Probability that one out of four players gets all four aces What is the probability that exactly one player gets all four aces if a randomly shuffled deck is dealt to four players? 
My attempted working: ${4\choose1}\cdot\frac{1}{4!}\cdot\frac{1}{52\cdot51\cdot50\cdot49}\cdot{48\choose 9, 13, 13, 13}/{52\choose 13, 13, 13, 13}$.
However this probability seems to be too small. What am I doing wrong?
 A: There is another quick way to do it: the probability in question is $4$ times the probability that the first player gets all aces, so let us compute the latter. Imagine that the cards are dealt but their values are not assigned yet. Then the chance that the ace of clubs goes to the first player is $\frac{13}{52}$. If that happens, we are left with $51$ unassigned values out of which $12$ are held by the first player, giving him the chance $\frac{12}{51}$ to get the ace of spades, etc. This gives the final answer 
$$
4\cdot\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}=\frac{4\cdot 11}{5\cdot 49\cdot17}
$$
A: There are ${52\choose 13,13,13,13}$ ways to deal the cards to players N,S, E, and W.  There are 4 ways to choose a player who gets all of the aces.  Say it's N.  He can have ${48\choose 9}$ ways to get the rest of his cards.  Now deal the remaining 39 cards to E, S and W. 
A: To sum it up: Given a hand with all four aces, there are ${48\choose9}$ ways to choose the remaining 9 cards. Since there are ${52\choose13}$ possible hands, the probability is $$\frac{{48\choose9}}{{52\choose13}} = \frac{11}{5\cdot17\cdot49} = \frac{1}{378.6}$$and since all players have equal probability, four times that that any player has all four aces.
