The [0, 1] Half-Street Fixed-Limit Game - Alpha, MoP, OE Value Betting
Here I'm going to solve a handful of important equations for this game. First I want to talk about an important value in poker math, one so important that you've seen it in tons of places, and one so important that it has its own name. This value is bet/(bet+pot). In our fixed limit games where the pot is P and the bet size is 1, it's 1/(1+P). Typically we refer to this value with the symbol alpha, but here for the sake of ease I'm going to just refer to it as a. So a = 1/(1+P), and that gives us 1-a = P/(1+P), both of which we'll be making use of.
This game is covered in less detail under example 11.3 on page 116 of Mathematics of Poker. In keeping with their naming conventions, we'll refer to x_1 as the worst hand Villain calls with, y_1 as the bottom of Hero's value betting range, and y_0 as the top of Hero's bluffing range.
When we're value betting, we always have a hand lower than x_1 by definition. Therefore, the EV of a value bet with hand y has three parts. When Villain holds a hand from 0 to y, we get called and lose our bet and profit -1. When Villain holds a hand from x_1 to y, we get called and win 1 bet for a profit of 1+P. Finally, when Villain holds a hand from 1 to x_1, Villain folds and we profit P. This gives us the following equation for the EV of a value bet:
EV of Value Betting: (y - 0)(-1) + (x_1 - y)(1+P) + (1 - x_1)(P)
If we check in the same situation, the EV equation has 2 parts. When Villain holds a hand from 0 to y, we lose and profit 0. When Villain holds a hand from y to 1, we win and profit P. This gives us the following:
EV of Checking: (y-0)(0) + (1-y)(P)
The worst hand Hero will value bet with is y_1. At that point, the EV of checking will be the same as the EV of value betting. Therefore, we can set the EV of checking equal to the EV of value betting and simply solve for y, and that will give us y_1.
(y - 0)(-1) + (x_1 - y)(1+P) + (1 - x_1)(P) = (y-0)(0) + (1-y)(P)
-y + x_1 + Px_1 -y - Py + P - Px_1 = P - Py
-y + x_1 - y = 0
-2y = -x_1
y = x_1 / 2
So given x_1, the optimal exploitative value betting range is going to be [0, y_1] such that y_1 = x_1 / 2.
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