Hypergeometric Distribution Calculator

• N is the whole population (52 cards, 50 units in the batch) and K is how many of them count as successes (4 aces, 5 defective units).
• n is how many you draw, and k is how many successes you ask about.
• Use this instead of the binomial distribution whenever each draw changes the remaining pool: that is what “without replacement” means.
• For lottery games, the lottery odds calculator applies this distribution to the main draw and adds the bonus draw, the odds of every prize level and the chance with many tickets.

The numerator counts the samples that contain exactly k of the K successes (and therefore n − k of the N − K non-successes); the denominator counts all possible samples of size n. Every draw is a plain count of equally likely hands, so the result is exact. The combination calculator computes the individual C(a, b) terms if you want to verify by hand. You cannot draw more successes than exist (k ≤ K) or than you draw (k ≤ n), and if the sample is larger than the number of non-successes some successes are forced in. For a k outside that range the calculator shows a probability of exactly 0 and lists the values that are possible. With very large populations (up to 1,000,000,000) the terms are evaluated with the saddle-point method used by R's dhyper rather than as huge factorials.

The mean is the same as the binomial's (np with p = K/N), but the variance carries the extra factor (N − n)/(N − 1), the finite population correction. It is less than 1 whenever you sample more than one item: each success drawn depletes the pool, pulling later draws toward fewer successes and shrinking the spread compared with independent draws.

When the sample is a small fraction of the population (a common working rule is n/N ≤ 0.05) the correction is nearly 1 and the binomial with p = K/N is an excellent approximation. Sampling 10 units from a batch of 10,000 barely dents the pool; sampling 10 from 50 dents it badly, and the exact hypergeometric answer is then meaningfully different.

A batch of N = 50 units contains K = 5 defective ones. An inspector draws n = 10 units without replacement. What are the chances of finding defects?

1. No defects: P(X = 0) = C(45, 10)/C(50, 10) ≈ 0.310563.
2. Exactly one: P(X = 1) = C(5, 1) × C(45, 9)/C(50, 10) ≈ 0.431337.
3. At least one: 1 − 0.310563 ≈ 0.689437: the batch fails a zero-tolerance inspection about 69% of the time.

The summary reads mean = 10 × 5/50 = 1 defect per sample and variance 10 × 0.1 × 0.9 × 40/49 ≈ 0.7347 (SD ≈ 0.8571). Compare the binomial approximation with p = 0.1: it gives P(X = 0) = 0.9¹⁰ ≈ 0.3487, about 12% too high, because drawing 10 from 50 is a 20% bite of the population. Load example reproduces these numbers.

Cards work the same way. In a five-card hand from a 52-card deck (N = 52, K = 4, n = 5) the chance of exactly two aces is 0.03993, and the chance of at least one ace is 0.341158.

Matching lottery numbers is sampling without replacement: the draw picks 6 balls from 49 and your ticket marks 6 of them as successes (N = 49, K = 6, n = 6). The chance of matching exactly 3 is 0.01765, about 1.77%, and the chance of at least 3 matches is 0.018638, about 1 in 54. The jackpot, P(X = 6), is 1/C(49, 6) = one in 13,983,816.

The argument order differs between programs, so match the letters above with care: R takes the number of failures in the population (N − K) as its third argument, and SciPy takes the population size first.

The combination calculator counts the samples behind the formula, and the dice probability calculator covers draws with replacement. To compare two groups on the same kind of table, the Fisher exact test is a hypergeometric test. Read binomial vs Poisson vs hypergeometric to choose the right model.