# Math Dopers: A little help?

**URL:** <https://boards.straightdope.com/t/math-dopers-a-little-help/252935>\
**Category:** Factual Questions\
**Created:** [June 30, 2004, 7:53pm UTC](https://boards.straightdope.com/t/math-dopers-a-little-help/252935 "2004-06-30T19:53:04Z")\
**Posts on this page:** 2\
**Page:** 2

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**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\
**Post date:** [July 1, 2004, 9:14pm UTC](https://boards.straightdope.com/t/math-dopers-a-little-help/252935/21 "2004-07-01T21:14:18Z")

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If all you care about is the most likely number of matches, you don’t actually have to calculate large exponentials. The probability that there are exactly n matches is hypergeometric, as already stated; we can write it as  
P(n) = (A! B! (N-A)! (N-B)!) / (n! N! (A-n)! (B-n)! (N-A-B+n)!)  
(here N=60000, A=6100, B=3500). P(n) increases to the most likely number of matches n[sub]0[/sub] and then decreases; so P(n)/P(n-1) is \>=1 for n\<=n[sub]0[/sub] and \<1 for n\>n[sub]0[/sub]. But  
1 \<= P(n)/P(n-1) = ((A-n+1) (B-n+1)) / (n (N-A-B+n))  
implies  
n \<= (A+1)(B+1)/(N+2)=355.981…  
so we have n[sub]0[/sub]=355.

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**Author:** ![Josh\_dePlume](https://avatars.discourse-cdn.com/v4/letter/j/bbe5ce/32.png) [@Josh\_dePlume](https://boards.straightdope.com/u/Josh_dePlume)\
**Post date:** [July 1, 2004, 10:14pm UTC](https://boards.straightdope.com/t/math-dopers-a-little-help/252935/22 "2004-07-01T22:14:38Z")

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Routines for calculating hypergeometric probabilities are found in many software packages. Excel even has such a function, called hypgeomdist.

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