Suppose you play NLHM heads up and are using a Nash equilibrium strategy.
1. Is everytime you have the same hand is in the same situation do you do the same action? Or do you consider the history (previous hand played) aswell?

2. Does this strategy differ from how you would play in a zoom heads up game with infinitely many opponents? Ie you only play 1 hand against each opponent.

3. Is there a pure Nash strategy which does not loose against a mixed Nash strategy?

4. Suppose you know the (crappy) strategy of your opponent completely, we call it S1, and you compute the optimal play against it, call it P1, if your opponent is given P1 and computes the optimal strategy in repsonse, S2, etc. Consider the sequence P1,P2,P3..., does this sequence converge to Nash equilibrium? Or does it oscillate?

I will have a go at the answers...

1. You have the same options each time, but should have a RANGE which behaves with some randomness. So sometimes you bet out and sometimes you check/call the exact same hand. GTO never remembers a previous hand for this reason, but humans may choose to do so.

2. GTO strat would work perfectly in Zoom, but other exploit methods may work also. From question 1, it follows that GTO works in zoom.

3. "Mixed" is undefined, but if you mean a mixture of methods, or an approximation of Nash with added exploit strategy, then pure Nash would not lose money to this, heads up, and would likely not lose money multiways unless both opponents are colluding specifically against you. However, pure Nash may not be the most profitable strat at a given poker table.

4. Yes, all logical exploit strategies arrive at Nash when level infinity is reached. It oscillates but each revolution is closer to Nash.




Sent from my iPhone using Tapatalk

I mean whether a strategy which doesnt involve any randomness (no RANGE as you call it) can break even against a Nash strategy which does?

A pure Nash strategy: Makes same decisions in same situation.
Mixed: Employs randomness to make its decisions. (Thats what wikipedia says atleast)

Since poker is all about incomplete information and randomness, GTO will crush ABC-TAG in any reasonable sample of hands. The TAG player must randomize, to have a chance. Most flops involve texture and usually neither player has much of a hand. You cant check/fold to the GTO player every time you miss.


Sent from my iPhone using Tapatalk

But it doesnt make sense to check some of the time and bet some of the time in zoom. Here you can take the same action everytime. So is zoom optimal strategy different then? But you say GTO doesnt depends on history so it should be the same? Its clear that you cant check fold every time, but why cant you bet everytime? Or check some weak made hands etc. Not clear to me. Is there a proof that you need randomness?

Nash likely starts out the hand the same way 2-bets, 3-bets preflop, but will always have a range response to flop texture. There are actually a small subset of runouts that would be considered "threshold" and would require a complete Nash solution. In theory a complete solution would respond the same way each time.


Sent from my iPhone using Tapatalk

I think it helps to change the term "pure Nash" to "completed Nash solution" in your discussion if you want Wikipedia to make sense. We are mere mortals and will never possess a 6-handed 100BB complete Nash solution. The bots are almost there heads up. Fkn bots! Lol


Sent from my iPhone using Tapatalk

I think I did not answer #3 correctly. I may not be able to answer, but I think there is a single equilibrium, and a complete Nash solution, which would not lose to the mixed solution. But I do not think humans can memorize the pure solution and should not try, therefor the mixed strategy involving rng generated by the randomness of 52! deck combinations to be the only strategy adaptable to human poker players.


Sent from my iPhone using Tapatalk

Or just double blind zoom with no avatars and no idea who the opponent is. For one hand against one opponent then the GTO would be to act the same each time. But, if it was a simple shove/fold toy game you would still shove with value 1/3 and shove with bluffs 2/3 , and by doing so earn profit equal to the size of the pot on the flop. So even in a pure strategy, you are random, it just depends wether you are memorizing which exact combos to bluff, or just using RNG.


Sent from my iPhone using Tapatalk

Just to clarify though, it is possible to play a pure strategy against Nash and break even, an example is if you pick rock every time against an opponent using 1/3 rock, 1/3 paper, 1/3 scissors. But throwing rock every time is just easily exploited by non-Nash strategies. It also doesn't matter if your opponent is making a conscious move to exploit you or not, you can still be exploited. A toddler who doesn't know how to play the game but just throws paper every time because he likes it will still seriously exploit you.

That is what I am saying, that a dominant strategy in toy games can also be equilibrium without the need for Nash, yet is also considered Nash. Nash proved there is an equilibrium, and that is the basis for game theory, but for any game that is interesting, and the stock market is certainly interesting, the pursuit of equilibrium is the rub, not just accepting that equilibrium is a thing.


Sent from my iPhone using Tapatalk

The Nash Equilibrium doesn't consider history from previous hands, but it may involve randomness / mixed strategies.

You could play a NE strategy there, yes. But even if you only play 1 hand against each opponent, you may still be able to find weaknesses that are frequently present in the overall population and then try to exploit these. So just because you only play a single hand against each opponent does not mean that playing NE is automatically the most profitable option.

There is almost certainly no pure Nash strategy for any kind of complex poker variant. But for every mixed decison in a NE you can simply play any one of the mixed actions with 100% frequency instead, and that change won't affect the EV of the strategy against the original NE. So it's trivial to find pure strategies that break even against the mixed Nash Equilibrium (if you know that equilibrium), but these pure strategies will generally no longer be part of a NE.

The process you describe will generally oscillate / cycle without ever finding a NE. Fictitious Play is an algorithm for finding NEs that works fairly similar though, but it adjusts strategies more gradually.

If you're playing a NE strategy, you never consider history. Whether you take the same action in the same spot depends on whether that spot requires a pure or mixed strategy.

No.

If a pure Nash strategy exists, then I think so, otherwise it wouldn't be a Nash strategy (Villain would have an incentive to adjust, a means to exploit you).

A Nash equilibrium describes a pair of strategies. Essentially what you're asking is whether, for any two equilibria (x1, y1) and (x2, y2), another equilibrium is achieved by playing x1 against y2. I think so for the reason I mentioned, but I may have a misunderstanding.

I've got nothing to add to Plexiq and heehaww's excellent answers.

If this is the case, wouldn't (x1,y2) and (x2,y1) also have to be equilibriums?

edit: oh wait that's what you're saying.

So how does one in principle compute the Nash equilbirum strategy?

As I mentioned above, Fictitious Play is somewhat similar to the process you described in your OP.

Fictitious Play works like this instead:
S1: Arbitrary Strategy
P1: BR(S1), the best response against S1
S2: BR(P2)
P2: BR(AVG(S1, S2)), the best response against the average of S1 and S2
S3: BR(AVG(P1, P2))
P3: BR(AVG(S1, S2, S3))
...
SN: BR(AVG(P1, P2 ,..., Pn))

The average of the strategies AVG(S1, S2, S3,..., Sn) and AVG(P1, P2, P3,..., Pn) converges to a NE.

How do you take the average of strategies? Is this computationally feasible to solve in practice, for smaller game, say you are on the flop already?

As for averages, exactly as you would think, e.g.:
S1 raises AA 100% of the time in a certain spot
S2 raises AA 0% of the time in that same spot
AVG(S1,S2) raises AA 50% of the time...

And yes, there are plenty of solvers around for smaller game variants. Fictitious Play is a bit out of fashion though, CFR+ is probably more popular these days (https://arxiv.org/abs/1407.5042).

Is there Nash equilibrium for more then 2 players games.For example 6max poker?

Yes, Nash Equilibria exist for multiplayer games.

The Nash condition only states that no single player can unilaterally increase their EV by changing away from the equilibrium strategy when all other players remain static. It's important to keep in mind that it does not give any guarantees if two or more players deviate.

Yeah, I believe before Nash it was known there was a Nash Equilibrium solution for 2 player zero sum games (obviously it wasn't called Nash Equilibrium at the time). Nash generalized this concept and proved that such an equilibrium exists for more than 2 person games and I also believe for any game with a finite set of actions not just a zero sum game. (Someone with a better understanding of this can correct me if I messed anything up)

I think I read in the Tiptons book that there is no mathematical Theorem that claims to exist NE for 3or more player games.It does not mean that it dosnt exist just it does not exist, but I'm not sure I remember it well.I ll try to find that part of book.
But its not obv that there is such state of game.For example if there 3 players on the river one with nut/air type of range and the other two with bluff catchers.What is NE for this spot?And lest say they need to call or fold in same time.

It looks to me that in NE polarized range should be indifferent between bluffing and checking and that bluff rest of players should call 50 % of the time combined equal fractions each,but if one of them calls every time then the other guy loose more money because they chop when polarized range bluffs and lose when he dose not.

From what I understand, there is definitely a mixed solution (RNG freq play) for all the preflop scenarios, and then on the flop based on texture. Once it gets to two players in the hand it might revert to pure solution (brute force solved) but not for sure.

But so long as the game is finite, there is equilibrium possible.


Sent from my iPhone using Tapatalk

So when Polk has two overs, an ace, and a gutterball oop, on the flop with pot of 20BB and 80BB stacks behind, he has a pretty good idea what lines he will take and how to wind up on the river. I would consider that mixed strategy Nash.


Sent from my iPhone using Tapatalk