H5N1
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Wed Jun-29-05 12:29 PM
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There is a formula for figuring (for instance) the odds of getting five of a kind, all at once, in one of three shakes of five dice.
It might be some kind of binomial distribution formula but I cannot seem to find a simple, spelled out version on the net.
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CBGLuthier
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Wed Jun-29-05 12:31 PM
Response to Original message |
1. Yahtzee probabilities here |
kick-ass-bob
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Wed Jun-29-05 12:33 PM
Response to Reply #1 |
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Good to see my alma mater is doing some good out there!
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H5N1
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Wed Jun-29-05 12:37 PM
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4. thanks, actually I am looking for the formula |
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a formula that lists and explains the variables, perhaps (n) = the number of dice or cards, perhaps (p) = the probability of a particular outcome per dice or card... things like that. I knew the formula years ago... but I forgot. Seems it may involve some factorials as well.
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CBGLuthier
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Wed Jun-29-05 12:43 PM
Response to Reply #4 |
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top of page 21 on this pdf file http://www.madandmoonly.com/doctormatt/mathematics/dice1.pdfLooks like a continuation of the same stuff from the other page but concludes with a formula.
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H5N1
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Wed Jun-29-05 12:47 PM
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8. must be in there somewhere, thanks |
hfojvt
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Wed Jun-29-05 12:45 PM
Response to Reply #1 |
7. except they did not do it all at once |
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also, I have never done it with matrices, but that is probably the way to formalize all of the adding that I would do.
For the straight probability of a Yahtzee it would be the sum of the odds of getting a Yahtzee on the first roll added to the odds of getting one dice that you save and then matching it to the other four in your next two turns, although you have to track the various choices made by the player and the matrix does that very well. For example if you roll 4,4,1,3,5 on your first turn, save the fours, but then roll 2,2,2 on your second turn, your odds of a yahtzee are better if you save the twos and you have a 1/36 chance of a yahtzee on your last turn.
Unless you are my little brother - otherwise known as Mr. Dice. Once when we were playing triple Yahtzee and we both needed two Yahtzees in our last three turns. He got them. I did not.
Another time we were playing risk and he rolled two ones on defense as I was attacking him. I then proceeded to roll three ones on offense, losing two armies. What are the odds of that? 1 in 216! No problem, when you combine his good luck with my bad luck.
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hfojvt
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Wed Jun-29-05 12:33 PM
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1/6 to the fifth power times 18 (or .0023148). But I got my BA in math from Minnesota a long, long time ago in a galaxy far, far away.
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H5N1
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Wed Jun-29-05 12:38 PM
Response to Reply #3 |
5. How about an all-purpose formula? |
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How did you come up with a name like hfojvt?
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hfojvt
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Wed Jun-29-05 01:03 PM
Response to Reply #5 |
10. same way the creators of 2001 came up with HAL |
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HAL = IBM - 111. I only wish I had gone with dmpxo.
I do not believe there is an all-purpose formula for these things.
You just have to add up the possibilities. Since there are six different possible yahtzees I multiplied by 6. Since there are three turns I muliplied by 3 (only I skipped a step and just multiplied by 18). There are 6 to the fifth power possible combinations of five six-sided dice - 11111, 11112, 11113, 11114, 11115, 11116, 11121, 11122, and so on up to 66666.
But the odds of a Yahtzee in three turns are different than getting it in one turn (an also much harder to calculate). For instance if your first roll is 22225, your odds of a Yahtzee in your next two rolls are 11/36 (a 1/6 chance of getting a two on the 2nd turn added to a 5/6 chance of a non-two multiplied by a 1/6 chance of a two on the third turn or 1/6 + 5/36).
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redqueen
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Wed Jun-29-05 01:09 PM
Response to Reply #10 |
Ready4Change
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Wed Jun-29-05 12:57 PM
Response to Original message |
9. My way of figuring it: |
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Compare possible results against possible successes.
For example: If you roll 2 dice, you might get 1 and 1, or 1 and 2, ..., 6 and 5, or 6 and 6. That's 36 possible results. Out of those, only 6 are successes. (1 and 1, or 2 and 2, or...)
So you have:
Chance of 2 of a kind: 6 of 36 = 1 in 6
So, roll a pair of dice 6 times and odds are you'll get a pair once.
Continue that method outwards:
Chance of 3 of a kind: 6 of 216 = 1 in 36 Chance of 4 of a kind: 6 of 1296 = 1 in 216 Chance of 5 of a kind: 6 of 7,776 = 1 in 1296
Now, if you roll 5 dice 3 times you have:
3 for 1296 = 1 in 432.
I'm assuming ANY 5 of a kind will do. (all 1s, or all 2s, etc...) If you insist on a SPECIFIC 5 of a kind (all 6s, for example) you'd have
3 tries at 1 in 7,776 = 3 in 7,776 = 1 in 2592
And if that's wrong, you now know why I don't gamble. :)
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hfojvt
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Wed Jun-29-05 01:07 PM
Response to Reply #9 |
12. same answer I got in post #3 |
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only you explicated more.
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Theres-a
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Wed Jun-29-05 01:05 PM
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Hope you're not here to mutate into something that's easily transferable between humans.:scared:
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H5N1
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Wed Jun-29-05 01:36 PM
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15. I didn't see any mention of H5N1 here |
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Suprisingly. So I thought I would add a little H5N1 to the mix.
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Theres-a
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Wed Jun-29-05 01:40 PM
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17. There's a user who posts in LBN |
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Pandemic 1918,who links to recombinomics articles,but they usually get moved to the science forum.
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H5N1
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Wed Jun-29-05 01:45 PM
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Oh well, I suppose that may change if the wrath of H5N1 starts clobbering the hell out of us. Most people (americans)never heard of it.
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redqueen
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Wed Jun-29-05 01:09 PM
Response to Original message |
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Edited on Wed Jun-29-05 01:10 PM by redqueen
:hi:
Welcome to DU!
on edit: creepy username... love it. :)
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H5N1
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Wed Jun-29-05 01:37 PM
Response to Reply #14 |
16. I might become very popular over the next several years |
redqueen
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Wed Jun-29-05 02:02 PM
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19. Well, if you mean YOU, then great! |
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If you mean what your name means... please don't! :scared:
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48pan
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Wed Jun-29-05 02:10 PM
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20. The calculation is very long. Use an existing table. |
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The chance of getting 5 of a kind on a single roll is 6 to the 4th power.
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