Picking a number nearest two thirds the average I am interested in the following game for $n$ players:
Each of the $n$ players chooses a random integer from the interval [0,100] and the average of these numbers (call it $a$) is computed. If we denote by S the set of chosen numbers that are greater than $\frac{2a}{3}$ then we call the player that picked min(S) as the winner of the game.
I am looking for strategies that would yield the highest expectation of winning the game.
One way would be to model the choices with some probabilistic distribution (which?) and compute the expected value of $\frac{2a}{3}$.
I would like to know if anyone happens to see any other tricks that could be useful in such a game.

Okay, here are some short remarks. I assume that if you pick the right number you win regardless of how many other that pick the same number and otherwise you lose. I will discuss the game-theoretical solution of the game. The question how one should play with real human opponents is more of a psychological question and off-topic for this site.  
I believe one can eliminate any strategy where you pick a number greater than $3$ by repeatedly removing weakly dominated strategies. Playing $0$ is also a dominated strategy. On the hand if everyone plays $n$, where $n$ is one of $1$,$2$ or $3$, then that is a Nash equilibrium. Just check that if $n$ is one of these numbers then the smallest integer greater than $\frac{2n}{3}$ is $n$. 
