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