Picking a number nearest two thirds the average

Question

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.

Answer 1

The problem with this game is that the only Nash equilibrium is 0. Given a prior distribution of answers from the other players, you should always guess lower. But if everyone does this, it changes the prior distribution until the "right" guess is again lower. It's more fun if you restrict play to [1,100] rather than [0,100].

In practice this isn't actually an exercise in mathematics, it's an exercise in psychology. The "obvious" average is of course 50. People who "understand" the puzzle will guess around 33. Of course, there are those who "understand" this first level of recursion and guess 22; and so on. The problem, for those who ACTUALLY understand the puzzle (because they studied it) is not to do the recursion, because it has no end. The problem is actually to guess: how well do you think your OPPONENTS understand recursion?

Experiments show that most people can figure out roughly 1.3 levels of recursion. So the typical average in this game is around 26-30, and you should guess 18-20. Repeating the experiment with different populations will give different results; in particular, people who have advanced training in recursion (computer programmers or some mathematicians, but in general NOT other kinds of engineers) will, on average, manage another level or so, making the "correct" play 12-13. But really, if all the players understand the game, the only correct play is 0.

Answer 2

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 $n$.