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If a matrix game has a Nash equilibrium strategy in which certain options are not used (i.e. chosen with 0% probability), does that mean that those options are strictly dominated? If so, is it also correct to assume that those options will not be used in all Nash equilibrium strategies?

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  • $\begingroup$ Not my field, but I know that strictly dominated strategies will never be possible in mixed strategy Nash equilibria. $\endgroup$ – Theo Bendit May 12 '19 at 13:40
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The answer to your first question is NO. Consider a simple battle of sexes example.

\begin{bmatrix} & L & R \\ T & 2,1 &0,0& \\ B & 0,0 & 1,2 \end{bmatrix} Then $(T,L)$ is a Nash equilibrium in which strategy $B$ or player $1$ and strategy $R$ of player $2$ are not used. However, $(B,R)$ is a Nash equilibrium. In particular $B$ is the best-response to $R$.

If a strategy $s_i$ is a part of an equilibrium than it means that it is a best-response to a strategy pursued by the other player in this particular equilibrium. This does not mean that the other strategies are dominated; it only means that in response to that particular strategy of player $j$ used in this particular equilibrium the unused strategies lead to weakly lower payoff than $s_i$.

To answer your second question, a strictly dominated strategy will never be used in equilibrium. So if you determine that a strategy is strictly dominated you may analyze the game as if that strategy was "not there" (you can delete it - which leads to the notion of iterative deletion of strictly dominated strategies).

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