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:
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!
