How do I find the conditional probability of a second person being dealt a 'full-house' hand after the first person has been dealt a 'four-of-a-kind'?

Question

Standard 52 card deck, so total number of possible hands is 52C5.

A four of a kind is a hand where 4 of the cards are of the same value and out of necessity, the last one is of a different value. Eg. 4♠️, 4♣️, 4♥️, 4♦️, 5♠️. Call this event B.

A full house hand is where 3 cards are of the same value and the final 2 are of the same value. Eg. 1♠️, 1♣️, 1♥️, 3♦️, 3♠️. Call this event C.

Now P(C|B) = P(C n B)/P(B)

How do I find P(C n B)?

If you have a different or more efficient method that's cool too.

Thank you

Answer 1

You can compute directly $P(C\mid B)$: You know that the first guy got a four of a kind. Without loss of generality, he got the four aces and a king.

How many cards are left in the deck ? What is the total number of hands the second player can have ? How many of them result in a full-house ? (Hint: Logically, the second player won't have any ace in his hand. You then have to separate cases whether he has three kings, two kings or none)