Lets say I have 3 players, call them $p_1, p_2, p_3$. All players $p_i$ can choose between either play $A$ or $B$. Denote the probability that player $i$ will play $B$ with $q_i$. How do I find the mixed strategy for this game?

I was thinking something in the line of, assume $p_1$ plays $B$ then with probability $q_2$ $p_2$ plays $B$ and with probability $q_3$ $p_3$ plays $B$ so the expected payoff should be (where $u_1(B)$ denotes the "reward for $p_1$ playing $B$)

$$ q_3(q_2u_1(B) + (1-q_2)u_1(B)) + (1-q_3)(q_2u_i(B) + (1-q_2)u_i(B)) $$

On the other hand if $p_1$ plays $A$ the expected payoff is

$$ q_3(q_2u_1(A) + (1-q_2)u_i(A)) + (1-q_3)(q_2u_1(A) + (1-q_2)u_1(A)) $$

In order for $p_1$ to mix between strategies, this two payoffs have to be equal. Am I on the right track?


1 Answer 1


Yes, this is on the right track. But you need to remember that for example $u_1(B)$ is dependent on the plays of the other two players, so I would write the first equation, for instance, as:

$$ q_3\left (q_2u_1(s_1=B; s_2=B, s_3=B) + (1-q_2)u_1(s_1=B; s_2=A, s_3=B) \right) \\+ (1-q_3) \left (q_2u_1(s_1=A; s_2=B, s_3=B) + (1-q_2)u_1(s_1=A; s_2=A, s_3=B) \right) $$

Let's illustrate with the following example game: Each player, if he plays $B$ must pay each other player one coin, and the payoff of the game is that if there is one "odd man out" (with the other two matching) that player then collects four coins from each other player.

Then the payoff for player $1$ choosing $B$ is $$ 0 q_3q_2 -5q_3(1-q_2) -5(1-q_3) q_2 + 6(1-q_3)(1-q_2) $$ and the payoff for player $1$ choosing $A$ is $$ 10 q_3q_2 -3 q_3(1-q_2) -3(1-q_3) q_2 + 0(1-q_3)(1-q_2) $$ So one of the indifference equations is $$ 0 q_3q_2 -5q_3(1-q_2) -5(1-q_3) q_2 + 6(1-q_3)(1-q_2)\\ = 10 q_3q_2 -3 q_3(1-q_2) -3(1-q_3) q_2 + 0(1-q_3)(1-q_2) $$ or $$ 20q_2q_3 + 8 q_2+8q_3 = 6$$ and the other two are $$ 20q_1q_3 + 8 q_1+8q_3 = 6 \\ 20q_2q_1 + 8 q_2+8q_1 = 6 $$ Subtracting these last two, you can see that either $q_3=0$ or $q_2-q_3= 0$ so (since the case of all three playing $b$ all the time is obviously not a Nash equilibrium point) all thre of the $q_i$ are equal.

So the Nash equilibrium point comes with each player choosing $B$ $\frac{\sqrt{46}-4}{10}\approx 0.278$ of the time.

That value comes from solving $20q^2 + 8q+8q = 6$.


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