Consider the following game: Player 1 can choose A or B. Player 2 can choose C or D. The payoff matrix is as follows:
(A, C): (2, 2)
(A, D): (4, 2)
(B, C): (1, 1)
(B, D): (3, 5)
Is (A, D) a Nash Equilibrium? Player 1’s best choice if 2 chooses D is A. However, if 1 chooses A, C and D are equally good to 2, so my intuition is that (A, D) is not a Nash Equilibrium.
Yes, it is still considered a Nash equilibrium, as long as no one has a strict incentive to unilaterally deviate.
First cite: Wikipedia
Second cite:
Recall that every n-player game of finitely many possible strategies has a Nash equilibrium in mixed strategies (due to Nash), or the weaker result that every two-player zero-sum game of finitely many possible strategies has a Nash equilibrium in mixed strategies (due to von Neumann and Morgenstern). Consider a game where there’s more than one possible strategy profile, but the payout is 0 for all players no matter what, anyway. By the referenced results, the game has a Nash equilibrium; by the design of the game, we know that this must mean Nash equilibria are allowed to have “ties”.