The whole "opponent needs to fold x% of the time..."

[Boredom]

I'm trying to get better with this, so if somebody could help solve this hand? Hopefully I can learn how to apply this, thanks in advance.

I have AQ on the turn on a board of AT98 rainbow. I have exactly one PSB left effective. Let's assume for simplicity I know my opponent has AK. If I shove (for $100 into the $100 pot), how often must my opponent fold to make this play breakeven?

Thank you very much.

[T50_Omaha8]

Your EV needs to be zero; that is, you need x such that

200x - 100(1-x) = 0
300x - 100 = 0
x = 1/3 or 33%

[djk123]

So if you shove you will win the $100 in the pot the x% of the time he folds. The (1-x)% of the time he calls, you will win $200 16% of the time and lose $100 84% of the time.

[(x* 100) + (1-x)*.16*200 ] - [(1-x)*.84 * 100]=0

x=.342105, so about 34% of the time.

[Pokerfarian]

Doesn't that math (wrongly) rely on the fact that you have 0 equity if you dont bet? The answer is surely it depends?

[Boredom]

[ QUOTE ]
Doesn't that math (wrongly) rely on the fact that you have 0 equity if you dont bet? The answer is surely it depends?

[/ QUOTE ]

I may be way off here, but does the fact that I'm not considering not betting in this particular example make this irrelevant?

[djk123]

[ QUOTE ]
Doesn't that math (wrongly) rely on the fact that you have 0 equity if you dont bet? The answer is surely it depends?

[/ QUOTE ]

EV of checking is $0 since we never win. EV of betting $100 if villain folds 34% of the time is $0. So if we think villain will fold 50% of the time, betting is clearly superior because our EV will be positive, which is greater than 0. But if we think villain will only fold 10% of the time, betting is clearly inferior because our EV will be negative, which is less than 0.

[craig1120]

This is the way I do it:

0= 100x + (1-x)[300(.16)-100]

Is this correct?

[zaphod]

[ QUOTE ]
[ QUOTE ]
Doesn't that math (wrongly) rely on the fact that you have 0 equity if you dont bet? The answer is surely it depends?

[/ QUOTE ]

EV of checking is $0 since we never win. EV of betting $100 if villain folds 34% of the time is $0. So if we think villain will fold 50% of the time, betting is clearly superior because our EV will be positive, which is greater than 0. But if we think villain will only fold 10% of the time, betting is clearly inferior because our EV will be negative, which is less than 0.

[/ QUOTE ]

It still depends. He did not state if we have position or not. If we are last to act and our opponent has checked to us, we have greater than 0 equity by checking.
If we don't have position we have greater than 0 equity if our opponent is not always betting with AK.

It turn goes check-check it also depends on how often we get paid of/ how often we can push him off his hand when we miss..

So what is the most +EV move clearly depends on what our opponent does on both turn and river..

[Idiotex]

Help me out here. I've never done these EV calculations. But ignoring what has happened on previous streets, given that your turn decision is independent of what has taken previously and is in relation to the pot only... I can't see how the answer is 33-34% (although clearly it is since everyone is coming into it).

To me, if you risk $100 to win a pot of $100, the break even % to me seems to be 50%. Since 50% of the time you will win $100, and 50% of the time you will lose $100. I am aware that in reality you would have lost money on the hand on average, but your turn decision should be independant of what has happened on previous streets.

Sorry to be a bit of a tard on this one, but help me out here.

[Idiotex]

What is wrong with me?

If the villain folds 33% of the time we win the pot of $100.

But in the 67% of the time that the villain calls we lose an additional $100.

So, .333 x $100 - .666 x 100 = 33 - 66 = -$33

What is wrong with me?