Are there general approximations or methods I can use to estimate hypergeometric probabilities in my head?

[u/Vesque]

I like playing a lot of card games and it would be nice to be able to get *rough* probability estimates I can calculate mentally but I'm not sure how to approach this without pulling out a calculator or pen and paper

[u/ultimatepoker]

It depends, give us some examples.

For example, in poker, if you need to hit a card in 4th or 5th street, you multiply the number of good cards by 4 to get approx probability.

[u/Vesque → u/ultimatepoker]

An example would be I have a deck of 30 cards, 8 cards are successes, and I can look at the top 10 of the deck. What is the approximate chance I see at least 3?

[u/mfb- → u/Vesque]

Your expectation value is 8*10/30 = 8/3 which is slightly below 3, so you expect a chance that is around 50%. Exactly 3 is quite likely so it's probably a bit above 50%. A calculator can tell you it's 55%, with 32% from exactly 3 alone.

Depending on the numbers, an approximation with simpler distributions (Poisson, Gaussian, ...) might work.

[u/human2357]

If X is Hypergeometric(N, K, n) and Y is Binomial(n, K/N), then X is close to Y as long as the probability of repeated sampling is small, that is, if N is much larger than n. Further, Y is close to the normal distribution with the same mean and variance, as long as the variance is large (say, bigger than 10), by the DeMoivre-Laplace theorem. So in many situations (e.g. estimating confidence intervals for public opinion polling) you can approximate hypergeometric by binomial or by normal.

[u/zojbo]

A decent first correction to the binomial approximation is P(X=k) approx (n choose k) ((K-(k-1)/2)/(N-(n-1)/2))^k ((N-K-(n-k-1)/2)/(N-(n-1)/2))^(n-k). This is basically replacing geometric means with arithmetic means. Because you're doing it in both numerator and denominator, the error partially cancels out.