# Counting poker hands

In this morning’s blog post, John D. Cook talks about poker hands and their probabilities. In particular, he says

For a five-card hand, the probabilities of 0, 1, or 2 pair are 0.5012, 0.4226, and 0.0475 respectively.

Upon reading this, I assumed there’d be an explanation of how these probabilities were calculated. But no, he just leaves us hanging. So I sat down and worked them out.

The probabilities are going to be calculated by dividing the number of ways we can get a certain type of hand by the total number of possible hands. So let’s start by working out the denominator. There are 52 cards, so if we deal out 5 cards, there are

$\left(\genfrac{}{}{0}{}{52}{5}\right)=2,598,960$

possible ways to do it. The symbol that looks sort of like a fraction without the dividing line represents the binomial coefficient,

$\left(\genfrac{}{}{0}{}{n}{k}\right)=\frac{n!}{k!\phantom{\rule{thickmathspace}{0ex}}\left(n-k\right)!}$

This is not only the coefficient of binomial expansions, it’s also the number of ways we can choose k items out of n without regard to order, also referred to as the number of combinations.

To figure out the number of one-pair hands, imagine the deck laid out in 13 piles: a pile of aces, a pile of kings, a pile of queens, and so on. We then do the following:

1. Pick one of the piles. There are 13 ways to do this.
2. From that pile, choose 2 cards. There are

$\left(\genfrac{}{}{0}{}{4}{2}\right)=6$

ways to do this.

3. Pick three other piles. There are

$\left(\genfrac{}{}{0}{}{12}{3}\right)=220$

ways to do this.

4. Pick one card from each of these three piles. Since there are 4 cards in each pile, there are ${4}^{3}=64$ ways to do this.

So the total number of one-pair hands is

$13×6×220×64=1,098,240$

and the probability of getting a one-pair hand is

$\frac{1,098,240}{2,598,960}=0.42257$

Moving on to two-pair hands, we start with the same 13 piles we had before. We then do the following:

1. Pick two of the piles:

$\left(\genfrac{}{}{0}{}{13}{2}\right)=78$
2. Pick two cards from each of these piles:

$\left(\genfrac{}{}{0}{}{4}{2}\right)×\left(\genfrac{}{}{0}{}{4}{2}\right)=6×6=36$
3. At this point, any of the remaining 44 cards will give us a two-pair hand.

Therefore, the total is

$78×36×44=123,552$

and the probability of drawing a two-pair hand is

$\frac{123,552}{2,598,960}=0.04754$

Counting the number of no-pair hands is a little trickier because we have to make sure we don’t mistakenly count better hands, i.e., straights and flushes. Starting again with our 13 piles, we do the following:

1. Pick 5 piles

$\left(\genfrac{}{}{0}{}{13}{5}\right)=1,287$
2. Eliminate all the straights from this number. The number of straights is most easily done through simple enumeration:

A–5, 2–6, 3–7, 4–8, 5–9, 6–10, 7–J, 8–Q, 9–K, 10–A

So 10 of our 1,287 possible pile selections will give us a straight. That leaves 1,277 non-straight possibilities.

3. Pick one card from each of the selected piles. That gives ${4}^{5}=1,024$ possibilities.
4. Eliminate the flushes from this set of possibilites. There are 4 flushes, leaving 1,020 possibilities.

These steps give us

$1,277×1,020=1,302,540$

hands that are worse than one-pair. The probability of getting such a hand is

$\frac{1,302,540}{2,598,960}=0.50118$

I kept one more digit in my answers than John did, but they all match.