Ok what is the underlying concept here

[Numbr2intheWorld]

So lets say you get to the river on some hand with some board and you decide to overbet twice the pot as a bluff. Now, if you decide to do this, you have to win the hand 2 out of 3 times to break even. Otherwise, you have made a mistake (in a vaccuum)

But looking at the other players perspective, he's faced with a 2 times the pot overbet and he has a bluff catcher. This bluffcatcher always beats a bluff from his opponent Now, he is calling a 2x pot bet to win 5x the pot. This means he has to be right 40% of the time for calling to be a breakeven play.

Now my question is, why isn't it 1/3 of the time? Is it just because when calling your considering a range and with the betting you are considering only with a bluff? What concept fills in this gap?

[mcatdog]

I don't understand your question. Are you expecting the two probabilities to add up to 100%? There's no reason why they should.

[bigspenda73]

your math is wrong, if I shove $10 into a $5 pot my opponent has to call $10 to win $15, not to $10 to win $25

[Knytestorme]

Wouldn't the issue be that you are not counting the times where you win 5x the pot after he calls but his bluffcatcher is < your bluff?

[BigPapi]

because the size of the pot has changed with your bet...

[Numbr2intheWorld]

I don't understand your question. Are you expecting the two probabilities to add up to 100%? There's no reason why they should.

No, i'm expecting that they = the same.

[jackvance]

They're not gonna be the same because of the dead money in the pot I think.

your math is wrong, if I shove $10 into a $5 pot my opponent has to call $10 to win $15, not to $10 to win $25

It's the same thing; the pot odds calculation is $10/$25=40%. Or the EV calculation you are referring to:

he has to risk $10 to win $15, breakeven (1-x)(-$10)+x(+$15)=0 => x=0.4

[IowaSkinsFan]

Hmmmm this is interesting. Still wrapping my head around it.

[mcatdog]

There's no reason for them to be the same. Also if you expect them to be the same then why did you want it to be 1/3 when the other one is 2/3?

Often times if you're trying to understand a concept it helps to think about the extreme situations. For example if you bet $1 into a $100 pot as a bluff then your bluff needs to work 1/101 = 1% of the time to be profitable, and villain's call needs to be correct 1/102 = 1% of the time to be profitable. If you bluff $100 into a $1 pot then your bluff needs to work 100/101 = 99% of the time, and his call needs to work 100/201 = 50% of the time.

[Parasurama]

Massimo, so the frequency that your opponent has to call for you to break even should be equal to the frequency that he has to have the best hand for a call to break even? This is what you're saying.

[IowaSkinsFan]

Yeah I'm pretty sure mcat is right. What was confusing to me is that the scenario has input for two variables so there is no solving to do. We know that opp is overbet bluffing so of course bluff catcher should call everytime. They are really just two seperate statements.
1. If we are unaware of opps range, how much does our overbet have to work to be profitable?
2. If we are unaware of opps range, how much does our call have to win to be profitable?

[noble007]

Yeah ISF is right they are two seperate statements.

Even though you think you are looking at the same situation from the other players perspective you aren't because in eg.1a the bettor is bluffing with a frequency of 100% and in eg.1b the bettor is bluffing at an unknown frequency.

If you want the concepts to link and the math to balance the first statement would have to read something like -

So lets say you get to the river with some hand on some board, what's the optimal bluffing frequency if you were going to overbet twice the pot?

& looking at the other players perspective, he's faced with a 2 times the pot overbet and he has a bluff catcher. This bluffcatcher always beats a bluff from his opponent. Now, he is calling a 2x pot bet to win 5x the pot. This means he has to be right 40% of the time for calling to be a breakeven play.

(now the concepts are linked and the numbers match - 40% is the optimal bluffing frequency for the bettor in this example as well.)

Probably not what you are asking but anyhoo...

[dsaxton]

The real question is why should they be equal?

One player is risking x to win y, whereas the other player is risking x to win x + y. As mcat pointed out, there's no reason for these numbers to be the same. In fact, the break even points are only equal in the totally trivial case where the player is bluffing $0. (If x / (x + y) = x / (x + x + y), some algebra shows that x = 0.)

[Numbr2intheWorld]

Yeah ISF is right they are two seperate statements.

Even though you think you are looking at the same situation from the other players perspective you aren't because in eg.1a the bettor is bluffing with a frequency of 100% and in eg.1b the bettor is bluffing at an unknown frequency.

If you want the concepts to link and the math to balance the first statement would have to read something like -

So lets say you get to the river with some hand on some board, what's the optimal bluffing frequency if you were going to overbet twice the pot?

& looking at the other players perspective, he's faced with a 2 times the pot overbet and he has a bluff catcher. This bluffcatcher always beats a bluff from his opponent. Now, he is calling a 2x pot bet to win 5x the pot. This means he has to be right 40% of the time for calling to be a breakeven play.

(now the concepts are linked and the numbers match - 40% is the optimal bluffing frequency for the bettor in this example as well.)

Probably not what you are asking but anyhoo...

This was actually what i was thinking, but yea my terms we're two seperate things. Thanks guys