[Theory] Do symmetrical deviations from mixed betting frequencies lose EV in GTO?

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[Theory] Do symmetrical deviations from mixed betting frequencies lose EV in GTO?

1. Problem Definition
I'm going back to the study of some underlying GTO principles currently and there is a question that has been bugging me for a while now. So in heads up play, a Nash equilibrium between two players is defined as a set of two strategies from which neither player can deviate in order to increase their EV. Often times, equilibrium play results in certain hands being indifferent between two actions (say betting and checking), which manifests in a mixed strategy for these hands. Let's say Player A has two different hands that are both supposed to mix between exactly 67% betting and 33% checking.

Scenario 1: In the first scenario, Player B is sticking to the equilibrium response no matter what Player A does. Player A can now freely choose any betting and checking frequency for the mixing hands without losing EV. Since Player B is not adapting and as betting/checking for these hands are equal in EV, there are infinitely many alternative strategies for this player that generate the exact same EV vs. the equilibrium response. Player A can choose any deviation from the optimal mixing frequencies.

Scenario 2: In the second scenario, Player B is allowed to adjust his strategy if Player A deviates. Now, let's say Player A chooses to bet one of the hands that are supposed to mix 80% and check it 20% of the time. This would be an asymmetrical deviation meaning that Player A deviated from optimal frequencies and the overall betting frequency of his whole range changed. This play should lose EV. Now there exists an exploit that Player A can choose as Player B was not sticking to the optimal mixing frequencies. This simply means that one player deviates from GTO and loses to the other player's exploit.

Scenario 3: Essentially, here comes my question: What if Player A deviates from the equilibrium by choosing different bet frequencies but deviates symmetrically? This means that if one hand bets 80% rather than 67% of the time (13% more betting) and the other hand bets 54% rather than 67% of the time (13% less betting), Player A deviates from the optimal mixing frequencies of each individual hand but keeps the overall betting frequency of the whole range fixed. Does an exploit exist now for Player A? My personal take is that there shouldn't be one. We often get taught that deviations from game theory optimal play allow for exploitation (like in Scenario 2). But wouldn't we need to refine this to saying that this relates only to asymmetrical deviations, whereas symmetrical deviations are no problem at all?

In other words: Do symmetrical deviations from mixed betting frequencies lose EV in GTO?

2. Experiments in GTO+

So I've played around a bit in GTO+ to investigate this and for simple toy games the intuition above indeed seems to be supported. The game is AQJ vs. KK (one street), so IP has a perfectly polarized range composed of nuts and air and OOP has a perfectly condensed range consisting of only bluffcatchers.

Equilibrium Solution
For the parameters that I've set, this is IP's range construction in equilibrium. The EV of the game for IP is 0.5555. The overall betting frequency is 55.6%. The value region always bets (A) and the required bluffing frequency is equally distributed among the two bluffing candidates Q and J (no removal effects). IP has a total of 4 bluffing combos (12 combos, each betting 33% of the time, 12 * 0.33 = 4).

Experiment for Scenario 1: OOP never adjusts
Here, IP commits mixing errors but OOP's response is node-locked to the initial equilibrium strategy. The EV of the game for IP does not change, still 0.5555.

Experiment for Scenario 2: Asymmetrical Deviations from Mixed Frequencies
Here, IP commits a slight mixing error (betting 63% of the time globally) and OOP is responding freely. As IP is bluffing slightly more than he should, OOP is calling all of his bluffcatchers (100% calling, 0% folding). The EV of the game for IP now now reduced to
0.4074, OOP is exploiting.

Experiment for Scenario 3: Symmetrical Deviations from Mixed Frequencies
Finally, IP is again deviating from the equilibrium but does so in a way that keeps the overall betting frequency exactly at 55.6%! J is now never betting, and Q is betting 4 out of 6 times. So as in equilibrium play, IP has exactly 4 bluffing combos again, but these arise as a consequence of symmetrically deviating from equilibrium play. The EV of the game for IP is now back to 0.5555. The symmetrical deviation (betting one hand more than he should and simultaneously betting the other hand less) did not lead to any EV loss.

3. Conclusion
So in this toy game, symmetrical deviations do not seem to lead to EV loss. I'm sure that in more complex games across several streets, they will do. My question essentially is when this will be the case and why... Any thoughts??

3 Comments

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Trinity 2 years, 3 months ago

My guess is whenever you change the range composition, you open yourself to being exploited on some degree.

In your experiment QQ and JJ are equivalent because they have the same equity (0), plus they don't interact with opponent's range in any meaningful way. But I think in real scenarios this would be very uncommon.

I'll add 2 more scenarios more similar to a real game.

Scenario 4: You're deciding whether to cbet or check in position. GTO wants to bet with a 30% frequency. The betting range is composed of 30% top pair+ and 70% no made hands. You decide to bet 30% overall but your betting range is 100% no made hands. Is this expoloitable? Of course.

Scenario 5: You're deciding whether to cbet or check in position. GTO wants to bet with a 30% frequency. The betting range is composed of 30% top pair+ and 70% no made hands. But each combination is betting at a different frequency based on whatever reason (equity, blockers, protection,etc.) You fail at picking the right frequency with each combination but your betting range still results in 30% top pair+ and 70% no made hands. Is this exploitable? Probably, but on a lesser degree and the exploit is harder to find.

Holonomy 2 years, 3 months ago

Yeah as pointed out by Trinity QQ-JJ are the same hand in your example. Is it possible there would be no exploit? Maybe, but I suspect unlikely. You certainly can't guarantee that there won't be one. Not sure if it was precluded by your question but if you had one of your value hands betting 50% and one of your bluffs betting 50% and you symmetrically deviated here you would obviously be exploitable as your B:V ratio is off.

tombos21 2 years, 2 months ago

I love a good poker experiment!

QQ and JJ are the same hand here. As long as the overall bluff:value ratio is correct the strategy is unexploitable. In practice, the reason the solver mixes many hands is to improve board coverage and minimize blocker weaknesses.

As an example, try giving the defender KQs and KJs instead of KK. These bluff-catchers interact with the QQ/JJ bluffs, forcing them to bluff the same proportion of each. If you only bluff QQ then the defender can exploit you by calling KJs and folding KQs.

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