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Fold Equity Hypothetical

  
 
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Jason
Old 03-08-2010, 05:40 PM     Post subject: Fold Equity Hypothetical #1 (permalink)  
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You're holding 32 and the pot has reached $100. The river comes and you missed your str8 draw. You missed your flush draw, too. You got the nut low with 3 high, but sadly, this isn't razz. This is no limit hold'em and all the cards are out there. For the sake of simplicity, we'll assume your lone opponent can beat 3 high and you're NOT chopping if it gets to showdown. You decide to make a pot sized $100 bet to TRY to take down the pot since that's the only way you can win.

Question #1
What % of the time does your opponent need to fold for this to be a BREAK-EVEN play?

Question #2
Hypothetically, if the villain had 3rd pair, 9's on a board that had a King and a Ten, what % of the time does the hero need to be bluffing with air (no pair) like we actually are in this example so that villain CALLING your pot sized $100 bet is a BREAK-EVEN play for him?

Question #3
What % of the time does your opponent need to fold for this to be a BREAK-EVEN play if hero bets $50 (1/2) pot instead of $100 (full pot)?

Question #4
Hypothetically, if the villain had 3rd pair, 9's on a board that had a King and a Ten, what % of the time does the hero need to be bluffing with air (no pair) like we actually are in this example so that villain CALLING your 1/2 sized $50 bet is a BREAK-EVEN play for him?
- Jason

 
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CBAT
Old 03-08-2010, 06:36 PM #2 (permalink)  
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Question #1
What % of the time does your opponent need to fold for this to be a BREAK-EVEN play?


Villian needs to fold 50% of the time. (100/200)

Question #2
Hypothetically, if the villain had 3rd pair, 9's on a board that had a King and a Ten, what % of the time does the hero need to be bluffing with air (no pair) like we actually are in this example so that villain CALLING your pot sized $100 bet is a BREAK-EVEN play for him?


Question #3
What % of the time does your opponent need to fold for this to be a BREAK-EVEN play if hero bets $50 (1/2) pot instead of $100 (full pot)?


Villian needs to fold 33.3% of the time. (50/150)

Question #4
Hypothetically, if the villain had 3rd pair, 9's on a board that had a King and a Ten, what % of the time does the hero need to be bluffing with air (no pair) like we actually are in this example so that villain CALLING your 1/2 sized $50 bet is a BREAK-EVEN play for him?


I'm not sure how to go about 2 and 4...I'm working on it though.
 
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dranger7070
Old 03-08-2010, 08:23 PM #3 (permalink)  
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1.) 100/(100+100) = 50%

2.) 100/(100+200) = 33%

3.) 50/(50+100) = 33%

4.) 50/(50+150) = 25%

Not 100% sure on 2 and 4, but I think I got them right...
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rpm
Old 03-09-2010, 12:34 AM #4 (permalink)  
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is this not a question of basic pot odds or am i confused?
in number two you bet pot, giving villain 2:1 pot odds, meaning he has to have the best hand 1 in every 3 times he calls you (33%).
so your bluffing frequency (assuming you don't try to value bet worse than his hand) has to be 33.333333% for it to be precisely breakeven.

in number four you bet half pot, giving villain 3:1 pot odds, meaning he has to have the best hand 1 in every 4 times he calls you (25%)

as for you, finding your required fold frequency for breakeven is done by bet/(bet+pot) so say you bet 100 into a pot of 100 (full pot), you do 100/(100+100)=0.5, translating to 50% of the time. hope this helps. i may be barking up the wrong tree here.
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Jason
Old 03-09-2010, 01:05 AM #5 (permalink)  
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Very nice. No, rpm, you all had it right. I would wager most players don't know or apply those concepts. I don't think it's talked about enough as it probably should be compared to other topics.

So, yeah, I've been doing fold equity calculations and musings as it was on my todo list because the demands for it in my games have increased more so than past games I've played and it's been a while since I've thought about it more deeply.

For anyone willing to indulge, let's pick it up one more notch:

Suppose all of the variables are the same except now we're playing the TURN with one card to come, so instead of 0% equity, we now have 25% equity ... not very realistic with one card to come, but just run with it ... to improve to the stone cold nuts, but if we miss, we're still stuck with 3 high.

Question 5
With 25% equity heading to the river, to be a BREAK-EVEN play, what % of the time do we need villain to fold if we make a pot sized bet of $100?

Question 6
With 25% equity heading to the river, to be a BREAK-EVEN play, what % of the time do we need villain to fold if we make a 1/2 pot sized bet of $50?
- Jason

 
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rpm
Old 03-09-2010, 01:40 AM #6 (permalink)  
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i like what you're doing here. i'll attempt these later this evening after i grind through some more readings for school. this is the kind of stuff i'm starting to grasp and be able to use in post-hand analysis, but need many more repetitions and refinements before it will become second nature during the actual play of a hand. it is also very important stuff i feel for understanding the theory which underlies every decision, and gives winning players their edge.
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JKDS
Old 03-09-2010, 01:43 AM #7 (permalink)  
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Question 5

20% of the time

Question 6

Heh, we dont need him to fold
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Originally Posted by OngBonga View Post
But no, jkds is lolvillager and anyone who wants to string him up is sighbad.
 
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Micro2Macro
Old 03-09-2010, 01:50 AM #8 (permalink)  
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for 5 and 6 we should assume we're betting all in right? otherwise we have implied odds when we improve..
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rpm
Old 03-09-2010, 06:13 AM #9 (permalink)  
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ok i admit question 5 and 6 are beyond my mathematical understanding of poker at the moment. i'm interested to know how to work them out. i THINK i can determine the EV of the bet if you know villain's fold frequency on the turn. but i'm not sure how to go about these ones. i'll try to do some research then come back...
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dranger7070
Old 03-09-2010, 12:10 PM #10 (permalink)  
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FUUUUUUUUUUUUU Jason lol I'm pretty tired atm and I haven't really given 5 and 6 that much thought, but its eluding me so far. I'll look at this after I wake up.
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eugmac
Old 03-09-2010, 06:48 PM #11 (permalink)  
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This took me some time to get the same answers as JKDS - a nice review of basic high school maths, heehee.

We have to add up the EVs of all the possible outcomes: 1. he folds, 2. he calls and we miss and lose, 3. he calls and we hit and win.

Breakeven means the target EV of this calculation is 0; the above three EVs must add up to 0.

So in the case of question 5, how often villain has to fold to our pot-sized bet in order for us to BE.

Let x be the probability of villain folding.

0 = $100x [the probability that he folds, and we win $100] + (1-x [the probability of the times he doesn't doesn't fold])($200*.25 [we win the whole lot when we hit] - $100*.75[we lose our bet when he calls and we miss])
0 = $100x + (1-x)($-25)
= $100x - 25 + 25x
125x = 25
x = 0.2

just change the appropriate numbers to solve question 6 - hint: when we 1/2 pot it, and villain calls, we've just contributed 25% of the total pot, where we have 25% equity. Do we need any fold equity to break even?
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dranger7070
Old 03-09-2010, 08:16 PM #12 (permalink)  
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Sweet ty eugmac. I wish I could have figured that out on my own tho.
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Jason
Old 03-11-2010, 06:42 PM #13 (permalink)  
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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
- Jason

 
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JKDS
Old 03-11-2010, 06:44 PM #14 (permalink)  
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Quote:
Originally Posted by Jason View Post

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
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But no, jkds is lolvillager and anyone who wants to string him up is sighbad.
 
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Jason
Old 03-26-2010, 09:29 PM #15 (permalink)  
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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.
- Jason

 
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JKDS
Old 03-26-2010, 09:56 PM #16 (permalink)  
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If you want, you could rescale the equation to make it a function of F, B, and E only by expressing B in terms of Q which makes the equation look a little nicer...but thats really just the math optimization nerd in me talking.
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But no, jkds is lolvillager and anyone who wants to string him up is sighbad.
 
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spoonitnow
Old 03-26-2010, 11:11 PM #17 (permalink)  
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Quote:
Originally Posted by JKDS View Post
If you want, you could rescale the equation to make it a function of F, B, and E only by expressing B in terms of Q which makes the equation look a little nicer...but thats really just the math optimization nerd in me talking.
When you solve for fold % it breaks down and you get the same thing either way. It ends up being just another application of the whole bet/(bet+pot) thing.
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Originally Posted by Ripptyde
I only have 2 simple rules when I am coaching a new student.

Rule # 1: don't ask questions

Rule # 2: don't ask questions

I have no interest in discussing strategy with a protege'. Your job is to remain quiet and listen. I have a very systematic approach that I will share with the right candidate and I promise that I will turn you into a force of nature and show you elements of the game of poker that you never knew existed.
 
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spoonitnow
Old 03-27-2010, 01:11 AM #18 (permalink)  
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Quote:
Originally Posted by Jason View Post
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.
Quote:
Originally Posted by Ripptyde
I only have 2 simple rules when I am coaching a new student.

Rule # 1: don't ask questions

Rule # 2: don't ask questions

I have no interest in discussing strategy with a protege'. Your job is to remain quiet and listen. I have a very systematic approach that I will share with the right candidate and I promise that I will turn you into a force of nature and show you elements of the game of poker that you never knew existed.
 
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