Be four players who get 9 cards each from 36 deck (swiss trick-game) with four different colors. Be A = {all cards from player 1 have same color} and B = {all cards from player 2 have same color}. What is $P(A ∩ B)$ with card(Ω) = $\binom{36}{9}\binom{27}{9}\binom{18}{9}\binom{9}{9}$.

  1. I know $P(A ∩ B) = \frac{card(A ∩ B)}{card(Ω)}$, so is this correct: $P(A ∩ B) = \frac{\binom{9}{9}\binom{9}{9}\binom{18}{9}\binom{9}{9}}{\binom{36}{9}\binom{27}{9}\binom{18}{9}\binom{9}{9}} = \frac{1}{\binom{36}{9}\binom{27}{9}}$
  2. Am I right to assume $P(A) = P(B)$?
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    $\begingroup$ I have taken that the cards have $9$ each of $4$ colors (suits) $\endgroup$ Sep 3, 2022 at 13:40
  • $\begingroup$ @trueblueanil yes, that is right. Thank you for your post. Am I right with my second assumption? $\endgroup$
    – Moe
    Sep 3, 2022 at 14:18
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    $\begingroup$ We are computing joint probability $P(A).P(B|A)\;\; for\;\; P(A\cap B)$ They are "equal" only in the sense that they are symmetric, i.e. it doesn't matter whether you distribute first to A or B. $\endgroup$ Sep 3, 2022 at 14:50

1 Answer 1


There is a question of which colors A and B have, so if you want to find the probability that each of A and B have only one color,
the correct expression would be$\quad\dfrac{\binom41\binom99\binom31\binom99\binom{18}9\binom99}{\binom{36}9\binom{27}9\binom{18}9\binom{9}9} =\dfrac{12}{\binom{36}9\binom{27}9}$


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