Computing Nash equilibrium strategy

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)

• Why the downvote? Might be a difficult question but I think it's clear enough. Jul 14, 2017 at 12:53

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.

• Nash's Existence Theorem proves that NLH has a Nash Equilibrium, assuming finite cash. There's a finite amount of distinct game states, meaning amount of players, stack sizes, pocket cards and boards. Jul 14, 2017 at 12:22
• @jollyJoker-> Nice comment, you are right. Since its finite, there's at least one (mixed) NE. Jul 14, 2017 at 12:34
• Whats a mixed strategy?
– user1457
Jul 14, 2017 at 12:53
• Mixed Strategies WIKI Jul 14, 2017 at 12:58
• @Paparazzi Note the "in principle" in the question. It's unsolvable in practice but there's a finite number of options so it's solvable in principle. Jul 15, 2017 at 14:50

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):

• Do we know that this sequence will reach an equilibrium, how do you know that they wont just switch back and forth in a cycle strategies?
– user1457
Jul 17, 2017 at 13:34
• That's a good question and I don't know. I did notice that if you try to solve a complicated strategy, the solver will usually end up getting down to a really low level of exploitability and then get stuck there for a long time. The level of exploitability at that point is usually so low that I just stop the solver. I'm sure most algorithms in these solvers are more complicated than just purely going back and forth but that's essentially whats happening. Jul 17, 2017 at 13:46