PDA

View Full Version : A little Hold'em math anybody?

12-16-2003, 05:52 AM
Apologizing ahead of time for the headache. How many possible deals and boards are there in texas Hold 'em? What percentage of these hands would give 3 or more players playable/bettable hands? (assuming a full 10 players)

Thanks

MrBlini
12-16-2003, 06:37 AM
Assuming position of hands and order of streets is important there are:
(52)(51)/2 x (50)(49)/2 x (48)(47)/2 x (46)(45)/2 x (44)(43)/2 x (42)(41)/2 x (40)(39)/2 x (38)(37)/2 x (36)(35)/2 x (34)(33)/2 deals,
(32)(31)(30)/6 flops for each deal,
29 turns for each flop,
and 28 rivers for each turn.

That is 52!/[27!(2^10)6]
= 1.206 x 10^36 according to Matlab.

The hands that are playable moreover depends on the hands that are dealt to other players, the characteristics of the other players, and what they decide to do with their hands, since there is often more than one correct decision for good players and additional ways that poor players will play a hand. There is no easy way to quantify this.

That said, the 2+2 table demonstrates that probably 19% or fewer hands should see the flop if everyone is skilled and playing as close to a good game as they can manage, and many fewer of these hands are playable beyond the flop.