I often hear people say limit holdem is solved whatever that means???

Isn't it impossible to "solve" a game with as many variables as poker?

So is PLO solved? I don't get how that is even possible with as many different combos and moves possible in the game and situations.

Sure the bad players and donks eventually lose it all, but i think there's always an edge with good players.

no, and won't be for a while

Limit Holden is solved because of the fixed betsize. Solved meaning that there is an unbeatable unexploitable game theory optimal strategy for the game.

NLHE is mostly unsolved because bet sizing can vary dramatically. For example, you can figure out some close to optimal valuebet/ bluffing frequencies in certain spots with a certain bet size or know your perfect frequencies when say bluff catching for a certain sizing but what if you take cases where you can bet 10, 100, 1000 times the pot?

PLO has its own difficulties. Sure the bet sizing is capped at pot but the combinatorics are so staggering that it will take a very long time if at all to figure out some unexploitable frequencies.

I still don't understand how there can be a strategy in limit holdem that solves the game.... with so many combos and strategy/counter strategy

Look up Cepheus.

Since the bet sizing is always fixed then you're able to break your play down into unexploitable frequencies, e.g. when you bet in a spot you're bluffing and value betting in a perfect ratio to the price youre laying your opponent so that your opponent is completely indifferent to either option of calling or folding, meaning you are unexploitable, which compares to exploitative strategy in the same game which relies on your opponent having imperfect frequencies (e.g. I know when my opponent bets he is bluffing too much for the price I am getting so I can exploit him by bluff catching wider or bluff raising more often). Hope this makes sense.

Only for HU limit holdem.

It makes sense but irrelevant.

unexploitable!=GTO

theoretically it could be solved, but wont be happening anytime soon.

How is it irrelevant? I basically just explained an example of how GTO works.

What you explained is not how GTO works.

All HU poker games can be solved and will have exactly 1 symmetric Nash equilibrium in which both players play a strategy that is unbeatable (And both will have a long term winrate of 0bb/100)

This follows easily from the works of John Nash and others (Beautifull Mind flick)


Going from 2 players to n>2 players screws things up. There will still be Nash equilibria, but the equilibrium wont necessarily be unique and it wont necessarily be symmetric. The notion that playing strategies that is part of a Nash equilibrium are unbeatable in poker games with players n>2 is a common misconception. Maybe it is, maybe it isnt.

Exactly 1???

Yes, in symmetric, zero sum, two man games, uniqueness is one of several neat qualities of equilibria, that does not extend to games with players n>2.

The key is the amount of players. As I alluded to before.

Disclaimer: I use to teach Game Theory at Grad school at U of Copenhagen and I was writing a PhD thesis in Game Theory before playing poker full time. But its almost 10 years since I left, so Im rusty and my knowledge is not up to date.

Even for mixed strategy?

Yeah. If you dont allow mixed strategies existence does not hold. Think Rock, Paper, Scissors. That game does not have N.E in pure strategies, only in mixed strategies.

Uniqueness does require that you remove redundant strategies tho. Again, think Rock, Paper, Scissors. If you add "Bowling ball" to the strategy set and have Bowling Ball beat Scissors, lose to Paper and tie with Rock, then Bowling Ball is a redundant strategy, as it will be exactly similar to Rock. A Rock, Paper, Scissors game with an extra redundant strategy as Bowling Ball, will have an infinite amount of N.E.

http://faculty.smu.edu/ozerturk/MixedNash.pdf

Was that directed at me?

Yes this battle of sexes appeared to have more than one NE. I'm just confused as I have always read there is AT LEAST one NE but never exactly one in all definition.

Disclaimer : I don't know anything about game theory except from Wikipedia.

Well, neither of those games are Zero Sum. The first game is not symmetrical either.

Two player, Symmetrical, Zero Sum games are a particular class of games. HU poker is among them. It is for those games that equilibria has all sorts of neat qualities. Poker with n>2 players is symmetrical and Zero Sum, but obv not a two player game.

Neither of those games in that paper are Zero sum, and thus does not fall into that category.

Thanks, that clear things up!

So is a HU subgame between CO and BN from a 6 max game,symmetrical and zero sum?

No!

But the game where we play one orbit and all players are in each position exactly once is a symmetrical zero sum game.

Thanks that's very helpful. Expect some PMs coming your way.

My understanding here is that if you have a three player game where one player plays "bad" (not cheating/colluding bad), then even if the other two play an identical NE strat. their win rates will differ depending on their position relative to the bad player. The effect of this goes up as you introduce more players, and it's not known if you will always have a positive win-rate in a 9 handed game regardless of position and other players skill.

Which is maybe what you said reworded, but seems more obvious and less scary.