Probability/Combinatorics Problem - Old Maid Cards > A special deck of Old Maid cards consist of 25 pairs and a single old maid card. All 51 cards evenly between you and two other players – 17 cards for each player. > > (a) how many different hands can be dealt to you? > > (b) what is the probability that your hand has exactly 2 pair (and 13 single cards)? (a) This is easy: $\binom{51}{17}$ (b) This one I'm having trouble. I thought about doing something like this: $\cfrac{\binom{25}{2}*\binom{47}{13}}{\binom{51}{17}}$ 25-C-2 = Choose 2 of 25 pairs 47-C-13 = There's 46 remaining cards (or 23 pairs) but you add 1 because of old maid card 51-C-17 = Total possibilities. I know this answer is wrong because its greater than 1. The solution is 0.30282. Any help is appreciated. Thank you.

Let’s split the calculation into two cases, hands with the Old Maid and hands without.

A hand with the Old Maid that has exactly two pairs contains the Old Maid, two pairs, and $12$ singletons. Such a hand therefore has cards from $12+2=14$ of the $25$ denominations. There are $\binom{25}2$ ways to choose the denominations of the two pairs, and there are then $\binom{23}{12}$ ways to choose the denominations of the $12$ singletons. For each singleton there are $2$ ways to choose which member of the pair we get. Thus, there are $2^{12}\binom{25}2\binom{23}{12}$ hands of this type.

Now see if you can modify that calculation to get the number of hands of the desired type that do not include the Old Maid.