Find the probability that when a deck of 52 cards is dealt…

Find the probability that when a deck of $52$ cards is dealt to four players, the first player receives exactly $n$ pairs "ace and king of the same suit".

Step 1: All outcomes

Only the first player's hand matters. It is a random set of $13$ cards:

\[ \binom{52}{13} = 635\,013\,559\,600. \]

Step 2: Sort the cards into three groups

• $4$ "AK pairs": for each suit, its ace and its king ($8$ cards);
• $44$ other cards (neither ace nor king).

For each suit, the hand contains both its ace and king (a pair), exactly one of them ($2$ ways), or neither ($1$ way).

Step 3: Count hands with exactly $n$ pairs

1. Choose the $n$ suits whose ace and king are both in the hand: $\binom4n$ ways. This uses $2n$ cards.
2. For the other $4-n$ suits, the hand must not contain both. Suppose exactly $j$ of these suits contribute one card (ace or king): choose the suits in $\binom{4-n}{j}$ ways and for each pick ace or king: $2^j$ ways.
3. The remaining $13-2n-j$ cards come from the $44$ others: $\binom{44}{13-2n-j}$ ways.

Adding over $j$:

\[ \boxed{P_n = \frac{\binom4n\displaystyle\sum_{j=0}^{4-n}\binom{4-n}{j}\,2^j\binom{44}{13-2n-j}}{\binom{52}{13}}, \qquad n=0,1,2,3,4.} \]

Step 4: Another form — inclusion–exclusion

The number of hands containing $j$ specified pairs (and anything else) is $\binom{52-2j}{13-2j}$. The standard "exactly $n$" formula of inclusion–exclusion gives

\[ P_n = \frac{1}{\binom{52}{13}}\sum_{j=n}^{4}(-1)^{j-n}\binom{j}{n}\binom{4}{j}\binom{52-2j}{13-2j}. \]

Both formulas give exactly the same numbers.

Step 5: The numbers

Most hands ($78\%$) have no ace–king pair of the same suit; getting all four such pairs happens in fewer than $2$ hands per million.

Step 6: A quick check for $n=4$

All eight aces and kings plus $5$ of the other $44$ cards: $\binom{44}{5}=1\,086\,008$ hands, and $P_4 = \binom{44}{5}/\binom{52}{13}\approx 1.7\cdot10^{-6}$. ✓