# Mixed Strategy Without an Option = Option is Dominated?

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?

• Not my field, but I know that strictly dominated strategies will never be possible in mixed strategy Nash equilibria. – Theo Bendit May 12 '19 at 13:40

$$\begin{bmatrix} & L & R \\ T & 2,1 &0,0& \\ B & 0,0 & 1,2 \end{bmatrix}$$ 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 $$s_i$$.