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?
Best Answer
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 $$ 4\cdot\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}=\frac{4\cdot 11}{5\cdot 49\cdot17} $$