Probability Each Player Receives an Ace

Question

A standard $52$-card deck is dealt uniformly at random to $4$ players, $13$ cards each.
What is the probability that \emph{each} player receives at least one ace?

Jeremy Wu · Accepted Answer

\textbf{Step 1: Count the total number of equally likely deals.}
A deal is an assignment of $13$ cards to each labeled player. The total number of deals is
$$
\#\Omega=\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}
=\frac{52!}{(13!)^4}.
$$

\par\textbf{Step 2: Count the favorable deals by distributing the aces first.}
There are $4$ aces. For each player to have an ace, the only possibility is:
$$
\text{each player gets exactly 1 ace.}
$$
The number of ways to assign the 4 distinct aces to the 4 labeled players is
$$
4! .
$$

\par\textbf{Step 3: Deal the remaining $48$ non-ace cards.}
After giving each player $1$ ace, each player still needs $12$ more cards from the $48$ non-aces.
The number of ways to deal these $48$ cards into 4 labeled hands of size $12$ is
$$
\binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}
=\frac{48!}{(12!)^4}.
$$

\par\textbf{Step 4: Multiply to get the number of favorable deals.}
$$
\#\{\text{favorable}\}=4!\cdot \frac{48!}{(12!)^4}.
$$

\par\textbf{Step 5: Divide favorable by total.}
$$
\mathbb{P}(\text{each player has an ace})
=
\frac{4!\cdot \dfrac{48!}{(12!)^4}}{\dfrac{52!}{(13!)^4}}
=
4!\cdot \frac{48!}{52!}\cdot \frac{(13!)^4}{(12!)^4}.
$$

\par\textbf{Step 6: Simplify factorial ratios.}
Note that
$$
\frac{48!}{52!}=\frac{1}{52\cdot 51\cdot 50\cdot 49},
\qquad
\frac{13!}{12!}=13.
$$
So
$$
\mathbb{P}(\text{each player has an ace})
=
4!\cdot \frac{13^4}{52\cdot 51\cdot 50\cdot 49}.
$$

Since $4!=24$ and $52=4\cdot 13$, one clean final form is
$$
\boxed{
\mathbb{P}(\text{each player has an ace})
=
\frac{24\cdot 13^4}{52\cdot 51\cdot 50\cdot 49}
=
\frac{6\cdot 13^3}{51\cdot 50\cdot 49}.
}
$$

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