Extending 'Guess 2/3 of the Average' Game In game theory, 'Guess $\frac{2}3$ of the Average' is a game where $n$ people are asked to choose a real number between $0$ and $100$ inclusive. The person with the closest answer to $\frac{2}3$ of the average value wins. It can be shown that there is a unique pure strategy Nash equilibrium where everyone picks the number $0$. 
(The reasoning is that the desired number can't be greater than $\frac{2}3 \cdot 100$ so everyone picks between $0$ and $\frac{2}3 \cdot 100$. Then iterate.)
Can we generalize this game. Specifically, say that you survey $n$ people who each pick a real number from a set $S \subset \mathbb{R}$. From their responses, you can form a vector $\vec{v}$ where the $i$ th entry is the $i$ th person's response. 
Now suppose you have function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ that is known to all of the $n$ people. The person whose value is closest to $f(\vec{v})$ is the winner. What strategy should each of the $n$ people employ?
Can we find essentially different functions and sets such that everyone ends up guessing the same number? For example, we could have chosen a different fraction $< 1$ rather than $\frac{2}3$ but in spirit, this is the same example. There are also examples of functions and sets where everyone does not end up picking the same answer. For example, if we restrict the people in our example above to choose only integers, then the strategy is pick $1$ or $0$ depending on what you think the majority of the people will pick.
 A: Lets choose $f: S^n\rightarrow S$ where $S \subset \mathbb{R}$ where for $v \in S^n$ the person who is closest to $f(v)$ wins. 
I think we should say that $f$ is continuous and $S$ is connected, in the 2/3-game $S$ would be $[0,100]$ and $f(x) = \sum_{i=1}^n\frac{2x_i}{3n}$
If this holds a rational player should pick only numbers in $f(S)$. For every number outside of $f(S)$ there is a better pick - Just like its more reasonable to pick 66,6 than it is to pick 70.
Now you take $f(f(S)) =: f^2(S)$ and continue with $f^3(S)$ ... The limit $n \rightarrow \infty$ of $f^n(S)$ gives you the range of numbers a rational player should pick. Although just like in the Guess-2/3-Game in a real world situation not every player is rational, so it might not be the smartest move to be rational.
A: Assume that $S$ is a compact interval $[a,b]$, and consider the "closest point" function $$r(x)=\begin{cases}x;&x\in [a,b]\\a;& x<a\\b;&x>b\end{cases}$$
Now, define $g:S\mapsto S$ as $g(x)=r(f(x,x,\cdots,x))$. If $g$ is continuous, then the Brouwer fixed-point theorem guarantees that everyone can simultaneously win by picking the same value (the fixed point). Note that requiring $g$ to be continuous is a weaker condition than requiring $f$ to be continuous.
