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.
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 $$3 \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48} + \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48}$$