<s>[INST] The probability that in a game of bridge each of the four players is dealt one ace 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?
 [/INST] Since the order doesn't matter, let's make hands one by one.


*

*For the first, there are 4 Aces and I need to choose 1 to give to this person, $\binom{4}{1}$. Then I need to choose 12 cards from the 48 non-Aces to complete this hand, $\binom{48}{12}$. Thus the number of ways to make this hand is
$$\binom{4}{1}\binom{48}{12}.$$

*I already gave the first person an Ace, so I have three left and I need to choose one to give to this person. I also used up 12 non Aces on the last person, so there are 36 non Aces left and I need to choose 12. The number of ways to make this hand is thus,
$$\binom{3}{1}\binom{36}{12}.$$

*and 4. follow the same logic and so I have
$$\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}$$
ways to make hands 3 and 4.
Finally, all the possible ways to make 4 13-card hands are
$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13} = \binom{52}{13,13,13,13} = \frac{52!}{13!\,13!\,13!\,13!}.$$
Lastly, putting it all together, the probability of each person getting an Ace 
is thus
\begin{align*}\frac{\binom{4}{1}\binom{48}{12}\binom{3}{1}\binom{36}{12}\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}}{\binom{52}{13,13,13,13}} &= \frac{4!}{1!3!}\cdot\frac{48!}{12!36!}\cdot\frac{3!}{1!2!}\cdot\frac{36!}{12!24!}\cdot\frac{2!}{1!1!}\\
&\qquad\times\frac{24!}{12!12!}\frac{1!}{1!0!}\cdot\frac{12!}{12!0!}\cdot\frac{13!\,13!\,13!\,13!}{52!}\\
&=\frac{24 \cdot 48! \cdot13^4}{52!}.
\end{align*}
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