SigmaCalc

Hypergeometric Distribution Calculator

The distribution behind card hands, lottery draws, and quality-control sampling from a fixed batch.

Population size (N)

Successes in population (K)

Sample size (n)

Successes in sample (k)

P(X = k)

0.274

P(X=k) = C(K,k)×C(N−K,n−k) ÷ C(N,n)

What you entered

Population size (N): 52Successes in population (K): 13Sample size (n): 5Successes in sample (k): 2

1
Ways to choose the successes from the population
C(13, 2)= 78
2
Ways to choose the failures from the rest of the population
C(39, 3)= 9139
3
Total ways to draw the sample
C(52, 5)= 2598960
4
P(X = k)
(78 × 9139) ÷ 2598960= 0.2743

Result

P(X = k): 0.274

Drawing 5 items without replacement from a population of 52 (13 successes), there's a 27.43% chance of getting exactly 2 successes — this is the distribution behind card-hand and lottery-style problems.

Hypergeometric probability - sampling without replacement

N is the population, K the number of successes in it, n the sample drawn and k the successes in that sample. Unlike the binomial, each draw changes what remains.

Population N	Successes K	Sample n	Successes k	Probability
52	4	5	1	0.2995
52	13	5	2	0.2743
20	7	5	2	0.3874
50	10	10	3	0.2178

The first row is exactly one ace in a five-card poker hand, at 29.95%. Row two is exactly two hearts in five cards. The distinction from the binomial is replacement: once you draw an ace, only three remain among 51 cards, so the probability of the next draw changes. Use hypergeometric whenever you sample from a finite population without putting items back - quality inspection from a batch, cards, or drawing names from a hat. When the population is very large relative to the sample the two distributions converge, and the binomial becomes a good approximation.

Without replacement is the key difference

Binomial probability assumes each draw doesn't affect the odds of the next (like flipping a coin). Hypergeometric probability is for sampling without replacement — like dealing cards — where each draw changes the remaining pool.

A classic example

The probability of drawing exactly 2 aces in a 5-card poker hand is a hypergeometric problem: population size 52, successes in population 4 (the aces), sample size 5, and you're solving for k = 2.

Frequently asked questions

What's the probability of drawing 2 aces in a 5-card poker hand?

Population N=52, successes K=4 (aces), draw n=5, want k=2. P = C(4,2)×C(48,3)/C(52,5) = 6×17,296/2,598,960 = 0.0399 = 4.0%. About 1 in 25 hands will contain exactly 2 aces.

How is hypergeometric different from binomial?

Binomial: each trial has the same probability (sampling WITH replacement, like repeated coin flips). Hypergeometric: each draw changes the remaining pool (WITHOUT replacement, like dealing cards). For large populations, they converge — drawing 5 from 10,000 is nearly binomial.

A batch of 100 parts has 8 defective. I inspect 10. What's the chance I find exactly 1 defective?

N=100, K=8, n=10, k=1. P = C(8,1)×C(92,9)/C(100,10) = 8×1.096×10¹⁰/1.731×10¹³ ≈ 0.383 = 38.3%. Finding exactly 1 defective is the most likely single outcome in this inspection.