This is a discussion on POKER MATH: Calculating Fold Equity within the online poker forums, in the Learning Poker section; We are on a holy mission in the name of the poker gawds to discover the lost and secret art of quantifying fold equity. Seeking 


#1




POKER MATH: Calculating Fold Equity
We are on a holy mission in the name of the poker gawds to discover the lost and secret art of quantifying fold equity. Seeking math wizards and numerological mages who have reached enlightenment in these arts.
I am using the formula EV = P(f) + (1f)[EV(P+R+R)  S] this has the same result as EV = P(f) + (1f)[EV(P+R) + (1EV)(S)] and we are trying to find the value for f (fold equity) where EV = 0 or break even. The following once in a lifetime scenario plays down at a poker table in some obscure and secret underground location. BB posts 400, SB posts 200, pot 600, Action folds down to BU BU raises 800, pot 1400, SB raises 800 allin, pot 2800, BB folds, BU calls 800 (maybe or maybe not), pot 3600 At some point the SB collected the following variables and proceeded to do a most simple mathematical calculation. Pot if fold (P) = 1400 (3.5BB) Stack if lost (S) = 1400 (3.5BB) Raise (R) = 800 (2BB) EV = 56.09 (66 vs K8o) = P(f) + (1f)[EV(P+R+R)  S] = 1400f + (1f)[0.5609(1400+800+800)  1400] = 1400f + (1f)[0.5609(3600)  1400] = 1400f + (1f)[2019.241400] = 1400f + (1f)619.24 = 1400f + 619.24  619.24f f = 619.24 / (1400  619.24) f = 619.24 / 780.76 f = 79% Eureka we have found the answer, fold equity is 79%. BOOM! Keeping his best poker face to not reveal any tells of his complete and utter confusion the SB quickly runs the calculation again, from a less than ideal spot this time. EV = 25% = P(f) + (1f)[EV(P+R+R) S] = 1400f + (1f)[0.25(1400+800+800) 1400] = 1400f + (1f)[0.25(3600) 1400] = 1400f + (1f)[9001400] = 1400f + (1f)500 = 1400f  500 + 500f f = 500 / (1400 + 500) f = 500 / 1900 f = 26% Snap! Dude on BU you should fold 1:4 times for us to break even, who's your daddy. But the BU called and he watches his chips get shipped away... Now, the question of the universe and everything can anyone PLEASE tell me what on earth 79% is supposed to mean? 
Similar Threads for: POKER MATH: Calculating Fold Equity  
Thread  Replies  Last Post  Forum  Thread Starter 
1015 buyin downswings  40  29th May 2016 7:00 PM  Cash Games  TheArnie 
#2




My Opinion
You really need to do all that. Its just allin preflop and you know these hole cards should hit. Because they lost the last two times you had them. Viceversa the hole cards just hit so fold them. Here's another angle. You know weak hole cards hit when the dealer button was on some spot. Wait for good hole cards and the dealer button in same spot. Give or take a spot for standard deviation. Remember, it hits 75 percent of time in a standard deviation with a coin flip. with 13 cards its even less of an error. Flushes and straights are a little exception to the rule. With those cards you need to see the flop. You want an open end straight or four cards to flush. Count it a third of the time. Your should hit your straight or flush a third of the time. If you have suited connectors and just hit a straight or flush best to fold them. Not even look at the flop.

#3




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What I really want to know, perhaps you did answer that and I missed it, my apologies, what does the answer 76% fold equity mean? Is it 4:3 or does it stay 3:4 and we just discard the minus, or do we read it as 1:4 instead? Is it even still relevant to folding or does it now favour calling? Is there anyone that knows how to interpret this? If my EV is >50% the answer tends to be negative. 
#6




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= 1400f + (1f)[0.5609(1400+800+800)  1400] = 1400f + (1f)[0.5609(3000)  1400] = 1400f + (1f)[1682.71400] = 1400f + (1f)282.7 = 1400f + 282.7  282.7f = 1117.3f + 282.7 f = 25.3% Quote:
You are solving: tEV = P(f) + (1f)[EV(P+R+R)  S] when tEV = 0. {The left side EV is total EV. I changed it to tEV make it obvious that the EV inside the equation is different. The EV in EV(P+R+R) is showdown EV, that percentage your hand has of the pot when you are forced to show it down.} Why? Because you want to find out how often your opponent must fold for you to make money by shoving. If you opponent folds exactly that percentage, you don't make any money (but you don't lose any either). If he folds more often, you benefit from that. If he calls more often, you lose. Now, in that light, do you see why it's negative? You picked a hand with a positive expectation when called. You benefit when he calls. If you had your way, you would force him to never fold. When he folds, you lose equity that you had from the showdown side of things. You want him calling 100% of the time, more even (if such a thing were possible outside mathematical equations). Compare that to the second hand, where your cards are not the favorite. In this case, if he calls 100% of the time, you will lose money on the shove. Now, you rely on him folding often enough that the money you earn from his folds compensates for the money you lose when he calls. He now needs to fold more than about 32% of the time, or else you lose money by shoving. I hope this makes sense and helps. When your hand is a favorite (greater than 50%), you prefer the call. You don't benefit from fold equity because it reduces your showdown equity. It hurts your total EV when the other guy folds. When your hand is a dog (less than 50%), you need a certain amount of fold equity to make up for the lack of showdown equity. 
#7




I made a mistake above and it's too late to edit it.
It's not true that we want him to call, even with a negative f. 100% of the 1400 that is already out there is better than 53% of the total pot at the end. But, it is true that we'll show a profit with the first hand no matter what. We don't need the person to fold to show a profit. If they call 100% of the time, we still profit. If they fold 100% of the time, we show more of a profit. It's like holding Aces preflop. Sure, it's great to play for stacks. But, if you could get you opponent to put in 95% of his stack and then fold for the remaining 5%, you'll make more money than showing it down. Basically, if f is negative, we show a profit even when called 100% of the time. If f is positive, we need them to fold that much or more to show a profit with the hand. 
#8




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with EV <50% we loose money when called but win money when they fold and f is a positive number that indicates how much folding will compensate for the deficit. is it then also correct to say that: with EV >50% we win money when called and when they fold so f is a negative number that indicates how much not folding will compensate for the excess. Not sure if that even makes sense.... Quote:
<50% we loose when called so we need them to fold enough to make up for what we loose when called, >50% we already win when called so the result is negative and there is no folding required to make up for anything. Quote:
I think a light may just have gone on.... you said if they call we still show profit but when they fold we show more profit. From this perspective "not folding" to reduce excess does start to make sense. Quote:
Would I be making a leap if I assume what you are saying is that we only equate positive results to fold equity and should consider any negative results to be 0%, no folding required? Thank you so much vinnie for taking the time to discuss this with such clarity and detail. I am still perplexed as to what the negative value actually means but am much more confident with the results thanks to you. You rox!!! 
#9




I totally messed things up with my first post by focusing on 50%. We could have a positive expectation even if we're not a favorite to win. If the pot is large enough. This happens when we have a hand that is 40% to win, but we're getting 21 on our bet. So, we can ignore the majority of that that mess. It was 2am and I was tired.
It would be more correct to day that if "[ EV(P + R + R)  S ]" is positive, we don't need the other player to fold to show a profit getting the money in. An example, say we have 40% chance to win a pot of $4000, with a raise of $1000 and a stack of $2000. [40% * (4000+1000+1000)  2000] [ 0.4 (6000)  2000] [ 2400  2000] [ 400 ] So, even though we aren't a favorite to win the hand, we get enough money from the pot that we don't need a fold to show a profit. Quote:
For the first example, assume they never fold. tEV = 1400 * 0 + (1  0) ($282.70) tEV = $282.70 In this case, we have a positive expectation even if we never get a fold. We don't show a negative expectation until they fold less than 0% of the time, which is impossible in the real world. In the second example, if they never fold: tEV = 1400 * 0 + (1  0) ( $650) tEV = $650 When they never fold, we lose money. In this case, we need them to fold a certain amount to show a profit. 
#10




re: Poker & POKER MATH: Calculating Fold Equity
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Your work here is done! Thanks again!!! Luv the sig, makes me smile every time... 
#11




Just for completion's sake and to conclude:
We calculated the fold equity required for a hand with 25% EV when raising allin by 800 into a 1400 pot risking a 1400 stack to break even (tEV = 0) at 31.7%. Pot (P) = 1400 Stack (S) = 1400 Raise (R) = 800 EV = 25% tEV = P(f) + (1f)[EV(P+R+R)  S] tEV = 1400f + (1f)[0.25(1400+800+800)  1400] tEV = 1400f + (1f)[0.25(3000)  1400] tEV = 1400f + (1f)[750  1400] tEV = 1400f + (1f)650 tEV = 1400f  650 + 650f 650 = 1400f + 650f  0 650 = 2050f f = 650 / 2050 f = 31.7% Quote:
tEV = 1400 * .32 + (1  .32) ( 650) tEV = 448  650 + 208 tEV = $6 
Similar Threads for: POKER MATH: Calculating Fold Equity > Texas Hold'em Poker  
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1015 buyin downswings  40  29th May 2016 7:00 PM  Cash Games  TheArnie 