Bet sizing to break even when stealing from the btn

SB folds x percent, BB folds y percent, our bet size is b in big blinds

Both players fold: x*y percent of the time (or just xy for short)

If we're not taking into account anything post-flop and assume we just lose the hand right then if the blinds don't fold:

EV of Stealing = (1.5)(xy) + (-b)(1-xy)

We want the EV of stealing to be zero:

(1.5)(xy) + (-b)(1-xy) = 0
1.5xy - b + bxy = 0
-b + bxy = -1.5xy
b(-1 + xy) = -1.5xy
b = -1.5xy/(-1 + xy)

Example: SB folds 75 percent, BB folds 60 percent

xy is 0.75 * 0.60 = 0.45

b = -1.5xy/(-1 + xy)
b = -1.5(0.45)/(-1 + 0.45)
b = -0.675/(-0.55)
b = 1.227

If you want to understand the math of this a little bit better, I wrote an entire series that teaches you how to do EV calculations using only basic addition and multiplication. The first part of the series is here.

Originally Posted by spoonitnow

Example: SB folds 75 percent, BB folds 60 percent

xy is 0.75 * 0.60 = 0.45

b = -1.5xy/(-1 + xy)
b = -1.5(0.45)/(-1 + 0.45)
b = -0.675/(-0.55)
b = 1.227

If you want to understand the math of this a little bit better, I wrote an entire series that teaches you how to do EV calculations using only basic addition and multiplication. The first part of the series is here.

Thanks, spoon. I'll read your article in a minute after I try to digest this. Your example is more realistic, but let me see if I can do the math using my more extreme scenario of two nits in the blinds using your formula. We are solving for b:

SB & BB each fold 80%, so 80% * 80% = both fold 64%

b = (-1.5 * .64) / (-1 + .64)
b = -96 / -.36

thus...

b = 2.66.

I'm guessing that theoretically I'd need the raise to be 2.66 to break even, no? (I understand this isn't necessarily in game thinking, but I'm trying to how the numbers break down.)