Fullhouse probability for first and third person my goal is to calculate the probability of first and third person getting a fullhouse at the start of the game (5 people and 52 cards). I understand that for the first person it would be: $$\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}}$$
But I just can't figure out how to deal with the second person in between since he can get a lot of possible combinations, so the cards to choose from for the third person will change. (for example if he also gets a fullhouse, there is one less rank for the third person fullhouse). 
Am I missing something? Right now I can't come up with any sollution, so some help would be really appreciated. 
 A: Completely ignore the second person!  Begin by dealing the cards to the first person and immediately after continue by dealing the cards to the third person.  The cards that the second person and any other people receive are completely and totally irrelevant to the problem and the calculations surrounding it.
Recall that $Pr(A\cap B) = Pr(A)Pr(B\mid A)$
Now, in calculating $Pr(B\mid A)$, you might want to consider two cases.  Either the pair of cards in our third player's hand match the pair of cards in our first player's hand, or not.  In either case, there are $\binom{47}{5}$ equally likely hands that our third player could have given a particular hand that our first player has.
For the third player to have the same pair as the first player, those two cards must be in his hand.  Pick what remaining three cards are used.  There are $11$ untouched ranks remaining, pick what the unused suit for our triple is.  there are thus $11\times 4 = 44$ possible continuations where our third player gets a full house sharing the pair for the first player.
For the third player to not have the same pair as the first player, we calculate very similarly to before.  There are $11\times 4$ ways to pick the triple for the third player and $10\times \binom{4}{2}$ ways to pick the pair for the third player.
This gives a probability of $\dfrac{11\times 4 + 11\times 4\times 10\times 6}{\binom{47}{5}}$ for $Pr(B\mid A)$ giving a final probability of
$$\frac{13\times 4\times 12\times 6}{\binom{52}{5}}\times \frac{11\times 4 +11\times 4\times 10\times 6}{\binom{47}{5}}$$
