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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?

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    $\begingroup$ 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). $\endgroup$ – afreelunch Mar 10 at 13:30
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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}

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