= 27d567bc2d48d9f40e9ccc20ea370b15)
Zack_M_Davis comments on 2013 Less Wrong Census/Survey - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (616)
The expected value of defecting is 4p/(p + 4(1-p), to within one part in the number of survey takers. Whether or not you defect makes no difference as to the proportion of people who defect.
The solution is to determine how likely it is that a random participant is going to defect, conditional on your choice of cooperate or defect. If you're playing with a total of N copies of yourself, you cooperate and get the maximal payout ($60/N). If you're playing against cooperate bots, you defect and get $60*4N/(N-1).
We can generalize this to partial levels. If you play with D defectors and C cooperators whose opinion you can't change, and X people who will cooperate when you cooperate (and defect when you defect), then the payouts are as thus:
C: (C + X)/(C + D + X) D: 4(C /(C + D + X)
You can solve for the break even point by setting C + X = 4 * C
So the answer is that you should defect, unless you think that for every person who is going to cooperate no matter what, there are at least three people who are thinking with similar enough reasoning to come up with the same answer you come up with (regardless of what answer that is).
I think you've got the denominators of your fractions wrong. There are 4 raffle tickets for everyone who defects. I get the values
C: (C + X)/(C + 4D + X) D: 4(C /(C + 4D + 4X)
which solves to a horrible quadratic surd.
If we wanted to we could combine your method with Zack's and assume that C people cooperate, D defect and X make the same choice I do, which is to cooperate with probability p. I think this gets kinda ugly though.
The fractions I wrote are payout * number of tickets, not the chance of winning. But you do have a point: changing many people from cooperate to defect does dilute the total pool of tickets, and not be an unnoticeable amount.
The corrected answer is Payout * Chance to win, which is:
C: (C + X)/(C + D + X) * (1 / (C + 4D + X)
D: (C/(C+D+C)) * (4 / (C + 4D + 4X))
And you don't want to combine my method with Zack's. You don't want a probabilistic strategy - you want to figure out what your beliefs are as far as "how many people do I expect to be in categories C, X, and D". Given your beliefs about how your choices affect others, there's exactly one right choice.
(By the way, the numbers I gave are the same as the ones you gave, only I cancelled a common factor of (C+D+X))
I think that your "one right choice" might sometimes be a probabilistic one. To make this more obvious, consider a game where the value of the prize is maximal when exactly half of the participants choose C, and the value goes down as the proportion gets further from a half (and any of the participants is equally likely to win the prize). Then I think it's obvious that the correct strategy is to estimate C, D, and X as before, and then cooperate with probability p so that C+pX=D+(1-p)X. Then because everyone else in X acts as you do you'll end up with exactly half the people choosing C, which is what you want.
Note that even some of the people in X who you are "acausally controlling" still end up choosing a different option from you (assuming that your random number generators are independent). This allows you to exactly optimise the proportion of people who choose C, which is what makes the strategy work.
I think the same thing applies in Yvain's game. In particular, if we thought that C=D=0 then I think that Zack's analysis is exactly correct (although I wouldn't have used exactly the same words as he does).
EDIT: I retract the last sentence. Zack's calculation isn't what you want to do even in the C=D=0 case. In that case I endorse cooperating with p=1. But I still think that mixed strategies are best in some of the cases with C or D non-zero. In particular what about the case with D=0 but C=X? Then I reckon you should pick C with p=0.724.
I think this is it. Suppose there are C CooperateBots, D DefectBots, and X players who Cooperate with probability p. The expected utility of the probabilistic strategy is (proportional to) (p(C + pX) + 4(1-p)(C + pX))/(C + 4D + pX + 4(1-p)X). Then (he said, consulting his computer algebra system) if C/X < 1/3 then p = 1 (Cooperate), if C/X > 3 then p = 0 (Defect), and p assumes intermediate values if 1/3 < C/X < 3 (including 0.7239 if C/X = 1, as you mention).