Computing Nash equilibrium strategy

Question

How does one in principle compute the Nash equilibrium strategy for NLH? Are there different methods?

Let us simplify the case and consider only 2 players and only 1 hand in isolation. (No history)

Answer 1

Your question is very general. Game Theory, which employs Nash equilibrium, is a set of methods to model situations where players are in conflict.

If your input is the entire game, the problem becomes unsolvable. First of all I am not sure if it exists a Nash solution, but even if it does, it would be nearly impossible to calculate, given all these parameters (players, stack_sizes, cards, position, gameplays, table-dynamics, history, meta-game, etc.)

You can use it partially for individual situation, though mainly to get an idea of where the game leads to. As an example, I would suggest looking at Nash Equlibrium for HeadsUP shortstack push/fold situations

Finally, bear in mind that even if you could magically calculate nash strategies for more complex situations, that wouldn't be necessarily winning in the long run. In gerenal, Nash equilibrium calculates a defensive strategy, so as not to get exploited by your rational opponent. In Holdem, most of your winning will result exactly from the opposite, i.e. finding the weakness of your opponent and exploiting it. As an example:

A well studied game is the paper-rock-scissors, where Nash mixed strategy equilibrium suggests you randomly play paper, rock, scissors, 33% of the time each. With this strategy, even a computer opponent will never be able to exploit you. The opponent will tend to play the same, and in the end you will reach Nash equilibrium, where no one wins or loses. If, however, you have an opponent who will ALWAYS play the rock, it is obvious that you want to exploit his "weakness" and play ALWAYS paper, thus gaining profit. This is also true in poker, of course in a more general and complicated way.

Answer 2

Nash Equilibrium is where two players don't have an incentive to change their strategy.

A simplified way of looking at how a solver works: If you have player1 and player2. You would start with some strategy. Then have player2 exploit player1. Then have player1 exploit player2's new strat. Then player2 exploit player1. And keep going back and forth until you reach an equilibrium or point where there is no way for either player to exploit the other.

Let us simplify the case and consider only 2 players and only 1 hand in isolation. (No history)

History will never matter for a Nash Equilibrium strat. Looking at 1 hand in isolation wont work because you have to look at the whole range. Unless you are referring to a toy game where each player always has the same specific hand. In which case:

• SB: AA
• BB: KK
• Blinds = 1 / 2

If we have 10bb stacks and SB is only allowed to shove or fold. And BB is only allowed to call or fold. A nash equilibrium will be SB shoves and BB folds. If the SB changes his strat he will make less $$ so he has no incentive to change. If the BB changes his strat he will make less $$ so he has no incentive to change.

EDIT: For fun solved this game with full ranges to 0 exploitabilty: (You can see that 96o is a mixed strat for SB and 97o is a mixed strat for BB)

SB Strat (red = shove):

BB strat (green = call):