Mixed Strategies for 3x3 matrix

Question

What are mixed strategies in this game?

Usually, I can find mixed strategies of 3x3 when there exist a dominant strategy that dominates another one and we eliminate dominant strategy. But in such case like this I don't know how to find.

Answer 1

One key consideration is that a strategy can be strictly dominated by mixed strategies as well. In other words, if we can assign a probability distribution of two actions such that they do strictly better than a particular strategy in expectation, than that strategy is strictly dominated.

For the first player, there are no strictly dominated strategies. However, for the second player, notice that a combination of $D$ and $F$ might potentially do better than $E$. Indeed, if we can find a probability distribution $p_D+p_F=1$ where $p_D$ (of $p_F$) denote the probability attached to $D$ (or $F$) that satisfy the following conditions, then $E$ does worse than this combination regardless of whether player 1 plays $T$, $B$, or $R$. $$ \begin{align} p_D\times 8 + p_F\times 2&>& 6\\ p_D\times 8 + p_F\times 2&>& 2\\ p_D\times -12 + p_F\times 4&>& -9\\ \end{align} $$ Here the left hand side of each inequality correspond to the average payoff of playing the mixed strategy when either $T$, $B$, or $R$ is played, and the right hand side corresponds to what happens if instead $E$ was played and we are wondering whether $E$ would be dominated, hence the inequality sign.

Note that $p_F= (1-p_D)$, which implies that the above inequalities would be required to satisfy $$ \begin{align} p_D\times 8 + (1-p_D)\times 2> 6&\iff& p_D>2/3\\ p_D\times 8 + (1-p_D)\times 2> 2 &\iff& p_D>0\\ p_D\times -12 + (1-p_D)\times 4> -9 &\iff& p_D<13/16.\\ \end{align} $$ Notice that there is a range of values for $p_D$ that would satisfy the above inequalities. One particular example would be $p_D=3/4$. As a result $E$ is strictly dominated in mixed strategies. Thus, it won't be played with positive probability in any Nash equilibrium, so we can eliminate $E$.

Once you eliminate $E$, then the row player would figure that playing $B$ would never be optimal because it would be strictly dominated by $T$, again under the consideration that $E$ would never be played. Then you will be left with your simple $2\times2$ game for which solving would be straightforward.