Find the probability in 52-card deck Find the probability that when a 52-card deck is distributed to four players, the first of them will receive exactly $n$ pairs of "ace and king" of the same suit.
I can calculate cout of possible cases: $\binom{52}{13}$ and count of positive cases for $n=4$: $\binom{4}{4}\binom{44}{5}$ but i dont know how to calculate count of positive cases for $n=1,2,3$
 A: Let $E_1$ be the event the first player receives both $\spadesuit\mathsf A$ and $\spadesuit\mathsf K$. Similarly, define $E_2,E_3$ and $E_4$ for the other suits. Using the principle of inclusion-exclusion,
$$
P(\text{no AK})=1-\binom41 P(E_1)+\binom42P(E_1 E_2)-\binom43P(E_1E_2E_3)+\binom44P(E_1E_2E_3E_4)
$$
(I am using the shorthand $P(E_1E_2E_3)$ to mean $P(E_1\cap E_2\cap E_3)$). Furthermore, we can determine
$$
\begin{align}
P(E_1)&=\binom{50}{11}\Big/ \binom{52}{13}\\
P(E_1E_2)&=\binom{48}{9}\Big/\binom{52}{13}\\
P(E_1E_2E_3)&=\binom{46}{7}\Big/\binom{52}{13}\\
P(E_1E_2E_3E_4)&=\binom{44}{5}\Big/\binom{52}{13}
\end{align}
$$
Then, using the generalized inclusion-exclusion principle, we can find the probabiltiy of getting each nonzero number of same suit $AK$'s.
$$
\begin{align}
P(\text{1 AK})
&=\binom41 P(E_1)-
2\binom42 P(E_1E_2)+
3\binom43P(E_1E_2E_3)-
4\binom44P(E_1E_2E_3E_4),
\\
P(\text{2 AK})
&=\binom42 P(E_1E_2)-
3\binom43P(E_1E_2E_3)+
6\binom44P(E_1E_2E_3E_4),
\\
P(\text{3 AK})
&=\binom43P(E_1E_2E_3)-
4\binom44P(E_1E_2E_3E_4),
\\
P(\text{4 AK})
&=\binom44P(E_1E_2E_3E_4)
\end{align}
$$
