TITLE: The probability that in a game of bridge each of the four players is dealt one ace
QUESTION [7 upvotes]: The question is to show that the probability that each of the four players in a game of bridge receives one ace is $$ \frac{24 \cdot 48! \cdot13^4}{52!}$$ My explanation so far is that there are $4!$ ways to arrange the 4 aces, $48!$ ways to arrange the other cards, and since each arrangement is equally likely we divide by $52!$.  I believe the $13^4$ represents the number of arrangements to distribute 4 aces among 13 cards, but I don't see why we must multiply by this value as well?

REPLY [4 votes]: After the cards have been shuffled and cut, there are $\binom{52}4$ equally likely possibilities for the set of four positions in the deck occupied by the aces. Among those $\binom{52}4$ sets, there are $13^4$ which result in each player getting an ace; namely, make one of the $13$ cards to be dealt to South an ace, and the same fo West, North, and East. So the probability is
$$\frac{13^4}{\binom{52}4}=\frac{4!\cdot48!\cdot13^4}{52!}=\frac{2197}{20825}\approx.1055$$