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?
 A: 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).
