spoonitnow's 5000th Post: Game Theory and Poker

[spoonitnow]

For 5 and 6, I didn't get it because the definition of unexploitative strategy is not 100% clear in my mind. I thought it was the strategy that prevents the opp from making any relatively +EV play? If so, how do you get to your criteria above?

Imagine both players are playing their own optimal exploitative strategy against each other at the same time. A consequence of this would be that neither player can do better by changing their strategy, because by definition, the optimal exploitative strategy is the best-performing strategy in terms of EV that you can have against a given opponent's strategy.

So neither player can improve their EV by changing their strategy, and this creates a certain equilibrium. If either player deviates from the equilibrium by changing their play, then they lose EV (since they will be performing worse than the optimal exploitative strategy, which was the maximum EV they could obtain).

In this equilibrium, neither player can be exploited by a change in strategy. Therefore, their strategies are both unexploitable.

So what is a change in strategy in this game? Villain has one strategic option, and that's where to place x_1. Hero has two strategic options, and those are where to place y_1 and y_0.

When Villain is playing his optimal exploitative strategy, the EV of folding x_1 will be the same as the EV of calling x_1. Similarly, when Hero is playing his optimal exploitative strategy, the EV of betting y_1 or y_0 will be the same as the EV of checking y_1 or y_0, respectively.

So we set the EV of folding x_1 equal to the EV of calling x_1, the EV of betting y_1 equal to the EV of checking y_1, and the EV of betting y_0 equal to the EV of checking y_0. When we do all of this, then both players are playing their respective optimal exploitative strategies, and therefore are both playing unexploitably. The solutions for x_1, y_1, and y_0 in this system of equations gives us each player's unexploitable strategy.

(2) in (3):
y_0=2y_1/(1-a) (4)
(4) in (1):
y_1=(1-2y_1/(1-a))/a
ay_1=1-2y_1/(1-a)
(a+2/(1-a))y_1=1
(a(1-a)+2)*y_1=(1-a)

y_1=(1-a)/(2+a(1-a))

so by (2):
x_1=2(1-a)/(2+a(1-a)) (5)

(5) in (3):
y_0=2/(2+a(1-a))

Not sure if there is a way to make them look better...

^^Some typos spoon, too many calling and folding, not enough betting and checking.

Thanks for pointing out the typos. And that's as good as you're going to make them look in text.