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In a $13$ card hand, given that a person has one ace, what is the probability that person has more than one ace?

In a $13$ card hand, given that a person has the ace of spades, what is the probability that person has more than one ace?

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closed as off-topic by Did, Yiorgos S. Smyrlis, JimmyK4542, Claude Leibovici, arjafi Sep 15 '14 at 6:43

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Those two probabilities will be the same. – TMM Sep 12 '11 at 0:42
Count the total number of hands in which a person has at least one ace, or, easier, count the number of ways the person can have exactly one ace. – gary Sep 12 '11 at 0:43
@TMM: One is about 50% larger than the other. – Brian M. Scott Mar 10 '12 at 11:14

This is just a matter of counting. Let’s look first at the second question. How many hands contain the ace of spades? The other $12$ cards may be any of the $51$ other cards, so there must be $\binom{51}{12}$ such hands. How many of them contain no other ace? To form such a hand, you must pick all $12$ of the other cards from the $48$ cards that aren’t aces, something that can be done in $\binom{48}{12}$ ways. Thus, $\binom{48}{12}$ of the $\binom{51}{12}$ hands that contain the ace of spades contain no other ace, and therefore $\binom{51}{12}-\binom{48}{12}$ of them do contain at least one other ace. Thus, the probability that the hand contains at least one other ace, given that it contains the ace of spades, is

$$\begin{align*}\frac{\binom{51}{12}-\binom{48}{12}}{\binom{51}{12}}&=1-\frac{\binom{48}{12}}{\binom{51}{12}}\\ &=1-\frac{48!}{12!\,36!}\cdot\frac{12!\,39!}{51!}\\ &=1-\frac{39\cdot38\cdot37}{51\cdot50\cdot49}\\ &=1-\frac{13\cdot19\cdot37}{17\cdot25\cdot49}\\ &=1-\frac{9139}{20825}\\ &=\frac{11686}{20825}\;, \end{align*}$$

which is a little more than $0.56$.

Now we can tackle the first problem. There are $\binom{48}{13}$ hands that contain no ace, so there are $\binom{52}{13}-\binom{48}{13}$ hands that contain at least one ace. To build a hand with exactly one ace, you must choose one of the $4$ aces and any $12$ of the other $48$ cards; this can be done in $4\binom{48}{12}$ ways. Thus, $\binom{52}{13}-\binom{48}{13}-4\binom{48}{12}$ hands contain more than one ace. Given that you have at least one ace, therefore, the probability that you have more than one is

$$\begin{align*} \frac{\binom{52}{13}-\binom{48}{13}-4\binom{48}{12}}{\binom{52}{13}-\binom{48}{13}}&=1-\frac{4\binom{48}{12}}{\binom{52}{13}-\binom{48}{13}}\\ &=1-\frac{\frac{4\cdot48!}{12!\,36!}}{\frac{52!}{13!\,39!}-\frac{48!}{13!\,35!}}\\ &=1-\frac{\frac4{12!\,36!}}{\frac{52\cdot51\cdot50\cdot49}{13!\,39!}-\frac1{13!\,35!}}\\ &=1-\frac{4\cdot13\cdot39\cdot38\cdot37}{52\cdot51\cdot50\cdot49-39\cdot38\cdot37\cdot36}\\ &=1-\frac{13\cdot19\cdot37}{17\cdot25\cdot49-19\cdot37\cdot9}\\ &=1-\frac{9139}{14498}\\ &=\frac{5359}{14498}\;, \end{align*}$$

or a little under $0.37$.

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You can argue indirectly for the first question, by finding the probability of having exactly one ace: after being given the first ace, throw away all the remaining aces, and then select the 12 cards from the non-aces. Then, if $X_n$ is the event of aving n aces,$ P(X_1)+P(X_2)+ P(X_3)+P(X_{4}) =1$ (I am assuming there are a total of four aces in the deck). Basically: the probability of having more than one ace, given that one already has one, is given by fixing one card to be the ace, and then seeing what the probability is that the remaining cards have exactly two, three, or four aces; for the event $X_2$, select a second card to be the ace, throw the two remaining aces; for $X_3$, select three aces, throw away the remaining ace, etc.

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