Comments: |
"closest to 1 bigger than the average of everyone else's guess" reduces to "name the largest number you can". If the number you name is more than a factor of n larger than the next contender, then your number is the only one larger than the average and you win by default.
"Bigger than the average of everyone's guess mod 100" confuses me. When does the mod get applied? Is a good strategy to play -1/(very large number), which is very large mod 100 and thus larger than average, but also very close to 1?
Ah, good point.
I had an earlier version of the post that I thought was equivalent and needlessly complex, but I see now that it's not equivalent, so I'll revive it:
Everyone votes a point on the unit circle, and your goal is to get closest to the average, rotated by some fixed small angle. And by "average" I mean "take all the points p1...pN as complex numbers, compute z = (p1 + ... + pN+1)/(N+1), and say the target is e^{i epsilon} z/|z| for some smallish epsilon, maybe like 0.1 or so.
I've tossed a +1 in the computation of the mean because it feels right to "ground" the recursion by putting in a virtual vote for 1, so the game isn't entirely symmetric under rotation of the unit circle.
*Edited at 2016-10-02 11:02 pm (UTC)*
There's a point on the unit circle, roughly (n+1)/n*epsilon degrees in front of 1. If everyone chooses that point, then the average-plus-epsilon lands on that point as well. If you think everyone but you is going to choose that point, you'd better choose it too, or the average-plus-epsilon will wind up closer to them than you.
I don't think I agree with this, even assuming you meant "radians" rather than "degrees". Let N be very big and let epsilon be pi/4. You seem to be saying there exists a point x such that if everyone chooses x, then x is the best point to play. But I'm still staring at this and believing that for all x, if everyone chooses x, my best move is quite close to x rotated by pi/4.
I specifically chose the space of moves to be not-simply-connected to dodge the consequences of Brouwer's fixedpoint theorem.
Sorry, I think you're right. The point I was thinking of was actually at (n+1)*epsilon radians, and it only works if (n+1)*epsilon is less than pi.
There's an old game that goes like this: you and your opponent choose a number 1 to 5, and you each get the number of points you bid, _unless_ your opponent's number is one less than your number, in which case your opponent gets the sum of the numbers picked and you get nothing. I think this game (or some continuous variant thereof) has the properties you're looking for.
This game can be solved: you can assume the answer is a probability distribution across the five choices, and solve for the distribution such that your expected score gain is zero regardless of what your opponent picks. If either player plays this distribution, it's like playing rock-scissors-paper using a random number generator.
(By "number" I mean "integer" in the above game.)
I am familiar with that game, and maybe you're right that the game I'm describing could be finitely approximated and would in that case just have a boring ol' mixed strategy Nash equilibrium. I don't think it's going to be "always play this specific point" as you seem to be suggesting above, but it could be something uninteresting and nonparadoxical like "pretty much play a random point on the circle, somehow smearily biased towards 1 since that the virtual vote for 1 is thrown in"
However, the whole point in asking the question hinges on the empirically observed fact that in other games that definitely do have deterministic nash equilibria, like "guess 2/3s of the average", where it's "Play zero", actual humans do something else. And I just vaguely wonder what actual humans would do in the unit circle game.
When economists play the 2/3 game with non-economists (Colin Camerer has some stuff on this that I quoted in that book you read) they bet on a pretty high number. Naive recursion usually peters out completely after about five rounds. Same as nesting thoughts, which the empirical game models: "He thinks she thinks Bob thinks Alice thinks Charles will bet on 33." | |