Quote:
Originally Posted by :::grimReaper:::
This sounds incorrect. In a complex game like poker, not playing GTO will be losing. The only way to prevent losing is to play GTO yourself, which will result in break-even play. Though this doesn't hold for all games, e.g. rock-paper-scissors as you mentioned.
Unless there is a unique, non-mixed (it always picks a single action) Nash equilibrium, there are value-maximising ways to play against an equilbrium that aren't GTO themselves.
It's also worth keeping in mind that there are usually an infinite number of GTO strategies: there will be free choices where you can choose to make an action from X% to Y% of the time, and the rest of the strategy shifts around to accommodate this choice.
If you have a mixed Nash equilibrium, it's easy to get non-GTO play that maximises value against the equilbrium player: any decision it makes, with any non-zero probability, can be played with any probability to get the maximum value. For example, if the GTO strategy sometimes folds and sometimes raises in a particular situation, I could always fold or always raise in that same situation, and break even against the GTO player. This is because a Nash equilbrium - GTO play - maximises value against itself. If an action it played (even with a tiny probability) got less value than some other action, it wouldn't be maximising its value, and it wouldn't be GTO.
More generally, you can arbitrarily mix and match actions that
any GTO strategy would ever play. All you have to do is avoid actions that
no GTO strategy would make, and you're all set to play a GTO player.
So, any hand where GTO bluffs, you can always raise that hand, playing against GTO, and you won't lose. Or you can always fold. That's not recommended play, against anyone else.