# Probability of dividing a deck of cards into 4 equal piles each containing an ace

After dividing a standard deck of cards into 4 equal sized piles, what's the probability that exactly one ace is in each pile? I've had a couple of ideas about how to set this problem up but nothing seems to come out correctly. For instance, I can look at the probabilities that each pile has a single ace and set this up as a multiplication of conditional probabilities. It also occurred to me that I could view the events as (the event that Ace of Spades and Ace of Hearts are in different piles), (the event that Ace Spades, Diamonds, and Hearts are in different piles), and (the event that all aces are in different piles). However, I'm having a hard time even determining what the first of these probabilities should be.

Is it correct that the probability of the first pile having a single ace is $\dfrac{\binom{4}{1}\binom{48}{12}}{\binom{52}{13}}$? It's been a while since I've done probability so I'm having a little trouble getting started, though I know that eventually I'm going to multiply a number of conditional probabilities together.

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I think that's the right way to go. Thanks. I hadn't thought of just subtracting out the piles as you go along. –  anon Sep 12 '11 at 0:20

You won’t need conditional probabilities: you can count the successful divisions all at once. Imagine building such a division by hand. First you lay out the four aces to start your piles; there are $4!$ different ways to do this. Now you have to add $12$ more cards to the first ace; in how many ways can you choose those $12$ cards from the $48$ that remain in your hand? Then you have to choose $12$ of the remaining $36$ cards to go with the second ace. And so on. When you’ve finally got the count of successful divisions, of course, you’ll have to divide it by the total number of possible divisions into four equal piles in order to get the probability. The calculation of that total number is much like the calculation that I just outlined, except that you don’t have to worry about any special cares.

Note: I’ve assumed that the order in which the piles are placed on the table matters. If not, there’s effectivly only one way to place the four aces on the table to begin with.

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I like this idea lot. I just assumed that it had to do with conditional probabilities since that was the section it was in. –  anon Sep 12 '11 at 0:23

The ace of spades always (with probability 1) goes into some pile.

After finding spades, what is the probability that the ace of hearts goes in a different one from the one spades got? It must be $\frac{3\times 13}{51}$.

Now assuming that spades and hearts landed in different piles, the probability for diamonds to end in a yet-unused pile is $\frac{2\times 13}{50}$.

Finally, the ace of clubs will hit the remaining pile with probability $\frac{13}{49}$. Multiply everything together and we get

$$1\times\frac{3\times 13}{51}\times\frac{2\times 13}{50}\times\frac{13}{49} = \frac{4! \times 13^4 \times (52-4)!}{52!}$$

(This result can also be justified by counting possibilities -- there's a factor $4!$ to decide which ace goes in which hand, four factors of $13$ to decide where in each hand each ace goes, and a factor of $(52-4)$ to place all of the non-aces afterwards. Then divide everything by $52!$, the total number of ordered decks).

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