mixed strategy nash equilibrium question! Suppose the game consists of only $2$ players, player $1$ and player $2$, and each of them has only $2$ strategies to choose between. This gives a $2$ by $2$ payoff matrix.
Player $2$ has no preference when choosing one of his strategies, while player $1$ chooses the strictly dominant strategy for himself. 
Questions are:


*

*In this case, is player $2$ going to mix her strategies? Given that whatever proportion player $2$ mixes her strategies, player $1$ would definitely gain positive earning.

*In this case, is there any mixed Nash equilibrium?
Thanks for help!
 A: Let's call player's $1$ strategies A and B, and player's $2$ strategies C and D. Let C be the dominant one. So, the payoff matrix is:
$$\small \begin{matrix}
& & \textrm{player } 1&\\ 
& & \textrm{A} & \textrm{B} \\ 
 \textrm{player } 2 &\textrm{C} & x_1, x_2 & y_1, y_2\\ 
& \textrm{D} & u_1, u_2 & z_1, z_2 \\ 
\end{matrix}$$
Since A is dominant, $x_2>y_2$ and $u_2 > z_2$.
Since player $1$ always plays A we can ommit strategy B and the game's matrix actually looks like this:
$$\small \begin{matrix}
& & \textrm{player } 1&\\ 
& & \textrm{A} \\ 
 \textrm{player } 2 &\textrm{C} & x_1, x_2\\ 
& \textrm{D} & u_1, u_2 \\ 
\end{matrix}$$
Assuming that player $2$ goes only for maximising her own profit (i.e. she doesn't get any happier when player $1$ gets less), player $2$ would play the strategy that gives her most profit, that is:
if $x_1>u_1$ player $2$ plays A, if $x_1<u_1$ player $2$ plays B.
If $x_1 = u_1$ then there are continuum different mixed stategy NEs since any pair $\{(1,0), (t, (1-t))\}$for $t\in[0,1]$ gives a NE.
