Maximizing cEV in check behind vs betting scenario.
 

Here's the scenario.

You get to the river of a cash game hand where the pot is $300. You have a hand with show down value. Ie, it's the best hand x% of the time. However, because there are weak hands in the villains range, you feel that betting will fold out weaker hands that beat yours y% of the time.

So, your cEV for checking behind is x*300.

Your cEV for betting (I'm assuming the bet is $200) is (300)(x+y)-200(1-x-y). I'm also assuming all worse hands fold so you lose every time you are called.

What I'm trying to do is figure out some sort of equilibrium between x and y that maximizes cEV. But, I'm not sure how to set the constraints either.

The more I think about this, the more I think it's probably a stupid question and you can't really find an equilibrium. But, I'll see what you guys think/suggest.

Re: Maximizing cEV in check behind vs betting scenario.
 

You should bet if 500y + 200x > 200 (x,y decimals)

Note that x being larger makes it better to bet. This may seem slightly weird but it comes from having a fixed y% of his whole range that is best & folds. That's what you're gaining irregardless of x & you're risking slightly less if x is larger.

Re: Maximizing cEV in check behind vs betting scenario.
 

[ QUOTE ]
You should bet if 500y + 200x > 200 (x,y decimals)

Note that x being larger makes it better to bet. This may seem slightly weird but it comes from having a fixed y% of his whole range that is best & folds. That's what you're gaining irregardless of x & you're risking slightly less if x is larger.

[/ QUOTE ]

Yea, I realize all this. I was just trying to figure out a way to determine if their are distinct values for x and y that yield the larges gains in cEV by betting rather then checking behind.

But, I realize this is a linear function, and the maximum is only constrained by how we define x and y. Obviously, the best scenario is when x and y are a maximum. However, the only relationship we can really say about x and y is that x+y<=1.