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Let's say, we have sequential finite game with 2 players. The first player is rational and given the other player is also rational, there is a Nash equilibrium. Now let's say that only the first player knows the Nash equilibrium.

Is it rational for the first player to play the nash equilibrium given that his oponent may not play the nash equilibrium?

I want to train 2 agents in tic tac toe against each other with reinforcement learning. I would like to know, if the agents will be able to play well against arbitrary oponent after we train them to play against each other, and thus to play one of NE.

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Let's answer the question for a one-shot game. I know that you're going to play pure strategy $s$. Then I should play a best response to $s$. If your choice is not a best response to what I am playing, then the choice of pure strategies is not a Nash equilibrium. A Nash equilibrium is a profile of $mutual$ best responses.

If you're programming a computer, then you need to forecast what the opponent will do, and you need to write a best response function for your computer. You could eliminate dominated strategies, but even in a simple one-shot two-person game, your program will need to have a part that forecasts and a function that computes the best response to that forecast.

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  • $\begingroup$ Makes sense. thanks :) $\endgroup$ – Jan Vainer Sep 19 '17 at 21:46

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