Let’s deal the cards 13 by 13. The number of possible hands (as a set rather than a sequence) for the first player is \(\binom{52}{13}\), for the second player is \(\binom{39}{13}\), for the third player is \(\binom{26}{13}\), and for the fourth player is \(\binom{13}{13}\). The product is the total number of ways to deal the cards to four players.

The deals that everyone has exactly one ace can be counted by first deciding which ace goes to which player. This can be settled in \(\)4!\(\) ways. Then we need to deal the remaing 48 cards, first in \(\binom{48}{12}\) ways for the first player, then in \(\binom{36}{12}\) ways for the second player, then in \(\binom{24}{12}\) ways for the third player, and finally in \(\binom{12}{12}\) ways for the fourth player. The product is the total number of ways to deal the cards to four players such that everyone has exactly one ace.

Thus the probability that each of the players gets an ace is,

\[\frac{4! \binom{48}{12} \binom{36}{12} \binom{24}{12} \binom{12}{12}}{\binom{52}{13} \binom{39}{13} \binom{26}{13} \binom{13}{13}}.\]