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.

  • $\begingroup$ This looks hard to me. The obvious problem is that your opponents may collude, in which case you're shafted. So you have to formulate some conditions that preclude this possibility. Even so, a solution may be hard to find. I suggest you start by trying to solve the case $n=2$, where collusion is not a problem. Or have you tried this already? $\endgroup$ – TonyK Mar 21 '11 at 18:44
  • $\begingroup$ a) "random" and "strategy" are contradictory -- I presume you mean "an arbitrary integer"? b) You'd need to say what you mean by "the highest expectation of winning the game". There's only an expectation if at least some of the strategies being followed involve random choices, and even then the expectation depends on the strategies. If you actually mean a dominant strategy, there isn't one for this game: for each strategy you could follow, there are strategies for the other players such that it would be better for you to follow a different strategy. $\endgroup$ – joriki Mar 21 '11 at 19:02
  • $\begingroup$ What happens if several players pick min(S)? Do they all get a full payoff for a win? $\endgroup$ – Johan Mar 24 '11 at 14:45

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.

  • 3
    $\begingroup$ Great answer. To expand only slightly, this is a classic game in behavioral economics called "p-beauty Contest". $\endgroup$ – anthony137 Mar 21 '11 at 19:39
  • $\begingroup$ "the only correct play is $0$": I don't think that can be right. If I know the other players are going to do that, then I pick $1$ to win. Also: did you say $1.3$ levels of recursion? As in $13/10$? That is a curious result. What were these experiments? $\endgroup$ – TonyK Mar 22 '11 at 16:04
  • $\begingroup$ @TonyK The result that 1 wins against a large collection of 0s is only because of the specific way the problem was stated: "strictly greater than" $(2/3){\bar x}$ rather than "not less than". As far as the value goes, check en.wikipedia.org/wiki/Guess_2/3_of_the_average $\endgroup$ – Paul Z Mar 22 '11 at 17:16
  • $\begingroup$ Is 0 actually a Nash equilibrium for the game? Why? $\endgroup$ – user8540 Mar 22 '11 at 18:44
  • $\begingroup$ @Paul Z: $0$ can never be right for this game, where the winning number has to be greater than $2a/3$. The game in your link is different: the winning number is the number closest to $2a/3$. And what do you mean, "as far as the value goes"? $\endgroup$ – TonyK Mar 23 '11 at 14:52

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 $n$.


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