Fold Equity Hypothetical

[Jason]

Good answers and explanations. But, there's still more questions to ask that I think would be useful to try to wrap this up. We've taken a look at some specific examples that examined fold equity. Suppose you wanted to create a mathematical formula to plug in #'s to easily figure what % of the time a player would need to fold to be a break-even play.

Question 7
How many other variables do we need to consider to formulate an equation to determine the % of the time a player will fold (we will define this variable as F) such that it is a break-even play?

Question 8
Define the varaibles from question 7 that need to be determined to calculate F.

Question 9
Express as a mathematical formula using the variables from question 8 to solve for F. F = ?

If you've already answered a question or you play regularly @ stakes you do not consider beginner, give others at least 24 hours or so to take a crack at some of these if they want

[JKDS]

If you've already answered a question or you play regularly @ stakes you do not consider beginner, give others at least 24 hours or so to take a crack at some of these if they want

awwww, but i already haz a formula for this

[Jason]

Ah, too bad no one else took a shot to finish this out.

When I was posing these questions to myself, I saw four overall variables to consider: my equity (E), my bet size (B), the TOTAL size of the pot BEFORE I bet (we'll call it Q). and the % chance the villain will fold to bet B (F). It actually took me longer to develop the equation than I anticipated as I would neglect a variable here or there. But, I had an Excel spreadsheet of different variables and finally came up with an equation that worked. It's mainly tedious algebra, but the start of the equation is that the sum of the times you are called versus the times you are not called is equal to 0. The times you are called is further divided into two parts based on the times you call and win and call and lose:

((Q+B)*E - (1-E)*B) * (1-F) + FQ = 0
(QE + BE - B + BE) * (1 - F) + FQ = 0
(QE + 2BE - B) * (1 - F) + FQ = 0
(1) * (QE + 2BE - B) - (F) * (QE + 2BE - B) + FQ = 0
QE + 2BE - B - QEF - 2BEF + BF + FQ = 0
From here, solve for F:
(F) * (B + Q - QE - 2BE) + QE + 2BE - B = 0
(F) * (B + Q - QE - 2BE) = B - QE - 2BE
F = (B - QE - 2BE) / (B + Q - QE - 2BE)

It also may be worthwhile to solve for B, because that is the one variable you are in direct control of. Most of your bets will likely be made somewhere close to 1/2 pot, pot, or the rest of your stack. So, if you have a good idea of your equity plus the % chance your opponent will fold, you could plug those in and see if it results with a reasonable bet size you could actually follow through with. So, picking up to solve for B instead of F:

QE + 2BE - B - QEF - 2BEF + BF + FQ = 0
From here, solve for B:
(B) * (2E - 1 - 2EF + F) + QE - QEF + FQ = 0
(B) * (2E - 1 - 2EF + F) = QEF - QE - FQ
(B) * (2E - 1 - 2EF + F) = Q * (EF - E - F)
B = Q * (EF - E - F) / (2E - 1 - 2EF + F)


Anyway, I just found this subject particularly interesting because I always thought if you equity was BELOW 50% heads-up that you had to have at least some fold equity when you bet, but apparently you DON'T always need that depending on the bet sizing and equity as was the case in Question #6. Also, it's sort of obvious, but worth knowing that no matter % chance they will fold, you will ALWAYS give them BETTER odds to make a hero call as in Question #1 and Question #2, we need them to fold 50% of the time but they only need to make a hero call to our air 33% of the time.

I'm still tinkering around with how to practically use this information. I think I'll play a little more aggressively in pots where I have good equity - in fact, it appears my EV depends on it. I'll probably plug numbers in some real life hand examples upon review to make the real calculations @ the table second nature.

Any other thoughts, tips, or questions about fold equity? I know from my experience moving up stakes, it becomes a much more important concept compared to lower stakes where you run into more calling stations and maniacs.

[spoonitnow]

Ah, too bad no one else took a shot to finish this out.

When I was posing these questions to myself, I saw four overall variables to consider: my equity (E), my bet size (B), the TOTAL size of the pot BEFORE I bet (we'll call it Q). and the % chance the villain will fold to bet B (F). It actually took me longer to develop the equation than I anticipated as I would neglect a variable here or there. But, I had an Excel spreadsheet of different variables and finally came up with an equation that worked. It's mainly tedious algebra, but the start of the equation is that the sum of the times you are called versus the times you are not called is equal to 0. The times you are called is further divided into two parts based on the times you call and win and call and lose:

((Q+B)*E - (1-E)*B) * (1-F) + FQ = 0
(QE + BE - B + BE) * (1 - F) + FQ = 0
(QE + 2BE - B) * (1 - F) + FQ = 0
(1) * (QE + 2BE - B) - (F) * (QE + 2BE - B) + FQ = 0
QE + 2BE - B - QEF - 2BEF + BF + FQ = 0
From here, solve for F:
(F) * (B + Q - QE - 2BE) + QE + 2BE - B = 0
(F) * (B + Q - QE - 2BE) = B - QE - 2BE
F = (B - QE - 2BE) / (B + Q - QE - 2BE)

Note first of all that this (above and below) is assuming that our bet can't be raised.

Since I've got a min or two I'll work it out from here. What Jason's missing here is that the value B - QE - 2BE is a very special number. If we call that number S, we get F = S/(S+Q) or F = S/(S+POT) which should look familiar. If you need a hint as to why this should look familiar, we find the fold % needed on a pure bluff as F = bet/(bet+pot).

It looks complicated, but S is just our true risk in the hand. If we're betting all-in B in a pot of Q, on average we're going to get back equity * total final pot = E * (2B + Q) = 2BE + QE. So if we subtract what we're getting back on average when we're called from our bet we get B - (2BE + QE) = B - 2BE - QE, ta da.