Reading this question about calculating approximate lottery odds inspired me to ask about approximating the values of a hypergeometric distribution. Specifically, the situation I often find myself in (for some definition of "often") is that I have a pool of $N$ items, of which $K$ are useful and the remaining $N-K$ are useless. I pick $n$ items from the pool, without replacement. The probability of getting at least one useful item is

$$\sum_{k = 1}^{n}\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} = 1 - \frac{\binom{K}{0}\binom{N-K}{n}}{\binom{N}{n}} = 1 - \frac{(N-K)!(N-n)!}{(N-K-n)!N!}$$

This is not so bad if I have a calculator, but I wonder if there is a way to approximate this result with mental math? The approximation can be very rough, perhaps valid to half an order of magnitude or so; simplicity of calculation is more important. If it helps, we may assume $K$ and $n$ are much less than $N$. (Sample values: $N = 45$, $K = 3$, $n = 2$) I will occasionally abuse this formula by setting $K \approx \frac{N}{2}$ but the accuracy can be worse in that case.

As an example of the kind of result I'm looking for, if I were picking with replacement, I could use the formula

$$1 - \biggl(1 - \frac{K}{N}\biggr)^n \approx \frac{nK}{N}$$

I guess this might be a reasonable approximation for picking without replacement as well, but is there something else that gives a better result without adding much complexity to the formula? Maybe adding or subtracting 1 somewhere, or squaring something? I'm mostly curious whether something like this is already known among people who study such things. (I'm not really expecting anyone to find an easy way to do much better than $nK/N$ that hasn't already been discovered by centuries of mathematically-inclined gamblers.)


1 Answer 1


If the population size is at least 20 times the sample size, calculating it as you would a binomial distribution gives a good approximation.

See: http://www.stat.rice.edu/~dcox/Stat305/Lessons/L0212/node2.html


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