Optimal strategy in three-player number-picking game Consider a game in which there are three players. Call them Player $1$, Player $2$, and Player $3$. Here are the rules:


*

*Each player is supposed to select an integer between $1$ and $100$. 

*Player $1$'s number is randomly generated. Player $2$ and Player $3$ both know that Player $1$'s number is randomly generated.

*The person with the largest number has to pay the other two people the number that each one of them said (i.e. say Player $1$ picks $5$, Player $2$ picks $70$ and Player $3$ picks $90$. In this case, Player $3$ pays $5$ to Player $1$ and $70$ to Player $2$).
Let's suppose you are Player $3$. Furthermore, suppose that Player $2$ plays optimally. What's the best strategy if you want to maximize profit?
I solved the $n = 2$ (two-player game) case here: Optimal strategy in probability-based game
I want to extend it to $n = 3$, but I can't figure it out. I would appreciate any help.
 A: Actually, there are only 2 players in this game. Player 1 does not choose a strategy between a set of strategies.
This game could be represented as $\mathcal{G}(X_1,X_2,H_1,H_2)$, where $X_i=[0,100]\cap\Bbb{Z}$, for i={1,2} and:
$$H_i(x_1,x_2)=
\begin{cases}
E(Z_i),  & \text{if $x_i> x_j$} \\
\frac{E(Z_i)}{2},  & \text{if $x_i= x_j$} \\\\
x_i, & \text{$x_i<x_j$}
\end{cases}$$
Now suposse $x_j<x_i$:  $ E(Z_i\mid x_i > Y)=-(\frac{x_i+1}{2}+x_j)$
$E(Z_i) = \left(\frac{1 - x_i}{100}\right)\left(\frac{x_i + 1}{2}+x_j\right) + \left(\frac{100x_i - x_i^2}{100}\right)=\frac{-3x_i^2-2x_ix_j+200x_i+2x_j+1}{200}$
The new optimum is $x_i=33-2x_j$, for  $ i,j=\left\lbrace1,2\right\rbrace$, $j\neq i$.
The solution to that system is $(11,11)$, which is the Nash equilibrium of $\mathcal{G}$.
Edit: Now assuming that if two of them tie, they will pay half. Notice that if they tie, the payment function had the same optimum as the original.
This solution is consistent with the previous game, if both of them tie, they will have to pay $\frac{x+x}{2}=x$ to each other, so the benefits will be $0$.
