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What's the probability of two players each holding two cards of the same suit in a 9-handed Texas Hold 'em?

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The probability of two players being dealt hole cards of the same suit can be thought of as follows: fix the suit ahead of time so, without loss of generality, assume it's hearts. The probability of player 1 being dealt one heart and then another heart is (13/52) times (12/51). Now there are theoretically two less hearts in the deck, so the probability of player 2 being dealt one heart and then another heart is (11/50) times (10/49). We multiply through: ((13*12*11*10)/(52*51*50*49))=0.002, which is about .2%.

Note that this result is independent of how many hands you play, so this holds for your 9 handed scenario, as well as a 6 handed game. This result is theoretical, and not completely independent, since in practice you glean information during a hand on whether or not they have hearts.

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