While studying the hypergeometric distribution, I came up with this formula below that yields the same numerical result of the standard formula, but looks completely different!

Why are the two formulas equivalent? HELP!

I don't get the intuition why they are equivalent, probably I don't understand the hypergeometric enough yet :/


$\ N$=total marbles in the urne

$\ G$=green marbles in the urne, among the $\ N$ total in the urne

$\ n$=marbles drawn for the urne

$\ g$=green marbles, among the $\ n$ marbles drawn for the urne

Standard formula:

$\ P(X=g) = \frac{{G \choose g} {N-G \choose n-g}}{{N \choose n}}$

Alternative formula:

$\ P(X=g) = \frac{{n \choose g} {N-n \choose G-g}}{{N \choose G}}$

I derived it starting from:

$\ P(X=g) = \frac{D \times R}{A}$


$\ A$=combinations of all $\ N$ marbles in the urne = $\ {N \choose G}$

$\ D$=combinations of $\ n$ drawn marbles, containing $\ g$ green marbles = $\ {n \choose g}$

$\ R$=combinations of $\ (N-n)$ remaining marbles, containing $\ (G-g)$ green marbles = $\ {N-n \choose G-g}$


$$ \frac{\frac{\color{blue}{G!}}{\color{red}{g!}(G-g)!}\frac{\color{blue}{(N-G)!}}{\color{red}{(n-g)!}(N-G-n+g)!}}{\frac{\color{blue}{N!}}{\color{red}{n!}(N-n)!}} =\frac{\color{red}{\frac{n!}{g!(n-g)!}}\frac{(N-n)!}{(G-g)!(N-n-G+g)!}}{\color{blue}{\frac{N!}{G!(N-G)!}}}. $$

  • $\begingroup$ thanks! I had made a mistake in the expansion of the two formulas, that's why I couldn't find the equivalence. Your clear expansion helped me find that mistake :) $\endgroup$ – elemolotiv Mar 21 '18 at 16:03

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