spoonitnow's 5000th Post: Game Theory and Poker

[daviddem]

y_1 = (1 - y_0)/a (1)
x_1 = 2y_1 (2)
y_0 = x_1/(1-a) (3)

(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...

That means x_1 will have the same EV for calling or folding, y_1 will have the same EV for calling or folding, and y_0 will have the same EV for calling or folding

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