The Hypergeometric Distribution

The hypergeometric distribution shows up whenever you reach into a finite pile and grab a handful without putting anything back. Deal five cards from a deck and count the aces; pull ten parts off a shipment and count the defects; audit twelve of a hundred trades and count the errors. Each item you remove changes what's left, so the odds shift with every draw, and that shifting is exactly what separates the hypergeometric from its more famous cousin, the The Binomial Distribution.

Set up the pool with three numbers: $N$ items in total, of which $K$ are "successes" (aces, defects, whatever you're counting), and you draw a sample of size $n$. The probability of getting exactly $k$ successes is a ratio of counting numbers:

$$P(X = k) = \frac{\dbinom{K}{k}\dbinom{N-K}{\,n-k\,}}{\dbinom{N}{n}} .$$

Read it as: (ways to pick $k$ of the $K$ successes) times (ways to pick the remaining $n-k$ from the $N-K$ non-successes), divided by (ways to pick any $n$ from all $N$). It's pure Counting: Permutations and Combinations: favourable arrangements over total arrangements.

With replacement or without: binomial vs hypergeometric

The single question that decides which model you need is: does each draw put the item back? Same experiment, one word of difference, two distributions.

	Binomial	Hypergeometric
Sampling	with replacement (or infinite pool)	without replacement (finite pool)
Draw-to-draw odds	fixed at $p$ every time	shift after each draw
PMF	$\binom{n}{k}p^k(1-p)^{n-k}$	$\dbinom{K}{k}\dbinom{N-K}{n-k}\big/\dbinom{N}{n}$
Mean	$np$	$n\,\tfrac{K}{N}$
Variance	$np(1-p)$	$n\,\tfrac{K}{N}\big(1-\tfrac{K}{N}\big)\tfrac{N-n}{N-1}$

Notice the means match if you read $p = K/N$: on average, the fraction of successes in your sample equals the fraction in the pool, replacement or not. The variance is where they part company.

The finite-population correction

That extra factor $\frac{N-n}{N-1}$ in the variance is the finite-population correction, and it's always at most 1, so sampling without replacement is less variable than sampling with. The intuition: once you've drawn most of a small pool, there's little left to be surprised by, in the extreme, if you draw the entire pool ($n = N$) the correction is zero and you know the count exactly. When the pool is huge relative to your sample ($N$ far bigger than $n$), the correction is nearly 1 and the hypergeometric becomes indistinguishable from the binomial. That's why "with replacement" is a fine approximation when you're sampling a tiny slice of a big population.

Worked example

A shipment has $N = 20$ items, of which $K = 5$ are defective. A quality inspector pulls $n = 4$ at random, without replacement. What's the chance exactly 2 of the 4 are defective?

$$P(X = 2) = \frac{\dbinom{5}{2}\dbinom{15}{2}}{\dbinom{20}{4}} .$$

Compute each piece: $\binom{5}{2} = 10$ (ways to choose 2 defectives from 5), $\binom{15}{2} = 105$ (ways to choose 2 good ones from 15), and $\binom{20}{4} = 4845$ (total ways to choose any 4). So

$$P(X = 2) = \frac{10 \times 105}{4845} = \frac{1050}{4845} \approx 0.217 .$$

About a 22% chance of catching exactly two defects. And the expected number of defects in the sample is $E[X] = n\,\tfrac{K}{N} = 4 \times \tfrac{5}{20} = 1$, the same one-quarter defect rate as the shipment, scaled to four draws.

Where you'll meet it

• Card and lottery problems. "Probability of exactly two aces in a five-card hand" is textbook hypergeometric.
• Quality control and auditing. Inspecting a finite batch without replacement is its native habitat.
• Fisher's exact test. The p-value for a $2\times2$ contingency table is computed from hypergeometric probabilities, the small-sample alternative to a chi-squared test.

The hypergeometric completes the counting family: it's the without-replacement partner of the The Binomial Distribution, built entirely from Counting: Permutations and Combinations, and it sits alongside the The Negative Binomial Distribution and the rest of the Common Distributions every quant should recognise on sight.