When do I use hypergeometric instead of binomial?

Whenever the draw changes what is left. Cards dealt from a deck, items pulled from a batch for inspection, balls drawn from an urn without putting them back. The binomial assumes the probability is identical every trial, which is only true if you replace what you took.

How different are the two answers?

It depends on how much of the population you draw. Take under about 5% of it and the difference is negligible — which is the standard rule for treating a survey sample as binomial. Take a fifth of it and the binomial overstates the spread noticeably, because it lets the same item be counted repeatedly.

How do I use this for lottery odds?

Population is how many numbers the machine holds, successes is how many are drawn, draws is how many you pick, and k is how many you want to match. For a 6-from-49 game, N = 49, K = 6, n = 6, k = 6 gives one in 13,983,816.

What does the mean mean here?

n·K/N — draws times the share of the population that is a success. It is the same as the binomial mean, which is the thing that surprises people: sampling without replacement changes the spread of the outcomes but not their average.

Hypergeometric Calculator

With population size 52, successes in the population 13, draws 5, successes wanted 2, hypergeometric comes to 27.4280 % — chance of exactly k. It is reached in 7 steps, the last of which is 0.2742797 * 100, and each one is printed on the page with its numbers filled in. The formula is the one published by NIST/SEMATECH e-Handbook §7.3.3, not an approximation fitted to it.

The probability of drawing exactly k of a kind from a finite population without replacement — the card and lottery case the binomial gets wrong.

Formula and sources checked 16 August 2026 · How we check

Population size 52, Successes in the population 13, Draws 5, Successes wanted 2

27.4280 %

Chance of exactly k for the example below. Editing a field recomputes the calculator below; this figure holds the answer the page was loaded with.

It is written into the HTML rather than drawn by a script, so a search engine reading this page without running JavaScript still finds an answer.

Population sizeA deck of cards is 52.Successes in the populationThirteen hearts in a deck.DrawsSuccesses wanted

Chance of exactly k

27.4280 %

Ways to pick k from the K successes: combinations(13, max(0, min(2, 13)))78 ways
Ways to pick the rest from the failures: combinations(52 - 13, max(0, min(5 - 2, 52 - 13)))9,139 ways
Ways to draw n from N at all: combinations(52, max(0, min(5, 52)))2,598,960 ways
P(X = k): (78 * 9139) / 25989600.274
P(X = k) as a percentage: 0.2742797 * 10027.428 %
Mean, n·K/N: 5 * 13 / 521.25 successes
What the binomial would say: combinations(5, max(0, min(2, 5))) * (13 / 52) ^ 2 * (1 - 13 / 52) ^ (5 - 2) * 10026.367 %

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Worked example

Exactly two hearts in a five-card hand: 78 ways to pick the hearts times 9,139 ways to pick three non-hearts, over 2,598,960 hands — 27.43%. The binomial, which assumes the deck is reshuffled between cards, says 26.37%. The gap is small here and grows as the sample takes a bigger bite out of the population.

How to work it out yourself

1.Count the ways to choose k successes from the K available: C(K, k).
2.Count the ways to fill the remaining draws from the failures: C(N − K, n − k).
3.Multiply those two — that is the number of favourable hands.
4.Divide by C(N, n), every hand that could be drawn.

The formula

Ways to pick k from the K successescombinations(13, max(0, min(2, 13)))
Ways to pick the rest from the failurescombinations(52 - 13, max(0, min(5 - 2, 52 - 13)))
Ways to draw n from N at allcombinations(52, max(0, min(5, 52)))
P(X = k)(78 * 9139) / 2598960
P(X = k) as a percentage0.2742797 * 100
Mean, n·K/N5 * 13 / 52
What the binomial would saycombinations(5, max(0, min(2, 5))) * (13 / 52) ^ 2 * (1 - 13 / 52) ^ (5 - 2) * 100

Source: NIST/SEMATECH e-Handbook §7.3.3 — comparing two proportions, which uses the hypergeometric, NIST DLMF §26.3 — binomial coefficients

Questions people actually ask

When do I use hypergeometric instead of binomial?: Whenever the draw changes what is left. Cards dealt from a deck, items pulled from a batch for inspection, balls drawn from an urn without putting them back. The binomial assumes the probability is identical every trial, which is only true if you replace what you took.
How different are the two answers?: It depends on how much of the population you draw. Take under about 5% of it and the difference is negligible — which is the standard rule for treating a survey sample as binomial. Take a fifth of it and the binomial overstates the spread noticeably, because it lets the same item be counted repeatedly.
How do I use this for lottery odds?: Population is how many numbers the machine holds, successes is how many are drawn, draws is how many you pick, and k is how many you want to match. For a 6-from-49 game, N = 49, K = 6, n = 6, k = 6 gives one in 13,983,816.
What does the mean mean here?: n·K/N — draws times the share of the population that is a success. It is the same as the binomial mean, which is the thing that surprises people: sampling without replacement changes the spread of the outcomes but not their average.

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