Probability that a five-card poker hand contains exactly two aces if it has exactly one face card

Question

We are dealt a five-card poker hand. What is the probability that this hand contains exactly two aces if it has exactly one face card?

Answer 1

Let $T$ be the event the hand has exactly $2$ Aces, and let $F$ be the event the hand has exactly $1$ face card. We want $\Pr(T|F)$. By the definition of conditional probability, we have $$\Pr(T\F)=\frac{\Pr(T\cap F)}{\Pr(F)}.$$

It remains to find the probabilities on the right.

For the probability of exactly one face card, note that there are $\binom{52}{5}$ equally likely poker hands. To count the hands with exeactly $1$ face card, note that the face card can be chosen in $\binom{12}{1}$ ways, and for each such way the $4$ non-face cards can be chosen in $\binom{40}{4}$ ways. Thus $$\Pr(F)=\frac{\binom{12}{1}\binom{40}{4}}{\binom{52}{5}}.$$

We now calculate $\Pr(T\cap F)$. The $2$ Aces can be chosen in $\binom{4}{2}$ ways, and then the face card in $\binom{12}{1}$ ways. The $2$ remaining non-face, non-Ace cards can be chosen in $\binom{36}{2}$ ways. Thus $$\Pr(T\cap F)=\frac{\binom{4}{2}\binom{12}{1}\binom{36}{2}}{\binom{52}{5}}.$$ Divide. There is a great deal of cancellation, and we end up with $$\Pr(T|F)=\frac{\binom{4}{2}\binom{36}{2}}{\binom{40}{4}}.$$

Remark: There are slicker ways to do the calculation. For given that there is one face card, the only thing that matters is the remaining $4$ cards.

There are $\binom{40}{4}$ ways to choose these $4$ cards from the $40$ non-face cards. And there are $\binom{4}{2}\binom{36}{2}$ ways to choose $2$ Aces and $2$ non-face cards/ That gives a quick path to the final answer,

However, it is useful to just let the full machinery operate.

Answer 2

Let A be the event of exactly 2 aces and let B be the event of exactly one face card.

Recall that $P(A|B)=\frac{P(A\cap B)}{P(B)}$

Now, $A \cap B$ is the event of having two aces and exactly 1 face card, which means that the remaining two cards must be of the numbers 2-10. So the number of ways to have this is $C^{12}_1 \cdot C^4_2 \cdot C^{36}_2$. Also, the number of ways to have exactly 1 face card is $C^{12}_1 \cdot C^{40}_4$. Thus $$P(A|B)=\frac{C^{12}_1 \cdot C^4_2 \cdot C^{36}_2}{C^{12}_1 \cdot C^{40}_4}=\frac{378}{9139} \approx .0414$$