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Suppose someone is continuously drawing one card each time (without replacement) from a deck of cards. He stops when he gets 3 of Hearts.

What's the probability that he gets any of the Aces before getting the 3 of Hearts?

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up vote 2 down vote accepted

If you are asking about a specific Ace, it is clearly $\dfrac{1}{2}$.

If any Ace will do, just look at the order our $5$ cards come in. (The others are totally irrelevant.) The probability our $3$ comes first is $\dfrac{1}{5}$. So the probability it does not come first is $\dfrac{4}{5}$.

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What if for the case that 3 of Hearts is both before all Aces and all Deuces? – CaptainObvious Oct 4 '12 at 1:34
All relative orders of the $5$ cards are equally likely. Another way to think of it is that once we choose the location for the set of $5$ cards (but not yet the individuals), for every such choice the $5$ cards can be arranged in $5!$ ways. If the $3$ or hearts is to be first, the rest can be arranged in $4!$ ways. So the probability is $\frac{4!}{5!}$, which is $\frac{1}{5}$. Therefore the probability of not first is $\frac{4}{5}$. – André Nicolas Oct 4 '12 at 1:40
I had misread the question as $3$ is first. – André Nicolas Oct 4 '12 at 1:43

Assuming you mean all the aces are drawn before the 3 of hearts, the answer is $\frac{4}{5}$ because the 3 can come in any of 5 positions as marked in Xs below:


Only the last position exhibits the condition in which all the aces have been drawn already.

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I mean any of the aces will count. – CaptainObvious Oct 4 '12 at 1:15

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