Heads-up, one preflop bet only. Perfect plays.

[well]

I think the different equities of a hand against different sets of hands is not as subtle as you state it is.
The difference between JTs against different pairs of sevens should in my opinion not be considered. An optimal strategy does not distinguish this difference.
I agree that for each of your vector there are/is (a) best countervector(s).
But those vectors will - I think - never consist of numbers other than 0 and 1. Only don't cares could occur otherwise.
But the search is not for a perfect countervector as they are easily found. I called it "perfect" strategies, whereas I actually meant optimal ones.

O, and there are 169 hands to consider, not 178.

Looking forward for the next posts,

Regards.

[AaronBrown]

Sorry, of course you're right about the number of hands. I don't know where the 178 came from.

Non-transitivity in hand values can get you into a rock-paper-scissors situation. I agree it doesn't affect the existence of equilibrium, there is a 1/3, 1/3, 1/3 equilibrium in rock, paper scissors. But it does complicate the analysis.

I disagree that the optimal vector weights would be zero or one for all hands. That's true in the 0,1 game; if the hands were transitive, that would be true in heads-up one-bet hold'em as well (except for the trivial case of using an intermediate value to get an exact probability).

But suppose we had a six card deck with Win, Rock, Paper, Scissors and Lose. Win always wins, lose always loses and RPS play like the game against each other.

Clearly the first player bets with Win, as there is nothing to lose. He will also bet with enough of his RPS hands, the fraction depending on the bet size, in order to induce calls when he holds Win and his opponent holds RPS. But he'll have to randomize among RPS, he can't play one always and another never.

[well]

I think there's a slight misunderstanding here.
What I wrote was

[ QUOTE ]
I agree that for each of your vector there are/is (a) best countervector(s).
But those vectors will - I think - never consist of numbers other than 0 and 1.

[/ QUOTE ]

That means, a strategy that does best against a certain strategy.
I think the optimal strategy however, will be mixed. And that's why we can't use the approach I stated in the original post. All strategies found with that algoritm will be fixed.

Oh - and what was the sixth card?

Regards.

[AaronBrown]

I seem to have developed some kind of numeric dyslexia in this thread. Dysnumia? I can't keep my numbers straight. Five cards it is.

Now that (I think) I understand you, I think I agree.