can someone please help me to solve this question:
Consider the normal form game:
i) For Player 1 the strategy B is strictly dominated by C.
ii) For Player 2 the strategy E is strictly dominated by F.
iii) The game has 2 Nash eq. in pure strategies and further equilibrium in mixed strategies. In this third eq. player 1 chooses between A,B, and C according to the probabilities:
σA= __________, σB= ___________, σC= _____________.
and player 2 chooses between D,E and F according to the probabilities:
σD= __________, σE= ___________, σF _____________.
It would be nice if someone of you can help me to find the solution for the probabilities (Step iii), with a quick explanation. I only can solve a 2x2 matrix but have never done this for a 3x3 matrix.
And are the solutions for i) and ii) correct?
Thanks for helping me!
Assuming that the numbers in the bimatrix are the payoffs to the players (as opposed to penalties on them), I believe your answers i) and ii) have the dominance the wrong way round. Each payoff to player $1$ in strategy B is strictly larger than the corresponding payoff in strategy C, so it's B that strictly dominates C, not the other way round.
Likewise, each payoff to player $2$ in strategy E is strictly larger than the corresponding payoff in strategy F, so E strictly dominates F.
Hint for solving iii):
Since strategy C is strictly dominated by strategy B for player $1$ it cannot appear with positive probability in any Nash equilibrium. Likewise, strategy F for player $2$ cannot appear with positive probability in any Nash equilibrium. You can therefore ignore those strategies entirely and obtain the Nash equilibria of the original $3\times3$ game by finding the Nash equilibria of the $2\times2$ game with bimatrix: \begin{array}{||c|c|} \hline &\text{D}&\text{E}\\ \hline \text{A}&6,12&4,8\\ \hline \text{B}&4,8&8,16\\ \hline \end{array}