The [0, 1] Half-Street Fixed-Limit Game - Unexploitable Solution
The moment I'm sure you've all been waiting for (well at least NightGizmo). Now we're going to find x_1 (the worst hand Villain calls with), y_1 (the worst hand Hero value bets), and y_0 (the best hand Hero bluffs) for the unexploitable strategy (also known as optimal strategy). Consider the following:
1. When Hero is playing unexploitably, x_1 will have the same EV for calling or folding.
2. When Villain is playing unexploitably, y_1 will have the same EV for betting or checking, and y_0 will have the same EV for betting or checking.
We get this from the definition of unexploitable strategy. It's like if we're playing rock-paper-scissors, and I'm playing unexploitably, then for you the EV of playing any of the 3 options should be the same. If that clears it up, cool. If not, then think on it.
So we're going to assume both Hero and Villain are playing unexploitably. 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. Now we just have to find all 6 EV equations and set them equal and figure out what to do with that information:
EV of calling x_1 = (y_1 - 0)(-1) + (1 - y_0)(1+P)
EV of folding x_1 = 0
Set them equal:
(y_1 - 0)(-1) + (1 - y_0)(1+P) = 0
-y_1 + (1 - y_0)(1+P) = 0
y_1 = (1 - y_0)(1+P)
y_1 = (1 - y_0)/a <--- Call this equation 1
EV of betting y_1 = (y_1 - 0)(-1) + (x_1 - y_1)(1+P) + (1 - x_1)(P)
EV of checking y_1 = (y_1 - 0)(0) + (1 - y_1)(P)
Set them equal:
(y_1 - 0)(-1) + (x_1 - y_1)(1+P) + (1 - x_1)(P) = (1 - y_1)(P)
-y_1 + x_1 + Px_1 - y_1 - Py_1 + P - Px_1 = P - Py_1
-y_1 + x_1 - y_1 = 0
2y_1 = x_1 <--- Call this equation 2
EV of betting y_0 = (x_1 - 0)(-1) + (1 - x_1)(P)
EV of checking y_0 = (y_0 - 0)(0) + (1 - y_0)(P)
Set them equal:
(x_1 - 0)(-1) + (1 - x_1)(P) = (1 - y_0)(P)
-x_1 + P - Px_1 = P - Py_0
-x_1 - Px_1 = -Py_0
x_1 + Px_1 = Py_0
x_1(1+P) = Py_0
x_1(1+P)/P = y_0
x_1/(1-a) = y_0 <--- Call this equation 3
We finally get a system of three equations with three variables (remember a is a constant). They are:
y_1 = (1 - y_0)/a
x_1 = 2y_1
y_0 = x_1/(1-a)
A little bit of simple substitution clears things up, but I'll let the reader go through that. At least until I feel like typing it out.
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