poker hand conditional probability

Question

Consider a deck of cards, where the backs are marked in such a way that you can distinguish a pip card (Ace, 2,...,10) from a face card (King, Queen, or Jack). You deal a poker hand to a friend. Given that the friend has three pip cards and two face cards, what is the probability that your friend holds a full house?

Answer 1

The three pips have to match, and the two face cards have to match. The first condition is the same as saying that three cards, chosen from $40$, are either all As, all 2s, . . ., or all 10s. That probability comes out to: $$10\cdot\frac{\binom{4}{3}}{\binom{40}{3}}=\frac{1}{247}$$ Similarly, the probability that the face cards match is:$$3\cdot\frac{\binom{4}{2}}{\binom{12}{2}}=\frac{3}{11}$$ The probability of both of these things happening is the product of those two fractions, $\frac{3}{2717}\approx0.11\%$, or about $1$ in $906$.

Answer 2

Without loss of generality, let the first three cards be pip cards and the remaining two cards be face cards. A full house would necessarily have three cards of the same rank as each other and another two cards of the same rank. In this case since three are pip cards and two are face cards it must be that the pip cards are all the same and the face cards are the same.

The first pip card can be anything, it doesn't matter.

The second pip card must be the same rank as the first. There are three remaining pip cards that match the same rank out of the 39 remaining pip cards to have chosen from.

This step happened with probability $\dfrac{3}{39}$

The third pip card must also be the same rank as both the first two.

This step happened with probability $\dfrac{2}{38}$

All of these steps happening simultaneously so far occurs with probability equal to the products of the probabilities of each step individually.

The next card can be any face card, it doesn't matter which.

The last card must be the same rank as the other face card which happens with it's probability.

Finish the calculations by completing the necessary arithmetic.