Poker combinatorics: chances opponent has a full house given that you have 2 pairs I've read the answers to some very similar problems to this, but I can't manage to apply the knowledge to this specific one, sorry for this redundance. Here's the question:
What is the probability that your opponent is dealt a full house, given that you have two pairs?
My attempt: A: You have 2 pairs
B: The opponent has a full house
$P(B|A) = \frac{P(BA)}{P(A)}$
P(A) is easy enough to calculate, but it's P(BA) that's causing me problems. I'm interpreting it as the probability of the dealer selecting 10 cards that will provide the opponent with a full house and you with 2 pairs. Right now I have:
$P(BA) = \frac{ {13\choose 2}{13\choose 1}{4\choose 2}^{3}{10\choose 3}{4\choose 3}{??\choose??} }{52\choose 10}$
The ${??\choose ??}$ is supposed to represent the final card in the 2 pair hand, but there seem to be so many different cases that I don't know how to calculate it.
I don't even know if I'm close. Thanks for any help!
A: The number of ways you can receive two pairs is 
$$\binom{13}{2}\binom{4}{2}^2\binom{11}{1}\binom{4}{1}$$
since we must select two of the thirteen ranks for the pairs, two of the four suits from each of those ranks, one of the other eleven ranks for the singleton (omitted from your calculation), and one of the four suits of that rank.
Let's count the number of ways the other person can have a full house if you have two pairs.  We consider cases.
The triple comes from the same rank as your singleton and the pair comes from the same rank as one of your pairs:  This can occur in 
$$\binom{1}{1}\binom{2}{1}$$
ways since there is only one way to select the rank of your singleton for the triple, two ways to select one of the ranks of your pairs for your opponent's pair, and one way to select all of the remaining cards of each of those ranks.
The triple comes from the same rank as your singleton and the pair does not come from the same rank as one of your pairs:  This can occur in
$$\binom{1}{1}\binom{10}{1}\binom{4}{2}$$
ways since the pair come from one of the ten ranks not used for your two pairs or your singleton and two suits must be chosen from that rank.
The triple does not come from the same rank as your singleton and the pair comes from the same rank as one of your pairs:  This can occur in 
$$\binom{10}{1}\binom{4}{3}\binom{2}{1}$$
ways since your opponent's triple must come from one of the ten ranks not used for your pairs or singleton, three cards must be chosen from that rank, and one of the two ranks for your pairs must be selected.
The triple does not come from the same rank as your singleton and the pair comes from the same rank as your singleton:  This can occur in
$$\binom{10}{1}\binom{4}{3}\binom{1}{1}\binom{3}{2}$$
ways since your opponent's triple must come from one of the ten ranks not used for your pairs or singleton, three cards must be chosen from that rank, there is only one way to choose the same rank as your singleton for your opponent's pair, and two of the three cards remaining from the rank of your singleton must be chosen from the same rank as your singleton.
The triple does not come from the same rank as your singleton and the pair does not come from the same rank as your pairs or your singleton:  This can occur in 
$$\binom{10}{1}\binom{4}{3}\binom{9}{1}\binom{4}{3}$$
ways since the triple must come from one of the ten ranks not used for your pairs or singleton, three cards must be chosen from that rank, your opponent's pair must come from one of the nine ranks not already used, and three cards must be chosen from that rank.
Thus, for each way you could receive two pairs, your opponent could receive a full house in 
$$\binom{1}{1}\binom{2}{1} + \binom{1}{1}\binom{10}{1}\binom{4}{2} + \binom{10}{1}\binom{4}{3}\binom{2}{1} + \binom{10}{1}\binom{4}{3}\binom{1}{1}\binom{3}{2} + \binom{10}{1}\binom{4}{3}\binom{9}{1}\binom{4}{2}$$
ways.  
Hence, the number of ways for you to receive two pairs and your opponent to receive a full house is 
$$\binom{13}{2}\binom{4}{2}^2\binom{11}{1}\binom{4}{1}\left[\binom{2}{1} + \binom{10}{1}\binom{4}{2} + \binom{10}{1}\binom{4}{3}\binom{2}{1} + \binom{10}{1}\binom{4}{3}\binom{3}{2} + \binom{10}{1}\binom{4}{3}\binom{9}{1}\binom{4}{2}\right]$$
