this should just be
(z-x)*2+(1-z)*-1+z=z
z = (x+1)/2

and then you have P2's indifference at x:
EV(fold) = EV(call)
0 = 2 * y/(y+(1-z)) + (-1) * (1-z)/(y+(1-z))
0 = 2y-(1-z)
y=(1-z)/2

and then, solving, x=5/9, y=1/9, and z=7/9

If the money amounts were continuous rather that discrete dollar amounts I can't help thinking that the solution to the simple no limit game reduces to some beautiful 3D bifurcation diagram with the number you're dealt along the x axis, amount to bet on the y axis and betting frequency along the z axis.

Given your number n on the x axis you can pick any point inside the region of the graph at x=n so you will get a discontinuous 2D region of possible strategies for every n.

Ok, so I simplified the original problem a bit. The goal isn't to diverge from it, but more to use the simplified one as a building stone toward understanding what's going on.

1) Player 1 can only bet 1$ (or check obv.). Of course, it changes the whole dynamic of the problem, but still, I guess we could learn something.

2) I clustered the real numbers to simplify my very simple simulation. I don't think it alter the nature of the problem. So [0; 0.1[ --> 0, [0.1; 0.2[ --> 1, ..., [0.9; 1] --> 9

So now, knowing that there 1$ of EV in every pot, what's 1st player GTO strategy? And what his 2nd player GTO response?

I found out this:

- If 1st bets a (0,1,6,7,8,9) range and checks (2,3,4,5), he will make on average at least 52 cents per hand, no matter how 2nd responds.

- Based on 1st betting range, 2nd can minimize 1st's EV (to 52 cents) by having this calling range: (6,7,8,9).

Note that it doesn't matter if 2nd calls or folds with (2,3,4,5), his EV will remain 48 cents.

mufaasa and i solved the case when only a potsized bet is allowed above. the bettor bets the top 2/9 of his range and the bottom 1/9. the bluffcatcher calls the top 5/9 of his hands.

Had errors in my simulation. It's actually top 2/10 (20%) and bottom 1/10 (10%) for the better and top 5/10 (50%) for the caller.

EV(better) = 55.5 cents
EV(caller) = 44.5 cents

cool, is it easy to increase the number of hand divisions and see if it gets closer to the analytical result?

fmp

I guess so. For now, I only have a little 10x10 matrix in Excel with some basic equations (lazyness prevails!), but I'll see what I can do. Might not be soon though as I'm busy these days, but will let you know.

http://www.math.ucla.edu/~tom/papers/poker1.pdf Here is a paper by Chris Fergusson on this subject.