I am studying Game Theory and have problems with solving the questions regarding the game down below:
Consider $v_1 > v_2 > v_3 > 0$ and the following pay off matrix
$$ \begin{pmatrix} A/B & R_1 & R_2 & R_3 & \\ R_1 & 0,0 & v_1,-v_1 & v_1,-v_1 \\ R_2 & v_2,-v_2 & 0,0 & v_2,-v_2 \\ R_3 & v_3,-v_3 & v_3,-v_3 & 0,0 \end{pmatrix} $$
First question: Does this game have an equilibrium where one player plays a mixed strategy and the other a pure strategy? I tried to answer this by using a "claim" in my book which said that this is not the case. However, the proof of this uses the fact that there are strictly dominant strategies in the pay-off matrix which is not the case in this particular game. Any help is appreciated.
Second question: Does this game have an equilibrium where both players randomize between two of their options? In this, I have calculated the probablities where player A is indifferent between player B picking $R_1$ or $R_2$, for an example, assigning the the probablity of choosing $R_3$ zero and then looking at all the cases. This gives me the following mixed strategy equlibria
$$ q = \frac{v_1}{v_1 + v_2}, q = \frac{v_1}{v_1 + v_3}, q = \frac{v_2}{v_2 + v_3}, p = \frac{v_2}{v_1 + v_2}, p = \frac{v_3}{v_1 + v_3}, p = \frac{v_3}{v_2 + v_3} $$
where I have assigned player A to have a probability of $p$ to pick $R_1$ and $1-p$ to pick $R_2$, for example, if he mixes between these two options and the same for player B with $q$. Is this enough?
Because then afterward I have to find all equilibria where both player A and B mixes between all three options, however, only if this is possible (and if is not I have to tell why). Here I would then assign the probabilities of, ie. for player A to pick $R_1$ with $p_1$, picking $R_2$ with $p_2$ and picking $R_3$ with $1-p_1-p_2$ and likewise for player B but how do I even know if such equilibria exist in the first case? And if so under which conditions do these type of equlibria exist?
TIA for any help.