TITLE: Probability of deck of cards such that each person receives one ace
QUESTION [6 upvotes]: Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace.

The answer to this is given as$$\frac{13^4}{\binom {52}4}$$
My doubt is the following:

The book justifies ${\binom {52}{4}}$ as number of possible different combinations of the four positions in the deck occupied by 4 aces. That sounds like a case of arrangements to me, so shouldn't we think about permutations and not combinations if we are concerned about how the aces are to be arranged in the deck ?.
Shouldn't the denominator be ${\binom {52}{13}}$ since you are choosing 13 cards for 4 people.

REPLY [7 votes]: All we really care about is the placement for the aces; not where the other cards may be in the deck, nor even the suits of the aces.  
So let us take 52 blank cards and 4 stickers with 'ace' written on them.   To 'deal the cards', place the blank cards in for lines of 13, then unbiasedly select four from them and put a sticker on each.
How may ways are there to stick the aces on 4 different cards in the deck of 52? That is a selection of 4 from 52.  $$\binom {52}4$$
How many ways are there to do this so that each ace is stuck on one card in each from the four lines of 13? That is a selection of 1 from the first line, 1 from the next line, and so forth. $$\binom{13}1\binom{13}1\binom{13}1\binom{13}1$$
Divide and calculate.$$\dfrac{13^4}{\dbinom{52}{4}}$$