Probability of Two Pairs in Five-Card Poker Hand

[alexmahone]

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 [math]{13\choose 3}[/math] 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 [math]={13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}[/math]

Total no. of ways of selecting a five-card poker hand [math]={52\choose 5}[/math]

[math]p=\dfrac{{13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}}{{52\choose 5}}[/math]

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

 

[Jameson]

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 [math]{13\choose 3}[/math] 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 [math]={13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}[/math]

Total no. of ways of selecting a five-card poker hand [math]={52\choose 5}[/math]

[math]p=\dfrac{{13\choose 3}\cdot {4\choose 2}\cdot {4\choose 2}\cdot {4\choose 1}}{{52\choose 5}}[/math]

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

If there are two pairs, then that already accounts for 4/5 cards in the hand. Let's break this up into the 4 cards that contain the pairs and the one card that doesn't.

There are 13 possible ranks, as you said, and we are picking 2 of them to match. We need 2,2 or 3,3, etc. Those kind of matches. Then for each of those 2 ranks they could be 1 of 4 possible suits. How would you express that?

Once you pick the two pairs, we need to pick the last card. How many ranks are left available?

(The $\binom{4}{2}$ terms and the $\binom{4}{1}$ term are all correct. You need to modify your first term and add one more term.)