A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards.

(a) If you have one ace, what is the probability that you have a second ace? (b) If you have the ace of spades, what is the probability that you have a second ace?

  • $\begingroup$ Probability in both parts would be the same. $\endgroup$ – Comic Book Guy Mar 15 '12 at 19:37

The number of hands with exactly one ace is $4{48\choose 12}$. The number of hands with no aces is ${48\choose 13}$. So the number of hands with at least one ace is ${52\choose 13}-{48\choose 13}$. The number of those with more than $1$ ace is: ${52\choose 13}-{48\choose 13}-4{48\choose 12}$.

So the first probability is: $$\frac{{52\choose 13}-{48\choose 13}-4{48\choose 12}}{{52\choose 13}-{48\choose 13}}$$

You can simplify this by multiplying the numerator and denominator by $\frac{39!13!}{48!}$ to get relatively small numbers. I got a probability of $\frac{11199}{28996}$, but I wouldn't trust my arithmetic.

There are $51 \choose 12$ hands with the ace of spaces. If those, there are $48\choose 12$ without another ace. I'll leave the rest to you.


(a) I interpret the question as asking: Given that you have at least one Ace, what is the probability that you have at least two Aces. Other interpretations are possible. We use the language of conditional probabilities, because there are problems with similar structure where that language becomes almost necessary.

Let $A$ be the event "at least one Ace" and let $T$ be the event "at least two Aces." We want $P(T|A)$. Recall that for any events $X$ and $Y$, $P(X|Y)P(Y)=P(X\cap Y)$. So $P(T|A)P(A)=P(T\cap A)$.

First we calculate $P(A)$. It is easier to find the probability of the complementary event "no Aces." There are $\binom{52}{13}$ hands, all equally likely. The number of no Ace hands is $\binom{48}{13}$. So the probability of a no Ace hand is $\frac{\binom{48}{13}}{\binom{52}{13}}$, and therefore $$P(A)=1-\frac{\binom{48}{13}}{\binom{52}{13}}.$$ Next we find $P(T\cap A)$, which is just $P(T)$. There are $\binom{4}{2}\binom{48}{11}$ two Ace hands, $\binom{4}{3}\binom{48}{10}$ three Ace hands, and $\binom{4}{4}\binom{48}{9}$ four Ace hands. Add up, divide by $\binom{52}{13}$.

Now we divide to compute $P(T|A)$. For structural reasons, I would like to give an inefficient expression for $P(A)$. This is the probability of one Ace, plus the probability of two Aces, and so on. When we divide to get $P(T|A)$, the $\binom{52}{13}$ cancel, and we get $$P(T|A)=\frac{\binom{4}{2}\binom{48}{11}+ \binom{4}{3}\binom{48}{10}+\binom{4}{4}\binom{48}{9}}{\binom{4}{1}\binom{48}{12}+\binom{4}{2}\binom{48}{11}+ \binom{4}{3}\binom{48}{10}+\binom{4}{4}\binom{48}{9}}.$$ Note that the denominator is just the number of hands with $1$ or more Aces, and the numerator is the number of hands with $2$ or more Aces. We have essentially restricted the universe to hands that have at least $1$ Ace.

(b) Let $S$ be the event that our hand includes the $\spadesuit$ Ace, and let $T$ be as before. We want $P(T|S)$, and use $P(T|S)P(S)=P(T\cap S)$.

To find $P(S)$, note that there are $\binom{51}{12}$ hands that contain the $\spadesuit$ Ace. To find $P(T\cap S)$, we count much like in (a). There are $\binom{3}{1}\binom{48}{11}$ hands that contain the $\spadesuit$ Ace and one additional Ace. There are $\binom{3}{2}\binom{48}{10}$ hands that contain the $\spadesuit$ Ace and two additional Aces. Finally, there are $\binom{3}{3}\binom{48}{9}$ hands that contain the $\spadesuit$ Ace and three additional Aces. Add up, divide by $\binom{52}{13}$. We obtain $$P(T|S)=\frac{\binom{51}{12}}{\binom{3}{1}\binom{48}{11}+\binom{3}{2}\binom{48}{10}+\binom{3}{3}\binom{48}{9}}.$$

Remark: There are computational shortcuts that can and should be used to get numerical answers. We chose to concentrate on structure.

  • $\begingroup$ Thanks so much for your help! $\endgroup$ – quantum Mar 15 '12 at 19:43
  • $\begingroup$ You are welcome. I have left simplification, computation to you. Note that it turns out that "given Ace of Spades" gives a larger probability than "given at least one Ace." $\endgroup$ – André Nicolas Mar 15 '12 at 20:19
  • $\begingroup$ Yup sure. I don't so much care for the computation aspect. Like you remarked it's the structure that's important here. $\endgroup$ – quantum Mar 15 '12 at 23:24
  • $\begingroup$ In the last formula, what you calculated is $1/P(T|S)$? $\endgroup$ – Jack Jan 24 '16 at 16:56
  • $\begingroup$ The last expression calculates $\Pr(T|S)$. $\endgroup$ – André Nicolas Jan 24 '16 at 17:16

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