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Cards are drawn one at a time, without replacement, from a deck of 52 playing cards. What is the probability that the first card drawn is an ace, given that the 15th card is an ace?

I have trouble calculating the probability of seeing aces on the first and the 15th draws and the probability of seeing $i$ aces in the first 14th draws, $i=0,1,2,3$. Any hint is much appreciated, thanks a lot!

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Is the 15th card in the deck an ace, or is the 15th card drawn known to be an ace? –  Jonathan Rich Mar 13 '13 at 16:54
    
@JonathanRich: the 15th card drawn is an ace. –  drawar Mar 13 '13 at 16:57
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5 Answers

up vote 1 down vote accepted

All sequences of cards are equally likely. So our conditional probability is the same as the probability that the second is an Ace, given that the first is an Ace.

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So the probability is just $\frac{{C_2^4/C_2^{52}}}{{4/52}}$? –  drawar Mar 13 '13 at 16:52
    
The probability is $3/51$. You are using the conditional probability formula, which is overkill. But you are using it correctly. –  André Nicolas Mar 13 '13 at 16:58
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Hint: Take one ace out of the deck and put it in position $15$. Then you have how many aces left out of how many cards left?

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Knowing the $15$th card is an ace, there remain $51$ unknown cards, containing $3$ aces. So drawing any single card (f.e. the first card) out of the remaining $51$, the chance getting an ace is $$\frac{3}{51} = \frac{1}{17} \approx 5.88\%.$$

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Correct; however, the question was tagged "homework". It's good form to try to answer these types of questions in such a way as to lead to the answer rather than just giving it away. –  KeithS Mar 13 '13 at 17:05
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If you want a bit more complex way, just for the sake of it, you know the position of one ace and it is not first. Thus you are left with $51$ cards and three aces. There are $\binom{51}{3}$ ways of positioning the three aces. Now you want the first to be ace. There are thus $\binom{50}{2}$ ways to position the remaining aces. Divide $\frac{\binom{50}{2}}{\binom{51}{3}} = \frac{1}{17}$.

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Correct; however, the question was tagged "homework". It's good form to try to answer these types of questions in such a way as to lead to the answer rather than just giving it away. –  KeithS Mar 13 '13 at 17:06
    
@KeithS It was already given away and the answer accepted... –  Valtteri Mar 13 '13 at 17:07
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A standard card deck has 52 cards, four of which are aces. You know the value and position of one of them. The actual position is immaterial, except that you know that this card isn't the first one. You therefore know that there is one fewer card that the first card could be, and that there is one fewer ace that would produce a successful result. The probability is therefore the chance that one of the remaining aces, out of all the remaining cards, is the one on top.

The question could just as easily have been stated "what is the probability that the first card in the deck is an ace, given that one ace has been removed from the deck and set aside?"

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