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I am struggling to understand why my calculation is off by a factor of ten. P(AAABB)= $4/52 * 3/51 * 2/50 * 13 * 4/49 * 3/48 * 12$

My thinking is like this: P(full house) = P(3 of same card) * (# of ways to get 3 of same card) * P(2 of same card) * (# of ways to get 2 of same card that is different than the first)

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You’ve calculated the probability of being dealt a full house in such a way that the first three cards are the three of a kind and the last two cards are the pair. That is, there are $13$ possible ranks for the three of a kind; once you’ve picked one of them, the probability that the first $3$ cards dealt you are of that rank is $\frac4{52}\cdot\frac3{51}\cdot\frac2{50}$. The overall probability of getting a three of a kind in the first $3$ cards dealt you is therefore $13\cdot\frac4{52}\cdot\frac3{51}\cdot\frac2{50}$.

Similarly, there are then $12$ possible ranks for the pair, and the probability of being dealt $2$ cards of a specific one of those ranks from the $49$ cards remaining in the deck is $\frac4{49}\cdot\frac3{48}$, so the probability of being dealt a pair after the initial three of a kind is $12\cdot\frac4{49}\cdot\frac3{48}$.

Put the pieces together, and your $13\cdot\frac4{52}\cdot\frac3{51}\cdot\frac2{50}\cdot12\cdot\frac4{49}\cdot\frac3{48}$ is the probability of being dealt a three of a kind followed by a pair.

However, a full house doesn’t have to be dealt in that order. In fact, there are $\binom52=10$ different ways to choose which $2$ of the $5$ cards (in order of dealing) make up the pair, and that’s your missing factor of $10$. Each of the $10$ possible ways of inserting the pair into the three of a kind has the same probability — the probability that you calculated — so the overall probability of a full house is $10$ times your figure.

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