Hey all,

So I've seen this formula, that specifies that to determine villain's max bet to keep our call EV is:

Bet <= Pot * equity / (1 -2 * equity)

Can someone please explain how we arrive to this? Why do we multiple equity by 2 etc.

Just getting back to basics here and brushing up on the math side which I'm uber rusty at.

Thanks !

Imagine if you want to know what equity (Eq) is required to call to at least break even when a bet of size B is made into a pot of size P. The EV equation for this would be:

EV = Eq*(P+B) - (1-Eq)*B

Now set it equal to 0 to solve for break even and solve for equity:

0 = Eq*(P+B) - (1-Eq)*B

0 = Eq*P+Eq*B-B+Eq*B

0 = Eq*P+2*Eq*B-B

0 = Eq*(P+2B)-B

B/(P+2B) = Eq

Trying to edit my post unssuccessfully but hopefully you can see it stems from the EV equation for the equity your hand is required to have to make a break even call.

Hey, thanks for replying, I am familiar with the EV formula and it's variations you've shown but that's exactly the thing - I seem to overlook the missing link to see how we arrive to:

Villain's max bet for us to BE or better <= Pot * equity / (1 -2 * equity)

I have not seen this equation before:

Bet <= (Pot * equity) / (1 - 2 * equity)


But it looks like tells you what fraction of the pot to bet given you know what equity you'd like to offer.

Now let's assume we want our opponent to at best be break even on a call:

EV <= 0

Substitute our EV equation from before

Eq*(P+B)-(1-Eq)*(B) <= 0

Now let's isolate the betsize (B) in terms of equity (Eq) and potsize (P)

Eq*P+Eq*B-B+Eq*B <= 0

Eq*P+2*Eq*B-B <= 0

Eq*P+(2*Eq - 1)*B <= 0

Eq*P <= -(2*Eq - 1)*B

Eq*P <= (1-2*Eq)*B

Now when we we multiply an inequality by a negative number it will flip the sign on the inequality. So when Eq < .5 , 1-2*Eq will be negative and we get:

(Eq*P)/(1-2*Eq) >= B

Which is the equation you shared.

Not sure how relevant/good it is anymore though. Seems weird if I am thinking about this right as your opponents equity is approaching 50% this equation says you should bet larger and larger but as it approaches 0 you should essentially bet nothing.

Maybe I am missing something or not interpreting the results correctly.

It just comes from solving for the bet size variable.

b = bet size
p = size of the pot before the bet
q = equity

EV(call) = q(p + 2b) - b

To find all bet sizes that can be profitably called, set this >= 0 and solve for b:

q(p + 2b) - b >= 0
qp + 2bq - b >= 0
b(2q - 1) >= -qp
b <= -qp/(2q - 1)
b <= qp/(1 - 2q)

This assumes that equity < 0.5, but in that case we would call any bet size.

Ahh yes interpreting it as a maximum betsize to call makes more sense thanks!

Ignore my inequality flipping comment it's not tru that 1-2*Eq is negative when Eq<.5

I'm pretty sure you meant to say

This assumes that equity < 0.5, for in the case equity > 0.5, we would call any bet size.
-----------------
Here is another look at bet sizing math and that is the case of hero betting when he has equity > 50%.

In that situation, he rather have villain call rather than fold. Since a call nets the pot, Pot, the relevant equation is:

Choose the value of Bet such that Bet > Pot*(1-eq)/(2*eq – 1), eq>0.50

Example: If hero eq is estimated to be 60%, Bet > 0.40/0.20 * Pot = 2*Pot

Thus, bet the largest amount greater than 2 * Pot that you think villain will call.

It is always wise to adjust these pure math-based results for such factors as opponent tendencies, future action and realized equity, stack sizes, position, etc.

Yes, that is what I meant, thanks. Poorly phrased by me.

Wow thanks for all the responses and the breakdown. Yes indeed this was referred to as the calculation for bet sizes made by villain for which we are +EV given equity and where, it was mentioned that when we have >0.5 we ofc call.

I should have provided more context. Anyway, def food for thought. Thanks !

Hey Stat, I know I'm asking noobish math questions, but can you share please how you came to this equation?

Sure. The EV equation for hero making a bet of Bet into a pot of Pot and villain calling is:

EV = eq*(Pot+ 2*Bet)- Bet.

Now if equity is greater than 50%, you generally want villain to call. If he folds, you win the pot. So, you want to find the bet size that gives you an expected win greater than Pot. Therefore, the solution inequality is:

EV = eq*(Pot + 2*Bet) – Bet > Pot

eq*Pot + 2*eq*Bet – Bet > Pot

Bet*(2*eq -1) >Pot*(1-eq)

Bet > Pot*(1-eq)/(2*eq-1), eq > 0.50

I showed in my previous post that with equity = 60% you need a bet of 2*Pot. Suppose you only made a pot size bet. Then

EV = 0.60*(Pot + 2*Pot) – Pot = 1.8 Pot – Pot = 0.8*Pot

This is less EV than if villain folded. A bet of 2*Pot gives EV = Pot so any bet greater than 2*Pot that is called does better than a fold.

Thanks !

Trying to make sense of the implications, suppose we do have 60% equity and we need to bet 2x pot for Bet EV > Pot , so Bet EV is higher than what we win when villain folds.

Am I getting it right so far ?

Something doesnt sit right with me tho as it seems getting any extra $ into the pot is better than villain straight out folding and putting no money when we are ahead

.

assuming strong strategies for the different betsizes as well as strong counter strategies, there's a curve that describes each situation. The curve may in fact look different than above, for example if limited by stack size the curve would end on a high point instead of turning downward as betsize increased as above. Also, since the assumption is that the strategies are strong, the negative ev betsizes are non existent.

Just to clarify, this math is only valid for the nuts/air vs bluffcatcher game, where the bettor wants to bet as much as possible. So in this case, 60% equity would equate to Hero having 60% nuts and 40% air in his betting range, and Villain having 100% bluffcatchers. If Villain has some fraction of nuts and is allowed to raise, then this math is not valid in terms of finding the optimal bet size.

The inequality

Bet > Pot*(1-eq)/(2*eq-1), eq > 0.50

gives the set of bet sizes that Villain cannot profitably call. If Villain plays well, he will always fold to these bets and Hero will never win more than the pot. If Hero bets less than this amount, then Villain will win some fraction of the pot in EV when he calls, and so Hero's EV drops below the size of the pot.

Although I did not state it specifically in my first post but did in the second, I think it is clear that I was only considering a call or fold to hero’s bet. A possible villain raise would add another dimension to the analysis as will other factors such as future betting, stack sizes etc.

I don’t agree with the statement that the math only applies to the nuts/air vs bluff-catcher assumption.

Stat, if I understand correctly, you are saying that we want Bet EV > equity in pot, so Bet EV is bigger than what we win when villain folds to our bet.

But isn't it true that if we bet and villain folds we win 100% of the pot and not just the eq*pot.

So maybe do you think it would be more precise to say that we want Bet EV > Check Behind EV , as this is the % of Pot we win when we check?

Would love this clarified.

Also, if you could please answer a much more noobish question -

I am really brushing up here and going back to square one with my algebra these days but isnt this supposed to be

Bet(2eq-1) > Pot / eq * pot ?

How did you arrive to Pot * (1-eq) on the right side of the ineqality?

And my final question -

This is fascinating, where would be a good resource to start to learn about this and how this works, specifically in poker with relation to stacks, strategies etc?

Thanks everyone, this thread just keeps on giving, I think I might become a reg here on the theory forums

eq*Pot + 2*eq*Bet – Bet > Pot is equivalent to A + X*Bet > B. Then X*Bet > B-A. Then Bet > (B-A)/X

I am considering two cases.

Case 1. Villain folds to a bet. Then EV = Pot

Case 2. Villain calls a bet. Then EV = eq*(Pot + 2Bet) – Bet. I then solve for Bet so that the EV > Pot, the Case 1 result

basically the optimal betsize is found at the point on the curve from which there is a loss of value in both directions. betting too big causes a loss of value betting ev(as fewer combinations can make such a bet, which brings down the average ev of the betting range). betting too small causes a loss of value betting ev(as the stronger hands are not receiving as much ev as they should, which brings down the average ev of the betting range).

I don't know of any good resources for learning the graphic implications of poker, but I have a small collection of stuff that you may be interested in if you happen to be a visual learner, or if you just think graphs are cool(I do lol).

https://forumserver.twoplustwo.com/1...alization+yaqh

thread delivers with profit maps like none other. very cool ty yaqh.

Thanks for clarifying stat, def food for thought.

Dude. This is f'in gold. Thread exceeds my expectations.

A. Thank You for the explanation, it is beautifully put.

B. Yes, I am a visual learner. And just now I am finally figuring out how to use it in studying poker and I can see it instantly on how different my learning curve is when I see and organize the material visually for myself.

Is it possible to re-post or send me the 1st and 2nd photo you posted in your last reply? They are abit blurry/pixelated and I can't read them clearly and they seem fascinating.

Also, if you could shed some light on how you build them that would be amazing.

Thanks !