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).