Re: AK
 

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I don't think we're good the 2/3 times you need to be to raise-call.

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I see this reference to how many times one has to have the best hand. Where does the math come from? How do you get to a 2/3's win percentage?

Thanks for an answer.

Re: AK
 

An approximation, if villain is behind and calls we win one, if he is ahead and raises and we call we lose 2.

vilain could also fold and we gain nothing, call with a better hand and raise with a worse one, but latter two are unlikely and cancel each other.

Re: AK
 

[ QUOTE ]
[ QUOTE ]
I don't think we're good the 2/3 times you need to be to raise-call.

[/ QUOTE ]

I see this reference to how many times one has to have the best hand. Where does the math come from? How do you get to a 2/3's win percentage?

[/ QUOTE ]

let P be the probability that we have the best hand.

when we raise with the best hand and get called, we win 1bb. when we raise with the worst hand and get reraised and call, we lose 2bb. breakeven is when our EV (the sum of these possibilities) = 0. so:

P - (2 * (1-P) = 0
P - 2 + 2P = 0
3P - 2 = 0
3P = 2
P = 2/3

hth!

Re: AK
 

Thanks to both of you!