Counting Two Pair 5-card Poker Hands

Question

I'm trying to understand combinatorics and I'm having some trouble with the popular 'Two Pairs' problem. The question asks how many Two Pair hands can be made from a deck of 52.

 

I know the answer is 13c2 * 4c2 ^2 * 11

 

I thought the answer was 13c3 * 4c2 ^ 2 * 4c1 (or 4)

 

My reasoning is a follows:

 

First, I figure, you have to 'choose' 3 unique ranks from the 13 available. And then, you 'choose' (ie count) the number of ways to create a pair from 4 suits (you do this twice for each pair). Finally, you multiply by the number of ways select a single card from a suit (which is just 4). Can someone explain what's wrong with my reasoning? Can someone explain how I can adjust my reasoning so that I don't make this mistake again?

 

As a followup question, how would the problem change if it were 'Three Pair' 6-card hands?

Would it be: 13c3 * 4c2 ^3?

 

Answer 1

 

What could the pairs be? Any two of the following: A K Q J 10 ...3 2 --so which kind of pairs it will be is ₁₃C_2;

 

Now, within the first pair (there are four possible cards--spades, hearts, diamonds, clubs) the choices can be done in ₄C₂ = 6 ways.

 

The same 6 ways applies to whatever second pair is chosen. We have now accounted for the pairings choices. There are 44 other cards that may be finally chosen for the first card of this poker hand.

 

The # of different such 5-card hands is 13C2•4C2•4C2•44 = 78(6²)(44) = 123,552 hands.

 

Dividing the above by 52C5, we have P(2 pairs in 5-card poker hand) = 0.04754