# Is the Nash equilibrum strategy probabilistic?

Suppose you play with a Nash equilbirum strategy in a heads up NLH match, can you play every equal situation in the same way? (Equal in the sense of the same cards on the board and in your hand and the same bet sequence). Or do you have to roll a dice and play different ways some percent of the time?

I am wondering if there is a proof or disproof of this, if not is there a good arguemnt for or against?

My question is, is there a pure (not mixed) nash equilbirum strategy in poker? Ie such that you dont loose money against a mixed nash strategy?

## 2 Answers

There is a surprising amount of mixed strategies in GTO poker. Generally if there is a mixed strategy then the EV of the two actions are very close. I don't know of any proofs but you can solve specific spots, and they almost always have a bunch of mixed strats. The guy talking about dynamics doesn't understand what a Nash Equilibrium is. Nash Equilibrium or playing GTO is different than a maximally exploitative strategy.

Edit: There is nothing dynamic about a Nash EQUILIBRIUM. The players are in an EQUILIBRIUM because both players don't have an incentive to change their strategies.

Here is an example of a GTO solution to a specific spot in NLHE BTN vs BB... You can see there are a lot of mixed strats. If you had a strong enough computer you could solve the whole game. For example HU limit holdem has been solved and there are plenty of mixed strats. There is also nothing dynamic about the GTO solutions. (I made the GUI with a lot of help from stack overflow programmers and was hoping to contribute in the poker community since that's what I know.)

• @lessharm:I didnt mention exploitation, I refer only to Nash strategies. Still, while there are surely spots when you can play GTO, these are extreme scenarios (like push-fold strategies, bubble in tournaments etc.). The way JustMe asked, it seems that he means to play GTO in poker in general as a whole. For No-Limit holdem as a whole, you cannot pre-define a play against all opponents , against every hand and taking into account every possible dynamic. Jul 17, 2017 at 6:41
• @koita_pisw_sou Yes you can, thats what Nash equilibrium is all about, thats the whole point of it.
– user1457
Jul 17, 2017 at 7:52
• Not insulted, just curious if that's the case as you say, can you provide with some evidence? I am talking specifically about NL holdem as a whole and not specific spots/scenarios Jul 17, 2017 at 8:02
• Just read wikipedia dude. Nash strategy existence in NLHM was proved many years ago. Scroll down to informal definition.
– user1457
Jul 17, 2017 at 8:07
• Wow, noticed the poker stack exchange didn't have much traffic and a lot of bad advice/answers. So tried to answer a couple questions. Someone went and down voted all of my other answers.. I guess I'll leave you guys to it. Jul 17, 2017 at 9:31

In poker, there are never identical situations, because even if you play with the same cards and the same opponent for days, you will end up creating history and dynamics between you.

However, you will end up facing decisions similar to what you have faced in the past. And yes, in the long run if you want to be balanced and unpredictable you will end up mixing your play with various moves/strategies/unorthodox plays and even random ones

• The fact that no poker hand is identical is true, due to the reasons I explained. There is no proof as to whether a mixed nash equilibrium exists, to "force" you to play in a random manner. You can sketch a proof by contradiction though (ex. if you always, 100% of the times bet the flop with a certain amount when you have top-pair, it is obvious that your opponent will "read" you, thus could benefit from it in the long run) Jul 14, 2017 at 11:18
• +1 To get you even. I just stay away from this guy. Jul 17, 2017 at 11:47