This analysis uses the same mathematics we used to compute the probability of 15 or more JFK witnesses turning up dead within a one year period. It's the SAME problem, but in this case the time interval is even shorter - 4 months, making it even more improbable.
We seek to calculate the probability of at least 15 microbiologists dying UNNATURAL deaths within a 4 month period. The deaths were a combination of homicides, suicides, accidents and undetermined origin.
The key VARIABLE is: How many world-class microbiologists are there? Not knowing this number, I have calculated the probabilties for the three assumptions: 1000, 2000, 5000 biologists.
For a 4 month period the probability is 1/3 the full one year probabilty, which is what I initially used for the computations since one year mortality data is availiable. To obtain the odds of unnatural deaths, I used 1999 mortality data:
http://www.nsc.org/lrs/statinfo/odds.htm IMPORTANT NOTE: THE PROBABILITY OF 16 OR MORE STRANGE DEATHS OCCURING IS MUCH LOWER THAN IT IS FOR 15 OR MORE DEATHS. I USE 15 DEATHS AND 5000 WORLD-CLASS BIOLOGISTS AS ASSUMPTIONS BECAUSE I USED THEM IN THE PRIOR JFK ANALYSIS AND I AM TOO LAZY TO DO ANY MORE WORK ON THIS TONIGHT.
FOR 16 STRANGE DEATHS, OUT OF 5000, TO OCCUR IN A 4 MONTH PERIOD, THE ODDS ARE ABOUT 1 IN 100 MILLION.
FOR 15 DEATHS, the probability is approximately 1 in 15 million.
Assuming there are 5000 world-class microbiologists in the world, the probability is 1/3 * 1/5,239,858 or 1 in 15.7 million.
Assuming there are only 1000 world-class microbiologists, the probability that at least 15 would die UNNATURAL deaths in a FULL calendar year is:
1 out of 21,230,606,601,227,800 or
1 in 21,230 trillion, 606 billion, 601 million, 227 thousand, 800).
For 1000 deaths to occur in a 4 MONTH period, the odds are 1/3 of this, or: 1 out of 63,690 trillion....
These are the mortality rates:
........................1 year...Lifetime
Probability of:
suicide.................0.000107 0.008197
homicide................0.000062 0.004739
accidental death........0.000359 0.027778
undetermined death......0.000014 0.001101
The probability of an unnatural death FOR ANY AVERAGE PERSON in A GIVEN YEAR is the sum of the probabilities of the four categories = 0.000542, or approximately 1 in 2000.
The probability of an unnatural death in a LIFETIME is the sum of the probabilities of the four categories = 0.041815, about 1 in 25.
The Poisson Distribution
Although the Normal (Gaussian) probability distribution is by far the most important, there is another which has proven to be particularly useful - the Poisson Distribution, which is derived from, and is a special case of the Normal Distribution.
The Poisson Distribution applies when the probability "P" for success in any one trial is very small, but the number of trials N is so large that the expected number of successes, pN, is a moderate sized quantity. The formula is: P(m) =a**m*exp(-a)/m!
In words, the Probability of EXACTLY m successes = a to the m'th power times the exponential function of (-a), all divided by m factorial.
If m= 15, m factorial = m! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1
Now lets use Poisson to determine the probability of a given number of individuals from ANY predefined group meeting unnatural deaths within a one-year period.
The only assumption we are making here is the number of biologists.
Assume for the sake of argument, there are N= 1000
Let p= Probability of any individual dying from UNNATURAL causes within a given year = 0.000542
Let a= Expected Number of deaths = pN= 0.542
Let m= Actual Number of UNNATURAL deaths = 15
The probability of exactly m=15 UNNATURAL deaths within a given year out of a predefined group of N = 1000 is:
P(m) =a**m*exp(-a)/m! or p(15)= 0.542**15*exp(-.542)/15!
Here are the probabilities for m=1 through m=15 deaths occuring within ONE year:
Prob(X=m) = probability of EXACTLY m DEATHS
Prob(X>=m) = probability of at AT LEAST m DEATHS (the one we want)
column 1 is n, the number of deaths,
column 2 is the probability of at least n deaths,
columm 3 is the actual odds (for example, 1 out of 5,239,859)
m.......Prob(X=m)........Prob(X>=m)
Assuming 5000 biologists:
n deaths..prob...1 out of..
1 0.933
2 .753 1.33
3 .508 1.97
4 0.288 3.47
5 0.138 7
6 .057 17
7 0.020 47
8 .00670 148
9 .00196 510
10 .00051 1,939
11 0.00012 8,070
12 0.000027 36,485
13 0.0000056 178,135
14 0.0000010 934,300
15 0.000000190 5,239,859
For a 4 month period (1/3 of a year), the probability is is 1/3 times 1/5,239,858 or approximately 1 out of 15 million.
FOR 16 DEATHS, IT IS ABOUT 1 OUT OF 100 MILLION.
Here are the probabilties for other cases (remember, the probabilities are actually 1/3 of those given, since the time interval is 4 months or 1/3 of a year):
-----------------
1000 biologists
n....prob..1 out of
1 0.418 2.39
2 0.103 9.69
3 0.0177 56.26
4 0.0023 427
5 0.0002 484,018
6 0.000022 45,091
7 0.00000169 588,306
8 0.000000114 8,752,118
9 0.00000000683 146,245,847
10 0.000000000368 2,712,122,977
11 0.0000000000180 55,278,020,364
12 0.000000000000814 1,228,276,488,499
13 0.0000000000000338 29,551,271,527,958
14 0.00000000000000130 765,351,111,903,523
15 0.00000000000000004710 21,230,606,601,227,800
-----------------------------
2000 biologists:
n.......prob....1 out of
1 .661 1.51
2 0.295 3.39
3 0.096 10.38
4 0.024 41
5 0.005 195
6 0.00089 1,113
7 0.000136 7,343
8 0.0000181 55,093
9 0.00000215 463,452
10 0.000000231 4,321,227
11 0.0000000226 44,239,588
12 0.00000000202 493,399,077
13 0.000000000167 5,954,899,756
14 0.0000000000129 77,331,852,139
15 0.00000000000093 1,075,228,365,705