# Probability that a five-card poker hand contains two pairs

What is the probability that a five-card poker hand contains two pairs (that is, two of each of two different ranks and a fifth card of a third rank)?

My attempt:

Let us first pick the 3 different ranks. There are ${13\choose 3}$ ways of doing this. Out of each rank consisting of 4 suits, we must pick 2 cards, 2 cards and 1 card respectively. So, no. of ways $={13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}$

Total no. of ways of selecting a five-card poker hand $={52\choose 5}$

$p=\dfrac{{13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}}{{52\choose 5}}$

This doesn't match the answer given in the textbook. Where have I gone wrong?

• You forgot to choose which of the three ranks would be the one with only a single card. Nov 14 '15 at 18:59
• Undercounting, which is unusual. If done in your style It should be $\binom{13}{2}\binom{11}{1}$. Nov 14 '15 at 19:00

You're pretty close, but there is a problem: you do have to choose 3 ranks, but they're not all going to be treated the same. One will be a single, and two others will be pairs. If you multiply by a factor of $\binom{3}{2}$ I think you'll have it.

In total, there are $$52\choose5$$ ways to draw a hand (this is our |S|).

We want to choose 2 out of four cards of one value, 2 out of four cards of another value, and one other card not of the first two values (This will be our |E|).

First we choose two values, there are 13 values (2 to A), so $$13\choose2$$.

Then we want to choose two cards of the first value out of four cards, $$4\choose 2$$

Again, we want to choose two cards of the second value out of four cards, $$4\choose 2$$

And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), $${52-8\choose1} = {44\choose1}$$

So we get: $${{{13\choose2}\times{4\choose2}\times{4\choose2}\times{44\choose1} }\over{52\choose2}} = {198\over4165} ≈ 0.0475$$