spoonitnow's 5000th Post: Game Theory and Poker

The [0, 1] Half-Street Fixed-Limit Game - OE Bluffing

By definition, the best hand Hero will bluff with (y_0) will be higher than the worst hand Villain calls with (x_1), so we have x_1 < y_0 < 1. We could prove that y_0 = 1 is never true, but that would take away from more important things, like the optimal exploitative bluffing frequency. So let's look at the EV of betting and checking y_0.

When we bet y_0: When Villain holds 0 to x_1 we get called, lose, and profit -1. When Villain holds 1 to x_1, Villain folds, we profit P.

EV of y_0 as a bet = (x_1 - 0)(-1) + (1 - x_1)(P)

When we check y_0: When Villain holds 0 to y_0 we lose and profit 0. When Villain holds y_0 to 1 we win and profit P.

EV of y_0 as a check = (y_0 - 0)(0) + (1 - y_0)(P)

To be bluffing all of the hands that will be +EV, we'll need y_0 to have the same EV as a check or as a bet.

(x_1 - 0)(-1) + (1 - x_1)(P) = (y_0 - 0)(0) + (1 - y_0)(P)

Now if we solve for y_0, we'll have the y_0 for the optimal exploitative bluffing percentage.

(x_1 - 0)(-1) + (1 - x_1)(P) = (y_0 - 0)(0) + (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
y_0 = x_1 (1+P)/P
y_0 = x_1/(1-a)

So given x_1, Hero's optimal exploitative bluffing range is [y_0, 1] such that y_0 = x_1/(1-a).

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.