I was thinking that this game would pretty much only measure (belief of) rationality, but now I see that it measures (belief of) honesty to a good degree as well. By guessing 100, one is being dishonest.
No it is not, especially since cousin_it has upfrontly told us what he values. You are assuming that everyone who submits has a utility function that highly values winning this game, which, given the comments around here, seems to not be true (or is at least widely believed to not be true).
Don’t confuse ‘has different values than I’ with ‘irrational’.
If we all take you at your word, that does indeed make it more interesting. If every other entrant acts rationally in choosing P, we must have P = (2/3)(PN + 100)/(N+1), if there are N other participants. This solves to give P = 200/(N+3).
But we don’t know N, we can only guess at it, or some probability distribution of N. N is at least 2, since Psy-Kosh has claimed to have entered, and anyone making this calculation must count themselves among the N. But Psy-Kosh posted before cousin_it revealed its guess of 100, so we might guess that Psy-Kosh voted 0 on the grounds given in the original post, which then modifies the calculation to give P = 200/(N+5).
But suppose of the N non-cousin_it entrants, K entered before knowing cousin_it’s entry, and all chose 0. Then we get P = 200/(N+3+2K).
Now, Aumann agreement only applies if the parties confer to honestly share their information. However, this has been framed as a competitive game, and someone who wants to be the exclusive winner would to better to avoid any such procedure, or to participate in it dishonestly.
A simpler analysis would be to point out that if Psy-Kosh voted zero (as would have been rational without cousin_it), then if everyone else votes zero, all will win except cousin_it. However, if someone votes slightly more than zero, then that one will be the exclusive winner. Someone who values an exclusive win above a tie might try to persuade everyone to vote zero and then defect.
Edit: I have entered, based on the above considerations. My entry was greater than zero.
a) I have successfully moved the average by ~2X compared to if I’d stayed silent.
b) If N is known and everyone acts rationally as you describe, a pair of colluding players can screw everyone over: one guesses 100, the other guesses the new average.
If I may attempt to contribute your brain matter’s outward velocity: When I was contemplating submitting a guess (which I didn’t do), I actually concluded (I have no evidence for this, sorry) that you in particular would probably guess 100. Had I acted on that, the effect would have been nearly the same, except for the possibility that others would be drawing the same inference.
Whether or not it’s boring is a matter of taste. But the point was to test a hypothesis, not to be interesting. Have you just subverted the point entirely; or does your claim to have sent a guess of 100 actually serve a purpose?
The “guess 2⁄3 of the average” game is quite well-studied and has been played many times in controlled environments and on the Internet. No point in running yet another simulation.
My claim serves a purpose of providing some simple mathematical exercises to all involved.
Except to potentially gain information specific to this community, which is what I assumed the point was.
(This is not to suggest that your modification is not interesting. It is. I just think it’s kind of poor form to hijack someone else’s post like this. If you wanted to play a different game, with a different point, I think the best course would have been to start your own separate game, with you taking the entries, and to have left Warrigal’s game as it was. There’s no reason we couldn’t have done both.)
If it’s any consolation, I’d assign high probability that some other merry prankster is going to intentionally screw with the results by not trying to win at all. Any future study of group rationality on LW would be advised, based on past experience, to throw out some outliers on principle.
Assuming you have actually entered a guess of 100, you may change it. If we didn’t let people change their entries, they might as well wait until the last second and then enter.
I believe the “correct” answer a is now given by the equation:
a = 2⁄3 (100 + ((n-1) a)) / n
where n is the total number of players, including cousin-it.
Except this doesn’t take into account people who didn’t see or ignored cousin-it ’s post. But if cousin-it’s guess was stated at the outset, I think the “correct” answer would be as above.
Edit: Richard Kennaway points out that this simplifies to a = 200 / (n+2)
Boring game. Let’s make it interesting! I hereby swear that I sent Warrigal a guess of 100. Use this information wisely.
I was thinking that this game would pretty much only measure (belief of) rationality, but now I see that it measures (belief of) honesty to a good degree as well. By guessing 100, one is being dishonest.
No it is not, especially since cousin_it has upfrontly told us what he values. You are assuming that everyone who submits has a utility function that highly values winning this game, which, given the comments around here, seems to not be true (or is at least widely believed to not be true).
Don’t confuse ‘has different values than I’ with ‘irrational’.
Or just plain wrong.
If we all take you at your word, that does indeed make it more interesting. If every other entrant acts rationally in choosing P, we must have P = (2/3)(PN + 100)/(N+1), if there are N other participants. This solves to give P = 200/(N+3).
But we don’t know N, we can only guess at it, or some probability distribution of N. N is at least 2, since Psy-Kosh has claimed to have entered, and anyone making this calculation must count themselves among the N. But Psy-Kosh posted before cousin_it revealed its guess of 100, so we might guess that Psy-Kosh voted 0 on the grounds given in the original post, which then modifies the calculation to give P = 200/(N+5).
But suppose of the N non-cousin_it entrants, K entered before knowing cousin_it’s entry, and all chose 0. Then we get P = 200/(N+3+2K).
Now, Aumann agreement only applies if the parties confer to honestly share their information. However, this has been framed as a competitive game, and someone who wants to be the exclusive winner would to better to avoid any such procedure, or to participate in it dishonestly.
A simpler analysis would be to point out that if Psy-Kosh voted zero (as would have been rational without cousin_it), then if everyone else votes zero, all will win except cousin_it. However, if someone votes slightly more than zero, then that one will be the exclusive winner. Someone who values an exclusive win above a tie might try to persuade everyone to vote zero and then defect.
Edit: I have entered, based on the above considerations. My entry was greater than zero.
Good, but it’s even more interesting.
a) I have successfully moved the average by ~2X compared to if I’d stayed silent.
b) If N is known and everyone acts rationally as you describe, a pair of colluding players can screw everyone over: one guesses 100, the other guesses the new average.
The combined effect is making my brain explode.
If I may attempt to contribute your brain matter’s outward velocity: When I was contemplating submitting a guess (which I didn’t do), I actually concluded (I have no evidence for this, sorry) that you in particular would probably guess 100. Had I acted on that, the effect would have been nearly the same, except for the possibility that others would be drawing the same inference.
He’s obviously lying
Whether or not it’s boring is a matter of taste. But the point was to test a hypothesis, not to be interesting. Have you just subverted the point entirely; or does your claim to have sent a guess of 100 actually serve a purpose?
The “guess 2⁄3 of the average” game is quite well-studied and has been played many times in controlled environments and on the Internet. No point in running yet another simulation.
My claim serves a purpose of providing some simple mathematical exercises to all involved.
Except to potentially gain information specific to this community, which is what I assumed the point was.
(This is not to suggest that your modification is not interesting. It is. I just think it’s kind of poor form to hijack someone else’s post like this. If you wanted to play a different game, with a different point, I think the best course would have been to start your own separate game, with you taking the entries, and to have left Warrigal’s game as it was. There’s no reason we couldn’t have done both.)
Now that you said it, I see how my comment has made the community worse off. I somehow didn’t see it then. I’m sorry.
Edit: please don’t upvote this.
If it’s any consolation, I’d assign high probability that some other merry prankster is going to intentionally screw with the results by not trying to win at all. Any future study of group rationality on LW would be advised, based on past experience, to throw out some outliers on principle.
Assuming you have actually entered a guess of 100, you may change it. If we didn’t let people change their entries, they might as well wait until the last second and then enter.
On the bright side, I suppose, you certainly have helped meet the goal of gaining information specific to this community.
I believe the “correct” answer a is now given by the equation:
a = 2⁄3 (100 + ((n-1) a)) / n
where n is the total number of players, including cousin-it.
Except this doesn’t take into account people who didn’t see or ignored cousin-it ’s post. But if cousin-it’s guess was stated at the outset, I think the “correct” answer would be as above.
Edit: Richard Kennaway points out that this simplifies to a = 200 / (n+2)