

[muriel_volestrangler]

146. The relevance from #83 was this:

"In scenario A, you cannot assign equal probabilities to "___B___ ___B___" and "___B___ ___G___". If both children were boys, then the choice between the revelation in Scenario A (that the child on the left is a boy) and that in B is equal, and so each scenario is equally likely to be followed. But if the child on the right was a girl, then Scenario A had to be followed, which makes that scenario twice as likely to come about from that combination of children. And similarly, if the child on the left was a girl, then Scenario B had to be followed. So the probabilities are:

Scenario A:
X or 2X
Scenario B:
X or 2X"

Not only am I not denying

"GG - 25%
BG - 25%
GB - 25%
BB - 25%"

, but those probabilities are central to my argument. GG is eliminated as a possibility. The other 3 possiblities remain, and thus that of BB is one third of the remaining possibilities. It is you who wants to unilaterally delete one of the possibilities, although it is still possible.

Because we don't know if the parent was forced into revealing the sex of the younger child as a boy (since the elder one is a girl).


Irrelevant. Nothing I just posted required we know that.

No, it's relevant (if you insist on proceding the way you want to). See the argument from #83.

We knew that BEFORE they identified one of the children as a boy, fixing the probability of the gender of one child at 100%. Being given that information CHANGES THE ODDS.

The information eliminates the GG option; it does nothing to change the odds of the others.

For your step-by-step calculation, I highly suspect you are going to go through part (d) of Exercise 1 from the Canberra presentation (page 14; while the New Scientist wording fits part (a), page 7). And, having examined that properly now, I think I may be able to explain it. Notice that this is a slightly different situation. You have introduced a 3rd random event, which happens after the birth of the children - "which child did I examine?" (by asking "what are the odds the boy has an older sibling, or a younger sibling?", you are saying "I found a boy, by random examination; what are the odds I examined the younger child, or the older one?"). Yes, the odds of an older sibling are 50%, and 50% for a younger one.

This has implications for how many possibilities are eliminated before we work out the odds. Under the original question, the only possibilities eliminated are the GG pair - a 25% chance. But, if we pick a child at random, see that he is a boy, and say "we picked a child at random, and found he was a boy", then while we still have the possibility of having picked the boy half of a mixed pair, we have already eliminated, by this time, the possibility that our random examination found the girl half of a mixed pair.

And these two questions are not the same. The setup in the New Scientist article is "I have two children. One of them is a boy", not "I have two children. I picked one of them at random, and he is a boy".

I'll try to illustrate. Let the '^' indicate the child examined.

pair        BB      BG      GB      GG     (probability each 25%)
           /  \    /  \    /  \    /  \
examine   BB  BB  BG  BG  GB  GB  GG  GG   (probability each 12.5%) 
          ^    ^  ^    ^  ^    ^  ^    ^
"he is
a boy"    Y    Y  Y    N  N    Y  N    N  

So, with this random choice of child, you have thrown away 4 out of the 8 possibilities; and 2 out of the 4 remaining are a pair of boys. But, as I have said, this is not what the original problem was.

In reply #138, I posted some javascript code that runs as an HTML page, to do 10,000 tests of the setup in the New Scientist question. I've now done the same for "I chose a child at random, and he is a boy". Here it is:

<html><head>
</head>

<body>
<script language=JavaScript>

function randInt(max)
{
 return 1 + Math.floor(max * Math.random());
 /* a random decimal number between 0 and 0.99999...., multiplied by max to be between 0 and 'not quite' max */
 /* then take the whole number part, and add 1 to be 1,2,...,max */
}

/* we will call a random result of '1' a boy, and '2' a girl */


mixedTotal = 0;
allBoysTotal = 0; /* starting values for the running totals */

for (i = 0; i < 10000; i = i + 1) /* do this 10000 times */
{
 elder = randInt(2);
 younger = randInt(2); /* make both random choices, before examining them */

 examinee = randInt(2);

 if (examinee == 1) /* look at the elder */
 {
  if (elder == 1)  /* if elder is a boy */
  {
   if ( younger == 1)  /* so we know both are boys */
   {
    allBoysTotal = allBoysTotal + 1; /* add one to the total of all-boy families found */
   }
   else /* elder is boy, younger is a girl */
   {
    mixedTotal = mixedTotal + 1; /* add one to the total of 'mixed' families found */
   }
  }
  else 
  {
   ; /* elder is a girl; we do nothing */
  }
 }
 else /* look at the younger */
 {
  if (younger == 1)  /* if younger is a boy */
  {
   if ( elder == 1)  /* so we know both are boys */
   {
    allBoysTotal = allBoysTotal + 1; /* add one to the total of all-boy families found */
   }
   else /* younger is boy, elder is a girl */
   {
    mixedTotal = mixedTotal + 1; /* add one to the total of 'mixed' families found */
   }
  }
  else 
  {
   ; /* younger is a girl; we do nothing */
  }
 }
}

document.write("mixed: "+mixedTotal+"; both boys: "+allBoysTotal);
</script>

</body></html>

Try them both, and you'll see that, for the NS question, you get about 5,000 mixed pairs, and 2,500 boy pairs; for the "I chose one at random", you get about 2,500 of each.

[gcomeau]

150. Sigh...

No, I have not introduced any "new event" with that question. I just asked a different question about the EXACT SAME INFORMATION we were presented with in the first place to force you to look at the odds involved in a different way.

The setup in the New Scientist article is "I have two children. One of them is a boy", not "I have two children. I picked one of them at random, and he is a boy".

The second statement is not what I said. I did not say, imply, or otherwise suggest that I picked the boy randomly. But if you prefer:

"I have two children. One is a boy."

There. Now given that statement I am just asking you a different question ABOUT the information than the question that was originally asked. No "new events" have been introduced. No new information has been provided. I just told you to look at it from a different angle. We are dealing with the EXACT same data set.

If you disagree, tell me what exactly the event or the information was that was introduced by asking you to calculate a probability. Asking you to figure something out about the information provided is not an "event".

[muriel_volestrangler]

155. Your question is asking me to figure the order the specified boy comes in

You tell me "I have two children. One is a boy."

You ask me "what are the odds he has an older sibling, and what are the odds he has a younger sibling?"

When you ask me that, you ask me "have I told you about a younger sibling or an older sibling?" If you did not choose which to tell me about at random, then you had reasoning behind your choice, and I cannot look into your mind, and I cannot give odds on a reasoning process of yours. I can only give odds on random events.

If you are going to use the answer for this for anything else, I must reckon the possibilities from now on as the permutations of "what sexes are the members of the family?" and also "what is the placing in age of the boy that was identified?" That is what I did in #146.

Have you read exercise 1, part (d) of the Canberra presentation? Can you see how this is the same setup?

Have you looked at the 2 javascript programs, and tried running them?

[gcomeau]

166. No. It. Is. Not.

It's asking you to calculate the odds given what you know. That's it.

When you ask me that, you ask me "have I told you about a younger sibling or an older sibling?"

No I bloody well do not. If I told you about an older or younger sibling asking you to calculate the odds of there being one would be purely stupid. The odds would be 100% or 0%, because I had just freaking told you if there was one or not.

Have you read exercise 1, part (d) of the Canberra presentation? Can you see how this is the same setup?

Yes I have. No it's not. Unless you want to argue the original question was also, since I'm presenting the EXACT SAME INFORMATION in the EXACT SAME FORMAT.

[muriel_volestrangler]

169. Here's why your question is "have I told you about a younger sibling or an older sibling?"

You tell me: "I have two children. One is a boy."

You ask: "what are the odds he has an older sibling, and what are the odds he has a younger sibling?"

"He has an older sibling" is exactly equivalent to "he is a younger sibling". And "he has a younger sibling" is exactly equivalent to "he is an older sibling".

So your question is exactly equivalent to:
"what are the odds he is a younger sibling, and what are the odds he is an older sibling?"

'He' is the boy you told me about. So the question is
"what are the odds I told you about a younger sibling, and what are the odds I told you about an older sibling?"

OK, I shouldn't have dropped 'what are the odds' from the question phrasing in #155, but I did include that in #146, when I first pointed out that was what your question really is - saying 'examine', because that's how the Canberra presentation puts it.

The main point is you have introduced the new random choice of which boy has been specified, and that is why you change the odds from the new Scientist question. And it's why part (d) has a different answer to part (a) in the Canberra presentation.

[gcomeau]

171. This is getting absurd..

You mean the question is "Is the boy I told you about the younger or older sibling"?

Fine. Yes. That's what we're calculating the probability of with that particular question. So what? That doesn't change any of the information provided! I cannot believe you are making me spell things out to this level of specificity, but here we go.

This was the original question, to the letter:

"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

Agreed?

Now, what is the information provided by the question? What does the question establish that we must take into account?

It's the bolded part. Agreed? The non-bolded part provides NO information. It is just an information query, telling you what you are being requested to determine FROM the information provided. Are we agreed at least that far?

[muriel_volestrangler]

173. Yes, I think we agree that bolded part is the information we are given

In the full question, "I have two children. One is a boy born on a Tuesday" is the information.

[gcomeau]

175. Progress! We have a point of agreement!

Alright, then if the only thing I change is the question part, without touching a single letter of the information part, and without introducing any new information in the content of the question, I'm not changing a single detail of what is being examined. I'm just asking you a different question about the exact same data.

Agreed?

(for example: I can ask "alright, then given those conditions what are the odds of Y?" and nothing has been altered. I cannot ask "then given those conditions what are the odds of Y if X is also true" because "if x is also true" introduces new paramaters.)

[muriel_volestrangler]

180. Yes, but a different question can, of course, have a different answer

so I'm not sure where you think this will get you. But carry on.

[gcomeau]

184. Yes, it can.

Carrying on.

Starting from our initial conditions:

"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

I am now altering the question, and only the question. The question will introduce ZERO new information or constraints.

"I have two children. One is a boy born on a Tuesday. What is the probability that he has an older sister?"

Repeat the question for probabilities he has an older brother, younger sister, and younger brother.

[muriel_volestrangler]

188. You have now introduced a selection

Remember that "one is a boy born on a Tuesday" means "at least one is a boy born on a Tuesday".* When you now say "he", you have selected a particular boy. What if both were boys born on a Tuesday? By asking about 'he', even though a particular boy has not yet been specified, you have made a selection.

We can say, from the original 196 permutations of boys and girls born on different days:
there is 1 permutation in which both are Tuesday boys
there are 7 permutations in which the elder child is a Tuesday boy and the younger child a girl
there are 6 permutations in which the elder child is a Tuesday boy and the younger child a boy born other than on Tuesday
there are 7 permutations in which the elder child is a girl and the younger child is a Tuesday boy
there are 6 permutations in which the elder child a boy born other than on Tuesday, and the younger child a Tuesday boy

There are, adding them up, 27 permutations that are still possible. 13 of them produce a combination of 2 boys.

* Since you have been arguing that "one is a boy. What is the probability I have two boys?" would give 50%, I think you have to be assuming, as I do, that "one is ..." means "at least one is ..."; because if it meant "exactly one is ...", then the answer to "one is a boy. What is the probability I have two boys?" would have to be zero.

[gcomeau]

191. I have introduced nothing but a new question.

Remember that "one is a boy born on a Tuesday" means "at least one is a boy born on a Tuesday".* When you now say "he", you have selected a particular boy.

No I haven't.

What if both were boys born on a Tuesday?

Then he would be ANOTHER boy born on a Tuesday. Nothing I asked rules that, or anything else, out. It was ONLY a question about what we had already been told.

I haven't introduced any selection. The criteria provided in the intital statement established the fact that AT LEAST one child is a boy, born on a Tuesday. We know that a boy, born on a Tuesday, is one of the two freaking children.

My asking what the odds are that that boy has an older sister selects NOTHING. It is just a question about information we have already been provided. I have introduced no new information, or criteria. Can that boy have an older brother? Sure. Can he have an older brother ALSO born on a Tuesday? Yep. Younger brother? Yes indeed. Younger brother ALSO born on a Tuseday? Affirmative.

I. Am. Not. Selecting. Anything. It is just a question.

[muriel_volestrangler]

195. You cannot say 'he' in a question without selecting a 'he'

You can say "what are the odds the younger child is a boy born on a Tuesday?" You can ask "what are the odds the elder child is a boy born on a Tuesday?"

When the information was "I have 2 children. At least one is a boy", that was exactly equivalent to "I have 2 children. They are not both girls". If you said "I have 2 children. They are not both girls. What are the odds he has an elder sibling?" , there could not be an answer. "At least one is a boy" has not specified a boy.

[gcomeau]

200. You can...

...when before asking the question ONLY ONE PERSON HAS BEEN DISCUSSED.

Remember the bolded part? The part we agreed was the information part? The part that says there is a boy born on a Tuesday who is one of two children and never specifies anyone else?

That "he". The "he" that the ORIGINAL STATEMENT, NOT MY QUESTION already selected for identification of his gender and birthdate and number of siblings.

I am not introducing new information by asking you a question about someone that has already been discussed. That there is a boy who was born on a Tuesday who has a sibling is information already introduced.

[Thor_MN]

201. Where this line of reasoning is going to go astray

The odds that the "identified" child has
an older sister 25%
a younger sister 25%
an older brother 25%
a younger brother 25%

Great. But that isn't the original question. That is talking about "the other" child. The orginal question is about both of them.


Here an equivalent game. We each mark 100 $10 bills, me with a G, you with a B. We shuffle them up and lay them down in pairs. All of the GG pairs get taken off the table and go to the house. We now know that each of the pairs has a B $10 bill. Now, since you think the odds are 50-50, you won't mind if I get to choose if I want the mixed pairs or the BB pairs. I'll take the mixed pairs every time and break even over time while you lose half your money.

[gcomeau]

207. It's not going astray at all.

It's just going to an answer that isn't the one you say it is.

The odds that the "identified" child has
an older sister 25%
a younger sister 25%
an older brother 25%
a younger brother 25%

Great. But that isn't the original question.

No it isn't. What it is however is providing an equivalent answer.

Those first two are the BG/GB combinations. The kid that was originally identified as 100%, definitely a boy (who was irrelevently born on a Tuesday) PLUS an older or younger sister.

And those second two are the two variants of the "BB" combination. The kid that was originally identified as 100%, definitely a boy (who was irrelevently born on a Tuesday) PLUS a younger or older brother) Or in other words... the combinations in which the original speaker who was providing us with the initial information HAS TWO BOYS. Now, what was that original question agan? Oh yeah, It was the odds the guy had two boys.

There they are. You just posted them. They're a 50/50 split with the BG/GB combos. I didn't calculate those odds, I made you do it. Did you get the math wrong? You know you didn't.

That is talking about "the other" child. The orginal question is about both of them.

Why yes, it was. You do realize that one child + "the other child" = BOTH children right?

Here is an equivalent game. We each mark 100 $10 bills, me with a G, you with a B. We shuffle them up and lay them down in pairs. All of the GG pairs get taken off the table and go to the house.

[Thor_MN]

216. You have proven nothing. I tossed you a cookie and agreed that your altered scenario has 50-50 odds

I calculated the odds for YOUR question.

And then dismissed it out of hand because it is NOT >THE< QUESTION.

In your question, you are correct, the odds are 50-50 because you have added the concept of identifying one of the children and then start asking about the other.

Problem is, the original question does not identify a child, just the sex of one of them.

My game is fully eqivalent to the original question. We have removed the GG, because we know that all the pairs must contain a B bill. Please explain why half of the mixed pairs of bills should be removed because they aren't possible to be in both orders. You should have lots of practice as you have said it over and over. Please, the bills are sitting right there in pairs, GB, BG, BB. 25 of each, on average. Tell us again why they can't exist? Or are you going to stick with the "LaLaLa, I can't hear you" line of reasoning?

[gcomeau]

219. The scenario is not, in even the tiniest little itty bitty detail....

...altered.

In your question, you are correct, the odds are 50-50 because you have added the concept of identifying one of the children

NO I HAVEN'T

See this? This right here?

"I have two children. One of them is a boy born on a Tuesday.

THAT identified one of the children AS a boy born on a freaking Tuesday. NOT ME. I simply asked you what the odds were of THAT ALREADY IDENTIFED BOY having various combinations of older and younger male and female siblings given ONLY what we were already told.

If you want to cvlaim otherwise, tell me what information I introduced that changed anything?

That there is a boy? We were told that.

That that boy has a sibling? We were told that.

That that boy was born on a Tuesday? We were told that.

Oh, look, we've run out of things my question mentioned, and we were already told every single one of them. NOTHING HAS BEEN ALTERED.

[Thor_MN]

224. Yes, Tell us WHICH boy in the familes with two boys?

You don't know, therefore the child is NOT identified by saying "a boy". To claim that you know the identity of one of the children, you have to be able to unambiguously identify the child in all cases. When the family has two boys, you can not know which one the parent is taking about. If you are quite done with your screaming, and want to claim that you know who the boy is, tell me how you know if he is the older brother or younger brother, "given ONLY what we were already told", one of them is a boy. If it isn't there, YOU introduced it.

Since it isn't there, your example over-represents the number of BB families by twice the amount possible. So we are left with: older brother with a younger sister, a younger brother with an older sister, or a boy with a brother. We can not say older or younger with two boys, because we simply do not know if "a boy" is the older or younger sibling. So there it is, again, 1 out of three for two boys when one does not alter the conditions by introducing additional information.

Oh, look, we removed the condition you added and once again come up with 1 in three for the the odds of two boys.

[gcomeau]

225. What the hell are you talking about?

Did my question introduce that information? Did it presume WHICH child the boy was? No it did not. It simply acknowledged that the boy was one of two children, which we had already been told.

So what did I CHANGE?

You don't know, therefore the child is NOT identified by saying "a boy".

How can you possibly not get this?

If I tell you I have two children one of which is a boy born on a Tuesday, I have just identified ONE OF my children as being a boy born on a Tuesday. I have just told you that a boy born on a Tuesday who is one of my two children exists.

If you are quite done with your screaming, and want to claim that you know who the boy is, tell me how you know if he is the older brother or younger brother,

Where the hell did I say which one he was? What are you even talking about? You're spouting nonsense. The entire damn point of the question is that we do NOT know so I'm asking the odds of it being either one. How do you get from there to claiming I said I know which one it is??????

[Thor_MN]

226. Right here.

"And it's because you insist on unnecessarily complicating the calculation by pretending it matters that you weren't told the birth order. But fine, I'll do it your way. We'll pretend as if birth order matters. In that case, let's use more precise terminology.

B(y) = Youngest Boy
B(o) = Oldest Boy
G(y) = Youngest Girl
G(o) = Oldest Girl"

Those are your very own words, no?


You keep insisting that you know the identity of the boy in a 2 boy family. With a mixed pair family, "a boy" is either the older or younger and is identifiable. With a 2 boy family, you have no information. By insisting on including a condition you can't glean from the given information, you are introducing new data. You are being forced to try to sneak it in, because it's the only way you can arrive at your 50% solution, other than your proposal wiping out half the mixed sex pairs, for some reason which you won't explain.

The only possibilities (again) are: older girl/younger boy, older boy/younger girl, a boy with a brother. So 2 boys are 1 in three. It logically follows from the normal distribution of sibling pairs, 25% GG, 50% mixed, 25%BB. Remove the GG, which I think we all understand, you are left with a two to one ratio of mixed siblings to 2 boys. How can you possibly not get this?



You also keep ignoring my questions. I've noticed that you have abandoned the thread where I ask why you have to remove half the mixed pair of cards or bills. Please explain, (again) why half the mixed pair siblings have to be removed, for a reason given in the original question, that does not also exclude the 2 boy families. Every explaination you have given, also excludes the 2 boy families. (reminder, you have maintained it can't be both BG and GB. (that works in any given family, so we will consider it) It follows that it also can't be both BG and BB, and GB and BB are impossible as well, so your 50% answer isn't right even by your own logic.)

If you want to go back to introducing "the other", a different question from the original, yes, the odds of a second boy is 50%. Left with just the original question it's one in three.

[gcomeau]

227. Uh-huh...

"And it's because you insist on unnecessarily complicating the calculation by pretending it matters that you weren't told the birth order. But fine, I'll do it your way. We'll pretend as if birth order matters. In that case, let's use more precise terminology.

B(y) = Youngest Boy
B(o) = Oldest Boy
G(y) = Youngest Girl
G(o) = Oldest Girl"

Those are your very own words, no?

Those are my words where I'm identifying ALL the possible individual positions of gender and birth order in which a child in a 2 child family could fall.

Now, where the HELL do you think I claimed to know which one of those this kid was, or required that to be known?

You keep insisting that you know the identity of the boy in a 2 boy family.

What I keep insisting is that the fact that one child is a boy, who has a sibling, and was born on a Tuesday has been identified.

Which it has. Or were we not told there was a boy?

Were we not told that boy was one of two children?

Were we not told that boy was born on a Tuesday?

The only possibilities (again) are: older girl/younger boy, older boy/younger girl, a boy with a brother. So 2 boys are 1 in three.

That's ridiculous.

You can either divide the siblings up by younger and older.... or not. I don't actually care since it will give you the same (correct) answer either way.

But you're just blatantly applying a double standard of categorization to double the odds of a sister. If the sibling happens to be female ... oh WELL THEN!!!! They could be younger, or they could be older, and that's TWO categories!!!

But if the sibling happens to be male... well then they're just a sibling. And just you never mind about that younger and older thing, "brother" is only one category!

Because, you know... the laws of probability are gender dependent...

[Thor_MN]

228. If you don;t understand probability, you are NEVER going to get it.

"What I keep insisting is that the fact that one child is a boy, who has a sibling, and was born on a Tuesday has been identified. "

not in the case of two boys, you can not know if the boy is the older or younger, he has NOT been identified.


"Because, you know... the laws of probability are gender dependent..."

The odds of a mixed pair are the same as having a same sex pair. I think we can agree on that? That the number of boys equals the number of girls? Are we OK with that? Except in this situation, because of the initial conditions, we remove the case of two girls, leaving the ratio of same sex (two boys) at half that of a mixed pair. 2 mixed pair to 1 pair of boys. One in three odds of having two boys. The odds in this question are altered from the odds for all families, because we have information that excludes some of the families.

There is not a damn thing it the world that is going to change that 2 to 1 ratio, short of environmental factors that alter the ratios of the sexes. If you can't see that, you are never, ever going to get it.

If your answer of 50% were correct, there would be three boys born for every two girls. Outside of the conditions of this question, in 100 families, on average, there would be 25 pairs of girls, 50 pairs mixed, 25 pairs boys. Do you agree?

Because the question gives information that removes the pairs of girls from possibility, leaves 50 pairs of mixed, 25 pairs of boys. That's one in three for a pair of boys.

You maintain the odds are 50%, so there would have to be 50 pairs of boys to balance the 50 mixed pairs.

Step back to all families, adding the pairs of girls back in (we really didn't kill them) we have in your world 150 boys and 100 girls. (50 girls from the 25 GG pairs, 50 girls from the 50 mixed pairs, 50 boys from the 50 mixed pairs, 50 boys from the 25 BB pairs and 50 more boys from out of your imagination) Granted the actual ratio of births is slightly higher for boys, but no where near your scenario. Wait, you aren't from China are you? There are a lot of missing girl children there (I think that they may really kill them), what with the 1 child rules and the desire for a boy, maybe that's where you come up with your skewed views of what the odds should be.

Oh, wait, we don't have to add boys to make the odds 50% we have to remove half the mixed pairs. For some reason that you can't or won't explain. So we are left with you being wrong, unable to explain why the odds should be 50%.

[gcomeau]

230. For the love of God...

"What I keep insisting is that the fact that one child is a boy, who has a sibling, and was born on a Tuesday has been identified. "

not in the case of two boys, you can not know if the boy is the older or younger, he has NOT been identified.

I asked this last time. You ignored me. I'm asking again.

Where did I either claim to know or require it be known that the boy was the older or younger?

The odds of a mixed pair are the same as having a same sex pair. I think we can agree on that?

If the only thing we know about the family is the existence of two children in it.

That is NOT the only thing we know about the family...

Oh, wait, we don't have to add boys to make the odds 50% we have to remove half the mixed pairs. For some reason that you can't or won't explain.

I have explained it a dozen times in this thread. I will now do it again.

I am going to go through this step by step. Each step will conclude with a question for you. Kindly answer them if you bother responding.
This is what we were initially told:

"I have two children. One is a boy born on a Tuesday."

Given those two statements this is what we know:

1. The child who is a boy born on a Tuesday is one of two children.

Do You Care To Dispute That Fact?

2. Being one of two children, he is EITHER the oldest, OR the youngest sibling. The scenarios in which he occupies ONE OF THOSE TWO categories total 100% of all possibilities.

Do You Care To Dispute That Fact?

3. Given that there are no biological biases on gender based on birth order, there is a 50% chance he is the youngest sibling, and a 50% chance he is the oldest sibling.

Do You Care To Dispute That Fact?

Assuming you have not yet denied any of the mind numbingly obvious things I have stated to this point, this is the current matrix of possibilities before us:

Scenario 1: (50%): B(discussed)/? - Where the boy we are discussing is the younger brother of an older sibling of unknown gender.
Scenario 2: (50%): ?/B(discussed) - Where the boy we are discussing is the older brother of a younger sibling of unknown gender.

Moving on...

4. The other sibling is either male or female.

Do You Care To Dispute That Fact?

Assuming you haven't introduced "hermaphrodite" as a statistically significant possibility we should be considering at this point, that now leaves us with 4 possibilities.

Scenario 1a: B(discussed)/B - Wherein the boy we have been discussing is the younger brother, and has an older brother.
Scenario 1b: B(discussed)/G - Wherein the boy we have been discussing is the younger brother, and has an older sister.

Scenario 2a: B/B(discussed) - Wherein the boy we have been discussing is the older brother, and has a younger brother.
Scenario 2b: G/B(discussed) - Wherein the boy we have been discussing is the older brother, and has a younger sister.

Do You Care To Dispute That Fact?

5. ONCE AGAIN, given that there are no biological biases on gender based on birth order, the 50% odds we are in scenario 1 are now evenly divided between 1a and 1b as follows:

Scenario 1a: 25%
Scenario 1b: 25%

And the 50% odds we are in Scenario 2 are now divided between 2a and 2b as follows:

Scenario 2a: 25%
Scenario 2b: 25%

Do You Care To Dispute That Fact?

Now, in Scenario 1a, and 2a, there are two boys. In 1b and 2b there is a boy and a girl. The question was what the odds of there being two boys was. 25% + 25% = 50%.

As for why, when you start from all the possible two child combination in a family where you do NOT know the gender of one child, you end up cutting the odds of a mixed boy/girl combo in half when you are told that one of the children is a specific gender... we actually just walked through that too but didn't call it out specifically when it happened. See, it happens in steps 3 and 4 when we recognize what possible scenarios still exist once we are told one of the children is a boy. Because when we are told that we are EITHER in a variant of scenario 1, OR a variant of scenario 2... but we cannot be in BOTH. It is one, or the other.

EITHER the boy is the youngest, OR he's the oldest.

IF he's the youngest, then the youngest sibling is NOT a girl. As in, there is a 0% chance of that being the case. And IF he's the oldest then the oldest sibling is NOT a girl (0% chance). And we know that it IS one of those two situations. WE DO NOT KNOW WHICH ONE (See that all caps statement? That one I just wrote? That's me NOT claiming to know WHICH kid is the freaking boy so don't even think about claiming I said we'd identified which one he was), but since either way the possibility of ONE of the children being a girl was reduced from 50% to 0%, the cumulative odds of it being a BG/GB combination were just cut in half.

There, I have now explained it yet again. And if you can't answer "Yes, I dispute this point" to any of the questions where you were asked to do so above, I also just proved it's right by independently calculating the odds of there being two boys without going through the "eliminate one of these mixed pairs" steps.

[Thor_MN]

234. Wow, absolutely stunning

You can not seem to grasp where I'm talking about real life and where I talking your fantasy world. You also have ignored all my questions, but I will answer you point by point and show you where you deviate from reality.

"I asked this last time. You ignored me. I'm asking again.

Where did I either claim to know or require it be known that the boy was the older or younger?"

I have multiple times pointed out that you can not know the identity of the boy in a two boy family, but you are either unable to understand me, or have chosen to ignore it. A little later, I'll lay it out again in more detail so maybe you can grasp that I'm not taking about the mixed pairs.


1. totally agree, never have disputed that.

2. Agree, but it's where you start to lose your grip. The boy does have to be ether the oldest or the youngest. In a mixed pair scenario, you can assume that "a boy" is either the oldest or the youngest. But in 2 boy pair, which is he? This is where you claim that you know his identity, but you can not possibly know. The person could be talking about either the older boy or the younger boy. So half the time in a two boy family, "A boy" is the older, half younger. Can you see how this cuts the 2 boy scenario in half, in relation to the mixed pairs? Wait, don't bother to answer that because you haven't answered any of my other questions.

3. absolutely correct, never once have disputed that.

4 and 5. Your downfall, taking about "the other". It not incorrect, mind you, just not relevant. If the question had anything to do with "the other" child, this would be important. But it does not, so this point is driving you crazy. Here is also where you claim that you can determine is "a boy" is the older or younger. In the 2 boy families, you can not gain that information from what is given, so both 1a and 2a are the same. In your "a" scenarios, half the time the 2 boy parent is talking about the older, half the younger, cutting it's frequency to half the "b" scenarios. These points are your biggest problem. And yes, I absolutely care to dispute your frequencies in 5, they show your lack of knowledge of probability and statistics. The actual frequencies are:

1a 12.5% BB
2a 25% BG
3a 12.55 BB
3b 25%. GB

And before you go all, "that doesn't add up to 100%, you're crazy" remember those two girl pairs we removed? forgot about them didn't you...

If you don't believe those numbers, please, for the love of whatever you hold sacred, answer this question: Do you understand that the normal distribution of pairs equally possible events is 25% AA, 25% AB, 25% BA, 25% BB? I'm using those numbers, you are whacking the hell out of them.


So with the 2 girl pairs removed, we are talking about 75% of the total. Let's refigure so we can calculate the odds in our reduced population (required in the initial conditions) Careful, watch your step, we are shifting frames here. Out of 75, we have 25 BG, 25 GB, (Summed) 25 BB. So one in three odds of a BB pair. Care to dispute that?


OK, onto the shambling mess at the bottom of your post..


"As for why, when you start from all the possible two child combination in a family where you do NOT know the gender of one child, you end up cutting the odds of a mixed boy/girl combo in half when you are told that one of the children is a specific gender..."

WHAT? Where are you going to bury half the mixed pairs? They are mixed pairs, we KNOW that there is one of each, It's the very freaking definition of mixed pair. Why the hell do we have to get rid of half of them? What knowing the sex of one of the children actually allows us to do is remove the matched pair of the opposite sex. That is all it allows us to do. It in no way, shape or form allows us to remove half the mixed pairs. There is a boy, so I must reduce the number of pairs of children where there is exactly one boy by half, because there is a boy? Does that make sense in your world?

"See, it happens in steps 3 and 4 when we recognize what possible scenarios still exist once we are told one of the children is a boy. Because when we are told that we are EITHER in a variant of scenario 1, OR a variant of scenario 2... but we cannot be in BOTH. It is one, or the other."

Sure we can, it is the very definition of possibility. We are not talking just one family, we are taking about populations of families. Notice that "S" at the end of the last sentence, it means plural, as in multiple, as in there are families in both scenarios, hence possible. In any GIVEN family, they can not be in more that one of the states. But that does not alter anything about the possibilities in other families. So it's really just an indicator that you are uncomfortable shifting frames from one family to all families.

"EITHER the boy is the youngest, OR he's the oldest.

IF he's the youngest, then the youngest sibling is NOT a girl. As in, there is a 0% chance of that being the case. And IF he's the oldest then the oldest sibling is NOT a girl (0% chance). And we know that it IS one of those two situations. WE DO NOT KNOW WHICH ONE (See that all caps statement? That one I just wrote? That's me NOT claiming to know WHICH kid is the freaking boy so don't even think about claiming I said we'd identified which one he was), but since either way the possibility of ONE of the children being a girl was reduced from 50% to 0%, the cumulative odds of it being a BG/GB combination were just cut in half."

I have never claimed that you could not identify the child in a mixed pair. I have, multiple times, told you that you can not identify the "a boy" in a TWO BOY family. The fact that a given family can not be both BG AND GB does nothing to the cumulative odds. A given family can not be BG and BB either, so now we have 0% chance!!! A GB famliy can't be BB!!! 0% !!! OMG!!! There are NO two boy families? Doesn't makes sense does it? So it doesn't make sense in your application either. As I pointed out above, your ratios are wrong, the one in three odds are a result of normal distributions of 2 outcomes, taken in pairs.


"There, I have now explained it yet again. And if you can't answer "Yes, I dispute this point" to any of the questions where you were asked to do so above,"

I have disputed your points, pointed out where you are wrong, answered your questions. Now kindly do the same.


"where you do NOT know the gender of one child, you end up cutting the odds of a mixed boy/girl combo in half"

"I also just proved it's right by independently calculating the odds of there being two boys without going through the "eliminate one of these mixed pairs" steps."


Your words, not mine. You removed one of the sets of mixed pairs and then claim that you didn't.

I'll repeat the questions that I'd like you to answer.

1. Do you understand and agree that the normal distribution of 2 outcomes, taken in pairs, is

one quarter AA
one quarter AB
one quarter BA
one quarter BB?

2. Do you see, that when we remove one of those, because we know that one of the outcomes was B, we are left with

one third AB
one third BA
one third BB?

Do you care to dispute either of those "mind numbingly obvious" assertions?

If so, please tell me why, now that we know that one outcome was B, that either AB or BA must also be removed, for a reason that does not also exclude BB.


I will take anything, other than your silence, that does not answer those questions to indicate that you either; really are clueless or are just trying to see how long I will continue to point out your errors.

[gcomeau]

235. Yes, stunning would be the word.

Ok... I just wrote a massive and somewhat derogatory rant ripping every individual thing you just said to tiny little pieces... then I deleted it as counterproductive.

Instead, we're just going to get to the heart of the points you disputed and I'm ignoring everything else. I'm not even going to curse you out for yet again ignoring something I explicitly told you.

Now, I told you in the last post I was calculating the odds using a different approach. There are TWO ways you can approach this problem.

1. Begin from the set of ALL possible 2 child combinations assuming we know NOTHING about the children in question. Then start introducing the information we have been given and playing the elimination game.

2. Begin from the set of all possible 2 child combination GIVEN WHAT WE KNOW.

Both approaches are valid if done correctly, but it's harder to do the first correctly because you have to make sure the implications of the information you are introducing is FULLY accounted for. And that isn't always obvious. This is NOT the approach I was using in the last post.

Now, when you said this:

And before you go all, "that doesn't add up to 100%, you're crazy" remember those two girl pairs we removed? forgot about them didn't you...

You made it clear you lost track of which approach we were using. There was NEVER a 2 girl combination present to be eliminated using the second approach. We constructed the initial matrix of possible child combinations using what we already knew from the beginning. And one of the things we knew was that one kid was a boy and that there being 2 girls was IMPOSSIBLE. The odds of there being 2 girls was 0%, not 25%. They occupied NO space in the total, 100% space of possible combinations we might be dealing with. None.

Do you understand this so far?

[Thor_MN]

236. Yes, I do understand that you can not answer my very simple questions

And are going to keep going LaLALa forever.

Fine, we will do it your way. We know that there are no 2 girls combinations so we have

33.33% BG
33.33% GB
33.33% BB

We can not have anything else, without creating a bunch of BB combinations out of thin air. Remember those questions of mine that you ignored? Do you understand what the normal distribution of 2 combinations, taken in pairs looks like? There are 2 times the number of mixed pairs than either of the matched pairs alone. Said another way, the second set of matched pairs brings us back to even odds. Remove it (because of the conditions of the question) and the mixed pairs outweigh the remaining matched pair 2 to 1.

There, done correctly, your way. Now I have proved it both my way and your way (several times), you have totally ignored my simple questions, so I take it you have absolutely no defense for your skewed numbers.

Or are you going to continue to say that you have to remove either BG or GB because it can't be both, because we know there is a boy in the set? Careful, remember that arguement also removes the BB, because if it's BG it can't also be BB and if it GB it can't be BB as well. Or are you going to insist that there are 3 boys born for every two girls? You have to go down one of those two wrong way streets in order to get your 50%


If you choose to reply, would you kindly do me the favor of answering my questions from my last post? I have answered you questions, you continue to ignore mine. Please, post your derogatory missive, I won't take offense. My money is on that it was counterintellgent rather than counterproductive. Did you find yourself backed into a corner with no escape but to admit that it is 1 chance in 3 for a two boy pair, knowing that a 2 girl pair was not possible? Would you even tell me what it was that I ignored even though you explictly told it to me? It was so important that you felt it was worth mentioning, yet not important enough to say what it was.


Before you go off on another rant, take a normal deck of 52 cards, shuffle them up, lay them down in pairs, set aside the pairs of red cards for the count, count the pairs of black cards, compare with the number of pairs of one red and one black. Restore the full deck, shuffle and repeat. Start a talley, it may take several runthroughs to average things out, do at least 4 or 5 trials. Keep repeating until you can achieve your 50% mixed pairs to 50% pairs of black cards. Come back once you have acheived that goal. See you never.

[gcomeau]

237. Sigh...

Fine, we will do it your way. We know that there are no 2 girls combinations so we have

33.33% BG
33.33% GB
33.33% BB

No, that is not doing it my way. That is doing it your completely incorrect way still. You are STILL drawing up the total possibilities based on not knowing "GG" is impossible, then weeding from there and doing a crappy job of it to arrive at that set of three possibles

I said to start with what you KNOW.

What you know is that there is a boy, with a sibling of unknown gender. Tell me how the HELL you get from there directly to BG, GB, and BB being the three equally possible outcomes?

"Well, he could have a sister... OR he could have a brother... OR he could have a sister?"

What is wrong with you?

As for your questions:

I'll repeat the questions that I'd like you to answer.

1. Do you understand and agree that the normal distribution of 2 outcomes, taken in pairs, is

one quarter AA
one quarter AB
one quarter BA
one quarter BB?

For the umpteenth time, yes, I understand those are the odds WHEN WE KNOW NOTHING ABOUT THE OUTCOME.

2. Do you see, that when we remove one of those, because we know that one of the outcomes was B, we are left with

one third AB
one third BA
one third BB?

Yes, I understand that when we INCORRECTLY, ONLY remove the AA combination and do nothing further to account for the full extent of the information we have been given, that is the wrong answer we get.

Did you want to ask me any more redundant questions I've answered 15 times in this thread?

If so, please tell me why, now that we know that one outcome was B, that either AB or BA must also be removed, for a reason that does not also exclude BB.

We don't remove either one because we don't know which kid is the boy. We adjust their CUMULATIVE odds of occurrence based on the knowledge that one of them became impossible even though we don't know which. The same way we adjust the odds of our holding a straight if we know one of the cards we need is out of play even if we don't know what freaking suit it is. We know ONE of the straights we would like to get has become impossible. Knowing which is unnecessary to adjust the odds to account for that knowledge.

We know ONE of the kids is not a girl. Knowing which is unnecessary to adjust the odds to account for that knowledge.

Before we knew anything about the children all we knew is that there were two of them, and BOTH had a 50/50 chance of being a boy, and a 50/50 chance of being a girl.

"Kid 1" : boy (50%) ... girl (50%)
"Kid 2" : boy (50%) ... girl (50%)

However, we were told one of the kids is a boy. Absolutely, definitely, with 100% certainty a boy. This means that ONE of those two kids just had the odds shift to boy (100%) ... girl (0%)

Now I want you to read this next sentence at least 20 times before replying, because so far in this thread it's like every time I type these words they're invisible.

WE DO NOT KNOW WHICH KID IT IS.

See? See that? That is something I KNOW. So don't fucking tell it to me. But we do not HAVE to know which one it was, we know it happened to one of the two. So we know that the cumulative odds now are:

1 boy (100%) and either 1 boy(50%) or 1 girl (50%)

Which gives us a 50% probability there are two boys.

Or, to once again do this the simple way:

The boy has a sibling. Since having one boy does not apply a biological bias to the gender of other children a mother gives birth to there's a 50% freaking chance that that boy has a brother, and that there are two boys.

[Thor_MN]

238. The word CUMULATIVE does not magically reduce the mixed pair condition.

"Yes, I understand that when we INCORRECTLY, ONLY remove the AA combination and do nothing further to account for the full extent of the information we have been given, that is the wrong answer we get."

So the number of mixed pairs are cut in half, because we know one of them is a boy? THAT MAKES NO SENSE. Having a boy and a girl is the very definition of a mixed pair. Making removing any of the mixed pairs absolutely ridiculous. I could hand my 5 year old niece 100 pairs of red and blue cards, tell her to pick out the pairs that had no blue cards and she would never once chose a mixed set. Why would you?


"Did you want to ask me any more redundant questions I've answered 15 times in this thread?"

This is the first time you have bothered to answer any of my questions.


"We don't remove either one because we don't know which kid is the boy. We adjust their CUMULATIVE odds of occurrence based on the knowledge that one of them became impossible even though we don't know which. The same way we adjust the odds of our holding a straight if we know one of the cards we need is out of play even if we don't know what freaking suit it is. We know ONE of the straights we would like to get has become impossible. Knowing which is unnecessary to adjust the odds to account for that knowledge.'

You seem to have missed an end tag there and launch into some sort of poker "analogy". Here you appear to claim that you must alter the chances of having a girl (card) in your family because you know that I have a girl in my (cards) famliy. Elswhere you have (correctly) argued that one outcome does not alter the odds of another. "given that there are no biological biases on gender based on birth order" Your words not mine. Yet you say that you must alter the odds, based on the knowledge of my (cards) family.


"However, we were told one of the kids is a boy. Absolutely, definitely, with 100% certainty a boy. This means that ONE of those two kids just had the odds shift to boy (100%) ... girl (0%)

Now I want you to read this next sentence at least 20 times before replying, because so far in this thread it's like every time I type these words they're invisible.

WE DO NOT KNOW WHICH KID IT IS."


Wait, YOU were the one going on and on about knowing the kids identity. Now you are shifting gears and saying you don't know which one? That was my argument from the beginning. Starting to agree with me, are you?


"See? See that? That is something I KNOW. So don't fucking tell it to me. "

Actually, I did have to tell it to you since you kept insisting that you knew the identity of "a boy" in the 2 boy families. You conveniently ignored that and tried to distract with the mixed pair, is which case, the identity can be assumed.


"But we do not HAVE to know which one it was, we know it happened to one of the two. So we know that the cumulative odds now are:

1 boy (100%) and either 1 boy(50%) or 1 girl (50%)

Which gives us a 50% probability there are two boys.

Or, to once again do this the simple way:

The boy has a sibling. Since having one boy does not apply a biological bias to the gender of other children a mother gives birth to there's a 50% freaking chance that that boy has a brother, and that there are two boys.

Correct, given that you are back onto "The other" which is not the original question. Given one child is a boy, the odds of the other child being a boy are 50%. THAT IS NOT THE QUESTION. I'm going to repeat that since you seem to keep forgetting it. The odds of a boy having a brother in a two child family are 50%, which is completely irrelevant to the original question. Do you not understand that the original question does not mention "the other"?


In your skewed example, you claim that there are is a 25% chance of BB, 25% chance BG, 25% chance BB, 25% chance GB. For that to be true, you have, with one set of each, 6 boys and 2 girls. You are also arguing that half the mixed pairs must be removed. Your sum arguement makes it even more whacked, saying that there must be 2 BB pairs and only one mixed pair, leading to five boys and one girl. So in total, your arguements calim the odds of 2 boys is 2 out of three, yet you say it is 50%. You are not even consistent with yourself.

[gcomeau]

240. When you're ready to actually read what I write, let me know.

Wait, YOU were the one going on and on about knowing the kids identity. Now you are shifting gears and saying you don't know which one? That was my argument from the beginning. Starting to agree with me, are you?

Did you see the bit where I said every time I say that it's like it's invisible?

Thanks for proving my point.

What I told you is that we had identifed one child as a boy born on a Tuesday but NOT WHICH ONE. And then I told you, and told you, and told you.

Post 225: "Did my question introduce that information? Did it presume WHICH child the boy was? No it did not." -- That's me telling you I wasn't claiming to know which kid the boy born on a Tuesday was. And if that was insufficiently clear later in the same post:

"Where the hell did I say which one he was? What are you even talking about? You're spouting nonsense. The entire damn point of the question is that we do NOT know so I'm asking the odds of it being either one. How do you get from there to claiming I said I know which one it is??????" -- That's me getting pissed off that you are continuing to claim I identified which boy it was and not-exactly-subtly telling you that was not the case.

Post 227: "Those are my words where I'm identifying ALL the possible individual positions of gender and birth order in which a child in a 2 child family could fall.

Now, where the HELL do you think I claimed to know which one of those this kid was, or required that to be known?" -- Does that look familiar? There I am again, telling you rather emphatically that no, I did NOT claim to have identified WHICH kid the boy was.

Post 230: "I asked this last time. You ignored me. I'm asking again.

Where did I either claim to know or require it be known that the boy was the older or younger? -- Oh looky there. There I am AGAIN after you once again ignored what I was typing and kept ranting about we not being able to know which boy it was telling you, AGAIN, that I had claimed no such thing. And then later in the same post:

"And we know that it IS one of those two situations. WE DO NOT KNOW WHICH ONE (See that all caps statement? That one I just wrote? That's me NOT claiming to know WHICH kid is the freaking boy so don't even think about claiming I said we'd identified which one he was),"

Was that unclear? I don't think so, but a lot of good it did me, because here you are... STILL apparently totally oblivious to the fact that I'm not now claiming, nor have I ever claimed to know which kid the freaking boy was.

And the only reason I didn't have to club you over the head with that AGAIN in post 235 in response to your YET AGAIN completely and totally ignoring it was because I deleted the response I wrote after it got extremely heated.

Are you seeing a pattern emerging here?

It's painfully obvious you are paying no attention to what I'm writing. At this point you still clearly don't even know what my argument actually is, you just keep making up your own in your head and arguing with that.

[Thor_MN]

241. Was I unclear? Your idiotic notion that that there are two BB possibilities implies that you know

I have told you many times, you are assuming that you can obtain the identity of "a boy" in a two boy scenario from the original conditions. You have totally ignored ME. You have ranted about that it has to be "the other". You have tried to argue the mixed pair when I clearly told you your flawed assumption was the matched pair. The problem of ignoring the blinding obvious lies within your own eyes. Try reading what I have written, just for freaking once.

Please, just try to obtain your odds with a deck of cards. You like cards, you brought up poker, I assume you have, or can borrow a deck of cards. Take 10 minutes, less time than it took to grab quotes and just attempt to get your 50%. Explain why you need to not count half the mixed pairs of colors because we know that there is a black card in the mixed pair. Or explain why you need to add black cards to the deck. Those are the only ways you are going to get to your 50%, by adding additional conditions, or answering a different question. Just try it.

In a mixed pair one can know the identity, in a 2 boy scenario one can not. You claim you do not care about the identity, but you are, since your 25% BB, 25% BG, 25%BB, 25% GB ABSOLUTELY, POSITIVELY IMPLIES THAT YOU KNOW WHICH ONE THE "A BOY" IS.

In a mixed pair, we know that "a boy" must be either the older or the younger. Make sense? BG, GB
In a matched pair, we have no clue if "a boy" is the older or younger. Understand? BB

3 possibilities, One in three.

My explanations have led to the same answer, the same way, every time. Your ridiculous notions double the number of boys and/or remove a matched pair in order to get to your beloved 50%. Which only works if you do just one of your unexplainable modifications to the conditions set in the question.


You seem to have truely run out of ideas at this point. All you have left is to claim that I am ignoring you, when in fact, you have ignored my point on the exact same thing. I see nothing in your post that claims 50%, so effectively your arguement has deteriorated to "I'm rubber, you're glue."

[gcomeau]

244. Listen genius...

Two POSSIBILITIES means I don't know which one it is. If I knew which one it was the other one would be impossible.

And now I'm done talking to you.

[Thor_MN]

246. That is EXACTLY what I have been telling you, but you refuse to listen

With a mixed pair, since they are not the same, you can infer older or younger. With 2 boys, you can not, so to use your phrasing
"other one would be impossible."

But enough of that, I have found a way to shoot down you method that I think even you can comprehend.

You have maintained starting from the "a boy"

Started with 4 boys, and determine that they had either an older or younger brother or sister.

25% BB
25% BG
25% BB
25% GB

Great, fantastic!!! We have your 50% answer.... except... that's 6 boys. Where did you get the extra 2 boys? Did you clone them? Conjure them? Where did they come from?

So by very simple math, you started with 4 boys and ended up with 6. Put a fork in it, that theory is done. It is impossible. It doesn't work without adding two non-existent boys.


Want to know why? BECAUSE 2 BOY FAMILIES HAVE 2 BOYS!!! Your BB is the same as BB. EXACTLY WHAT I HAVE BEEN TRYING TO TELL YOU. The possibilities are and always will remain, BG. GB. BB. One in three for two boys, knowing that there is a boy, and every arguement you have used is dead. gone. been proven ridiculous. Next time, try to not work towards an answer you have already decided to be true. Let the answer be the last thing to be determined, rather than jiggering the numbers in the middle to achieve the answer you want.

[gcomeau]

247. Have you taken a blow to the head?

First, what part of "I'm done talking to you" did you not get?

Second... a list of just a few possible straights in poker. (h) = hearts, (d) = diaamonds, (s) = spades, (c) = clubs)

2(h),3(h),4(h),5(h),6(h) (yay, straight flush)
2(h),3(h),4(h),5(h),6(d)

...

OMG! I just claimed there was more than one 2 of hearts in a deck!!!!!! AND more than one 3 of hearts! AND more than one 4 of hearts... Clearly one of those two hands is IMPOSSIBLE!!!!! :eyes:

Now go away. I'm not arguing this with you anymore. It's pointless.

[Thor_MN]

248. OK, have you got anything relevant? Or the will to do what you said you were going to, Grant?

5 card poker hands don't tell you anything about 2 children, you are not drawing them from a deck. You can use them to simulate distribution of families, but "adjusting odds" in one hand based on information from a another hand? Ridiculous.

[gcomeau]

250. Already did,. Bye. -nt

[Thor_MN]

251. OK then, bye.

[Thor_MN]

242. And just to add to your pain

"I said to start with what you KNOW.

What you know is that there is a boy, with a sibling of unknown gender. Tell me how the HELL you get from there directly to BG, GB, and BB being the three equally possible outcomes?

"Well, he could have a sister... OR he could have a brother... OR he could have a sister?"

What is wrong with you?"

I know that the the ratio of mixed pair siblings is twice that of a matched pair of boys. You agreed with me...

"For the umpteenth time, yes, I understand those are the odds WHEN WE KNOW NOTHING ABOUT THE OUTCOME"

So, yes, "he could have a sister... OR he could have a brother... OR he could have a sister?"

it's right there, you agreed with it, what is wrong with you?

Does knowing there is a boy somehow alter the families that have boys in them? We are looking at them precisely because they have a boy in the family, it doesn't change the family in any way. When you walk through a grocery store, do half the mixed pair siblings disappear once you recognize one of them is a boy? Back off the dosage of whatever meds you are on, they are messing with your grip on reality.

[qazplm]

189. with all due respect

for the love of God, please stop referring us back again and again to your program.

We understand your program. We disagree with the logic you use to set up your program.

Obviously, you agree with it, so citing that the program you set up with the parameters and logic you have advocated, when we disagree with that logic and those parameters, tells us absolutely zero.

[muriel_volestrangler]

190. But neither of you has come up with an adequate explanation

Between you, you have invented terms like "collapsing probabilities" with no meaning, and have misused the term 'possible' when you mean 'true'. You have not attempted any model we can check, nor have you even reported the results of you tossing coins to model the situation in the problem. You have asserted, in sloppy language, that your calculations are right, and when the errors in them have been pointed out, you have just reasserted them.

You are claiming to have overturned the mathematics of probability, with an argument in which the hols have been shown. The logic in the program is the logic of the situation in the question. You have made no attempt to say where you think the program is wrong. This is the first time either of you has actually said you've read through the program.

[qazplm]

202. again

1st of all, the logic in the program is not THE logic of the situation, it is your logic of the situation.

Now it MAY be correct, but that's for you to prove and saying some mathematicians agree with me (and a brief look online shows not all agree, and a brief look online shows that wording makes a huge difference to some mathematicians including whether or not you use the words "at least" which you've said is irrelevant) is not proof of anything.

So no, we haven't "claimed to overturn the mathematics of probability" we have said that the wording in this word problem asks a different question than you believe it asks, which leads us to a different answer than you prefer, while still using all of the acceptable "mathematics."

I don't have a problem with your use of the words sloppy language or misusing terms, not because I agree with them, but at least that is an attempt to refute our point with your own points. But telling us that we should run your program is absolutely pointless. We know what your program will do, ok, tell you what I will stop here and run your program....

I'm back, ran your program, does what you say it does, to the number, so what? I don't buy the logic put into the program because your program addresses a different question, as I've pointed out before, then is presented IMO and confuses what we are actually looking for. I believe the variables are different than you are.

So running your program is not some panacea to the logical issues here. As a matter of fact, if you "ran your program" looking for the question we propose is being asked, you'd undoubtedly come up with our answer, not yours. Which of course doesn't by itself validate our position because you disagree with what we think the question is.

[muriel_volestrangler]

208. You have in no way explained how you think the program addresses a different question

The logic if the program is very simple: 2 children are each assigned a sex. If at least one of them is a boy, then (if both are boys, add one to the 'all boys' total, otherwise add one to the 'mixed' total'). Repeat that 10,000 times, and show the answer.

That corresponds directly to the puzzle. If you think it doesn't, then point out the step or steps you think are wrong.

Show me any mathematician online who agrees with you, or disagrees with me.

Why do you think that I (or some others on this thread) have not proved our solution? We have talked, many times, about how the information allows you to eliminate just one of the 4 possible permutations (the GG one), and how the others still have chances equal to each other, and how thus the BB permutation has odds of 1 in 3. I have talked about how the 2 permutataions of BG and GB are one combination of a mixed pair, but that combinations do not have equal chances to each other - and have given examples of throwing 2 dice.

I'd remind you that the problem setter does not agree with you about "what question is being asked".

I know the program is not a panacea; but, when you continually refuse to apply the established mathematical methods of working out odds, and insist you own, different, answer is correct, it's worth doing a test to see which method actually comes out with the correct answer.

[qazplm]

212. I absolutely have

and in fact it was in response to you, which you have still not responded to, and I'm not going to repeat it simply because you can't be bothered to find it. I'm not going to keep typing the same answer to you in various places in this thread.

The "logic" of the program requires that there be four variables. We reject that there are in fact four variables being sought. It's that simple. We have told you "many times" how we disagree tha tonly one possible permutation can be eliminated. We have told you many times how we disagree that GB is not equal to BG. We have told you many times even if GB isn't equal to BG, one of them is still necessarily eliminated. We disagree that GB and BG do not have "equal" chances to each other. They absolutely do have equal chances to each other if nothing else.

If you want to talk about the problem setter, I'd remind you that in the very first link you provided, even that "expert" said that the original problem setters language was problematic at best. I can intend to say the weather is cold and use the word lukewarm, that doesn't mean that others should simply ignore the plain meaning of the word lukewarm.

No, it is not worth doing a test.

I. HAVE. ALREADY. TOLD. YOU. I. HAVE. DONE. YOUR. PROGRAM.

I. AGREE. IT. PUTS. OUT. THE. ANSWER. TO. THE. QUESTION. YOU. ENTERED.

I. DISAGREE. THAT. THE. WORDING. OF. THE. PROBLEM. EQUALS. THE. QUESTION. YOU. ENTERED.

Your own link shows that such a disagreement is recognized by other experts due to the wording of the problem. A quick look on wiki also shows the very thing.

I've even said, and I don't know if others have said this, that even if the rules of probability say this is true, you don't check logic at the door and say, well, if probability says it, it must real life. You have two completely independent binary variables, neither of which has a thing to do with the other. You've determined one. The other one is going to be binary regardless.

This is DIFFERENT from the monty hall problem. The chances were always 1/3rd you picked the right door. The chances were always 2/3rds one of the other two doors was the right door. And there is only one variable involved, which one has the car. So it makes sense that switching gives you better odds (depending on the way the question is asked and the knowledge of the host. Other versions the odds are different. Kinda like in this case so to that extent they are similar).

Otherwise, you don't have one variable but two in this case, and they are not related to each other like the variables in the MHP are.

[dnbn]

137. The analogy between genders and cards is wrong.

You can have two girls but you cannot have two 5 of spades.

[gcomeau]

140. No it isn't.

Who said anything about 2 fives of spades? I said 4 fives, different suits.

And 2 girls DIFFERENT BIRTH ORDERS.

One of those girls is "removed from play" when you are given the information that either the youngest or the oldest sibling is a boy. Thus.. NOT a girl. You don't have to know which one, you just have to know it was one of them. At that point, no, you can NOT have 2 girls. There's only one freaking kid left. It can't be twins or something.

Just like you don't have to know which 5 it was, you just have to know A 5 is out of the deck.

[dnbn]

144. Going back to the genders (don't want to spend time on the cards)

You are saying in post #129:

When we were told one of the kids was a boy, that did NOT just tell us we could eliminate that last one. It also told us we are now in one of two possible situations:

1: The boy that has been identified is B(y). If that is the case G(y) G(o) AND G(y) B(o) are impossible. Knock them both off, odds of having two boys is 50/50.
2: The boy that has been identified is B(o). If that is the case G(y) G(o) AND B(y) G(o) are impossible. Knock them both off, odds of having two boys is 50/50.

The problem is that you don't know which boy was identified and you cannot just assume that a boy was identified. It is a subtle difference but it is a significant one. The same way that saying that a boy was born on Tuesday is significant. In that case there was a specification which changed the the apriori ensemble. When the birth day is not specified we can assume an un-conditional (i.e. before saying "one is a boy") ensemble has 10,000 pairings (you can pick any number). When saying "one is a boy" the apriori ensemble is reduced to 7500 and only 2500 of them are of 2 boys (probability of 1/3=2500/7500). If he said "my first child is a boy" then the apriori ensemble would have bee reduced to 5000 and the probability of having two boys would be indeed 0.5. The same for saying "my second child is a boy". But assuming he didn't say either you are still stuck with 2500 out of 7500.

Splitting the case into two and adding their probabilities must be weighted by the proper conditional probabilities. Those probabilities must be mutually exclusive.

I am willing to continue this with proper formulation using Bayes theorem.

[gcomeau]

145. Still not getting it.

You are saying in post #129:

When we were told one of the kids was a boy, that did NOT just tell us we could eliminate that last one. It also told us we are now in one of two possible situations:

1: The boy that has been identified is B(y). If that is the case G(y) G(o) AND G(y) B(o) are impossible. Knock them both off, odds of having two boys is 50/50.
2: The boy that has been identified is B(o). If that is the case G(y) G(o) AND B(y) G(o) are impossible. Knock them both off, odds of having two boys is 50/50.

The problem is that you don't know which boy was identified

And just why, exactly, do you think there are TWO scenarios up there? Because we don't know that. The fact that we don't know that is fully accounted for by those two scenarios. So no, not knowing that is not "the problem".

and you cannot just assume that a boy was identified.

What? When the guy tells you "I have two children. One is a boy." That's not identifying one of the kids as a boy? REALLY?

Let's do this another way. I'm going to make a statement then I'm going to ask you some odds:

"I have a boy. He has a sibling".

What are the odds he has an older sibling?

What are the odds he has a younger sibling?

[dnbn]

148. It is not identifying the boy as first nor second.

Neither of these two scenarios happened in the sense that you don't have any information assuming the one or the other.

The new problem problem is not equivalent to the pair of boy/girl.

[gcomeau]

149. I. Know. That.

Please stop telling me things I clearly already know as if it makes a difference. Not one single thing I wrote required that whether the boy was first or second be identified. Not one.

And I'm not concerned if you think the new problem is equivalent. It's not intended to be equivalent... YET. Please just answer the question.

"I have a boy. He has a sibling"

What are the odds he has an older sibling?
What are the odds he has a younger sibling?

[dnbn]

152. I don't want to open another front.

With respect to the "I have a boy problem" (without Tuesday) I can show you in one line that the probability for having 2 boys is 1/3, but you may be unfamiliar with probability theory. I will try therefore the statistical approach:

There are 10,000 fathers each with 2 children. You will agree with me, I assume, that 2500 of them have 2 daughters, 2500 of them have 2 boys and 5000 of them have a boy and a girl.

Out of those 10,000 fathers only 7,500 fathers can come to you and say "I have a son".

Those 7,500 fathers include all of the 2,500 of those who have 2 boys.

The percentage of those who have two boys among the ones that can claim "I have a son" is 2500/7500 = 1/3.

This is the direct way of looking at this statistically. It has the equivalent probabilistic aprroach using Bayes theorem. Your way of doing it is in correct by ignoring the normalization factors that Bayes theorem requires.

The main point of the opening post is that any specification, such as "a boy born on Tuesday", affects the groups (or probabilities) of a single boy and two boys differently.

[gcomeau]

160. It is not another front!

It is exactly the same situation. I am just forcing you to look at it from a different angle so you see what the probabilities involved actually are. And for the love of all that is holy, if one more person tries telling me what the breakdown of "x" number of 2 child families is in terms of boys and girls as if I obviously don;t already know that I'm going to lose it. I KNOW.

And my point is that no, the specification of one of the boys having been born on a Tuesday makes absolutely zero difference to the odds of the other sibling being a boy, and there thus being two boys. It's slight of hand. It's misdirection. It's an irrelevant piece of trivia that has no impact on the odds of the specific question being asked... that being the odds of there being two boys given what we've been told.

Now would you mind answering the question I asked so we can move on to the next step and prove it?

[dnbn]

168. I still won't open another front. But I will explain why Tuesday matters.

Specifying Tuesday as the day on each A boy was born reduces the original group of people having one boy by a factor of 7. It does, however, reduce the group of people that have two boys by a smaller factor. The reason for that is that the latter group has two chances of meeting the Tuesday requirement with their two boys.

[gcomeau]

172. Nobody is asking you to open another front.

It's the exact same freaking scenario. I'll even make it to the letter if you want.

"I have two children. One is a boy born on a Tuesday."

There happy?

Now, all I'm doing is asking you a different question ABOUT that situation. Unless my asking a different question about it bends reality this is not a "different front", it's just forcing you to look at the EXACT SAME INFORMATION from a different angle. I'll even leave the "born on a Tuesday" in there and it still won't change a single thing.

Now, what is the probability that boy has an older sibling, and what is the probability he has a younger sibling? Is it, or is it not, 50/50 given what you have been told?

[qazplm]

186. that's exactly my point

logically, one of those two GB/BG is eliminated.

Now we cannot say which one is eliminated, we don't know, but we know the boy cannot simultaneously be the first boy and the second boy, one of those two outcomes no longer will happen.

If birth order doesn't matter, then effectively we can stop saying BB, GG, BG, and GB, and simply say, BB, GG, and Mixed.
Get rid of GG and we are left with two categories, BB and Mixed.

But the reality at the end of the day is, you are dealing with two, completely separate, completely binary probabilities. One of those binary probabilities has been set to 100%. The other, independent probability is 50/50. I see this as different from the Monty Hall problem. I understand the concept of that problem.

Your odds were always 1/3rd in picking the right one of three doors. Those odds don't change because one of the two doors you didn't pick has a goat behind. The odds were always 2/3rds that one of the doors you didn't pick has the car behind it. So therefore 2/3rds of the time, picking the other door works.

One variable, one probability, makes sense. This is two separate probabilities, completely independent of each other, and a different problem.

[dnbn]

103. BG and GB have the same outcome with respect to having 2 boys or not.

However, they are distinct events when you give a probability of 0.25 to each combination of two children. If you don't separate GB and GB you would deal with only BG, BB and GG and you could wrongly claim that these 3 combinations have a probability of 1/3 in the general population.

[gcomeau]

112. No...

...you would assign 0.25 probabilities to each of them if you were calculating the probability of any possible combination of two children any family could possibly have.

Which. We're. Not. Doing.

Edit: But if you insist on doing it the hard way, see post 113.

[notesdev]

229. I'm with you on this

according to the original statement of the question, BG and GB are the same exact case.

The "explanation" apparently requires that one make a number of assumptions not explicitly stated in the original question.

The commonsense approach is to realize that the gender of the second child is not determined in any way by the gender (or day of week of birth) of the first child, and if we are applying a simple 50/50 standard (any child is equally likely to be a boy or a girl, slightly different from actual demographics but an implied assumption by the question) then the chance of the second child being a boy is 1 in 2.

Anyone arriving at any other answer ought to step back, apply common sense, and ask themselves, how did I overthink this so much that I got a different answer?

This question holds a great lesson in both common sense and how to discard useless information when analyzing a question.

[hunter]

93. Okay, I'm taking bets....

I have two kids, one of them a son.
.
.
.

Is my other kid a boy or a girl? Place your bet.

.
.
.

Here we go....

.
.
.

The answer is...

.
.
.


Hah, hah. Everybody who bet the other kid is a girl is wrong. Pay up.


The smart bet was that Hunter was gaming you. So maybe you bet boy.

The boring bet was to flip a coin. You bet boy or girl based on that. But I don't believe any alternate universe opened up where the sex of my kids was undetermined just because I asked the question.

But if you voted "girl" because the "probability" seemed to be in your favor, well sorry, you were gamed. The goofiness of the problem is in our language and our games. You can sell anything to people who don't recognize the grotesque inadequacies of our language. Meaningless questions get meaningless (and oftentimes harmful...) answers.

Next I will be taking bets for the 2005 World Series. Who will win, the White Sox or the Astros?

[CHIMO]

107. Interesting

In fact very interesting.

One is a boy born on a Tuesday.

I read that as a unique identity of one of the two children. Then if the other child is a boy, it was not born on a Tuesday.

[muriel_volestrangler]

109. If you do read it that way, the calculation for that has also been made

See #34 (and the diagram in #36, which may help).

[CHIMO]

121. Yes

Following the solution logic that has been given brings one to that point.

However, I still find it an interesting question. I have not isolated the flaw yet.

I have to relate the problem to the following analogy.

Two paper bags. In one bag there is a piece of red paper with the word Tuesday written on it. This represents a boy for a Tuesday birth.

In the other bag we have seven pieces of red paper, each one with a different day of the week. This represents a boy and the day of the week that he will be born. Also in this bag, seven pieces of white paper, each with a different day of the week printed on it. This represents a girl and the day of the week that she will be born.

Leaving out the question of only one boy being born on a Tuesday, then I come up with a probability of one in two that there are two boys.

Taking one paper from each bag gives us one in two that we have two red pieces of which at least one is Tuesday.

[muriel_volestrangler]

122. Because in that description, you have specified the red Tuesday paper for one bag

Try using the diagram in #36. When you say "one of the children is a boy born on a Tuesday", you are limiting the possibilities to either the row or the column that corresponds to a Tuesday boy - so our possibilities are all those in row number 2 or column number 2. So we end up with the (lop-sided) cross of those two - which has a total of 27 possibilities. 7 of those in the column involve a girl, 7 in the row involve a girl, and the other 13 (which include the intersection of the row and column) are '2 boys'.

If you say "in this bag we know we have a red 'Tuesday' piece of paper", then you are limiting yourself to just row number 2. That has 7 options with a girl in, and 7 with 2 boys (if you accept 2 boys both born on Tuesday as fulfilling "one is a boy born on Tuesday"; if you don't accept that, it would be 6 possibilities of 2 boys, and a 6 in 13 chance). So in that method, it is 7 in 14, or 1 in 2. But that's because doing it that way, we've taken a little bit more information than the question actually gave us - we've said "this child is a Tuesday boy" rather than "one of the children is a Tuesday boy".

[CHIMO]

123. Yes

But let's set aside whether it is interpreted as at least one.

In the published solution the result is approaching one in two.

If I follow the published solution then I walk down a predetermined path. My intrinsic sense is that the incidents are not connected. They are all independent.

The analogy that I gave was mine and the solution I gave was mine. I choose to stay on the analogy. I guess that the hiccup comes from future and present cases. I would prefer to stay in future possibilities.

If one were to assign the second in the day that a child was born, it would seem to me that it would approach one in two much more closely.

The point that a child was born on a Tuesday does not change the probabilities. The child had to be born on one of seven days. Building a solution around that point may be the error?

Also in the published solution it begins by accepting one in three. This may also be an error?

[muriel_volestrangler]

125. The "1 in 3" result is a well-accepted and understood one

It's well known that the possible 2 child families are split 25% girl-girl, 25% boy-boy, and 50% girl-boy. So when girl-girl is eliminated by another piece of information, the possibilities left are in the proportion 2 (for girl-boy) to 1 (for boy-boy). So the odds are 2 in 3, and 1 in 3.

What we do is then apply the same logic to the case with Tuesday involved. We start with 196 possible permutations, all of equal probability, work out which are eliminated by the information of "one is a boy born on Tuesday" (it leaves 27 permutations), and then see how many of those are boy-boy (it's 13). Caraher, in reply #116, shows it for a similar question with intermediate numbers involved - whether the child is born in the AM or PM. There, it works out at 3 in 7.

"If one were to assign the second in the day that a child was born, it would seem to me that it would approach one in two much more closely." Quite right. The more specific the information given, the closer the odds approach 1 in 2.

Yes, it seems very strange that specifying 'Tuesday' makes a difference to the odds, even though the final question does not involve the day of birth. That's what makes this so interesting.

[CHIMO]

126. And

That is why I think there is a flaw in the logic.

So how can the day of birth change from one in three to approach one in two?

There is an error in the logic.

Following the given solution one must be accepting the flaw.

A unique solution is one in a hundred ninety six. What are the probabilities that something might occur and what are the probabilities once something has been defined.

I need to have a few nights sleep on the question and the logic.

The well known part is self evident in the probabilities.

The probability that one will have two boys with a least one born on a Tuesday is one in 0.06633 or roughly 1 in 15.

[krispos42]

130. It seems to me...

...that you are assuming that because a boy was already born that makes it more likely that a girl will be born next time around. I think the phrasing of the question and the logic box (BB/BG/GB/GG) was done to make you make that assumption.





Or I could be full of shit.. :dunce:

[muriel_volestrangler]

133. No - both children have already been born

What this is about is the amount of information that has been revealed about the children. We, as people with incomplete information about the children, are working out the probabilities of the full information about them.

We start from each child being 50% likely to be a boy, 50% a girl. For each child, that is independent from another child.

When all we knew was "there are 2 children", we could say "there is a 25% chance it is 2 girls, 50% chance it is a girl and a boy, 25% it is 2 boys". If we are told "one of the children is a boy", we know we can eliminate the case where both children are girls. One of the other possibilities must be the reality, and the proportion they are in to each other is not affected by the new information we received - we just heard about "one of them", so we cannot pick one of them and say "we know this one is a boy" - that would be assuming a little bit more than we have been told. So the odds of a pair of boys are now 1 in 3 - meaning that if, at that point, we always guessed "they are 2 boys", we would turn out to be right 1 time in 3.

[krispos42]

136. There are three possibilities, but they are not equally possible

There are only two genders to pick from, and the odds of a baby being a certain gender is entirely independent of previous babies being born. There's a 50% chance the unknown child is a boy, and thus a 50% chance we get BB, and there's a 50% chance the unknown child is a girl, thus giving us either a GB or BG. Remember, the order doesn't matter.

I think there's a 50% chance it's a BB,a 25% chance it's a GB, and a 25% change it's a BG once we know that a boy is born already.

[muriel_volestrangler]

138. No, the 3 permutations are equally possible

The thing is, 'the unknown boy' is not an independent object - we don't know which child it is that is unknown. The child is, to quote Rumsfeld, 'an unknown unknown'.

Each permutation (ie when order is specified), such as 'BG', 'GB' or 'BB', is equally likely (as was 'GG', but we know that is impossible, so we tkae that out of the calculation). The 'combination' (ie when order is not important) of boy-girl, which corresponds to 2 of the permutations, is twice as likely as the combination of 2 boys. For an explanation of permutations and cominbations, see http://www.mathsisfun.com/combinatorics/combinations-permutations.html , or http://www.mathsisfun.com/pascals-triangle.html (section 'Heads and Tails') for a sample calculation of the chances of a combination (in that case, 2 heads in 4 tosses).

Here's a short program, in Javascript, to pick 2 'children' at random, and then work out if this produces a boys-only pair, or a boy-girl mixed pair, and repeat that 10,000 times; and then print the totals.

<html><head>
</head>

<body>
<script language=JavaScript>

function randInt(max)
{
 return 1 + Math.floor(max * Math.random());
 /* a random decimal number between 0 and 0.99999...., multiplied by max to be between 0 and 'not quite' max */
 /* then take the whole number part, and add 1 to be 1,2,...,max */
}

/* we will call a random result of '1' a boy, and '2' a girl */


mixedTotal = 0;
allBoysTotal = 0; /* starting values for the running totals */

for (i = 0; i < 10000; i = i + 1) /* do this 10000 times */
{
 chicken = randInt(2);
 egg = randInt(2); /* make both random choices, before examining them; order is not important */

 if ( (chicken == 1) || (egg == 1) ) /* if either, or both, are boys, we record this in a total */
 {
  if ( (chicken == 1) && (egg == 1) ) /* if both are boys */
  {
   allBoysTotal = allBoysTotal + 1; /* add one to the total of all-boy families found */
  }
  else
  {
   mixedTotal = mixedTotal + 1; /* add one to the total of 'mixed' families found */
  }
 }
 else 
 {
  ; /* this is when they are both girls; we do nothing */
 } 
}

document.write("mixed: "+mixedTotal+"; both boys: "+allBoysTotal);
</script>

</body></html>

Save that as a HTML file, and then open it in a browser. Each time you reload it, you will see the mixed total at around 5,000, and both boys at around 2,500.

[gcomeau]

176. Seriously...

...you need to stop being hung up on insisting that the intial odds of boy/girl combinations before we know any information about the gender of one child is the most important thing here. It isn't.

The dispute is over how you take into account the additional information you have been given, that one of the children is 100% definitely a boy and only one of the childfren is of unknown gender.

Your response to this, that it is:

GB - 33%
BG - 33%
BB - 33%

Is the verbal equivalent of saying:

"Well... the boy could have an older sister... or he could have a younger sister.. or he could have a brother. Therefore it is twice as likely that he has a sister."

In reality you can either state it this way:

"He either has a sister or a brother.. with a 50% chance of each"

Or this way:

"He either has an older sister, an older brother, a younger sister, or a younger brother, with a 25% chance of each."

But you can't just double the number of sister possibilities by declaring that dividing them into older and younger doubles the "sister" search space and not do the same thing for the brother possibilities!

Stated verbally that should be ridiculously obvious. But when we start trying to do it just with the numbers and beginning from the set of all possiblle 2 child ombinations you're getting lost, because you're not recognizing the extent of the effect of the information provided when you're told that one of the kids is absolutely, definitely a boy. It not only completely eliminates the possibility there are two girls, it cuts the odds of it being a "boy/girl" combo in half because the odds of ONE of the kids being a girl just dropped from 50% to 0%. Now all there is, is a 50% chance that the OTHER kid is a girl, and a 50% chance that the other kid is a boy.

[muriel_volestrangler]

181. I am not "doubling the number of sister possibilities"; those possibilities are already there

It is you who wants to throw one away even though they are both still possibilities.

the verbal equivalent of saying:

"Well... the boy could have an older sister... or he could have a younger sister.. or he could have a brother. Therefore it is twice as likely that he has a sister."

No. We cannot say "the boy". We have not yet selected a boy. All we know is that at least one child is a boy; or, put alternatively, that the children are not both girls.

The problem is that when you say "the boy", you have selected one of the children. That is another step in the problem, and thus you change the odds that have to be calculated.

It not only completely eliminates the possibility there are two girls, it cuts the odds of it being a "boy/girl" combo in half because the odds of ONE of the kids being a girl just dropped from 50% to 0%. Now all there is, is a 50% chance that the OTHER kid is a girl, and a 50% chance that the other kid is a boy.

No, it does not 'cut the odds'. The only odds that drop to 0 are the odds that both children are girls. That outcome is now impossible; the other outcomes are still possible.

Have you actually run an experiment with this? I've given you the programs; all you have to do is read them, and run them. They show that the odds I calculate are correct. Try running them. Then you'll see that your method must be wrong (and I've pointed out why - because you are still choosing a child, and thus introducing a new step), and then you will see that you have to think again.

[gcomeau]

183. Yes we can say "the" boy

We've been told the gender of one kid. That would be "the boy". And no, that is not an act of selection. That is an act of recognizing what we have been told.

And yes, I know you've given me the programs that do what you keep telling me to do that I keep telling you is the wrong way to do it. What would running them prove?

[muriel_volestrangler]

185. Well, you can hold out against my arguments, against those of 2 mathematics professors,

against those of the New Scientist writer, and against those of the person who actaully set the problem (because he said "Tuesday has everything to do with it"). Perhaps if you actually tested things, you'd see that reality agrees with all of these people, and not you.

:shrug: Your choice. Running them would prove that your result is wrong, and thus you might think again.

[gcomeau]

187. Your program is not "testing" the disputed point.

What you want me to do is to do the calculation your way, which is how you've programmed it to run, and then see it gives your answer.

I already know that.

[muriel_volestrangler]

193. The programs do not calculate probabilities; they total outcomes of random events

And the one in which the only 2 random events are the sex of the 'chicken' and 'egg', when at least one is a boy, shows that the situation of 2 boys occurs half as frequently as a mixture of boy and girl.

The outcome of that program is not from a probability calculation. It is from many trials of the scenario.

[qazplm]

204. its not "at least"

its one of them are boys. At least would change my opinion.

It's pretty clear that to use your own argument even mathematicians find that a meaningful distinction and I've already linked you to an explanation of why that is so with little substantive response from you.

[muriel_volestrangler]

209. "I have two children. Exactly one of them is a boy. What are the odds that both are boys"?

Clearly, the answer to that must be '0'. If exactly one of them is a boy, then they can't both be boys. But that would be a trivial and boring question. So the interpretation of that must be "at least one of them is a boy".

And, in the question with "one is a boy born on a Tuesday", if you want to interpret that as "exactly one is a boy born on a Tuesday", then the answer woud be non-trivial - it would be 6/13. Now, I believe the only time that answer in this thread has been given has been by me, in #34 (which I repeated, in #122).

And we have not seen that "even mathematicians find that a meaningful distinction ". The comment you quoted about Wilfred/Winifred etc. gets us nowhere. It is selecting one of the children to tie the 'boy' information to. it applies the name, just to make the selection extra-clear. The question does not selection a child.

[qazplm]

214. it is not "exactly" either

you keep introducing different words.

At least is not exactly. They have two very seperate meanings.

As I've shown elsewhere and which you have not bothered to look, so since you are being lazy let me link it AGAIN:

Here is my take on the difference between stating, “at least one is a boy” and “one is a boy.”

“At least one is a boy” is a statement about the set of two children. It is not a statement about any one of the two children.

“One is a boy” (after you eliminate the “exactly one is a boy” interpretation) is a statement about one of the two children.

In the former case, I believe the only way to interpret “other child” without extra assumptions is as the child left over after the “at least one” requirement is satisfied.

In the latter case, the “other child” is the one not referred to in the “one is a boy” statement.

A solution to the former cannot at any point identify one of the children as a boy without changing it to the latter, even though we know one of the two must be a boy.

Example:
“At least one is a boy”
“One, Fred, is a boy.” (I add the name in there, because 1) it makes the demonstration clearer without changing the problem and 2) as a nod to one of the other posts)

If you say, “Fred is a boy,” it certainly follows that “At least one is a boy.” A statement about a member of the set allows one to make deductions about the set as a whole.

If you say, “At least one is a boy” it does not follow that “Fred is a boy.” Fred could be a girl (Winifred) and the statement could still be true.

So the two can’t be equivalent.

“Fred is a boy”
the possibilities are
Fred – brother
Fred – sister
and the probability is 50%. Every solution getting 50% is, at some point, identifying a member as a boy, and reducing it to the above problem.

“At least one is a boy”
Wilfred – brother
Wilfred – sister
Winifred – brother
Here, at no point, are we specifying the gender of either member. Using the (to my mind) only meaningful definition of “other child” given above, the answer is 1/3. Two observations: 1) the fact that either Wilfred or his brother could be used to satisfy the “at least one” requirement does not mean you count it twice. 2) The statement “Fred is a boy” amounts to eliminating option 3, and reducing it to the 50% solution.

I suppose the confusion comes down to whether we’re abiding by the unwritten law of solving math problems — that you use only the information given you in the math problem. I accept the “at least one” statement as true because it’s given. If I start asking how it was established to be true, then it’s very easy to change the problem to the 50% one, as demonstrated repeatedly in these posts.

This was from one of the comments to the NY Times article. There is another after that explains it even better, but I'm not going back to cut and paste that one as well.

[muriel_volestrangler]

220. As I said, the 'Fred' comment is explicitly selecting a child

Its entire purpose in naming the child 'Fred' is to say 'we have selected a particluar child, Fred, about whom information will now be given'. But the point of the problem is that a child has not been selected, about which to give the information. The information tells us one of the children is a boy (or a born on a Tuesday), but it does not specify which child. We can only apply the information to the pair of children.

[qazplm]

256. no, it's not

which child is fred, you don't know.

Fred is no different a label than boy.

[qazplm]

192. you've just committed two logical fallacies

argument from authority and ad hominem.

[muriel_volestrangler]

194. I'm giving reasons to actually try running the program

Both of you are claiming you have worked this out better than people with experience in the area, and yet it seems you're unwilling to check with reality on what happens. We've tried telling you the problems in your arguments, but you don't listen to that. It would do you good to check the reality before restating your arguments again.

[qazplm]

203. again

arguing that you are right because other smart people say you are is a logical fallacy, argument from authority.

AGAIN, we do not dispute that running your program will lead to the results you say.

We do not dispute this because at the end of the day, you and we disagree on the question being asked and more fundamentally the variable being sought. So if we think you should be testing for x, and you think you should be testing for y, then why in god's name would you running a test for y, and finding it, somehow makes us think that you shouldn't be testing for x?

The "reality" is that you have put in initial conditions that we believe are "garbage" and we have put in initial conditions that you believe are "garbage" and so garbage in, garbage out would seem to apply.

IF we both believed that the values to be sought were GB, BG, and BB and THEN we said, no it still isn't one/third, then you'd have a point. But we DONT believe those are the same values, we effectively believe it should be BB and Mixed, two values, not three. So tell you what, plug in our two values to your program, what do you get? You wont get 2/3rds.

Now, you may be right, or we may be right. But the fact that you can't even understand that we are talking about different questions shows you don't understand what the issue is. The issue is not in the mathematics, the issue is in the presentation of the question, the value sought, and it's an issue that to use your logical fallacy back at you, many mathematics have addressed (a simple perusing of wiki shows this) and that the numbers change based on the information you have and how you got that information because the question asked changes.

[caraher]

206. I do agree that running the program will not be persuasive

But the point at issue is unambiguously and correctly resolved by the program (and the question is simple enough that writing and running a simulation is massive overkill!).

The entire problem is that some of us refuse to accept as correct the equivalence between the statements

"I have two children. One is a boy."

and

"I have two children. They are not both girls."

Both are verbal expressions ruling out the possibility GG.

Provided we leave aside details such minor deviations from 50% for the odds of having a boy rather than a girl, everyone here agrees that the probabilities for each of the following outcomes (taking account birth order) are 25%:

BB, BG, GB, GG

If we cull from our population all the GG outcomes, we're left with three possibilities, each of which we've already established are equally likely:

BB, BG, GB

Since these are equally likely, and since probabilities need to add to 100% for our new subpopulation (all non-GG 2-child families), each of those outcomes has a 1/3 probability.

This is not ambiguous or even mildly controversial!

[gcomeau]

211. Oh?

The entire problem is that some of us refuse to accept as correct the equivalence between the statements

"I have two children. One is a boy."

and

"I have two children. They are not both girls."

How about this?

"I am one of two children. I am male."

Is that equivalent too? In that it establishes the existence of two children, one of whom is a boy?

[qazplm]

213. so here's the thing

you think those statements are equivalent and we dont.

We think BG and GB are equivalent and you dont.

The idea that this isn't ambiguous or controversial absolutely isn't true.

I've done enough reading up to know that despite a few folks on here claiming it to be so, there is all sorts of controversy about this, folks who disagree with each side presented, there are folks who claim a third side (anywhere from 0% to 100%, including one who argues the correct answer is 75%, etc.).

Obviously they can't all be right. Then again, they can all be partially right, depending on the phrasing of the question.

[notesdev]

233. The birth order is not relevant

So GB and BG are not two different cases.

Applying basic algebra and knowing that the first boy is a constant, not a variable, there are only two cases for the variable we are trying to solve for: B or G.

Since the first child's gender is a constant, it drops out of all valid equations that contain the genders of both children.

[notesdev]

232. A program with bad logic gives an incorrect answer

The proper logical approach of the problem is to reduce it to solving for the single relevant unknown - the gender of one child.

Whether it is the first child or second child is irrelevant - so breaking up the problem into BB/BG/GB/GG is an error in two respects:

1 - the gender of the first child is irrelevant to the gender of the second child
2 - which child was born first is also irrelevant to the gender of the second child

There are ONLY two cases here - either the second child is a boy, or the child is a girl.

With that established, there is absolutely no other information in the puzzle that tells us to favor either one as a solution.

Therefore, we are left with two equally likely possibilities, each with a 50% probability of being true.

This 'problem' and its suggested solution are a great example of arrogance born of people thinking they are a lot smarter and/or more clever than they really are.

[krispos42]

128. So the question is... given two offspring, what are the odds they are the same gender

Well, then 1 in 4.



There is a 1 in 4 change that, if anybody procreates twice, they'll have two boys.





Since he's procreated once already and gotten a boy, then there's a 50% chance that after then 2nd child is born it will be a boy as well.



I guess it depends on which specific point in time he's referencing.



Now I'll go read the article...

[pokerfan]

132. But he's procreated twice already

and all we know is that at least one is a boy, eliminating the possibility of two girls, hence: BB, BG and GB are all equally probable. Had he specified order or otherwise identified the boy (e.g. my eldest is a boy) then it becomes 50%.

[ismnotwasm]

139. Cool

Since it deals with probabilities that *is* a neat trick. I'm not very good with math, nor intuitive about it, so I love stuff like this; both entertaining and easy to understand.

As an aside, I was reading it out loud to my husband before reading the solution and he 'got' it quicker than I did. He's no math whiz either, but he was a bit more intuitive; "Well there's seven days in a week" he started to say, so he was on the right track.

[hansberrym]

142. The puzzle is much less counter-intuitive if we substitute abstractions such as coin flips and dice

rolls.

In the original example each trial is Boy/Girl plus day of week with each parent getting 2 trials. These real life examples can confuse and bring in assumptions and biases, but if we substitute to an abstract case we can avoid the disbelief that creeps in.

Substitution:
Participant for Parent
H/T Coin flip for Boy/Girl Child
Dice roll for born on certain day of the week

So that each trial is now a Coin flip(Heads or Tails) plus a dice roll(1,2,3,4,5, or 6) with each participant getting two trials.

If we consider just those participants that scored at least one H/2 combination and then ask what percentage of that subset of participants scored 2 Heads we have no pre-existing assumptions on what the outcome should be, and so the "counter-intutive" result of the original case and corresponding non-beief in the calculations does not creep in.

In order to score a H/2(Heads on coin flip and 2 on the dice roll) combination, a participant has to have at least one H, so we go to the well known coin flip possibilities involving two trials HT, TH, HH, and TT then continue with the same sort of calculation as in the New Scientist article.

For myself, this substitution eliminated much of the counter-intuitive aspect of the puzzle leaving only some doubt concerning the special case where a particpant scores H/2 on both trials and how that should affect the probability that any particapant who scored at least one H/2 combination would also have scored 2 heads. That remaining doubt was resolved by considering the number of successful trials is not neccessarily the same as the number of successful partipants. The participants scoring 2 heads (as a group) will indeed score 50% of the successful (H/2) trials which goes along with the intuitive result, but since some of the participants scoring 2 Heads will have 2 successful trials(H/2), the number of succssful participants scoring 2 Heads will be less than half the number of successful participants scoring 1 Head.

[Ready4Change]

143. Illusions due to mathematical perversity, imo.

I've followed the logic train on these, and I see the arguement.

Still, I reject them. Both on the Monty Hall question and on this 'Son born on a Tuesday' question.

It is, again imo, akin to concluding that, since you've thrown a coin 10 times and gotten 8 heads vs 2 tails, that you must be 'due' a few tails results.

What if the son were born Wednesday? You'd get the same result, which would appear to nullify the 'Tuesday' results, (or rather, set them into a super set over the Thur-Monday sons.) That is, until you figure the results for all those other weekdays. Then what do you have? Shock, even odds.

[pokerfan]

147. Try playing the Monty Hall problem

here:

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

Try one strategy (always staying or always switching) and see how you do in the long run.

[CHIMO]

151. Baloney

The solution given is wrong.

The fact that a person was born on Tuesday, Monday or any other day does not remove the speaker from the whole population. The analysis is incorrect. Period.

On the point of 1 out of two or one out of three, I would like to rephrase the scenario. Instead of the individual person giving sex of one half of the offspring, look at it this way. If a poll was done on a large population and the following questions asked:
Do you have children?
How many children do you have?
Do you have at least one boy?

From those that answered that they had at least one male and two children, what is the proportion of that population that would have two boys?

[muriel_volestrangler]

153. One third (nt)

[CHIMO]

154. Correct

One third. And one third for the Tuesday part as well.

[muriel_volestrangler]

156. Your last sentence is a non sequitur

"And one third for the Tuesday part as well." You have not given any calculation for that.

Did you see these comments to the New Scientist piece, by 'wtl'? I think they explain it quite well:

Most of these comments appear to have missed the point about the probability calculation. Note that:

(a) Probability of getting a second boy if you already have a boy is 1/2

(b) This is not the same as the probability of having two boys if you already have two children. That probability is 1/4.

(c). But if you state that at least one of the children is a boy, the probability becomes 1/3. This is greater than 1/4 because you have to exclude all those who have two girls from the population of parents in question. In effect, you have enriched the population for parents who have two boy by excluding all those who have two girls. This probability is less than 1/2 (see point a) since you have to include those whose first child is a girl but second is a boy.

(d) The probability of having two boys when you have at least one boy born on a Tuesday approaches 1/2. This occurs you have again enriched the population for those who have two boys: there is a higher chances that those who have two boys have at least one who is born on Tuesday, and so that is a greater proportion of parents with two boys in the population in question. You have excluded all those who do not have at least one boy born on Tuesday from the population of interest (i.e you have excluded those with two girls, those with one boy but not born on Tuesday, as well as those with two boys but neither born on Tuesday). As stated in the article, if you include a rarer piece of information (e.g. a boy born on 1 Jan), you enrich the population further and get even closer to 1/2. You can not exceed 1/2 because there is always going to be at least as many parents who have one boy given X as those who have two boys given one is X.

While the probabilities calculations in the article are correct, one a valid criticism remains: What is the point of asking such question? Why do you care how many parents have two boys among those who have at least one boy on a Tuesday? In real life, I can't see much point to these probabilities - you are applying the question to a very small population based on seemingly arbitrary information. But, I guess the point of the article is that it's fun! :)

...

For those who are still having trouble visualising the situation, perhaps this simpler explanation might help:

Among all parents who have two children, one quarter of them will have two boys.

Among all parents who have at least one boy, one third of them will have two boys.

Among all parents who have at least one boy born on Tuesday, close to one half will have two boys.

You are not changing the probability of having a boy, nor are you changing the probability of having two boys once you already have two kids. What you are changing is the group of parents we are talking about. We aren't talking about all parents who have two kids, we are only talking about parents who have two kids and one of which is a boy born on a Tuesday. When the speaker says "I have two children. One is a boy born on a Tuesday.", he is narrowing the population of parents down so that it is more likely he has two boys. That is, he is not just any parent with two kids, he is a parent with two kids but at least one boy born on a Tuesday.

While there aren't as many parents with a boy born on a Tuesday as those parents with any two kids, a greater proportion of parents with a boy born on a Tuesday have two boys compared to parents with any two kids.

Still a little confusing I guess, buy hopefully more easily understood.

[CHIMO]

159. Not Correct

The fact that a child is born on Tuesday does not increase the probability that another child will be born on the same day!

We are talking about all parents that have two children, one of which is a boy.

Because you wish to follow the given solution you are coming to the same conclusion.

You can not have a correct solution if you accept false assumptions.

[muriel_volestrangler]

164. I never said anything about "the probability that another child will be born on the same day"

and neither did those comments. :shrug:

[CHIMO]

162. The Tuesday

Part doesn't require a calculation.

It is a Euclidean deduction.

[muriel_volestrangler]

165. In your words, 'baloney'

There's nothing 'Euclidean' about it. You're not even trying to make an argument here. "It is a Euclidean deduction" is an empty phrase.

[dnbn]

157. The latter is incorrect.

The property of having a boy born on a Tuesday is higher among the people that have two boys than it is among the people that have only one boy (the former group has 2 chances of meeting this requirement versus only one for the former). The ratio between the two boys group and the only one boy group is therefore higher once a Tuesday requirement is added. 13/27 is the correct answer.

[CHIMO]

158. No

It can't be.

If one works from the universal part then we would have a discrepancy.

Let's call this the CHIMO axiom, two things can not be true when one is wrong.

If you find the given solution to be correct, then do the same calculation for Monday, Wednesday, Thursday, Friday, Saturday and Sunday. Your analysis then would say that there were no girls born.

So prove that my position is wrong. Not that the other is correct.

[muriel_volestrangler]

163. Consider the diagram in post #36

When the specification is "at least one boy, who is born on a Monday", then the possibilities are shown by the leftmost column and topmost row (ie the column and row marked '1'). There are 27 possibilities (each side of the grid is 14 long, and the top left position is common to both), and 13 of them are '2 boys'. If we ask about "at least one boy, who is born on a Tuesday", then the numbers again come out as 13 and 27 (and that's the 2nd diagram I drew).

If we ask about "at least one boy, who is born on a Monday or a Tuesday", then we have the union of those two - but there's some overlap, and the total number is 52, of which 24 are '2 boys'. Add 'or Wednesday', and it's 33 out of 75; add Thursday and it's 40 out of 96; Friday makes it 45 out of 115; Saturday 48 out of 132; Sunday (ie just 'at least one boy', with no date at all specified) 49 out of 147 (which is 1 in 3).

Each time we allow another day of birth, we find some overlap with the cases we've already considered. As the day of birth becomes less specific, the odds tend towards 1 in 3.

[CHIMO]

170. The

Question is not relevant.

One child was born on a cloudy day. Has no relevance.

The failure is in not recognizing that the Tuesday part is not relevant.

The population is still all the children born. The day they were born has no relevance to the question. Trying to incorporate that as part of the solution is the point of departure.

By counting the grains of sand in a spoon and relating that to the outcome would be just as relevant. The day the child is born has no relevance to the calculation. One has to take the whole population.

Unfortunately I can not follow a schematic that I haven't had time to think about.

I have done a spreadsheet on the possibilities and haven't been able to solve it.

Therefore I have to approach it from another angle. At least for the time being.(Perhaps I can get some sleep now.)

[dnbn]

167. The Monday, Wednesday, etc

probabilities, and obviously all of them equal to the same value, apply to different populations. The different populations are mutually exclusive for the group that have one boy but not so for the group that have two boys - another indication that that group is relatively larger than its original percentage.

[CHIMO]

161. Incorrect

13/27 is wrong.

The statement is a red herring.

The probability that he has two boys is 1 in three.

It is the same as the normal population with two children.

[dnbn]

174. If you start from normal populations, that ...

we agree have a ratio of 2 to 1 between the one boy camp and two day camp, and eliminate from them the ones that don't have a boy born on Tuesday, we will eliminate relatively more from the one boy camp. The reason for that is that any parent in the two boy camp has two chances to meet the Tuesday requirement while a parent in the one boy camp has only one chance of meeting that. The ratio between the two population will therefore be smaller after the elimination of non-Tuesdays.

This kind of effect will happen for any specification that the two boys have a higher chance in meeting in unison than a single boy.

[eShirl]

223. "girl or boy being equally likely, and any birth equally likely on all different days of the week."

50%

[boston bean]

231. It could never be GB if there are only two children.

[tinrobot]

239. Here's the answer in tabluar form

The trick to this is that we have a fixed set of data, almost like a deck of cards. We are not creating new children (where the possibility is 50/50) Instead, we already have two children and are simply handing out information about them.

It's similar to giving out hints, and each hint removes possibilities. For example, if I said I'm holding a red card, you know that there's one less red card in the deck, so the possibility of drawing red goes from 26/52 to 25/52.

It's the same here. Like a deck of cards, we have a fixed set of permutations and are given hints about the solution. By hinting that one child is a boy, we eliminate all possibilities of two girls. By hinting that the boy is born on a Tuesday, we eliminate more possibilities.

Most people "get" that if you say that one child is a boy, that hint gives information that reduces the possible solutions from four (gg gb bg bb) to three (gb bg bb)

The same happens with the day of the week, but it's less intuitive. As with the gender of the child, the information that the boy was born on a tuesday reduces the number of possible solutions. Here's a chart with all 27 possible permutations of two children, one of which matches up with the hint -- a boy born on a Tuesday.



The cases with two boys are highlighted. As you can see, 13 of the 27 permutations produce two boys.

Hope this helps people to visualize the solution.

[eShirl]

243. no, there are only 2 possibilities, not 3: the other child is a girl OR the other child is a boy.

roughly 50/50

[tinrobot]

245. Birth order plays a part.

Two children times two possibilities each = four combinations (bb/bg/gb/gg)

It's twice as likely to have one of each gender (gb/bg) as it is for two boys (bb).

Eliminating one of four possibilities (gg) leaves three. That's the math, not really that hard to grasp. If you can't subtract one from four, then I don't have much to say beyond that.

[RoyGBiv]

249. I just want to say ...

... this thread has been AWESOME.

I know some of the participants have gotten themselves all riled up, but I would like to note that through it all, following this has been fascinating, and I mean that in a positive way. I've been a participant in discussions among historians that get at least as heated, but they're generally both informative and entertaining.

And, FWIW, I forwarded this entire thread and the original article to a group of mathematicians and statisticians who deal with advanced problems on a daily basis, and *they* ended up having a similar kind of discussion. I won't bother revealing what they determined. It'd just irritate a certain contingent, and I have absolutely no business entering this debate myself.

Good to make the brain work. Thank you.

[OneTenthofOnePercent]

254. 50%... If the sex of one entity is known then that entity is not random anymore.

In statistics, previous occurances do not effect future outcomes.
By stating "one is a boy" that places into the category of previous occurance.

[muriel_volestrangler]

258. But you don't know the sex of 'an entity'

You know that one is a boy, or both are boys. But you do not know the sex of a particular child. You know that it is definitely not the case that both are girls.

[OneTenthofOnePercent]

261. I have flipped two coins, one in each hand...

in one hand, I will not tell you which hand, the coin is heads. Based on the known current information, what is the probability that the coin in the other hand (therefore both coins) are heads?

It's 50%. You KNOW one coin is heads and the other coin only has two possibilities.

[muriel_volestrangler]

264. But for all I know, you had to look at both coins before you could say "one is heads"

If you tell me that you have looked at only one coin, and it is heads, then it is indeed 50% probability that the other coin is heads. But you have not told me that a particular coin you have looked at is heads; you have just told me that one of them is.

When you had flipped 2 coins, and told me nothing, then the information tells me there's a 25% possibility of 2 heads, 50% of a head and a tail, and 25% of 2 tails. When you tell me one of the coins is a head, I can eliminate the last option (of 2 tails); but the information I have - that at least one of the coins is heads - tells me nothing about the relative probability of 2 heads v. a head and a tail. A head and a tail are twice as likely, still - so the chances of 2 heads are now 1 in 3.

[OneTenthofOnePercent]

266. If you are treating discreet permutations as separate results... that might apply.

But to say the situation of having one head & one tail or one tail & one head, respectively, is a different scenario is all too often a moot point.

IMO, unique permutations do not matter unless the order of the outcome matters.

[muriel_volestrangler]

268. Each permutation, before any further information is given, has an equal probability

and so, though there is not 'order' to think of in the result, it's highly relevant that a mix of a boy and a girl is twice as likely as 2 boys, when you are calculating the relative odds of the combinations. If you were betting on the toss of 2 coins, you'd need to remember that a head and a tail is twice as likely as 2 tails. If you were throwing a pair of dice, you'd need to remember that a 1 and a 2 is twice as likely as a double 1.

[Thor_MN]

267. I can see where some might argue that "one is a boy" COULD mean that the other is a girl.

To me, that is assuming something that shouldn't be assumed, but I can see where one might argue that the odds are 0% of having two boys. If the question is sloppy in any matter, it hinges on that.

If I were brought in the back door of a house and asked someone who could see out the front window what kind of wall the house across the street had and they replied brick, I would not assume that the house across the street had 4 brick walls. There are many houses in my neighborhood that have brick fronts, but different materials on the sides and back. To me, saying you have "a boy" tells me something about one of their children, but nothing about the other, especially if the next thing is a question about the odds on the pair of them.

I really don't see where anyone argues 50% beyond realizing that a snap answer is wrong. I can see arguing the 0%, but 50% makes no sense at all.

[Jim__]

262. Delete - posted in the wrong spot.

[drm604]

269. Can someone post how Foshee reached his conclusion?

The full article is only available to subscribers and I'm curious how he arrives at his answer (which I think was 13/27, from reading this thread).

[muriel_volestrangler]

271. Yes, 13/27, for the 'boy born on a Tuesday' problem

Sorry the article is now subscription only - it was generally available when first published, I think.

Try the explanations in #34 and #36. #34 is a slight modification of the wording from the article (for when you say 'one of them is a boy born on a Tuesday' means 'one, and only one, is a boy born on a Tuesday' - which changes the odds from 13/27 to 12/26), but the argument is basically the same. And #36 is my attempt for a diagram to illustrate the possibilities, and how the information rules some of them out (2 cases - one where both children being boys born on a Tuesday is held to satisfy the wording 'one is a boy born on a Tuesday', and one where only precisely one can be a boy born on a Tuesday).

Or try #239, where tinrobot has set out a table of the possibilities, and those that are '2 boys'.

[saras]

272. write 'em all down

It's really straightforward WHY it's 13/27 if you draw the whole table, instead of starting with a reduced one - I used a spreadsheet.

If all you know is that they have two children, you can conclude that...
(Child 1 M/F) * (Child 1 7 days) x (Child 2 M/F) x (Child 2 7 days) = 2*7*2*7 = 196 possibilities
And each possibility is equally likely (let's pretend).

If they tell you one of them (you don't know if it's child 1 or child 2) is a boy born on Tuesday then you go through the table and scratch out every entry that doesn't have at least one boy born on Tuesday (at this point, you don't care what the other one is, even if it's another boy born on Tuesday), and watch that you don't scratch out entries with child 2 born on Tuesday just because child 1 wasn't. The crossed-out ones are irrelevant.

Count how many entries are not crossed out - you'll find 27. Out of the original 196 possibilities, this is how many of them have a boy born on Tuesday. This is the part that's tricky until you see how it happens.

Now go and cross out every entry that isn't two boys, and count what's left - you'll find 13. This is how many of the above small number also have the other child being a boy. And hopefully you've all noticed that 13/27 is (14-1)/(28-1)