# 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 $$n=1,2,3$$

• 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 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 Sep 14 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 at 18:36
• $\binom{4}{3}(\binom{44}{7}+2*\binom{44}{6})$ ? Sep 14 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 at 19:12

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}