<p>It depends on whether the other players' strategies in the initial equilibrium are still best responses to the deviator's new strategy. For example, <span class="math-container">$(T,L)$</span> is the unique NE in the game below. Player 1 (the row player) can deviate from <span class="math-container">$T$</span> to <span class="math-container">$B$</span> without hurting his own payoff, but <span class="math-container">$(B,L)$</span> is not a NE, since <span class="math-container">$L$</span> is not a best response to <span class="math-container">$B$</span> in this game. </p>

<p><span class="math-container">\begin{array}{|c|c|c|}\hline
&amp;L&amp;R\\\hline
T&amp;1,1&amp;2,0\\\hline
B&amp;1,0&amp;0,1\\\hline
\end{array}</span> </p>

<p>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 <span class="math-container">$T$</span> to <span class="math-container">$B$</span> is still without loss of profit, but it changes the NE from <span class="math-container">$(T,L)$</span> to <span class="math-container">$(B,L)$</span>. There are also infinitely many other possible deviations for player 1 if you consider deviations in mixed strategies.  </p>

<p><span class="math-container">\begin{array}{|c|c|c|}\hline
&amp;L&amp;R\\\hline
T&amp;1,1&amp;2,0\\\hline
B&amp;1,2&amp;0,1\\\hline
\end{array}</span> </p>

CC BY-SA 4.0
2019-03-10T15:12:52.483Z
2019-03-10T15:12:52.483Z
Herr K.
<p>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?</p>

<game-theory><nash-equilibrium>
Non-negative deviations from Nash Equilibrium
David
# Non-negative deviations from Nash Equilibrium

(Asked by: username_0 on 2019-03-10)

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?

**Tags:** game-theory, nash-equilibrium


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## 1 answers

### Answer by username_1 on 2019-03-10. Score: 3 | Accepted answer

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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