Conditional Probabilities in Bridge The question reads as follows:
Each of 4 bridge players has been dealt 13 cards from a 52-card
deck.
(a) Given that your partner has at least one ace, what is the probability that she has at least two aces?
(b) Given that your partner has the ace of spades, what is the probability that she has at least two aces?
(c) Compare your answers. Do they make sense? Explain
For both a and b I used P(A | B) = P(intersection of A and B)/P(B)
For a) I got .37 by doing ($\binom{4}{2}$$\binom{48}{11}$+ ... $\binom{4}{4}$$\binom{48}{9}$)/($\binom{4}{1}$$\binom{48}{12}$+ ... $\binom{4}{4}$$\binom{48}{9}$) = 0.37
For b) I got .56 by doing ($\binom{1}{1}$$\binom{3}{1}$$\binom{48}{11}$+ ... $\binom{1}{1}$$\binom{3}{3}$$\binom{48}{9}$)/$\binom{1}{1}$$\binom{51}{12}$ = .56
I am not sure that I did A and B correctly, or how to explain how it makes sense in the context of part c.
 A: Your working for $(a)$ and $(b)$ is correct. Here is another way to look at it -
a) Sets of $13$ cards where at least one ace is present $ =  
\displaystyle {52 \choose 13} - {48 \choose 13}$
Probability of exactly one ace, given at least one ace is present $ = \displaystyle {4 \choose 1}{48 \choose 12} \ / \ \bigg[{52 \choose 13} - {48 \choose 13}\bigg] = \frac{9139}{14498}$
Subtracting from $1$ gives probability of at least two aces, which is same as your answer.
b) Probability of only one ace, given ace of spades is present
$ = \displaystyle {48 \choose 12} / {51 \choose 12} = \frac{9139}{20825}$
Subtracting from $1$ gives you the desired probability of at least two aces, which is same as your answer.
(c) is asking you to explain why it makes sense that probability in (b) is higher than in (a), basically compare the results with explanation.
A: (a) Using hypergeomentric PDF in R:
a.2 = sum(dhyper(2:4, 4, 48, 13))
a.1 = sum(dhyper(1:4, 4, 48, 13))
a.1; a.2; a.2/a.1
[1] 0.6961825   # P(A >= 1)
[1] 0.2573349   # P(A >= 2)
[1] 0.3696372   # P(A >=2 | A >= 1)

