Well, after reading through some of the "Programming Poker" thread below, I thought I'd wade in with something I figured out recently.


I notice at one point that Vince asks to see a proof that an optimal strategy can be discovered in a poker game. I can't give a mathematical proof of this (though David has repeatedly stated that one exists). I can, however, give you some analysis I did of a simplified heads-up game, to show how this type of optimal strategy would be found.


Basically, the game I'm using for a model (which I posted about in the general theory forum) is a two player game where each player antes one chip, the first player can then bet 2 chips or fold. If the first player bets, the second can either call or fold. The winner is then determined.


I define the following variables:


N = the total number of possible hands.


Cn = the probability of getting hand n.


Fn = the percentage of the time I will bet with hand n.


Xij = my opponent's expectation if I bet with hand i and he calls with hand j.


Yj = my opponent's total expectation if he always calls with hand j.


P = the probability I will bet. P = sum over n from 1 to N (sum (n),1,N) of Cn*Fn


Since my opponent's expectation if he always folds with a specific hand is -1, the optimal strategy is one for which all the Yj's are equal to -1.


Yj = (sum((n),1,N) of Cn*Fn*Xnj)/P


Since Xij and Cn are known once you specify the game, this leaves a system of N equations in N unknowns (any hands we can't make a profit against we can safely bet every time we have them, so this works out). Solving the equations gives you the optimal betting strategy for the first player to act.


I went through the calculation for a game with 3 possible hands:

AA, AK, KK. I'm imagining that these hands are dealt from an infinite deck of half aces and half kings, and that the winner is determined by dealing 2 more cards from the same deck and comparing hands. No straights or flushes, quads beat trips, which beat 2 pair. It is not possible to get worse than 2 pair.


Cn = (.25, .5, .25)


Xij = ( 0 -1.5 -1.5)


(+1.5 0 +1.5)


(+1.5 -1.5 0 )


This gives us:


(.25)(0)(F1) + (.5)(+1.5)(F2) + (.25)(+1.5)(F3) = -P (this one isn't possible, btw)


(.25)(-1.5)(F1) + (.5)(0)(F2) + (.25)(-1.5)(F3) = -P


(.25)(-1.5)(F1) + (.5)(+1.5)(F2) + (.25)(0)(F3) = -P


P = .25(F1) + .5(F2) + .25(F3)


So ignoring the case where our opponent has aces and setting F1, which represents when we have aces, equal to one, we have:


(-.375) - (.375)F3 = -((.25) + (.5)F2 + (.25)F3)


(-.375) + (.75)F2 = -((.25) + (.5)F2 + (.25)F3)


This all simplifies to:


-.125 - (.125)F3 = (-.5)F2


-.125 + (1.25)F2 = (-.25)F3


Substituting .25 + (.25)F3 for F2 in the second equation, we get:


-.125 + (5/16) + (5/16)F3 = (-.25)F3


(5/16)F3 = (3/16), so F3 = .6


Therefore F2 = .4, so if we always bet with aces, bet 60% of the time with kings and bet 40% of the time with AK, we have the optimal strategy. If our opponent folds any hand but aces, and calls with aces, we lose 1.5 5% of the time, an additional 1.5 3.75% of the time, and we lose 1 40% of the time (when we fold). So are total losses (in a hundred hands) are 7.5 + 4.625 + 40 = 52.125, and we win 1 45% of the time (when we bet and our opponent doesn't have aces). So overall we lose 7.125, but if our opponent mixes up his calling with hands other than aces, we have the same result. So acting first is a disadvantage.


I'd love to hear anyone's thoughts on this. Especially Vince's.


Lenny