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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).

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I guess your intuition is that, when player #1 chooses $U$ (with probability $1/3$) you want player #2 to choose $L$ simultaneously, and when #1 chooses $M$, #2 chooses $C$ or $R$. But in fact player #1 has no control over #2. For instance, when #1 chooses $U$, #2 may also choose $C$, which gives #1 zero payoff. – GWu Apr 7 '11 at 5:51

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.

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Hint: If he chooses U, the average payoff is (10+0+3)/3 = 13/3 to player #1 and (0+10+3)/3 = 13/3 to player #2. – Carl Brannen Apr 7 '11 at 2:42
I am trying to find the best "mixed" response -- meaning that just as player two's moves are given in probabilities, I should give player one's moves in probabilities too, in a way that would maximize the pay-offs. What you are suggesting is picking one "pure" strategy for player one (sorry for my miscommunication). – user9233 Apr 7 '11 at 2:56
@Daniel: The utility one gets from choosing a mixed strategy is a weighted average of the utilities one would get for choosing the corresponding pure strategies. Since an average of some numbers is always between the max and min, no mixed strategy can ever achieve higher expected payoff than any pure strategy. So best responses are always either pure strategies, or if there are several pure strategies tied for best, then mixtures among these pure best responses will also be best responses. In your case there is no tie, so any non-pure mixture would be strictly worse than a pure strategy. – Noah Stein Apr 7 '11 at 12:25

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 $M$.

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The way I'd put it is that the only time a mixed, (probabilistic) strategy is best is when you can assume that your opponent is using a probabilistic strategy. – Carl Brannen Apr 8 '11 at 21:05

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