How this instrument works
The hypergeometric distribution answers a very specific question: from a finite population of N items, K of which are 'successes' and the rest failures, if you draw n items without putting any back, what is the probability that exactly k of them are successes? The defining feature is 'without replacement' — each draw changes the composition of what's left, so the odds shift as you go, unlike a binomial experiment where every trial faces the same fixed probability.
The classic illustration is a deck of cards: draw 5 cards from a standard 52-card deck (N = 52), of which 4 are aces (K = 4), and ask for the probability of exactly 1 ace among the 5 (n = 5, k = 1). The PMF, P(X = k) = C(K,k) × C(N−K,n−k) / C(N,n), counts it directly: C(K,k) ways to choose the k successes from the K available, times C(N−K,n−k) ways to choose the remaining n−k draws from the N−K non-successes, divided by C(N,n), the total number of equally likely ways to choose any n items from the population.
The binomial distribution is the hypergeometric distribution's close cousin for the 'with replacement' case — when each draw is put back before the next, the probability of success stays fixed at K/N every time, and the binomial formula applies instead. As the population N grows very large relative to the sample size n, removing a few items barely changes the remaining proportion, and the hypergeometric distribution converges toward a binomial distribution with p = K/N — which is why survey sampling from a huge population is often modeled with a binomial approximation even though, strictly, no replacement is happening.
- Enter the total population size into Population size (N) — the full pool you are drawing from.
- Enter how many successes exist in that population into Successes in the population (K) — for example, 4 aces in a 52-card deck.
- Enter your sample size into Number of draws (n) — how many items you pull out, all at once or one by one without replacing them.
- Enter the count you want the probability for into Observed successes (k) — how many of the draws you want to land as successes.
- Read P(X = k) — probability of exactly k successes among the draws — the resulting hypergeometric probability.
- k cannot exceed n, K, or leave more non-successes needed than remain in the population — the instrument flags any combination that is not physically possible.
Worked example — exactly 1 ace in a 5-card poker hand
Enter N = 52 into Population size, K = 4 into Successes in the population, n = 5 into Number of draws, and k = 1 into Observed successes — the classic question of drawing exactly 1 ace in a 5-card hand from a standard deck. The instrument computes C(4,1) = 4 ways to choose the 1 ace, C(48,4) = 194,580 ways to choose the remaining 4 non-ace cards from the other 48, and C(52,5) = 2,598,960 total possible 5-card hands.
P(X = k) reads 0.299473635608, computed as (4 × 194,580) / 2,598,960 = 778,320 / 2,598,960. So a little under 30% of all possible 5-card hands contain exactly one ace — the single most likely ace-count in a random hand, ahead of zero aces, two aces, or more.
Questions
How is the hypergeometric distribution different from the binomial distribution?
The binomial distribution assumes each trial is independent with the same fixed success probability p throughout — as when drawing with replacement, or flipping a coin repeatedly. The hypergeometric distribution instead draws without replacement from a finite population, so each draw changes the odds for the next one: pulling an ace out of a deck makes the next card slightly less likely to be an ace, since there are now only 3 aces left among 51 cards. Use hypergeometric whenever items are removed permanently from a finite pool as they're drawn.
When can I approximate a hypergeometric probability with a binomial one?
When the population N is large relative to the sample size n — a common rule of thumb is n less than about 5% of N. In that regime, removing a handful of items barely shifts the remaining proportion of successes, so the binomial distribution with p = K/N gives a close approximation. Drawing 5 cards from a 52-card deck does not qualify (5 is nearly 10% of 52), which is why the exact hypergeometric formula matters for the worked example rather than a binomial shortcut.
Why do I need four separate inputs instead of just two?
Because the hypergeometric distribution is fully defined by the population's composition (N total, K successes within it) and the sampling plan (n draws, k of them observed as successes) — all four numbers interact in the formula. Leaving any one out would make the probability impossible to pin down: knowing there are 4 aces in a deck (K = 4) tells you nothing about the probability of drawing exactly 1 unless you also know the deck size (N) and how many cards you're drawing (n).
What does it mean if the instrument flags my inputs as invalid?
It means the combination is not physically possible: you cannot draw more items than exist (n > N), observe more successes than you drew (k > n), observe more successes than exist in the population (k > K), or need more non-successes than remain after removing the K successes (n − k greater than N − K). Any of these makes C(N,n), C(K,k), or C(N−K,n−k) undefined for the scenario, so the instrument asks for a consistent set of numbers instead.
Does the order of drawing matter for this calculation?
No — the formula counts unordered combinations (via C(N,n) and its components), so it treats a sample as a set of n items regardless of the order they were drawn in. Whether you deal 5 cards one at a time or grab a fistful of 5 at once, the probability of exactly k successes among them comes out the same.