Will Your Opening Hand Have the Combo?

A 60-card deck holds 4 copies of a key card. Draw 7: what are the odds of at least one? Cards don't go back once drawn, so every draw changes the odds. That's the hypergeometric distribution, and deck builders live by it.

How it works

The hypergeometric distribution: sampling without replacement

In one sentence

If $n$ items are drawn without replacement from $N$ , of which $K$ are successes, the number of successes $X$ is hypergeometric: $P(X=k) = \binom{K}{k}\binom{N-K}{n-k}\big/\binom{N}{n}$ , with mean $nK/N$ and variance $n\frac{K}{N}\frac{N-K}{N}\frac{N-n}{N-1}$ .

General formula

P(X = {\htmlData{sym=3}{k}}) = \frac{\binom{\htmlData{sym=1}{{K}}}{\htmlData{sym=3}{{k}}}\binom{{\htmlData{sym=0}{N}}-{\htmlData{sym=1}{K}}}{{\htmlData{sym=2}{n}}-{\htmlData{sym=3}{k}}}}{\binom{\htmlData{sym=0}{{N}}}{\htmlData{sym=2}{{n}}}}

Count the hands that work and divide by all possible hands. Each card drawn leaves the deck slightly different, which is what separates this from flipping coins.

What the symbols mean

$N$
Deck size
How many cards are in the deck.
$K$
Copies
How many copies of the key card the deck holds.
$n$
Cards drawn
The size of the hand.
$k$
Copies in hand
How many of the key card end up in the hand.

Count hands: there are $\binom{N}{n}$ equally likely hands of $n$ cards from a deck of $N$ .
A hand with exactly $k$ copies picks $k$ of the $K$ copies and $n-k$ of the other $N-K$ cards, in $\binom{K}{k}\binom{N-K}{n-k}$ ways.
Dividing gives the hypergeometric probability. The expected number of copies is $nK/N$ , the same as for the binomial.
Unlike the binomial (drawing with replacement), each card drawn changes what's left, so the distribution is a little narrower. With a big deck and a small hand the two nearly agree.

Learn the math

Geometric and Hypergeometric DistributionsStats · Lesson 16, Discrete Random Variables