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From a bridge deck of $52$ cards, we draw $13$. What is the probability that we have $5$ spades in our hand?

I think that there are $\dfrac{52!}{13! \cdot 39!}$ ways we can choose $13$ cards. There are $\dfrac{13!}{5! \cdot 8!}$ ways to have $5$ spades in our hand.

$$P = \frac{\left(\dfrac{13!}{5! \cdot 8!}\right)}{\left(\dfrac{52!}{13! \cdot 39!}\right)}.$$

Am I doing something wrong?

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Henri has already posted a solution below. In many problems like this, it is helpful to do a simple sanity check. Your answer is $P\approx 2 \cdot 10^{-9}$. This means that we should almost never see a hand with 5 spades in real life. But of course that's not the case! It is quite common that a player has 5 spades. – Yury Dec 28 '12 at 19:11
up vote 5 down vote accepted

There are rather more than $\dfrac{13!}{5! \cdot 8!}$ different ways to have five spades and eight non-spades in your hand.

There are $\dfrac{13!}{5! \cdot 8!}$ ways of choosing five spades from thirteen spades. But there are $\dfrac{39!}{8! \cdot 31!}$ ways of choosing eight non-spades from thirty-nine non-spades.

So you want $$P = \frac{\dfrac{13!}{5! \cdot 8!} \cdot \dfrac{39!}{8! \cdot 31!}}{\dfrac{52!}{13! \cdot 39!}} $$

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You are correct that there are $\dfrac{52!}{13! \cdot 39!}$ possible hands.

To determine the number of hands with 5 spades, treat the spades and the rest of the cards separately. First, choose 5 spades from among the 13: $\dfrac{13!}{5! \cdot 8!}$ . Next, you want to fill the other 8 cards in the hand. If you want hands with exactly 5 spades then there are 39 cards left to choose among; however if you want at least 5 spades then there are still 47 cards to choose among. So form that number the same way as the others, and the total number of hands meeting your criteria is the product of the ways to form those two groups. And you are correct that the probability is just the ratio of those hands to all possible hands.

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For any 6 specific spades, your method of choosing can choose them in 6 different ways rather than just 1. Similarly, you get the count wrong for any number of spades greater than 5. – Hurkyl Dec 28 '12 at 19:06
You're right. That's what I get for trying to generalize the problem on the fly. A correction factor is needed because in the case of allowing more spades in, a card can get in the hand either in the initial selection or in the selection to fill out the hand. – half-integer fan Dec 28 '12 at 19:27

I think an easy way to see it is like this:

$\binom{13}{5}$ = number of ways to choose 5 of 13 spades
$\binom{39}{8}$ = number of ways to choose 8 cards from the remaining 39 cards (39 = 52 cards total - 13 spades)
$\binom{52}{13}$ = choose any 13 cards from 52

$\cfrac{\binom{13}{5} \cdot \binom{39}{8}}{\binom{52}{13}}$

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