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?
 A: 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.
A: 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:
$${{{13\choose2}\times{4\choose2}\times{4\choose2}\times{44\choose1} }\over{52\choose2}} = {198\over4165} ≈ 0.0475$$
