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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 $n=1,2,3$

voder228
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  • 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. – lulu Sep 14 '21 at 18:19
  • 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 – voder228 Sep 14 '21 at 18:34
  • 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. – lulu Sep 14 '21 at 18:36
  • $\binom{4}{3}(\binom{44}{7}+2*\binom{44}{6})$ ? – voder228 Sep 14 '21 at 18:46
  • 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. – lulu Sep 14 '21 at 19:12
  • Should say: In principle, you could do $n\in {1,2,3,4}$ and then deduce $n=0$ by subtraction. I recommend not doing this, however. If you do all the cases $n\in {0,1,2,3,4}$ then you can use the sum as a check. Sadly, though, if the check fails, all you will know is that at least one of the cases is wrong, you won't know which. – lulu Sep 14 '21 at 19:14

1 Answers1

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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.
$$ \begin{align} P(\text{1 AK}) &=\binom41 P(E_1)- 2\binom42 P(E_1E_2)+ 3\binom43P(E_1E_2E_3)- 4\binom44P(E_1E_2E_3E_4), \\ P(\text{2 AK}) &=\binom42 P(E_1E_2)- 3\binom43P(E_1E_2E_3)+ 6\binom44P(E_1E_2E_3E_4), \\ P(\text{3 AK}) &=\binom43P(E_1E_2E_3)- 4\binom44P(E_1E_2E_3E_4), \\ P(\text{4 AK}) &=\binom44P(E_1E_2E_3E_4) \end{align} $$

Mike Earnest
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