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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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This one is difficult for me to calculate directly. Any suit can be the matching one, and any pair of seats can have the matched suits. So I just tried Monte Carlo.

For two handed, the problem is simple. There are only 4 cards dealt, so the 2nd, 3rd, and 4th have to match the suit of the first one dealt.

P = 1 * (12/51) * (11/50) * (10/49) = 0.0106

So it's about a 1% chance for this to happen 2 handed and it matches the values I got for a 100000 deal simulation. The chances increase with each seat, since there are more opportunities for the matches. The following are the values calculated for 2-9 seats.

Chances that at least 2 hands at the table together hold 4 cards in a single suit:

  • 2 seats: 1.0%
  • 3 seats: 3.1%
  • 4 seats: 6.0%
  • 5 seats: 9.7%
  • 6 seats: 14.1%
  • 7 seats: 18.8%
  • 8 seats: 23.8%
  • 9 seats: 29.3%
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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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