# What is the reasoning behind calculating the probability of getting a two pair in poker?

A two pair is a five card hand with two of one value, two of another value,and one of a third value. Ex: AABBC. Looking online shows that the correct formulation is [(13c2)(4c2)(4c2)(11)(4)]/(52c5). I do not understand why you multiple (4c2) twice though.

The reasoning I tried to apply (incorrectly) was as follows: There are 13 ranks, 4 cards in each rank. Then there are 4c2 ways to get 2 cards of the same rank, for each of the 13 ranks. So for the first pair, we have 13*(4c2) ways to get the desired pair. Once we have that pair one of the ranks is now off limits, so there are only 12*(4c2) ways to get the second pair. And once we have that only 11 ranks remain for the fifth card, each with 4 possibilities, giving us 11*4 options for it. Thus, the total number of ways to get a two pair is 13*12*(4c2)(4c2)(4*11)/(52c5). This is a different value from the one above though. Also, while my method multiplies (4c2) twice as well, the reasoning for it seems to be different than in the method above.

Can someone please explain why the correct method works, and where my thinking has gone wrong?

EDIT: Corrected my final formula.

• What are $13$c$2$ and the like? Aug 28, 2016 at 15:32
• 13c2 is another way of writing $\binom{13}{2}$ Aug 28, 2016 at 15:32
• @Bernard: $13c2$ means $13$ choose $2$. Aug 28, 2016 at 15:33
• In your reasoning, you mention 4*11, but you didn't include them in your formula. Aug 28, 2016 at 15:36
• Also, internet have 52C5 in the denominator, not 52C2. See en.wikipedia.org/wiki/… Aug 28, 2016 at 15:37

We wish to construct a 2-pair hand. First, choose which two ranks are to be pairs; there are $\binom{13}{2}$ ways to do this. Next, choose which two of the first rank we have in our hand - $\binom42$ - and then choose which two of the second rank we have as our second pair - $\binom42$ again. Finally, choose the last card from among the remaining $11$ ranks, so there are $11\cdot 4$ cards to choose from.
By choosing the two ranks for the pairs at the same time, which can be done in $\binom{13}{2}$ ways, you avoid this problem; but you can also just divide your answer by 2 to get the right answer.