Consider a two-person game with payoff matrices defined by \begin{equation} P= \left( \begin{array}{ccc} 0 & 4 & 1 \\ 2 & 2 & 4 \\ 3 & 2 & 2 \end{array} \right) \quad \text{ and } \quad Q= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 3 & 1 \\ 3 & 4 & 1 \end{array} \right). \end{equation} Determine if this game is non-degenerate. (I don't know how to apply the definition regarding the number of best responses to check.)
Remark:
This is actually the second part of a question. The first part asks us to prove the following result (which I am not sure whether it is related to this):
Let $G$ be a bimatrix game with actions $M=\{1, \ldots, m \}$ for the row player and $N= \{ m+1, \ldots, m+n\}$ for the column player. Let $a_1, \ldots, a_k \in M \cup N$ be a sequence of actions such that for all $i=1,\ldots,k$, $a_i$ is weakly dominated with respect to $A \setminus \{a_j : 1 \leq j <i\}$. Then $G$ has an equilibrium $(x,y)$ with the support of $x$ in $M \setminus \{a_j : 1 \leq j \leq k\}$ and the support of $y$ in $N \setminus \{a_j : 1 \leq j \leq k\}$.
Since the games has $3$ pure strategies for both players, the only chance for degeneracy is either:
It is easy to see that there is no pure strategy with more than one best response.
Now you could try to find probabilities for exactly two strategies that equalize the expected payoff of all three pure strategies of the other player.
However, you will not be able to do this. As Stefanos' comment shows, we get to a $2 \times 2$ game by iterated elimination of strictly dominated strategies. Since the $3 \times 3$ has no pure strategy with more than one best response, nor does the $2 \times 2$ game, since we only eliminated strictly dominated strategies, which can never be best responses (and so the fact we eliminated one strategy for each player means we can't find a mixed strategy for point 2. above either) . Thus the game is non-degenerate.