Poker Math : Required Fold Equity

This is a pretty basic poker equition, but it will eventually become part of a larger project. I’m putting it here because I managed to mess it up twice and would really like to have the correct version in the future.

EV = our expected value
f = probability our opponent folds to our bet
p = current pot (before our bet)

Note: I’m completely ignoring any further play
EV = (fold equity) + (SD equity given call)
EV = f*p + (1-f)*(SD equity)
EV = f*(p – (SD equity)) + (SD equity)
We of course want EV > 0, so
f*(p – (SD equity)) + (SD equity) > 0
f > -(SD equity)/(p – (SD equity))

(SD equity)/(SD equity – p)” width=”147″ height=”41″ />

To make this equation complete, let
w = probability of winning at showdown when called
b = our bet
then,

SD equity = w(p + 2b) - b

These equations make sense because:

  1. When we have 0% chance to win if called, our showdown equity is –b, so we have:
    f > b/(b+p)
  2. On the other hand, when we have a lock (100% chance to win), our equity is p+b (the pot plus opponent’s call), so we have:
    f > (p+b)/(p+b-p) = (p+b)/b, which means we never want the villain to fold
  3. This also tells us that the result “f > 1” lies somewhere between 0% and 100% to win. This is obviously because of pot odds, since we win (p+b), but lose only b, we don’t actually need a lock to want the villain to call every time (ex: when you have straight vs set, you never want opponent to fold, even though a set will suck out quite often)

Related Post: Chart of Required Fold Equity