Poker Math: Extending the 2/4 Rule

Three more math books arrived today, now I have a stack of 9 to get through, maybe 1-2 years worth if I learn fast. That means even less hope for poker, so I might as well pass on another idea I’ve had that I probably won’t have time to work on. Continue reading

Breakeven Fold Equity

EDIT: It has come to my attention that I defined “break even” incorrectly. Instead of defining it to be 0 EV, it should actually be defined as our EV if we go to showdown. This is because we need to compare the two best lines (checking vs betting) and see which one is better (ignoring getting raised off hands and having to c/f to a bluff). I will correct the chart soon.

This summer I did a little analysis of required fold equity (FE). I had high hopes for the project, but nothing great came of it. What did come out of it, however, was a chart that might be of use to you guys. It’s not what I wanted to release, but I don’t really have the time to do more meaningful analysis atm.

Required Fold Percentage

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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