Hypergeometric distribution

Cameron Laird claird at lairds.us
Tue Jan 3 12:08:03 EST 2006


In article <1136201733.518543.95060 at f14g2000cwb.googlegroups.com>,
Raven <balckraven at gmail.com> wrote:
>Well, what to say? I am very happy for all the solutions you guys have
>posted :-)
>For Paul:
>I would prefer not to use Stirling's approximation
>
>
>The problem with long integers is that to calculate the hypergeometric
>I need to do float division and multiplication because integer division
>returns 0. A solution could be to calculate log(Long_Factorial_Integer)
			.
			.
			.
This thread confuses me.

I've lost track of the real goal.  If it's an exact calculation of
binomial coefficients--or even one of several other potential
targets mentioned--I echo Steven D'Aprano, and ask, are you *sure*
the suggestions already offered aren't adequate?  Also, I think you
might not realize how accurate Stirling's approximation (perhaps to
second order) is in the range of interest.



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