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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?

2 Answers 2

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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?2012-10-04
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    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 $\frac{4}{5}$.2012-10-04
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    I had misread the question as $3$ is **first**.2012-10-04
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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:

X A X A X A X A X 

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.2012-10-04