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My problem is this:

A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?

Here is my work so far:

probability that there is at least one ace in the 13: 1-(C(52,13)-C(48,13))/C(52,13) = .3038

probability that there is an ace left at the end of dealing 12 cards: 12!/13! = .076923

probability of both of those happening: .3038*.076923 = .02337 = 2.34%

does this look correct? Thank you so much in advance. Nolan

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1 Answer

up vote 2 down vote accepted

You are making it too complicated. Just shuffle the deck and look at the $13^{\text{th}}$ card. It will be an ace with probability $\frac 4{52}$. Just like the first card. One error is that the chance that an ace is last among 13 cards given that there is at least one ace is greater than $\frac 1{13}$ because you might have more than one ace.

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Wow yeah, Cant believe I didn't realize that. Half of this seems to be making sure you're thinking of the problem in it's simplest terms. Thank you so much –  BadAtGraphs Nov 16 '12 at 8:21
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