# Non-negative deviations from Nash Equilibrium

I know that in a Nash Equilibrium, no player can profitably deviate from the equilibrium strategy assuming that the strategies of the other players remain the same. My question is, what if a player can deviate to a different strategy that generates an equivalent payoff? Would this no longer be a Nash Equilibrium?

• It would still be a Nash equilibrium. This concept requires the each player's strategy is weakly optimal given the strategies of the other players (so ties are OK).
– user17900
Mar 10, 2019 at 13:30

It depends on whether the other players' strategies in the initial equilibrium are still best responses to the deviator's new strategy. For example, $$(T,L)$$ is the unique NE in the game below. Player 1 (the row player) can deviate from $$T$$ to $$B$$ without hurting his own payoff, but $$(B,L)$$ is not a NE, since $$L$$ is not a best response to $$B$$ in this game.
$$\begin{array}{|c|c|c|}\hline &L&R\\\hline T&1,1&2,0\\\hline B&1,0&0,1\\\hline \end{array}$$
Of course you can easily construct a game where such a unilateral deviation still leads to a NE. As the game below shows, player 1's deviation from $$T$$ to $$B$$ is still without loss of profit, but it changes the NE from $$(T,L)$$ to $$(B,L)$$. There are also infinitely many other possible deviations for player 1 if you consider deviations in mixed strategies.
$$\begin{array}{|c|c|c|}\hline &L&R\\\hline T&1,1&2,0\\\hline B&1,2&0,1\\\hline \end{array}$$