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lavalamp comments on The Aumann's agreement theorem game (guess 2/3 of the average) - Less Wrong
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Comments (149)
I think it's entirely rational to submit a non-zero answer.
I would prefer to win outright, rather than tie, and I think it's safe to assume this is true of more people than just me.
If everyone does the "rational" thing of guessing 0, it will be a big tie.
If anyone guesses above 0, anyone guessing 0 will be beaten by someone with a guess between 0 and the average.
Therefore, a small non-zero guess would seem superior to a guess of zero, to those who value outright wins above ties (EDIT: and don't value a tie as being much better than a loss).
Perhaps I'll write a program to simulate what the best guess would be if everyone reasons as above and writes a program to simulate it...
This is not a test of individual rationality. It is a test of how well the Aumann assumptions apply to the group.
OK, but I don't see why the assumptions listed above should result in everyone guessing 0 (which is my point), and 2 minutes with google hasn't told me about any further assumptions.
If one of the assumptions is that a tie is as valuable as a win, then I see your point, but I haven't seen that made explicit.
You do see that zero is the only Nash equilibrium, right? If everyone plays zero, you gain nothing by defecting alone, because 1/N is still better than nothing (and your guess will always be greater than 2/3 of the average).
So you're arguing that it's not rational, under the assumption of common rationality, to play the unique Nash equilibrium?
Is making an "assumption of common rationality" really a rational choice, even here?
With the stakes as low as this, I would assign a very high likelihood to someone getting greater utility from throwing a spanner in the works for the lulz than from a serious attempt at winning, even if at least one such person hadn't already announced their action.
FWIW, I never suggested it was. Lavalamp claimed that zero was not rational under the assumptions in the OP's original justification, one of which was common rationality. It was the validity of that argument I was defending; not it's soundness.
Everyone playing 0 is only better than everyone playing 67 because it corrects individual defectors. It doesn't correct multiple coordinated defectors. If we know that there are no defectors, 67 is as good as 0, and if we know that there is some number of defectors who could conspire to play something else, 0 is not much better than 67. This becomes more interesting if the payoff on the non-equilibrium choices is greater.
Nash equilibrium is not a universal principle, it's merely a measure against individual uncoordinated madmen, agents unable to cooperate.
Actually, I think it does rather better against uncoordinated rational agents than it does against crazy people. I'm not sure why it should have any traction at all against the latter.
More generally, you're right, but: (a) that didn't seem to be the nature of lavalamp's argument; and (b) unless it's also incentive compatible in the standard sense, I tend to consider the possibility of coordination as changing the rules of the game (though that's just a personal semantic preference).
By madmen I meant "rational" agents who refuse to consider an option or implications of coordination (the kind that requires no defection).
Impossibility of coordination is a nontrivial concept, I don't quite understand what it means (I should...). If everyone follows a certain procedure that leads them to agree on 0, why can't they agree on 67 just as well?
Because given what others are doing, no individual has an incentive to deviate from 0 (regardless of whether they've agreed to it or not). In contrast, if they're really trying to win, every individual agent has an incentive to deviate from 67.
ETA: You can get around the latter problem if you have an enforcement mechanism that lets you punish defectors; but that's adding something not in the original set up, which is why I prefer to consider it changing the rules of the game.
Coordination determines the joint outcome, or some property of the joint outcome; possibility of defection means lack of total coordination for the given outcome. Punishment is only one of the possible ways of ensuring coordination (although the only realistic one for humans, in most cases). Between the two coordinated strategies, 67 is as good as 0.
What I wondered is what it could mean to establish the total lack of coordination, impossibility of implicit communication through running common algorithms, having common origin, sharing common biases, etc, so that the players literally can't figure out a common answer in e.g. battle of the sexes.
I'm sure I'm missing your point, but FWIW my original claim was only about the (im)possibility of coordination on a non-Nash equilibrium solution (i.e. of coordinating on a solution that is not incentive-compatible). Coordinating on one of a number of Nash equilibria (which is the issue in battle of the sexes) is a different matter entirely (and not one I am claiming anything about).
The purpose of this game, admittedly, is to test just how complacent / obedient the Overcoming Bias / Less Wrong community has become.
Think about your assumptions:
First you've got "common rationality". But that's really a smokescreen to hide the fact that you're using a utility function and simply, dearly, hoping that everybody else is using the same one as you!
Your second assumption is that "you gain nothing by defecting alone".
There's no meaningful sense in which you're "winning" if everybody guesses zero and you do too. The only purpose of it, the only reward you receive for guessing 0 and 'winning', is the satisfaction that you dutifully followed instructions and submitted the 'correct' answer according to game theory and the arguments put forth by upper echelons of the Less Wrong community.
In fact, there is much to gain by guessing a non-zero number. First of all, it costs nothing to play. Right away, all of your game theory and rationalization is tossed right out the window. It is of no cost to submit an answer of 100, or even to submit several answers of 100. Your theory of games can't account for this - if people get multiple guesses, submitted from different accounts, you'll be pretty silly with your submission of 0 as an answer.
"But that would be cheating." Well, no. See, the game is a cheat. It's to test "Aumann's agreement theorem" among this community here. It's to test whether or not you will follow instructions and run with the herd, buying into garbage about a 'common rationality' and 'unique solutions', 'utility functions' and such.
You see, for me at least, there's great value in defecting. You of course will try to scare people into believing they're defecting alone, but here you're presupposing the results of the experiment - that everybody else is dutifully following instructions. So anyway, I would be greatly pleased if the result turned out to be a non-zero number. It would restore my faith in this community, actually. And to that end, I would submit a high number...
If I were to play.
I think that you are profoundly mistaken about the attitudes and dispositions of the vast majority here. You appear to be new, so that's understandable. As you look around, though, you'll find a wide array of opinions on the limits of causal decision theory, the aptness of utility functions for describing or prescribing human action, and other topics you assume must be dogma for a community calling itself 'rationalist'. You might even experience the uncomfortable realization that other people already agree with some of the brilliant revelations about rationality that you've derived.
I was an avid visitor of Overcoming Bias, but yes I am new to Less Wrong. I had assumed that the general feel of this place would be similar to Overcoming Bias - much of which was very dogmatic, although there were a few notable voices of dissent (several of whom were censored and even banned).
Obviously. But there wouldn't be a point to my lecturing them, now would there? No, conchis made the canonical argument and I responded. And if you weren't so uncomfortable with my dissent you might have left a real response, instead of this patronizing and sarcastic analysis.
That's the problem with the internet: "I'm witty and incisive, you're sarcastic and sanctimonious". I'll admit the tenor of my last sentence was out of line; but I stand by the assertion that your psychoanalysis of this group is well off the mark.
Also, what exactly is so awful about a group norm of playing certain games seriously even when for zero stakes, in order to gather interesting information about the group dynamics of aspiring rationalists?
Pretty much nails it. pswoo's initial comment was fairly patronizing itself, so it seems a bit rich to criticise you (orthonormal) for playing along. But whatever.
By way of substantive response. Um, yeah. So, patronizing bits aside, I agree with much of your (pswoo's) comment. I just don't think it was especially relevant to the particular conversation you (pswoo) intervened in, which was about the validity of the standard argument rather than its soundness.
I will be very surprised if more than half of the answers are 0.
Agreed. This is why I specified that I think there are others who also would value a unique win, and why, in another comment I mentioned that of those of us who value a unique win, someone has to guess high.
This leads to quite a nice dilemma (as we'd all prefer for someone else to guess high), unless we believe cousin_it, who says he guessed 100.
Assuming that the rewards (and/or penalties) were adjusted such that everyone greatly prefers a tie to a loss, then I would have to agree that 0 is the Nash equilibrium (and would guess 0).
However, given that the only available reward here is social capital (if even that), I'd rather win outright, even if it brings a risk of losing, and I don't see why I would be alone in that order of preferences.
And I think I may be distorting the game as much as cousin_it, and equally unintentionally. Sorry...
I think we've basically resolved this, but just to clear up loose ends, I'm pretty sure it will be a Nash equilibrium provided everyone strictly prefers a tie to a loss; as far as I can tell the preference shouldn't need to be "great".
It has to be great enough to make me unwilling to risk a loss for the possibility of an outright win, which is why I said "greatly." But I suppose it's relative.
Nash equilibrium doesn't work like that. Each player's strategy must be optimal given perfect knowledge of others' equilibrium strategies. Your probabilistic reasoning only applies if you don't know others' equilibrium strategies (or if they're playing mixed strategies, but that isn't relevant here).
Sorry, context switch on my part-- I wasn't thinking about Nash equilibrium when I wrote that.
But I still don't see your point--if I assume that everyone's utility function is exactly like mine, I don't see how my probabilistic reasoning would differ from an equilibrium strategy, if I'm using the term right.
Did you just switch context again? My claim is about what happens if everyone strictly prefers to tie rather than to lose. In this case, given others' strategies, any individual's optimal strategy is to answer 2/3 of the average. The only way everyone can answer 2/3 of the average is if everyone plays 0, and this is the only strategy that nobody has an incentive to deviate from.
I think a reasonable assumption is that a tie between N people is worth 1/N of a unique win, to each of the winners. This would makes sense if there were a prize that was to split between the winners (and your utility function is linear with respect to the prize).
The problem is that the game is not well-defined, since we are not told how to value ties. If everyone is allowed to value ties as they please, the game is far more complicated. Instead the OP should say something explicit, and probably it should be "ties are exactly as valuable as wins". But it's hard to enforce that, isn't it?
Yes, because I would continue to prefer to win without a tie. :)
By allowing you to choose a real number rather than just an integer, Warrigal has made it easy to avoid ties. Since I have myself played a non-integer, it is extremely unlikely that anyone will guess the exact value of the answer, so there is no penalty to picking a probably unique real number very close to your actual guess.
EDIT: On second thought, there is a penalty in that you could tie for the right answer with someone else if you chose a salient number, but you would roughly halve your chances of being the winner if you added or subtracted an epsilon.
To improve on in this point, so that it's possible to move either way from anyone else's choice, we could make the choice to be from the open set (0,100) rather than from [0,100]. ;-)
if the average is less than 3/4 then the zeros will still win
It depends where in the range of 0-average the guess is.
But of course I see what you mean; I meant between 0 and (average * 3/4), sorry.
EDIT: (average * 3/4 + average * 3/8) is the upper bound, unless I forgot something or you're not allowed to go over.
EDIT 2: The point being, there's a lot more winning non-zero answers than zero answers.
Sure, but what are the odds of you getting the right one. If they're too low, then you could still better of with zero.
If I rank a tie and a loss the same, then I'm don't risk anything by guessing a non-0 value for the chance of winning outright.
If everybody reasons as you describe then everyone will guess 1/∞ and everyone will tie. You can't get closer to 2/3 of an infinitesimal than an infinitesimal, so it's stable.
Disclaimer: I'm not mathy. Maybe you actually can get closer to 2/3 of an infinitesimal than an infinitesimal.
The question required us to provide real numbers, and infinitesimals are not real numbers. Even if you allowed infinitesimals, though, 0 would still be the Nash equilibrium. After all, if 1/∞ is a valid guess, so is (1/∞)*(2/3), etc., so the exact same logic applies: any number larger than 0 is too large. The only value where everyone could know everyone else's choice and still not want to change is 0.
No, everyone who prefers to win outright will do the logic you just did and decide that 1/∞ is too small, for the reason you state.
Also, those of us who prefer to win outright will note that someone has to take one for the team and vote high to give the possibility of an outright winner, unless we believe cousin_it...
Reminds me of yet another semi-interesting game: everyone chooses a positive integer in base-10 notation, and the second highest number wins.
If the game as I've understood it is played over and over, I think it'd be much like rock, paper, scissors: http://www.cs.ualberta.ca/~darse/rsbpc.html
Really? I'd expect the numbers to spiral higher and higher.
After the first round, I wouldn't think there's much reason to guess higher than the previous highest number (or lower than the previous third number), which suggests convergence. If everyone updates based just on what other people did last time, then won't they cycle progressively closer around the initial second number? (Actually, I'm pretty sure the same holds even if people anticipate other's responses, provided they all reason forward the same number of steps.)
ETA: What happens If there is a tie for the highest number? Does the third highest guess win, or the two highest together? What if everyone guesses the same thing?
Eh? Why would anyone take one for the team? If it's bad to tie, surely it's worse to lose for the purpose of helping someone else win.
I should have explicitly stated this (you've caused me to realize that it's necessary for my whole line of reasoning), but it's also the case that losing is not a worse outcome for me than tying.
Ah. OK. I think that's generally supposed to be excluded by the set up (though in this case it was admittedly ambiguous). But given that you have those preferences, I agree that your reasoning makes sense.
There's no prize offered, but in theory, these people could collaborate to share whatever prize makes winning better than tying. Since there's no prize except winning, losing to help someone else does seem like it would be bad.