...I don't think this is a very wise offer to make on the Internet unless the "coin" is somewhere you can both see it.

When I heard about Yvain's PD contest, I flipped a coin. I vowed that if it came up heads, I would Paypal the winner $200 (on top of their winnings), and if it came up tails I would ask them for the prize money they won.

It came up tails. YOUR MOVE.

(No, not really. But somebody here SHOULD have made such a commitment.)

The second word in the winning secret phrase is pony (chosen because you can't spell the former without the latter); I'll accept the prize money via PayPal to main att zackmdavis daht net.

(As I recall, I chose to Defect after looking at the output of one call to Python's random.random() and seeing a high number, probably point-eight-something. But I shouldn't get credit for following my proposed procedure (which turned out to be wrong anyway) because I don't remember deciding beforehand that I was definitely using a "result > 0.8 means Defect" convention (when "result < 0.2 means Defect" is just as natural). I think I would have chosen Cooperate if the random number had come up less than 0.8, but I haven't actually observed the nearby possible world where it did, so it's at least possible that I was rationalizing.)

(Also, I'm sorry for being bad at reading; I don't actually think there are seven hundred trillion people in Europe.)

(donated $1,500)

Speaking of the sidebar, is there any way to make it optional whether a post goes up there? Maybe by including [LW] in the title or something? I enjoy blogging about rationality, but I also enjoy blogging about random things that go on in my personal life, and it's kind of embarrassing to have those show up on the LW sidebar.

Fanfanfiction idea: during the months of CelestAI's world takeover, a foresightful Hasbro VP races to get My Little Pony Equestria Girls made and released so that humans will be canon and people don't have to be ponies for eternity. Title: Friendship Is Optional.

(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))

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.

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.

Rot 13'd, but I'm confused - thought it would be taken for granted that this thread would contain lots of discussion of the survey, so you shouldn't read the thread before taking the survey.

conflicts with the Bay Area Solstice event

For the Prize Question, you should use a random number generator and cooperate with probability 0.8. Why? Suppose that the fraction of survey-takers that cooperate is p. Then the value of the prize will be proportional to p and there will be p + 4(1 - p) raffle entries. The expected value of Cooperating is p/(p + 4(1-p)) and the expected value of Defecting is 4(1-p)/(p + 4(1-p)). In equilibrium, these must be the same: if one choice were more profitable than the other, then people would switch until this was no longer the case. Thus p = 4(1 - p) and thus p = 4/5.

Addendum 29 November: Actually, this is wrong; see ensuing discussion.