Probability of Full House in 5 Cards Given a Pair of Aces What is the probability of a full house in a 5 card hand, given the first 2 cards are aces?
My thought process is that there are 4 other events that could occur: A: Ace
p: any card of given face
App —> (2/50) * (4/49) * (3/48)
pAp —> (4/50) * (2/49) * (3/48)
ppA —> (4/50) * (3/49) * (2/48)
ppp —> (4/50) * (3/49) * (2/48)
Am I approaching this incorrectly? I believe these are all the possible combinations for which this could occur.
A: Method 1:  If the first two cards selected from the deck are aces, then of the remaining $50$ cards, two are aces.  There are $$\binom{50}{3}$$ ways to select three cards from the $50$ that remain.  To obtain a full house, either one of the two remaining aces and two cards from one of the remaining $12$ ranks must be selected or three cards from one of the remaining $12$ ranks must be selected.  Thus, the number of favorable cases is
$$\binom{2}{1}\binom{12}{1}\binom{4}{2} + \binom{12}{1}\binom{4}{3}$$
where the first term counts the number of ways of selecting one of the two remaining aces, one of the remaining $12$ ranks, and two cards of that rank and the second term counts the number of ways of selecting one of the remaining $12$ ranks and three of the four cards of that rank.  Hence, the probability of obtaining a full house given the first two cards selected are aces is
$$\frac{\dbinom{2}{1}\dbinom{12}{1}\dbinom{4}{2} + \dbinom{12}{1}\dbinom{4}{3}}{\dbinom{50}{3}}$$
Method 2:  We correct your attempt.
You did not take into account the rank of the non-aces.  Observe that there are $48$ cards which can be selected as the first non-ace.  Choosing that card determines the rank of the cards which are non-aces.  Hence, there are three choices for the second non-ace and, if there is a third non-ace, there are two choices for that card.  The probability of selecting an ace, then two cards of one of the other $12$ ranks in that order is
$$\frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48}$$
Since there are three possible positions for the remaining ace, the probability of obtaining a full house with three aces given that the first two cards are aces is
$$\frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48} + \frac{48}{50} \cdot \frac{2}{49} \cdot \frac{3}{48} + \frac{48}{50} \cdot \frac{2}{49} \cdot \frac{3}{48} = 3 \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48}$$
The probability of obtaining a full house containing three cards of another rank given that the first two cards are aces is
$$\frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48}$$
Hence, the probability of a full house given that the first two cards selected are aces is
$$3 \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48} + \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48}$$
