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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 $n=1,2,3$

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  • $\begingroup$ I suggest trying $n=3$ next. You have to break it into cases, of course, according to whether or not the player has either the Ace or the King (or neither) in the unique unmatched suit. $\endgroup$
    – lulu
    Sep 14 at 18:19
  • $\begingroup$ in my opinion for $n=3$ count of positive cases: $\binom{4}{3}\binom{46}{7}$ - $\binom{4}{4}\binom{44}{5}$ but i think its wrong $\endgroup$
    – voder228
    Sep 14 at 18:34
  • $\begingroup$ As I say, you need to separate out three cases. The first, wherein the player has neither $A,K$ of the unmatched suit. The second (resp. third), wherin the player has the $A$ (resp. the $K$) but not both $A,K$ of the unmatched suit. The second and third cases have the same number (clearly) so you really only have to do two counts. $\endgroup$
    – lulu
    Sep 14 at 18:36
  • $\begingroup$ $\binom{4}{3}(\binom{44}{7}+2*\binom{44}{6})$ ? $\endgroup$
    – voder228
    Sep 14 at 18:46
  • $\begingroup$ That's what I get too. Now do $n=2$ the same way. Each subcase is trivial, but you need to be careful to make sure you get all the possibilities. $\endgroup$
    – lulu
    Sep 14 at 19:12
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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.
$$ \begin{align} P(\text{1 AK}) &=\binom41 P(E_1)- 2\binom42 P(E_1E_2)+ 3\binom43P(E_1E_2E_3)- 4\binom44P(E_1E_2E_3E_4), \\ P(\text{2 AK}) &=\binom42 P(E_1E_2)- 3\binom43P(E_1E_2E_3)+ 6\binom44P(E_1E_2E_3E_4), \\ P(\text{3 AK}) &=\binom43P(E_1E_2E_3)- 4\binom44P(E_1E_2E_3E_4), \\ P(\text{4 AK}) &=\binom44P(E_1E_2E_3E_4) \end{align} $$

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  • $\begingroup$ thank you very much $\endgroup$
    – voder228
    Sep 15 at 8:29

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