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 03-09-2012, 07:04 PM #1 veteran     Join Date: May 2011 Location: P(G) = 0.02% Posts: 3,260 Does GTO breakeven against non-GTO opponents in poker? (Starting this thread so as not to derail the thread where this discussion came up.) It's generally accepted that a GTO/balanced strategy does not maximally exploit non-GTO opponents, but it often is also claimed that such a strategy is inherently breakeven against any opponent. Can this be proven or disproven?
 03-09-2012, 07:25 PM #2 journeyman   Join Date: Feb 2010 Location: Greater Worcester, MA Posts: 303 Re: Does GTO breakeven against non-GTO opponents in poker? For certain games and certain GTO strategies, it is true. For example, the GTO strategy for rock-paper-scissors is break even against any strategy. But for complicated games like poker, if there were such a thing as a human or bot playing GTO, all you would know is that no strategy could beat it, but there are certainly strategies that will lose to it. For example, folding to any raise pre-flop with anything other than AA and then folding to any raise post-flop where AA is no longer the nuts. That strategy is going to lose to GTO.
 03-09-2012, 07:29 PM #3 enthusiast   Join Date: Feb 2011 Posts: 88 Re: Does GTO breakeven against non-GTO opponents in poker? The claim: "a GTO strategy is not the maximal exploit strategy agaist any other strategy" is true. The second claim ("a GTO strategy breakeven agaist any other strategy") is false. Can be true for some simplified game (eg. rock-scissor-paper), but generally is false. Take for example the 0-1 game (each player have a random number between 0 and 1, the higher win. the first player can push or check, if he push the second player can call or fold), here every deviation from the equilibrium for the first player lead to lose equity against a GTO strategy. In a complex game like poker, I espect almost every mistake to be punished vs a GTO strategy (the only one that i can think that are not punished are call or fold on the river with a bluff catcher). EDIT: beat by Mat the Gambler... we use the same example, but i think is the obvious one... i'm not the fastest at writing...
03-09-2012, 08:14 PM   #4
grinder

Join Date: Jul 2011
Posts: 406
Re: Does GTO breakeven against non-GTO opponents in poker?

Quote:
 Originally Posted by zumby (Starting this thread so as not to derail the thread where this discussion came up.) It's generally accepted that a GTO/balanced strategy does not maximally exploit non-GTO opponents, but it often is also claimed that such a strategy is inherently breakeven against any opponent. Can this be proven or disproven?
Yes such a strategy is break even because it is a zero sum game.
Imagine you are playing GTO in rock paper scissors. You are throwing rock 1/3 paper 1/3 and scissor 1/3. Imagine villain is also playing GTO.
Then if you play 18 times you will win 6 lose 6 and tie 6.
Example:
You throw rock 1/3 chance of winning 1/3 chance of losing
Play 6 times you win 2 lose 2 tie 2
You throw paper 1/3 chance of winning 1/3 chance of losing
Play 6 times you win 2 lose 2 tie 2
You throw scissor 1/3 chance of winning 1/3 chance of losing
Play 6 times you win 2 lose 2 tie 2

Now imagine villain is not playing GTO but instead throwing rock 2/3 paper 1/6 and scissor 1/6
You throw paper you have 2/3 chance of winning and 1/6 chance of losing.
SO if you play 6 times you win 2 and lose 1.
You throw rock you have a (1/6) chance of winning and a (1/6) chance of losing.
So if you play 6 times you win 1 and lose 1.
You throw scissors you have a (1/3) chance of winning and a (2/3) chance of losing.
Play 6 times you win 1 and lose 2.

So you won 4 times and lost 4 times and tied 10 times.

All that changed was the number of times you won/lost and tied.
However, you can easily see that your opponent is exploitable. All you have to do is throw paper 2/3 of the time rock 1/6 and scissor 1/6

In this scenario you would start crushing your opponent, but I don't feel like doing the math.

As you can see in a zero-sum game playing GTO is only important if you feel you are being exploited. All GTO does is make you IMPOSSIBLE to be exploited. However, you gain absolutely nothing from playing it and cannot profit. This is why you must find a weaker opponent and exploit them, that is how you make money at poker.

Last edited by sloppy ftw; 03-09-2012 at 08:15 PM. Reason: If both opponents were truely playing GTO this is ALWAYS TRUE. However, as poker is complex there is no unified GTO strategy

03-09-2012, 08:45 PM   #5
veteran

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Re: Does GTO breakeven against non-GTO opponents in poker?

I see that the GTO for rock-paper-scissor is always breakeven, but my question is whether the same applies to poker (or other games).

Quote:
 Originally Posted by sloppy ftw Now imagine villain is not playing GTO but instead throwing rock 2/3 paper 1/6 and scissor 1/6 [...] All you have to do is throw paper 2/3 of the time rock 1/6 and scissor 1/6 In this scenario you would start crushing your opponent, but I don't feel like doing the math.
Shouldn't we be throwing 100% paper?

03-09-2012, 09:25 PM   #6
Pooh-Bah

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Re: Does GTO breakeven against non-GTO opponents in poker?

Quote:
 Originally Posted by zumby Shouldn't we be throwing 100% paper?
yes. GTO rock/paper/scissors is called a "brittle" strategy, because once your opponent deviates from optimal strategy, your best response is radically different. If he mixes 33.33333333334% rock and splits the remainder equally between paper and scissors, you maximize by throwing 100% paper, not with any mixed strategy, so long as your opponent sticks to his non-optimal rock preference.

The same is true in poker for calling all-in bluffs. If you call at the optimal frequency, your opponent cannot gain by bluffing more or less often. If he bluffs even slightly more often than optimally, your best response is to fold 100% of the time. If he bluffs the slightest bit less than perfectly, you maximize by always folding.

Of course, if your opponent adapts to your new calling or r/p/s strategy, he can punish you for not mixing. R/P/S tournaments actually hinge on this, I understand. One player will play non-optimally, trying to tempt the other into an exploitable strategy.

 03-09-2012, 11:16 PM #7 journeyman   Join Date: Feb 2010 Location: Greater Worcester, MA Posts: 303 Re: Does GTO breakeven against non-GTO opponents in poker? No one has yet solved poker to the point of developing a GTO strategy. Theoretically, such a strategy might be so complicated that you'd need a computer a trillion times more powerful than the most powerful current computer to run it. But I'm sure you can imagine the basics of what such a strategy would look like in general. Imagine playing against the nittiest nit of all time. Preflop he'll fold anything other than AA to any raise, and he'll never reraise. Because, you know, he's only got one pair. On the flop, if an ace doesn't hit then he no longer has the nuts so he'll fold to any bet. If the board pairs, he'll fold to any bet because he's afraid of Quads. Such a player is obviously going to lose to any GTO strategy. He'd lose even more to a maximally exploitative strategy, but you already knew that part.
03-10-2012, 12:24 AM   #8
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Re: Does GTO breakeven against non-GTO opponents in poker?

Quote:
 Originally Posted by sloppy ftw Yes such a strategy is break even because it is a zero sum game. Imagine you are playing GTO in rock paper scissors. You are throwing rock 1/3 paper 1/3 and scissor 1/3. Imagine villain is also playing GTO. Then if you play 18 times you will win 6 lose 6 and tie 6. Example: You throw rock 1/3 chance of winning 1/3 chance of losing Play 6 times you win 2 lose 2 tie 2 You throw paper 1/3 chance of winning 1/3 chance of losing Play 6 times you win 2 lose 2 tie 2 You throw scissor 1/3 chance of winning 1/3 chance of losing Play 6 times you win 2 lose 2 tie 2 Now imagine villain is not playing GTO but instead throwing rock 2/3 paper 1/6 and scissor 1/6 You throw paper you have 2/3 chance of winning and 1/6 chance of losing. SO if you play 6 times you win 2 and lose 1. You throw rock you have a (1/6) chance of winning and a (1/6) chance of losing. So if you play 6 times you win 1 and lose 1. You throw scissors you have a (1/3) chance of winning and a (2/3) chance of losing. Play 6 times you win 1 and lose 2. So you won 4 times and lost 4 times and tied 10 times. All that changed was the number of times you won/lost and tied. However, you can easily see that your opponent is exploitable. All you have to do is throw paper 2/3 of the time rock 1/6 and scissor 1/6 In this scenario you would start crushing your opponent, but I don't feel like doing the math. As you can see in a zero-sum game playing GTO is only important if you feel you are being exploited. All GTO does is make you IMPOSSIBLE to be exploited. However, you gain absolutely nothing from playing it and cannot profit. This is why you must find a weaker opponent and exploit them, that is how you make money at poker.
what makes you think that the value of a solution of rock-paper-scissors is equal to the value of a solution to poker?

Last edited by RainbowBright; 03-10-2012 at 12:30 AM.

03-10-2012, 12:25 AM   #9
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Re: Does GTO breakeven against non-GTO opponents in poker?

Quote:
 Originally Posted by gedanken yes. GTO rock/paper/scissors is called a "brittle" strategy, because once your opponent deviates from optimal strategy, your best response is radically different.
No. rock paper scissors is a flexible strategy because as long as your opponent is playing a GTO strategy, it doesn't matter what you do because your EV is the same with all actions.

 03-10-2012, 12:28 AM #10 veteran     Join Date: Jun 2007 Posts: 2,344 Re: Does GTO breakeven against non-GTO opponents in poker? here's a proof. you're playing heads up, the SB antes and the BB antes. We know that a GTO strategy will include the SB raising some % of the time, if the BB folds 100% of the time, then he will be losing money. And since it's a zero sum game, then his losses will be your gains. whoops... mattthegambler obviously beat me as well. i'm trying to think of things to add. i've seen this statement made in other threads but people never back it up. i feel that someone well respected probably made a statement once which was misunderstood or something. Last edited by RainbowBright; 03-10-2012 at 12:33 AM.
 03-10-2012, 12:41 AM #11 old hand   Join Date: Mar 2008 Posts: 1,582 Re: Does GTO breakeven against non-GTO opponents in poker? GTO is NOT breakeven vs non-GTO opponents. It is breakeven vs GTO opponents when not accounting for rake. GTO is breakeven in Rock Paper Scissors, but that is because the game is set up in a way that each decision has equal and breakeven values vs a random action. However, poker is not set up in such a symmetrical way as there are options which are clearly -EV, for example folding 100% or shoving 100%. Both would lose a LOT vs a GTO strategy.
03-10-2012, 01:06 AM   #12
Pooh-Bah

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Re: Does GTO breakeven against non-GTO opponents in poker?

Quote:
 Originally Posted by RainbowBright i've seen this statement made in other threads but people never back it up. i feel that someone well respected probably made a statement once which was misunderstood or something.
I must not be explaining it well. The "someone well respected" is Bill Chen and Jerrod Ankenman in Mathematics of Poker.

In rps, if villain randomly throws each of the options. it doesn't matter what we do, we'll win 1/3, lose 1/3 tie 1/3.

but if villain throws rock 40%, paper 30%, and scissors 30%, what is our maximally exploitive strategy? It's not to mix slightly more paper in, it's to throw paper 100% and not change until villain adapts.

By throwing paper 100%, we win 40%, lose 30%, and tie 30%.

Say we tried paper 50%, scissors 25%, and rock 25%:

we throw paper
.5 * .4 = .2 paper beats rock
.5 * .3 = .15 paper ties paper
.5 * .3 = .15 paper loses to scissors
we throw rock
.25 * .4 = .1 rock ties rock
.25 * .3 = .075 rock loses to paper
.25 * .3 = .075 rock beats scissors
we throw scissors
.25 * .4 = .1 scissors lose to rock
.25 * .3 = .075 scissors beat paper
.25 * .3 = .075 scisors tie scissors
wins = .2 + .075 + .075 = .35
losses = .15 + .075 + .1 = .325
ties = .15+ + .1 + .075 = .325

so we aren't doing better than if we had simply thrown paper every time. So long as villain is not playing his GTO strategy, our maximally exploitive strategy is pure, not mixed.

taking your example of a villain that plays 2/3 rock, 1/3 scissors, 1/3 paper, my solution is to play only paper, and win 2/3. If you play your suggested mixed strategy of 2/3 paper, you win only 5/9.

Last edited by gedanken; 03-10-2012 at 01:13 AM.

 03-10-2012, 05:11 AM #13 adept     Join Date: May 2006 Posts: 1,058 Re: Does GTO breakeven against non-GTO opponents in poker? no. think of a jam fold game. gto faces a strat that plays gto but folds AA pre. almost every step away from gto is losing value, only a few threshold hands are breakeven.
03-10-2012, 08:48 AM   #14
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Re: Does GTO breakeven against non-GTO opponents in poker?

Quote:
 Originally Posted by gedanken I must not be explaining it well. The "someone well respected" is Bill Chen and Jerrod Ankenman in Mathematics of Poker.
I understand what they mean, but I don't think you do.

"This means that as a practical matter, if Y plays optimally, X has some latitude to make mistakes near his threshold values without it costing much. We call this type of strategy flexible." (MoP pg 229)

So Rock-Paper-Scissors is a flexible strategy... As a practical matter, if Villain plays optimally, Hero has complete latitude to make mistakes near his threshold values (he doesn't need to play 1/3 rock, 1/3 paper, 1/3 scissors... he can play 100% rock) without it costing Hero anything (his EV is exactly the same in all cases 0). We call this type of strategy flexible.

fwiw, it wasn't my example.

 03-10-2012, 10:23 AM #15 Pooh-Bah     Join Date: Jul 2006 Posts: 4,338 Re: Does GTO breakeven against non-GTO opponents in poker? a simple proof to show that a gto poker strategy is not breakeven against any other strategy is considering a "strategy" of always folding. clearly someone folding 100% of hands is losing against a gto strategy. the blinds make poker asymetric and therefore vastly different than a simple game like RPS. im wondering if the difference in blinds is the only thing causing this. would a heads up game with equally sized blinds be symetric in a game theory sense?

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