[dschatz]

Head's Up Hold'em has a nash equilibrium. Therefore there is at least one mixed strategy (I do X with probability P in Y situation) which cannot be negative expected value to any other strategy. In this sense, there is an optimal strategy. It doesn't mean that it is the maximum expected value against a particular opponent, but no opponent can win by playing (which is largely the goal of a casino).

Opponent modeling is purely advantageous, but not necessary

[ianstallings]

I found this artile by Bryce Paradis that elaborates on using a Nash equilibrium for optimal play. He is known for bringing advanced mathematics to the game of limit poker and winning a small fortune because of it. Here is his take:

* Q: What’s a Nash Equilibrium or “game theory optimal” strategy? – Failed Math, Port Perry, Ontario A: An equilibrium strategy is one that wins the most money possible against a perfect opponent (this does not mean an opponent who can see your cards, but one who always knows your range whenever you take an action and makes the best choice against that range). In the game “rock, paper, scissors,” the equilibrium strategy is to randomly choose between the three options, choosing each one a third of the time in the long run. Finding equilibriums in poker is much more complicated, but the concept can be useful when you’re playing lots of hands against tough opponents. For example, if your opponent bets half the pot on the river after a particular series of actions, the pot is offering him 2-1 on his bluff. If he were a perfect player, the right thing to do would be to call his bet a third of the time, since if you called more he’d exploit you by never bluffing and if you called less he’d exploit you by always bluffing. In reality, of course, our opponents are never perfect, and so the idea of playing an equilibrium strategy at the table is usually pretty academic. *

http://pokerpromagazine.com/proscorner/bryce-paradis/

[hannibal5]

I can see the possibility for confusion here.

You can certainly use Nash equilibrium when you have figured out the strategies your opponents are using. This is what Bryce Paradis is talking about. It can have practical value when playing Heads Up.

But If we are talking game theory and "solving poker", there is no single winning strategy that works against all other strategies and you can't calculate single Nash equilibrium that would be optimal in actual game against specific strategies.

[iopq]

Yes, you can.

Your opponent has 1352 different hand combinations. Assume he is playing the Nash equilibrium strategy. Make the perfect plays based on this. If he plays worse than the Nash equilibrium strategy, you beat him. If he plays perfectly, you tie.

Assuming your opponent plays perfectly works in chess. Chess programs are stronger than the best humans now.

[hannibal5]

>Assume he is playing the Nash equilibrium strategy.

You you can't do that assumption because you don't know what the strategy is. You can calculate Nash equilibrium only if you know the strategy your opponent is using. In full no limit hold em there is no single strategy winning strategy, so you don't know the strategy your opponents are using.

[jib]

No you don't need to know what the opponents strategy is. You calculate based on worst case (ie op playing perfect) and worst case is you break even. There is no way to maximize profit but you can play unexploitable, ie at a minimum not lose and possibly win assuming op doesn't play perfect.

[hannibal5]

In poker optimal strategy is not winning strategy.

An optimal strategy’s goal is to loose the least against any arbitrary strategy. It is a strategy that is impossible to exploit in poker because poker has antes.

Poker players must seek maximal strategy. A maximal strategy’s goal is to win as much as possible against a specific strategy.

[ianstallings]

Yeah in heads up play using a Nash equilibrium to "balance your range" and play a more unbeatable strategy while also using it to calculate probable hands your opponent holds might be helpful, but it's far from a solution to the problem of devising a winning strategy that always works. I think that's why he calls it academic. Because it won't work in the real world.

[hannibal5]

Nash Equilibrium assumes that each player knows equilibrium strategies of the other players. It does not apply to No Limit Hold Em.

You can solve equilibrium for simplified poker games like just Heads Up with only shove or call an all-in options though.

[tedsanders]

Nash equilibrium certainly applies to No limit hold 'em. It's a zero-sum game with finite choices over finite time. Could you explain why you think otherwise? Are you just saying it's practically impossible to calculate?

[hannibal5]

You can calculate Nash equilibrium only when you know the strategies of your opponents. There is no single winning strategy in complete No limit Holdem, so you don't know how your opponent is going to play.

It's theoretically possible to find Nash equilibrium over all possible strategies but that's not winning strategy. You just lose as little as possible. You lose against most/all strategies.

Take for example Kuhn poker (https://en.wikipedia.org/wiki/Kuhn_poker). It's very simple but first player has several optimal strategies.

[iopq]

No, you will tie or beat all strategies because your opponents will make huge mistakes like calling when you are almost never bluffing and folding when you are frequently bluffing.