# What is the probability that all four aces will be received by the same player?

A deck of 52 cards contains four aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 13 cards, what is the probability that all four aces will be received by the same player?

Answer is $$\dfrac{4\binom{13}{4}}{\binom{52}{4}}.$$

Why is $\binom{13}{4}$ taken in numerator?

• Because each player gets 13 cards? – tabstop Jan 25 '14 at 5:38

Before dealing out the cards, give each player 13 coins. Now there are 52 coins on the table, and we will deal cards on top of each coin. We will choose four of the coins to receive aces, and all ${52\choose 4}$ ways of choosing which four that is are equally likely.
There are four ways of choosing who the lucky player is to get all four aces, and ${13\choose 4}$ ways of choosing which of that player's coins will get the aces.
First choose one of the players - there are 4 ways to do this. Then the probability that this player gets all four aces is the probability that 13 cards chosen at random from 52 contain four aces, which is $\dfrac{\binom{48}{9}}{\binom{52}{13}}$. The required answer is therefore $\dfrac{4 \cdot \binom{48}{9}}{\binom{52}{13}}$, which is equivalent to the vadim123's answer. Notice that it doesn't matter who gets which of the rest of the cards.
Alternatively, again there are four choices for the player getting all the aces, and for each of these there are $\binom{48}{9,13,13,13}$ ways to deal the remaining cards. There are $\binom{52}{13,13,13,13}$ ways to deal four bridge hands, again giving the same result.