Comment by heyitsnick | Hacker News Reader

Comment by heyitsnick | Hacker News Reader

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> If you're playing the Nash strategy and everyone else isn't, you can still lose.

I'm not sure if we're talking in the same language here; you cannot lose (beyond your share of the rake) playing a GT strategy, regardless of what your opponent is doing. This is the basis of what GT is.

> But the main thrust of my post was that heads up (two players) is mostly solved and you don't seem to address that point. Is that not true?

No, it is not true. Even for FL hold'em, which has a significantly simpler game tree, the best attempts are still far from a complete solution. Large simplifications/assumptions to the game tree need to be made to make it a manageable size computationally. Much simpler representations for HUFLHE have been solved.

For all other heads up games, researchers are barely scratching the service; I don't even think there's a clear understanding of how to even go about simplifying the game tree to reduce it to a manageable size to even consider solving it.

[I put in the caveat that i (a) haven't read up on the latest in the last couple of years and (b) have not been involved directly with any research projects. I'd love to be corrected from someone who's involved in the latest research.]

> you cannot lose (beyond your share of the rake) playing a GT strategy, regardless of what your opponent is doing. This is the basis of what GT is.

What does GT mean here? Not game theoretic, since general Nash equilbria don't have this property. Nash is a strategy where no individual has a motivation to change from the equilibrium.

There are multi-player games that have this property that optimal play always results in statistical winnings, but I doubt that poker is one of them. For example, that would imply that collusion isn't effective against an optimal player.

Heads up is a different matter. There it's fairly elementary game theory that there exists an unexploitable strategy. A brief search says you're probably right, though, about heads up not being solved.

[edit: Rereading your comment, I think we are talking cross-purposes. I'm talking about heads up games only. You said "Heads up, where there is an optimal strategy, is essentially solved," I replied that "this is wrong, although there is a solution, it's far from solved..." and the conversation went from there. It seems at some point you switched to talking about ring games? On that I obviously agree, Nash does not apply.

So I maintain everything i said before; HU games are far from solved. Much the same that chess is far from solved.]

The game of poker is symmetrical and zero-sum (ignoring rake).

As you say, "Nash is a strategy where no individual has a motivation to change from the equilibrium." If I were playing a GT optimal strategy, the best you can do yourself is play the same strategy - you are not motivated to deviate. This will be EV neutral to both of us in a symmetric, zero-sum game.

Any deviation you make will either be EV neutral and therefore indifferent, or EV negative. If it is EV negative to you, I gain.

Of course none of this applies to multiway games, we're talking heads up.

Ah, I see what happened. You responded to my initial comment with:

> Firstly, although it is proven that there is an optimal (game theoretic) strategy to play (Nash) [...]

The reference to Nash here made me think you were claiming there to be an optimal strategy in the multiway case. I've been talking mostly about multiway ever since. I associate Nash equilibrium with multiway games and wouldn't use the term "Nash" to describe GTO play in a heads up game, even though a Nash equilibrium would be GTO. But maybe this is standard lingo?

Question: So is it the case that there are human players that are measurably better (in a statistically significant way) than the best AI players, heads up?

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