The Puzzle

You deal a standard 52-card deck evenly to four players (13 cards each).

Question: What is the probability that each player gets exactly one Ace?

Total Number of Deals

The total number of ways to deal the entire deck to 4 players with 13 cards each is:

\[ \binom{52}{13,13,13,13} = \frac{52!}{(13!)^4} \]

This is the size of the sample space.

Favorable Outcomes

We want each player to get exactly one Ace.

Step 1: Distribute the 4 Aces

There are:

\[ 4! = 24 \]

ways to assign one Ace to each of the 4 players.

Step 2: Distribute the Remaining 48 Cards

We now deal the remaining 48 cards such that each player receives 12 more cards (to make 13 total). The number of such distributions is:

\[ \binom{48}{12,12,12,12} = \frac{48!}{(12!)^4} \]

Final Probability

Combining both steps, the probability is:

\[ \frac{4! \cdot \frac{48!}{(12!)^4}}{\frac{52!}{(13!)^4}} = \frac{24 \cdot 48! \cdot (13!)^4}{52! \cdot (12!)^4} \]

This simplifies numerically to:

\[ \boxed{\frac{778169210}{12493275315}} \approx \boxed{0.0626} \]

So, there’s about a 6.26% chance that each player in a Bridge deal ends up with exactly one Ace.

Reference