Probability that two players have same color in trick game

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 $$P(A) = P(B)$$?
• I have taken that the cards have $9$ each of $4$ colors (suits) Sep 3, 2022 at 13:40
• @trueblueanil yes, that is right. Thank you for your post. Am I right with my second assumption?
– Moe
Sep 3, 2022 at 14:18
• 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. Sep 3, 2022 at 14:50

the correct expression would be$$\quad\dfrac{\binom41\binom99\binom31\binom99\binom{18}9\binom99}{\binom{36}9\binom{27}9\binom{18}9\binom{9}9} =\dfrac{12}{\binom{36}9\binom{27}9}$$