# What is the probability of two-pair poker hand?

What is the probability that a five card poker hand will have two pairs (with no additional cards)?

Example of two-pair poker hand: 2${\heartsuit}$, 2${\spadesuit}$, 3${\clubsuit}$, 3${\diamondsuit}$, A${\spadesuit}$

We have to pick three types of cards from {2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A}:
Number of ways of choosing three different numbers from set of 13 numbers = ${13 \choose 3}$
Number of ways of choosing suits for two of the pairs = ${4 \choose 2}^2$
Number of ways of choosing a suit for the side card = ${4 \choose 1}$

I believe the above way is correct and comprehensive way of counting all possible combinations. Am I wrong?

For context, the solution to one of the MIT questions isn't dividing by the extra 3: Question 1 and Answer 1

• I wanted to say, how the question has been asked before. My question in exactly what I wanted to ask, I'll edit my question. Thanks! – alice_in_wonderland Jul 22 '16 at 20:17
• your mistake is with the line: "Number of ways of choosing suits for two of the pairs" -- you have to multply this by $3 \choose 2$ because, you already chose the 3 numbers -- you have to chose the 2 with the pair – d_e Jul 22 '16 at 20:17
• Now I get it. Thanks! – alice_in_wonderland Jul 22 '16 at 20:18
• Edited again. I hope I am clearer now. – alice_in_wonderland Jul 22 '16 at 20:20
• I'm going to edit the Question, moving the "problem" stated in the title so that the body is more self-contained. Please review to see if I'm changing your intended meaning. – hardmath Jul 23 '16 at 1:02

Pick the two ranks of the pairs ${13 \choose 2}$, then pick the suits of the cards in the pairs (${4 \choose 2}^2$). Pick the fifth card from the remaining $11$ ranks ($44$).
Then divide by ${52 \choose 5}$.
You have to distinguish between the pair ranks and the non-pair rank. As you have it now, you're multiple-counting the hands. If you pick, say, $A,K,Q$, you have three choices for which two are pairs, but in your question you don't account for this. By choosing only two ranks, you leave no doubt as to which two ranks are the pairs. Now you have $44$ cards left (or, if you like, $11$ ranks and $4$ suits) from which to populate the fifth card of the hand.
• @alice_in_wonderland Suppose you have a deck of just $\{A\heartsuit,A\spadesuit,2\heartsuit,2\spadesuit,3\heartsuit,3\spadesuit\}$ and wish to select two pairs and a singleton. Clearly there are six ways to do this as you just have to discard one card to leave the required hand. That is equal to $\binom{3}{2}\binom{1}{1}\binom{2}{2}^2\binom{2}{1}$ and not to $\binom{3}{3}\binom{2}{2}^2\binom{2}{1}$ So it is apparent you do have to distinguish between pairs and singleton when selecting faces. – Graham Kemp Jul 22 '16 at 23:12