Edit: nevermind, i prob should read the entire thread cause it looks like this has already been answered, or attempted to be answered.

So even if you have the lock nuts on the river and you fold it, that's still a zero EV fold? You haven't lost any value by folding, nor have you gained any? How does that make sense? You just gave away an entire pot because you're an idiot! LOL

Maybe using the river is a bad example. Let's say you have AKs and the flop comes QJT of that suit on the flop... and you fold to a bet. How can that possibly be a 0EV fold?

Ah yes of course.

Folding in combination with whatever other action you take may be non-zero EV, but when you isolate the action of folding alone its always 0 EV.

No, I disagree, unless you mean "if you measure all other EVs relative to folding then folding is always zero EV", which is obviously true.

When you use the frame of reference of before-hand to after-hand, you're still just considering the outcome of the single action. If you don't see why, consider that the after-hand amount is centered around decisions made up to this point (i.e. the amount of money put in and the size of the pot)

I can work through a simple example if you want. Or you can try it yourself - you'll see that the relative EVs of all the decisions are the same.

If we could isolate the value of a whole strategy depending on our stack size, wouldn't that value be relative to the ev of folding preflop?

grunching but it depends what you mean by EV

if EV means "expected (i.e. average) _change_ in your stack size between right now and the end of the hand" then obv EV(fold)=0 because if you fold right now then your stack size isn't going to change at all by the end of the hand.

but there are other conventions for doing EV calcs, as well

Really not sure what you're getting at because obviously there are multiple actions over the course of a hand.

Using the conventional EV method you can do a before hand/after hand EV calculation, but what it amounts to is the EV of your entire decision tree. You aren't isolating a decision in any way.

It seems like what you're getting at is a before hand/after hand method that is also unconventional. I'm not going to rule out that this exists and I am curious as to what it would look like.

OK, let's take a simple example. It's on the river. The pot is 100, your opponent bets 50. You figure you have a 1/3 chance of winning if you call.

The whole hand was played HU, so up to this point you've each put in $50. Let's say you just have $50 left, so you will have to fold or call.

So, by the EV(fold) = 0 method, we have
EV(fold) = 0
EV(call) = 1/3*(100+50) - 2/3*50 = 16.66
very simple and straightforward.
EV(call) - EV(fold) = 16.66 which means that calling is 16.66 better than folding

By the pre-post method,
When you fold, your stack will decrease by 50 from pre to post hand.
When you call, your stack will either:
* be 100 larger than before the hand (1/3 of the time)
* be 100 less than before the hand (2/3 of the time)
EV(fold) = -50
EV(call) = 1/3*100 - 2/3*100 = -100/3 = -33.33
EV(call) - EV(fold) = -33.33 - -50 = 16.67 (rounding error)
which means that calling is 16.67 better than folding

... which means that both forms return the same answer.

Why would you use the 2nd method? Normally you wouldn't. I can't really think of a reason to do it off hand.

I first saw it myself in Sklansky's book "The Theory of Poker". I can't find my copy, I probably sold it in a fit of degeneracy. It was in the section describing GTO bluffing on the river. This example has actually caused me trouble from time to time, because people who are more familiar with the EV(fold)=0 school of thought get hung up on it. I don't remember if there was a reason to use the pre-post method there or not.

For example we use EV of a line convention in the solver (internally). So if you call 100$ on the flop and fold on the turn that fold is -100$.
While this is now how most people think this is the same thing but makes programming and comparing lines much easier because you don't need to track stack changes and thing about relative values so it's all come together nicely when summing up EVs of various lines.

I wouldn't say "EV of folding is always 0" although that's a good default assumption if you don't see asterisks around.

If you tried doing calculations by hand you would quickly find out that way (the EV from the start of the hand) is way easier to keep track as well.

I think this might be true in some circumstances, but you have to be 100% consistent. Since I don't do it often, I sometimes have to convince myself I'm doing it right by doing it the other way to check my answer.

I can't say this is wrong its just real ugly. Yeah, a bad fold is -EV under this convention, but a good call is also -EV.

But I mean if it makes calculations easier for some people and they take the same actions as a result I really can't argue.

I think the takeaway here is that
* fold=0ev is a good way to look at it. It's pretty simple and it reinforces the right mindset
* but it's important to remember that when you are calculating EV, you are *always* just considering the delta between your actions
* if you are finding a calculation to be complicated, consider other points of view and see if they simplify things

The call actually IS -ev. If you had the option to neither fold or call, that would be the best option.

The flip side of this is why it's always +ev to bet even when your opponent can profitably call. Because even though from his viewpoint calling is better than folding, he still loses more money than if you didn't bet at all.

Both answers are equivalent as long as you're consistent. If your stack is 85 bb after folding, you could also say that EV=85 bb. What matters is your EV relative to your other strategic options.

Yes

Hey guys, completing the analysis of RustyBrooks agree on the following ?.

1)

Suppose a game of HU ocn 150 big blinds each player.

On the river the pot of 100 big blinds is formed, and we go with the 100 big blinds remaining. Our equity is 34%. See if it is better to play XF or go all in. The villain always pay our bet.

Considering the EV (fold) = 0:

EV (XF) = 0
EV (ship) = 0.34 * 200 to 0.66 * 100 = 2

Considering pre-post hand:

EV (XF) = -50
EV (ship) = 0.34 * 100 to 0.66 * 100 = -32

In both cases the better the ship, right ?.

2)

I saw in different places, assertions that need to have an EQ of at least 50% to make a value bet if the villain never plays fold.
An advanced player says that in this situation I bet all EQ <51% will be negative EV.
a) Why not consider the money invested?
b) Obviously at her answer, I understand that analyzes the hand only at the moment of taking the decision, where EV (XF) = 0. Where obviously need at least 51% of EQ.

I hope an opinion on this !,

Greetings!

You don't come up with the same difference between your options, so I conclude that you've made a mistake somewhere. I don't have time to look at it now, though.

Your model is also incomplete unless you're claiming that if you check villain will always bet. If he won't then your potential actions are
check-fold
check-check
bet-call
(assuming he never folds)

So I have problems with how you've set this up. If you check and he goes all in, and you call, then it's the same as if you go all in and he calls. I think you have to include the option for you to check-call.

EV(check-fold) = 0
EV(ship) = .34*200 - .66*100 = 2
if he is betting the full 100 then
EV(check-call) = EV(ship) = 2
if he bet less, say, 50, then you'd have
EV(check-call) = .34*150 - .66*50 = 18
since 100 is the most he can bet, check-fold really isn't even an option, we can always call.


By the other method you'll have this
f(check-fold) = -50
f(ship) = .34*150 - .66*150 = -48
and so the difference between then, f(ship) - f(check-fold) = 2
(same as before)

If he'll always check behind then checking is clearly better than shipping, because
EV(check-check) = .34*100 = 34

If he checks behind, like, 10% of the time, even then checking is better, even if we assume we'll fold if he bets
EV = .1*.34*100 = 3.4

Yes, if i check the villain will always bet.

Thanks, i have a mistake here.

Yes!, your explanation is great.

In a while I write another situation to complete the conclusion

Hello RustyBrooks, I'll put another situation to finish making my question.

Now suppose in the previous example, our only alternative is to ship or play XF.
Besides the villain always bet when we play check.

Do you agree that we should bet all-in with each hand, with at least 33% higher equity ?.
With hands that meet the above condition, we would be making a value bet.

If you agree on this, why do you think that is considered a value bet those hands with equity greater than 50%? You think it's something similar to the case when we assume EV (fold) = 0 at the time of taking the decision without considering the money invested?

This situation I have seen in GTO strategies.

This question was necessary and / or explanation, in order to maintain the topic of the thread.

If he always bets when checked to, and always calls when you bet, check-calling = betting
In this case if equity > pot odds then betting > folding

If he sometimes checks, and your EV < 50%, checking is usually better than betting

This may be the reason for which, in the case of not knowing the% check back the villain, it is better to have an equity> 50% to bet?
In this way we ensure that the best option is to bet?

The real problem here is that you insist on folding to his bet. If you don't fold to his bet then checking is clearly superior when equity < 50%