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

Question

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

A two pairs is a hand where 2 of the cards are of the same value and also the next 2 cards are also of the same value but the last card is of a different value. Eg. 2♥️, 2♠️, 3♦️, 3♣️, 5♦️. Call this event, A.

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.

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

How do I find P(B n A)?

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

Thank you

Answer 1

After the first person gets a hand with two pairs:

• There are $52-5=47$ cards left in the deck
• There are $13-3=10$ foursomes left in the deck

Therefore, at this point:

• The total number of hands is $\binom{47}{5}=1533939$
• The number of hands with four-of-a-kind is $10\cdot(47-4)=430$

Hence the probability that the second person will get a hand with four-of-a-kind is $\frac{430}{1533939}$.