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hrishimittal comments on The Aumann's agreement theorem game (guess 2/3 of the average) - Less Wrong
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Comments (149)
I don't understand how the average guess will be 0. Can you please explain?
You will pick 100. I know that, so I'll pick 66. You know that I know that, so you'll pick 44 instead. But I know that you know that I know that, so I'll pick 29 instead. But you know that I know that you know that I know that, so you'll pick 20 instead. But I know-
This continues to infinity until both of our guesses approach 0.
There is no "infinity" to be considered here.
We are given a single equation
P = (2/3)P
with the unique solution P=0.
P = (2/3)P
P-(2/3)P = 0
P(1-2/3) = 0
P(1/3) = 0
P=0
QED
As a general rule, you shouldn't even mention infinity except in very select circumstances. Especially not when the solution is so simple!
But the correct equation is Pwin = (2/3)Pavg.
Naively appealing, but if the third step is "so you'll pick 44 instead", the first step claiming that "You will pick 100, and I know that" is incorrect.
But it can be rewritten in a different way:
The correct answer cannot possibly be above 66. So everyone knows that nobody will answer above 66, and thus the correct answer will not in fact be above 44. But everybody knows that, and so the correct answer will not in fact be above 29... etc...
Of course, where it breaks down is that we know some people will not reason as above.
If 4/5 of the players conspire to answer 100, they win over the rest of the players who answer 0, so it's not always a good idea to abide by the above argument. See also this strategy.
Edit: I fixed the reply; this is the original mistaken/confused argument on which cousin_it replied below (although the conclusion remains the same):
Given that a sole player answering 100 wins when all others answer 0, it's not a failure of reason to not abide by the above argument.
Wrong. If we have two players, one says 100, other says 0, average is 50, 2/3 of average is 33, player 2 wins. Add more players saying 0 and it gets even worse.
You are right; fixed.
The problem (with the edited scenario) is that, without an enforcement mechanism, all the players in the conspiracy have an incentive to defect (more precisely, they're indifferent between defecting and not defecting if they don't care how many winners there are; they strictly prefer defection if they would rather be the only winner.)
Conspiracy is a strategy that leads to winning, while "defecting" is something magical, a change that doesn't exist in isolation. If it was possible for one player to jump to a winning position, while other players remain where they were, then this is obviously preferable to that player, but that's not really the case.
Adhering to the conspiracy is a strategy that leads to losing if anyone else answers slightly below 100. It's not a strategy that exists in isolation either.
ETA: More generally, if you're going to try to undermine a basic concept in game theory as being "magical" (whatever that is supposed to mean), I think you owe more of an argument than the one you've given.
Vladimir is just following the footsteps of Aumann, who in 1959 proposed the notion of Strong Nash Equilibrium, which requires that an agreement not be subject to an improving deviation by any coalition of players. Other game theorists then realized (like conchis) that this requirement is too strong, since agreements must be resistant to deviations which are not themselves resistant to further deviations. (I'm mostly quoting from http://www.u.arizona.edu/~jwooders/cpcethry.pdf here.)
I propose that nobody should be downvoted for making a mistake that Aumann made. :)
What about mistakes that he continues to make? ;)
More generally, although I didn't vote Vladimir down (I have a general policy against voting down comments in conversations I'm actively involved in) I'm perfectly happy to vote down mistakes regardless of whether someone smart has made them before.
Okay.
You will pick X - a number whose value I don't know. I know that, so I'll pick X*2/3. You know that I know that, so you'll pick X*4/9 instead. But I know that you know that I know that, so I'll pick X*8/27 instead. But you know that I know that you know that I know that, so you'll pick X*16/81 instead. But I know-
This continues to infinity until X is multiplied by 0. At this point the value of X doesn't matter.
Nope, it's the same argument. You can't know that I pick X and at the same time know that I pick X*4/9 instead. From the outset, you can't assume to know precisely what I pick, and considering all possible values that I pick and you know I pick (a set of situations indexed by X) doesn't fix that.
It's the only Nash equilibrium. The only way everyone can win (and thus, the only way no-one would want to change their guess if they knew all the other guesses) is for all of us to guess a number that is 2/3 of itself: i.e. 0.
ETA: CannibalSmith's explanation is better.
ETA2: AllanCrossman's is even better.
Then I don't see the point of the game.
For this to apply in the real world, the players not only have to be rational, they also have to have common knowledge of each others' rationality. E.g. even if you're rational, if you think I'm stupid, and will guess 5, then you should no longer guess zero. Even if I am rational, and everyone else has common knowledge of everyone else's rationality, if they know that you think I'm irrational, then they know that you'll guess higher than zero, so they'll all guess higher than zero, and so on...
In general, the more "stupid" people there are, or the more "stupid" people we think there are, or the more "stupid" people we think others think there are, or... the further the average guess is likely to be from 0. So (I assume) the point is to test the assumption of common knowledge of rationality: i.e. how stupid people are, how stupid we think other people are, how stupid we think other people think other people are, etc.