Hypergeometric distribution

[Paul Rubin]

"Raven" <balckraven at gmail.com> writes:
> The problem with Stirling's approximation is that I need to calculate
> the hypergeometric hence the factorial for numbers within a large range
> e.g. choose(14000,170) or choose(5,2)

Stirling's approximation to second order is fairly accurate even at
low values:

    from math import log,exp,pi

    def stirling(n):
        # approx log(n!)
        return n*(log(n)-1) + .5*(log(2.*pi*n)) + 1/(12.*n)


    >>> for n in range(1,6): print n, exp(stirling(n))
    ... 
    1 1.00227444918
    2 2.00065204769
    3 6.00059914247
    4 24.0010238913
    5 120.002637086
    >>> 

To third order it's even better:

    from math import log,exp,pi

    def stirling(n):
        # approx log(n!)
        return n*(log(n)-1) + .5*(log(2.*pi*n)) + 1/(12.*n) - 1/(360.*n*n*n)


    >>> for n in range(1,6): print n, exp(stirling(n))
    ... 
    1 0.999494216712
    2 1.99995749743
    3 5.99998182863
    4 23.9999822028
    5 119.999970391
    >>> 

Reference: http://en.wikipedia.org/wiki/Stirling's_approximation