Hello. I am trying to solve GTO-play for a small poker-like game.

The game is as follows: Player A and Player B both post an ante.
Then they are both dealt a hand which is a number between 0-100.
Then player A may either bet (potsize) or check. If he checks the hand goes to showdown. If he bets Player B may either call or fold. If he calls it goes to showdown. Highest hand wins.

Algebraicly their ranges can be noted as follow:
Player A checks hands in the range 0 -> a
Player A bets hand in the range a -> 100

Player B folds hands in the range 0 -> b
Player B calls hands in the range b -> 100


Anyway, the trouble i have is to find a general formula for any players win probability in the scenario where A bets and B calls. Has anyone got a clue on this?


For the checking scenario i have concluded player A's winning probability to be a/100. And hence Player B's winning probability is 1-(a/100). Is this correct?

By "number" , do you really mean integer ? Also, you need to be precise: are both 0 and 100 included as possibilities?

The continous version of this has been solved about seventy years ago; you can find the solution in either MOP by Chen and Ankenman or Theory of Games and Economic Behavior by von Neumann and Morgenstern.

I'm going to assume the numbers are continuous, as it's simpler to solve this way. If they're integers the basic method applies, but you have to check the close cases to get it exactly right.

You work backwards from the last decision. B has to win 1 time in 4 to break even. So knowing that A bets on a to 100, B will call on 25 + .75*a. For example, if a = 60 then b = 70. If B bets on a number less than 70, he loses more than 1 time in 4. If he fails to bet on a number over 70, he gives up a chance to bet with better than 1 in 4 probability.

Call the numbers dealt n_a and n_b.

If A checks, his EV is 2*n_a, which on average is a/200. If he bets, there are three possible outcomes:

n_b < 25 + .75*a, A wins 2
25 + .75*a < n_b < n_a, A wins 3
n_b > n_a, A loses 1

To compute his EV given n_a > a, consider two cases:

25% of the time n_a < 25 + .75*a, so A always loses if called. In this case his EV is 2*(25 + .75*a)/100 - [1 - (25 + .75*a)/100] = 3*(25 + .75*a)/100 - 1.

75% of the time n_a > 25 + .75*a, so A wins half the time if called. Since he wins 3 half the time when called and loses 1 the other half, his EV if called is 1. That makes his overall EV 2*(25 + .75*a)/100 + [1 - (25 + .75*a)/100] = (25 + .75*a)/100 + 1.

25% of the first plus 75% of the second gives 1.5*(25 + .75*a)/100 + .5 = .01125*a + .875.

A's overall EV is his chance of checking (.01*a) times his average gain when checking (.005*a) plus his chance of betting (1 - .01*a) times his average gain when betting (.01125*a + .875). This is .00005*a^2 - .0001125*a^2 + (.01125 - .00875)*a + .875
= -.0000625*a^2 + 0.0025*a + .875.

To maximize this with respect to a, take the derivative, -0.000125*a + .0025 and set to zero. You get a = 20, which makes b = 40.

You should check my math, I just typed it in without paper or double checking. But the method is sound.

Given the restraints on player As strategy, the correct strategy is to always check.

We will show that the strategy "bet with a or higher" is worse than the strategy "always check".

It is easy to see that both strategies turn out the same if either player was dealt a card below "a" (if A has it, he checks and the strategies are literally the same. If A doesnt but B does, B folds to As bet but A was going to win anyway).

So we assume both players have a card higher than "a". Clearly player As EV for checking here is still 1.

Lets say he bets:
1/4 times, player B has a card below "b" (Bs calling range, 3/4 of As betting range as shown above by Aaron Brown) and folds. A wins 2
3/16 times, Player A has a card below "b" and Player B has a card above "b". Player A loses 1.
9/16 times they both have a card above "b". Player A wins 3 half the time and loses 1 the other half for an average of winning 1 (obviously, because both players have an even chance of winning, betting and being called cannot be better than just checking down)
So the EV of betting is 2/4 - 3/16 + 9/16 = 14/16 < 1

So the player betting with "a" or higher wins 2/16 less than the player who always checks whenever both player have a or higher, and wins the same otherwise.

The more common problem I believe is the same situation where Player A bets with anything in the range (0,a1) and (a2,100) (Betting with very small numbers can be advantageous since your chance of winning the showdown can be very small compared to the chance of them folding).

You worked out the problem for potsize P = 2 bets; work it out again for P = 1 !

This is the [0,1] game - it's completely analysed in the book "the maths of poker".

OP may wish to consider including a buffing range for the first better.

EDIT: Slow river_tilt, just saw bigpooch said the same thing.

Banzai is correct, I made a rookie mistake. I computed A's best value of a, but didn't check to see if it gave a positive net expectation. It didn't.

Assuming my original solution had been correct, A always loses if called if he bets with numbers between 20 and 40. So he might as well use 0 to 20 instead, that is bet with 0 to 20 or 40 to 100. That costs him nothing when he bets, and it doesn't change B's optimal strategy. But if when A checks, he now wins 30% of the time instead of 10%. This is enough to make it a winning strategy.

It says "potsize" bet after A and B ante (I set the ante at 1).