# 13 cards dealt. Probability of x aces and y clubs?

I know that p(x=4 and y=11) = 0 since if you have 4 aces, you cant have 11 spades as well in the same hand, but is there any relationship that can be written for p(x=i and y=k) for i=1,..,4 and k=1,...,13?

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- M3Lv 77 years agoFavorite Answer
i will instead break it up into i = non-club aces, 1 to 3, and j = ace of clubs, 0 to 1, k = non-ace clubs, 0 to 12 to get the relation

P(i,j,k) = C(4,i+j)*C(12,k)*C(39,13-i-j-k)/C(52,13)

i = 0 to 3, j = 0 to 1, k = 0 to 12

eg P(2,1,6)

= C(4,3)*C(12,6)*C(39,4)/C(52,13) = .0479%

or P(1,0,4)

= C(4,1)*C(12,4)*C(39,8)/C(52,13) = 19.18%

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