[jason-johnson]

>”Game theory optimal" (as poker players like to call it) is not really the optimal strategy. It's the nash equilibrium strategy assuming the other players are also playing a nash equilibrium strategy. As soon as one player deviates from the nash equilibrium it's not optimal anymore :).

This is not really what most people mean when they say “optimal strategy”. It’s true that exploitative play will make more money if you know exactly what your opponent is doing and if they keep doing it despite it not working. Neither of these will generally hold in an actual game.

The reason it’s called “optimal strategy” is because it works no matter what your opponent is doing. It will not make as much as a strategy tailored perfectly to your opponent but it will never lose to anyone under any circumstances, assuming infinite games (assuming infinite games just so we can ignore variance). The worse case the strategy has is break even to anyone else using it.

[pfortuny]

You are assuming there are no collusions among players (agreements to play against another one). That perturbs the Nash optimality.

[im3w1l]

I'm kind of curious how accidental collusion could work out. Like imagine multiple players playing in such a way that they help each other - but purely out of ignorance on how best to play the game!

[8note]

Isn't it pretty standard, and how professional poker players make money?

All of the good players identify the worst, riches player at the table, and they all take that players money.

Once they're out, you leave the table too

[brigadier132]

> Neither of these will generally hold in an actual game

These absolutely hold, 90% of casual poker players have the exact same strategy problems of not bluffing enough and calling too much.

[drdeca]

It sounds like you are maybe describing a dominant strategy? Oh, wait, no, you are saying that if you play the Nash equilibrium, then no other strategy among opponents will do better against it than the Nash equilibrium would, and therefore (as the game is zero sum) the worst their choice of strategy can make you get on average, is breaking even?

Ok, now that I (I think) understand your comment: I don't think you have to know exactly what non-Nash strategy someone is playing in order to exploit it. I don't think trying to estimate how someone is likely deviating from the Nash equilibrium, in order to try to exploit it, is necessarily a mistake. I think it could be feasible for someone to get better returns on average by noticing off-Nash patterns of play in other players, than playing Nash regardless would? (Not that I could win this way. I couldn't.)

[chii]

But isnt the definition of the nash equilibirum the strategy where no deviation from it is going to afford you a better outcome?

[drdeca]

A Nash equilibrium is a pair of strategies (or, one strategy for each player) such that no player can get a better result on average from deviating from it.

If one player isn’t playing a strategy that is part of any Nash equilibrium, then the best response might also not be part of any Nash equilibrium.

If all other players are playing a strategy from a given Nash equilibrium, then you can’t do better (in expectation) than you would if you were to play the strategy for you in that Nash equilibrium.

(A game may have multiple Nash equilibria. Possibly one such equilibrium could be better for you (or for everyone) than another.)