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: (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)).
