Conditional Probability in Poker

Question

I'm thinking of a ten person Texas hold'em game. Each person is dealt 2 cards at the start of the game. The question is:

GIVEN that you have been dealt 2 hearts (Event B), what is the probability that no one else has been dealt a heart (event A)?

I calculated P(B)= (13/52)(12/51) = 0.05882...

and P(A)=(39/52)(38/51)....(21/34) = (39!33!)(52!20!)=0.0009026... the probabilty of 18 cards being dealt (2 for each of the other 9 players) from a deck of 52, none of them being hearts.

I am unsure how to calculate P(A and B) to complete the conditional probability equation. Would it be = [(13/52)(12/51)][(39/50)(38/49)....(21/32)] ?

Answer 1

You don't have to compute $P(B)$. Just deal everybody else out of a $50$ card deck with $11$ hearts. There are $39$ non-hearts. Dealing $18$ cards with no hearts is then a probability of $\frac{39\cdot 38\cdot 37\cdot \dots 22}{50\cdot 49 \cdot 48 \cdot \dots 33}=\frac {39!32!}{50!21!}$

Answer 2

The value would be $$\frac{39}{50}\times \frac{38}{49}\times \frac{37}{48}\times \dots \times \frac{22}{33}=\frac{\frac{39!}{21!}}{\frac{50!}{32!}}=\frac{(39!)(32!)}{(21!)(50!)}$$

Just the probability of dealing no hearts to 18 spots from a deck of 50 cards, 11 of which are hearts.