optimal bluffing frequency

[spaminator101]

no such thing bluffing should be psycological not mathematical

[cats...]

its not that the 2nd player always folds regardless of his hand, its that he always folds given that he has this specific hand. In that case, if we claim 1/(p+1) to be the first guy's optimal bluffing frequency, the argument that he should increase this frequency if he knows that the 2nd player will always fold with hand X doesnt hold true.

is it not a fact of game theory that in a "one street game" that the first player's optimal bluffing frequency is such that the odds against him bluffing are the same as the pot odds the opponent receives? Also, that regardless if the 2nd player always calls or always folds with a certain hand, hand X in this situation, his EV will be the same? it's claimed to be true in the game theory section of TOP, but there's no proof, just a numerical example. if it is true can you lead me to a link with a proof?

"If either player can improve their equity unilaterally by changing their strategy, then the strategies aren't optimal."

i understand this, but is it true at the first player's optimal bluffing frequency, given hand X, where X is a specific inbetween hand, it doesnt matter what the second player does, he will have the same EV regardless?

thanks.

[AaronBrown]

There's no reason for the last actor in any game to randomize his strategy. He should take the action with the best expected value. In early stages, you sometimes randomize to affect future decisions by your opponent. But never at the end. That's why it's usually easiest to solve game theory problems backward from the end.

[AaronBrown]

I agree with that. The point of the "two pots" rule is that you have taken down a pot without showing, then it happened again. Sooner or later, they have to call you or you have to start bluffing. The trick is to get them out of phase, so they call when you're not bluffing and don't call when you are.

[cats...]

in the problem im working on, i didnt have the 2nd player randomize his strategy. In the game im working on there's only 3 possible hands, and it's a one-street game. So i made the assumption that given the 2nd best hand, the 2nd player always calls or always folds, and he will pick the higher of the two EVs, Knowing this, the first player will try to minimize that EV, which occurs when the two EVs are equal. I want to know if my original assumption that the second player always calls or always folds with the second best hand can be justified?