MPMK awkward river

[Oink]

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I wanted to check/fold, but I bet/folded because that's what 2+2 would tell me to do. I never realized half the people didn't know all the rules to a "blocking" bet. [img]/images/graemlins/tongue.gif[/img]

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Ok, thought more about this. Please tell me where I am wrong, but to me it seems that this example prooves Gehrig and Danzasmack wrong!!!!!!!

Say we are at the river OOP against a villain where we estimate our eq to 30%

Estimated actions by villain

- If bet to:
Ahead: call 60% raise 10%
Behind: call 25% fold 5%

- If checked to:
Ahead: bet 70% check 0%
Behind: bet 10% check 20%


This gives

EV(c/f) = 0.2*5 = 1

EV(c/c) = -0.7*1 + 0.1*6 +0.2*5 = 0.9

So we cant c/c because we are not good more than 1 in 7 (only 1 in 8). On the other hand we will be folding the best hand with 10% probability.

EV(b/f) = -0.6*1 -0.1*1 + 0.25*6 + 0.05*5 = 1.05

The value of the bet in it self is negative: -0.7 + 0.25 = -0.45, but the overall EV of bet/folding is better than c/f'ing because it prevents us of being folded of the best hand when we c/f.

WHAT AM I DOING WRONG HERE..?

EDIT: Oh and I cant see why EV(c/f) = 0 by def. When he checks behind a worse hand Hero will win the pot, how doesnt that give positive EV..?

[Oink]

EDIT: potsize = 5BB!

One more EDIT:

About EV of c/f equalling 0. To me that doesnt make any sense.

Think about it this way: Say we are at the river in the same scenario as the posted example but this time VILLAIN CANT MAKE AN ACTION IF CHECKED TO.

Then the EV of checking is

EV(c) = 0.3*5 = 1.5

since Hero will win the pot with 30% probability.


Now when vilain is allowed to make an action when checked to and assuming he will

Ahead: bet 70% check 0%
Behind: bet 10% check 20%

Then the value of folding is

EV(fold | if checked) = -0.1*5 = -0.5

And the value of checking

EV(call | if checked) = -0.7*1 + 0.1*1 = -0.6

To me it doesnt make any sense defining EV(c/f) = 0. Who the hell came up with that? Wannabee game theoretician Skalansky?

[Gurravasa]

I haven't read the whole thread but c/f is two actions right? There is a certain EV for the check and another for the fold. The fold will never earn us any money and therefor EV=O, but the check could earn us some money if nobody bets behind us and we pick up the pot. EV of a check is by definition always positive since it never cost us anything but could earn us money if noone bets. If you want to define a EV for the combination of a check and fold on one street it must always be positive, since it will never cost us anything but sometimes can gain some money.

EV(check)+EV(fold)=EV(c/f)

gives

>0 + 0 = >0

Don't understand why this is of any importance in reality but right should be right...

[zalazane]

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Now when vilain is allowed to make an action when checked to and assuming he will

Ahead: bet 70% check 0%
Behind: bet 10% check 20%



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Given that villain will be checking behind 20% of the time (and every time he does that you beat him), the EV of c/f should be .2*5

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EV(fold | if checked) = -0.1*5 = -0.5



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This is wrong, you can't give folding a negative value (you don't lose anything and you don't win anything)

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To me it doesnt make any sense defining EV(c/f) = 0. Who the hell came up with that? Wannabee game theoretician Skalansky?

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I don't know if anybody has defined EV(c/f)=0. but EV(f)=0

[Oink]

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EV(fold | if checked) = -0.1*5 = -0.5



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This is wrong, you can't give folding a negative value (you don't lose anything and you don't win anything)

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To me it doesnt make any sense defining EV(c/f) = 0. Who the hell came up with that? Wannabee game theoretician Skalansky?

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I don't know if anybody has defined EV(c/f)=0. but EV(f)=0

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When you say EV(f) = 0 by definition you ignore the times you fold the best hand. In the example above Hero will be folding the best hand with 10% prob. That negative value has to enter the equations somewhere when you compare actions.

I am sorry but defining EV(f) = 0 has to be the biggest misapplication of expected payoff in the history of game theory.

Again. Like they do it in math of poker.

if you play a game where villain cant make an action when checked to then Hero will win 1.5BB. If villain can bet if checked to then Hero will only win 1 BB because he will fold when villain bets. THIS MAKES THE VALUE OF THE FOLD = -0.5!

YOU CAN NOT IGNORE THE LOST VALUE WHEN YOU FOLD THE BEST HAND!

[zalazane]

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When you say EV(f) = 0 by definition you ignore the times you fold the best hand. In the example above Hero will be folding the best hand with 10% prob. That negative value has to enter the equations somewhere when you compare actions.



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It is entered to the calculation, because the pot size affects EV of other decisions.

Let's say the pot is 1 million BB and we win 50% of the time.

EV(fold) = 0
EV(call a bet) = ~500 000

[Gurravasa]

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EV(fold | if checked) = -0.1*5 = -0.5



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This is wrong, you can't give folding a negative value (you don't lose anything and you don't win anything)

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To me it doesnt make any sense defining EV(c/f) = 0. Who the hell came up with that? Wannabee game theoretician Skalansky?

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I don't know if anybody has defined EV(c/f)=0. but EV(f)=0

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When you say EV(f) = 0 by definition you ignore the times you fold the best hand. In the example above Hero will be folding the best hand with 10% prob. That negative value has to enter the equations somewhere when you compare actions.

I am sorry but defining EV(f) = 0 has to be the biggest misapplication of expected payoff in the history of game theory.

Again. Like they do it in math of poker.

if you play a game where villain cant make an action when checked to then Hero will win 1.5BB. If villain can bet if checked to then Hero will only win 1 BB because he will fold when villain bets. THIS MAKES THE VALUE OF THE FOLD = -0.5!

YOU CAN NOT IGNORE THE LOST VALUE WHEN YOU FOLD THE BEST HAND!

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Oink! Usually you count EV as your expectation from this point forward. And then EV for a fold is always 0. You are taking a wider angle of this wich is also correct since the EV for a fold is never 0 for the entire street. Just define what kind of EV you're taking about and we could all be friends.

[sharpie]

FWIW here are Oinks calculations if you make the EV of c/f = 0. I think either way works, but this way is easier to read:

Again, pot is 5BB before bets, we're ahead 30% of the time, villain will bet always when he's ahead if we check, will call 25% of the time when he's behind if we bet, and will bluff 10% of the time if we check:

EV of c/f = 0
EV of c/c (.7 * -1 + .1 * +6) = -.1
EV of b/f (.7 * -1 + .1 * +6 + .15 * +1) = +.05

I believe there's a chapter in poker essays 2 about this concept, but I only browsed through it in the bookstore a year or so ago and didn't really understand it at the time.

Edit: Another way to think about it, we're a .1BB deficit when we c/c the river, but we make .15BB out of him when we b/f against the 15% of hands that would've checked behind, thus b/f is +EV. Can anyone see any flaws in my thinking?

[sharpie]

For anyone who thinks betting is never correct if we wouldn't check/call, look at pages 212-213 of TOP. The idea is even though we don't have the winning hand often enough to call if we check, villain will call with enough worse hands, and those combined with the worse hands that would've bluffed us out and caused us to lose the pot had we checked make betting better than check/folding in that specific case.

[gehrig]

oink the value of folding the best hand is 0. the valuing of calling with the best hand is higher than 0, so folding isnt correct in that situation