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In a two player static game with a discrete strategic space that permits each player to chose one of the four possible strategies what is the maximum number of NASH Equilibrium for the game

  • $\begingroup$ Suppose any combination of the the four strategies of both players yields the same payoff. That is both players are indifferent between all four strategies. Then any combination and any randomization over them is a NE. Hence, you have 4x4=16 pure-strategy NE and infinitely many mixed-strategy NE. $\endgroup$ – Bayesian May 28 '19 at 20:00

Firsly, understand what a Nash Equilibrium (NE) is.

Simply put, a Nash Equilibrium is a state, where neither player has a reason to defect. So neither player can become better off by playing a different strategy.

Maximum number of NE is probably an extreme case, where it may help to draw the game in normal form and understand why there are multiple nash equilibria.

                    Player 1
                 A      B     C    D
              A (a,b) (c,d) ...
  Player 2    B   ...

In this case, the maximum number of NE's exists, when payoffs in all outcomes are the same. Then all outcomes are a NE. So the maximum number would be 16, because you have 16 possible strategy pairs in pure strategies and infinitely many in mixed strategies.


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