Finding best response function with probabilities (BR) given a normal-matrix representation of the game

Question

We are given players 1, 2 and their respective strategies (U, M, D for player 1, L, C, R for player 2) and the corresponding pay-offs through the following table:

$\begin{matrix} 1|2 & L & C & R\\ U & 10, 0 & 0, 10 & 3, 3 \\ M & 2,10 & 10, 2 & 6, 4\\ D & 3, 3 & 4, 6 & 6, 6 \end{matrix}$

Player 1 holds a belief that player 2 will play each of his/her strategies with frequency $\frac{1}{3}$, in other words $\alpha_2$=($\frac{1}{3}, \frac{1}{3}, \frac{1}{3}$). Given this, I need to find best response $BR_1(\alpha_2)$ for player 1. I am wondering how to do this mathematically. I have an intuitive way, and am not sure if it is correct. I think that if player 2 chooses $L$, player 1 is better off choosing $U$, anf if player 2 chooses $C$ or $R$ player 1 is better off choosing $M$, so best response for player 1 given his/her beliefs about player 2 would be $(\frac{1}{3}, \frac{2}{3}, 0)$, but I do not know if this is correct and how to get an answer mathematically (though I think it could involve derivatives which I would have to take to see what probability value would maximize the pay-off, just can't think of a function).

Answer 1

Why don't you compute the average payoff for player #1 for each of his three choices? Clearly he's going to choose the one that is largest.

Answer 2

I think Carl already gave the right answer. Even though mixed strategies may look better than pure ones, actually they are not. Suppose that player 1 choose a mixed strategy $\alpha_1=(a,b,c)$. Then the probability of each scenario is given by $$\begin{matrix} 1|2 & L & C & R\\ U & a/3 & a/3 & a/3 \\ M & b/3 & b/3 & b/3 \\ D & c/3 & c/3 & c/3 \end{matrix} $$ If you consider the payoff of player 1 in each of the 9 possible scenarios and compute the expected payoff, that will be $13a/3+18b/3+13c/3$. The maximum is achieved at $(a,b,c)=(0,1,0)$ which is $18/3$. Of course, the best response for player 1 is a pure strategy --- always choosing $M$.