Hypergeometric Draws From a Deck

If you flip a coin ten times, the number of heads follows the binomial distribution — each flip is independent, so getting a head doesn't change the odds of the next flip. But if you deal five cards from a deck and ask "what's the probability I got exactly two aces," the deck doesn't reset between cards. Every card you draw changes the composition of what's left. That's a different distribution, called the hypergeometric, and it's the one that governs essentially every "how many of a certain kind of card ended up in my hand" question you'll be asked.

The setup

You have a deck of $N$ cards, of which $K$ are "special" (say, aces, so $K=4$), and you draw $n$ cards without putting any back. The probability of getting exactly $k$ special cards among your $n$ draws is

$$P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} .$$

In plain English: the numerator counts the ways to pick $k$ aces out of the 4 available and the remaining $n-k$ cards out of everything that isn't an ace, and the denominator is the total number of ways to pick any $n$ cards from the deck at all. It's the same "count the favorable ways, divide by the total ways" combinatorics as any poker-hand question — the hypergeometric distribution is just this ratio written as a formula you can reuse instead of re-deriving each time.

Worked example 1: exactly two aces in a five-card hand

With $N=52$, $K=4$ aces, $n=5$ cards drawn, and $k=2$:

$$P(X=2) = \frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}} = \frac{6 \times 17{,}296}{2{,}598{,}960} = \frac{103{,}776}{2{,}598{,}960} \approx 3.99\% .$$

So roughly 1 in 25 five-card hands contains exactly two aces. Notice this differs from what a (wrong) binomial approximation would give if you treated each of the 5 draws as an independent $4/52$ chance — that ignores that the deck shrinks and the ace count changes with each card drawn, and it would overstate how "spread out" the count of aces should be.

Worked example 2: the expected number of aces

A useful shortcut: for a hypergeometric draw, the expected count of special cards is

$$E[X] = n \cdot \frac{K}{N} .$$

This matches the intuitive "proportion" answer even though the individual draws aren't independent — linearity of expectation doesn't need independence. For a 5-card hand, $E[X] = 5 \times \frac{4}{52} \approx 0.385$ aces on average. If instead you ask "how many aces do I expect in my hand given I know 3 of the other players already showed no aces in their 15 combined cards," you're really asking a hypergeometric question on a shrunk deck of $52 - 15 = 37$ cards with the same 4 aces still unseen — the same formula applies, just update $N$ (and possibly $K$) to reflect what's already been revealed.

Drag $n$ and $p$ in the explorer above to see how the shape compares to a true hypergeometric count of aces in a 5-card hand (here approximated with $p = 4/52 \approx 0.077$) — for a deck this large relative to the hand size, the binomial approximation is visually close, which is why it's often used as a fast mental shortcut even though the exact answer is hypergeometric.