# pure strategy equilibria and mixed strategy equilibria

Three players simultaneously pick a point on the interval $$[0,1]$$.

The player closest to the average of the three points wins $$1$$ dollar.

If there is a tie, then the dollar is split equally among them.

More formally, the players simultaneously choose strategies $$s_i ∈ [0,1].$$

The average of their choices is $$S = (s_1 + s_2 + s_3)/3.$$

Player $$i$$’s payoﬀ function is

$$U_i(s_1 , s_2 , s_3) = \begin{cases} 1/t, & \text{if i ∈ arg min_j |s_j − S| } \\ 0 & \text{otherwise} \end{cases}$$

where t is the number of players who tie (their choices are equally close to the average).

(a) What are the pure-strategy equilibria of this game?

(b) What are the mixed-strategy equilibria if the possible strategies are limited to playing $$0$$ or $$1$$, rather than $$[0,1]$$?

This question is about game theory, related with nash equilibrium. I have no idea where to start.

(a) Suppose all three choose the same number, $$x$$. The average is $$x$$ and the payoff is $$\tfrac{1}{3}$$. Is there a profitable deviation? If one player deviates, he can affect the average and move it only third of the way towards his newly chosen value. The average is still closer to $$x$$ than to him, so he'll get $$0$$. Thus, for each $$x$$ in the segment, the strategy in which all players choose $$x$$ is a pure equilibrium.
Is there an equilibrium where 2 players choose the same $$x$$ and someone else chooses $$y\neq x$$? No - from the argument above, the $$y$$ player gets $$0$$ and he should switch to $$x$$.
Is there an equilibrium where all three players choose different numbers? No, since in such a scenario at least one of the players will get $$0$$ (the one that is far away from the average) and he has a profitable deviation - get closer. For example: choose the same as one of the others and get $$0.5$$ with him.
(b) Suppose they choose 1 with probability $$x,y,z$$ (respectively). Write the expected payoff of player 3 when he chooses $$1$$ (as a function of $$x,y$$) and when he chooses $$0$$. From indifference, they are suppose the same - this gives you an equation for $$x,y$$. You can repeat the process to get $$3$$ equation with $$3$$ variables, and the solution is the desired equilibrium (again, $$x=y=z=0$$ and $$x=y=z=1$$ are pure equilibria, as stated in (a)).