Hypergeometric Distribution Calculator (Without Replacement)
By Santaji GadepatilinUpdated November 4, 2026
When you draw items without putting them back — cards from a deck, parts from a production lot — each draw changes the odds for the next one, and the binomial distribution no longer fits. This hypergeometric distribution calculator gives the probability of exactly k successes in n draws, the chance of at least one, and the expected number of successes.
How to use the hypergeometric distribution calculator
Enter the population size N, the number of successes K in it, the number of draws n and the number of successes k you want in the draw. The calculator returns the probability of exactly k successes, the probability of at least one success, and the mean and standard deviation of the number of successes.
Hypergeometric formula
P(X = k) = C(K, k) × C(N − K, n − k) ÷ C(N, n)
μ = n K ÷ N · σ² = n (K/N) (1 − K/N) (N − n) ÷ (N − 1)
Example
The chance of exactly one ace in a 5-card hand (N = 52 cards, K = 4 aces, n = 5, k = 1):
- C(4, 1) × C(48, 4) = 4 × 194,580 = 778,320 favourable hands
- C(52, 5) = 2,598,960 possible hands
- P = 778,320 ÷ 2,598,960 = 29.95%
- P(at least one ace) = 1 − C(48, 5) ÷ C(52, 5) = 34.12%
- Expected aces = 5 × 4 ÷ 52 = 0.3846, standard deviation = 0.5720
Aces in a 5-card hand
| Aces k | Hands | P(X = k) | P(X ≤ k) |
|---|---|---|---|
| 0 | 1,712,304 | 65.8842% | 65.8842% |
| 1 | 778,320 | 29.9474% | 95.8316% |
| 2 | 103,776 | 3.9930% | 99.8245% |
| 3 | 4,512 | 0.1736% | 99.9982% |
| 4 | 48 | 0.0018% | 100.0000% |
Notes
- k cannot exceed K or n, and n − k cannot exceed N − K; such combinations have zero probability.
- The counts use combinations — see the combinations calculator.
Frequently asked questions
What is the hypergeometric distribution formula?
P(X = k) = C(K, k) × C(N − K, n − k) ÷ C(N, n) — the ways to choose k of the K successes and n − k of the failures, divided by all ways to choose n items.
When should I use hypergeometric instead of binomial?
When sampling without replacement from a finite population. If the sample is less than about 5% of the population, the binomial distribution with p = K/N gives nearly the same answer.
What is the probability of being dealt exactly one ace in 5 cards?
C(4,1) × C(48,4) ÷ C(52,5) = 778,320 ÷ 2,598,960 ≈ 29.95%.
How is it used in quality control?
In acceptance sampling, a lot of N parts contains K defectives and you inspect n. P(X = 0) is the chance the sample shows no defects, so 1 − P(X = 0) is the chance of catching at least one.