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.


1 Answer 1


(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)).


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