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Does a nash equilibrium strategy always beat a non nash equlibrium strategy  in head ups spots? Does a nash equilibrium strategy always beat a non nash equlibrium strategy  in head ups spots?

01-27-2019 , 06:11 AM
Let's assume that I use nash equilibrium preflop ranges against an opponent who deviates from nash equilibrium with his preflop ranges, will I always beat him?

I'm asking this because in rock paper scissors, if I use all three at the same frequency and he only use one, I will still won't be able to beat him, so there must be something wrong here.

Are solvers being used in order to achieve a break-even result against competition where it's not possible to perform better and you would get exploited by it if you deviated?

Or is it a superior strategy against those who deviate on head ups spots?

Do players prefer not calling it gto because gto is indeed the best possible strategy where you will lose against no matter what while nash equilibrium doesn't always perform like that?

Last edited by Garpthefist24; 01-27-2019 at 06:21 AM.
Does a nash equilibrium strategy always beat a non nash equlibrium strategy  in head ups spots? Quote
01-27-2019 , 12:57 PM
I think the general thrust of your questions are answered in the sticky threads for this forum.
Does a nash equilibrium strategy always beat a non nash equlibrium strategy  in head ups spots? Quote
01-27-2019 , 12:58 PM
Any deviation will result in an ev transfer that is greater than or equal to zero.
Does a nash equilibrium strategy always beat a non nash equlibrium strategy  in head ups spots? Quote
01-28-2019 , 11:02 PM
Quote:
Originally Posted by Garpthefist24
Let's assume that I use nash equilibrium preflop ranges against an opponent who deviates from nash equilibrium with his preflop ranges, will I always beat him?
Heads-up, the EV transfer is greater than or equal to 0. You might be able to improve your EV from nash by exploiting your opponent's deviation.

Quote:
Originally Posted by Garpthefist24
Are solvers being used in order to achieve a break-even result against competition where it's not possible to perform better and you would get exploited by it if you deviated?

Or is it a superior strategy against those who deviate on head ups spots?
No human can play the exact strategy produced by solvers since they involve mixed frequencies for different actions with each hand. These strategies are hard to remember for humans as well as approximate. Plus since no one plays perfectly so you make more money by exploiting the other players.

Players use solvers to plug their own leaks, to generalize some of the solver results to exploit players in their population or generally improve their "default" strategies to preserve EV in unknown situations.



Quote:
Originally Posted by Garpthefist24
Do players prefer not calling it gto because gto is indeed the best possible strategy where you will lose against no matter what while nash equilibrium doesn't always perform like that?
Not sure what your question is but one clarification GTO and nash equillibrium are generally considered the same thing in poker circles.
Does a nash equilibrium strategy always beat a non nash equlibrium strategy  in head ups spots? Quote
01-29-2019 , 03:15 AM
It is a mistake to think that GTO in poker just breaks even.

As you understand, a GTO strategy always has an EV greater than or equal to 0 in the long run (in zero-sum games).

In rock paper scissors, a GTO strategy always has an EV of 0 in the long run.

The good thing about poker is that a GTO strategy does not always have an EV of 0 in the long run.
An example where a GTO strategy makes a lot of money is when our opponent's strategy is to fold every hand.
A GTO strategy probably makes a lot of money even against the best humans (Libratus did very well).
Does a nash equilibrium strategy always beat a non nash equlibrium strategy  in head ups spots? Quote

      
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