comment by Manfred ·
2013-09-10T02:48:28.293Z · LW(p) · GW(p)
I would claim that the correct probability to hold is somewhere around 0.999 in favor of composite if you take both boxes [...]
EDIT: Looks like I was right about probabilities, but too hasty about thinking that meant you should two-box. Omega can be malicious:
Suppose we do this Primecomb + lottery experiment a jillion times. What algorithm maximizes payout over those jillion times?
One-boxing sure seems like a good plan - usually the lottery will pay out, sometimes not, but no biggie since you can't affect it. And since there aren't that many prime numbers, the lottery and the box don't share numbers very often, though when they do you always lose the lottery.
But suppose you decide to two-box every time you see the lottery and the box have the same number. Now Omega's action is undefined - if the lottery number is composite Omega can basically choose whether you're to two-box or one-box. If we think in terms of the jillion trials of the same game, one-boxing would still be better, since when Omega undefinedly decides to make you two-box, you were going to win the lottery anyhow and could have gotten more money if Omega had decided to make you one-box.
However, if you two-box every time the numbers are the same, every time the numbers are the same you'll win the lottery. So if you see the numbers the same, it certainly sounds reasonable to try to be part of the lottery-winning group, right?
Hold up though. Suppose we get to program Omega a little bit. One version we make nice - call it Nicemega. It never makes the numbers be the same. so I always one-box and get lots of money. Another version we make mean - Meanmega. It chooses the number on the box to minimize the money it has to pay out. If you two-box when the numbers are the same, it makes you two-box whenever it can. If you are willing to two-box when the numbers are the same and you start seeing a lot of same numbers, you should switch plans, because you're probably getting Meanmega'd! So why should you two-box when you see the numbers the same, if it just means you're getting Meanmega'd?
In other words, the optimal strategy really can be globally optimal, even though sometimes it requires you to take locally bad actions. Seem a little familiar?
My favorite example for this is the Unexpected Hanging paradox.
A judge tells a condemned prisoner that he will be hanged at noon on one weekday in the following week but that the execution will be a surprise to the prisoner. He will not know the day of the hanging until the executioner knocks on his cell door at noon that day.
Having reflected on his sentence, the prisoner draws the conclusion that he will escape from the hanging. His reasoning is in several parts. He begins by concluding that the "surprise hanging" can't be on Friday, as if he hasn't been hanged by Thursday, there is only one day left - and so it won't be a surprise if he's hanged on Friday. Since the judge's sentence stipulated that the hanging would be a surprise to him, he concludes it cannot occur on Friday.
He then reasons that the surprise hanging cannot be on Thursday either, because Friday has already been eliminated and if he hasn't been hanged by Wednesday night, the hanging must occur on Thursday, making a Thursday hanging not a surprise either. By similar reasoning he concludes that the hanging can also not occur on Wednesday, Tuesday or Monday. Joyfully he retires to his cell confident that the hanging will not occur at all.
The next week, the executioner knocks on the prisoner's door at noon on Wednesday — which, despite all the above, was an utter surprise to him. Everything the judge said came true.
The question is - where was the flaw in the prisoner's logic? The answer: the only flaw is that the judge really was prepared to hang the man on Friday, even though by then it's not a surprise. Every prisoner you hang without surprise on Friday buys you four to hang with genuine surprise on the other days of the week. If you're unwilling to pay in the coin of failed surprises, you cannot buy genuine ones. If you're the judge, you roll a five-sided die and hang the prisoner on the corresponding day. If you roll Friday, then you damn well hang them on Friday to no surprise, or else you're not even really trying.
A similar logic leads to the rejection of two-boxing when you see that the numbers are the same. If you aren't willing to one-box when the lottery has rolled a Friday, er, a prime number, (and Omega has decided to be be a jerk and rub it in) then you're not ever actually one-boxing.
Replies from: Nate_Gabriel
↑ comment by Nate_Gabriel ·
2013-09-10T17:26:09.961Z · LW(p) · GW(p)
This post almost convinced me.
I was thinking about it in terms of a similar algorithm, "one-box unless the number is obviously composite." Your argument convinced me that you should probably one-box even if Omega's number is, say, six. (Even leaving aside the fact that I'd probably mess up more than one in a thousand questions that easy.) For the reasons you said, I tentatively think that this algorithm is not actually one-boxing and is suboptimal.
But the algorithm "one-box unless the numbers are the same" is different. If you were playing the regular Newcomb game, and someone credibly offered you $2M if you two-box, you'd take it. More to the point, you presumably agree that you should take it. If so, you are now operating on an algorithm of "one-box unless someone offers you more money."
In this case, it's just like they are offering you more money: if you two-box, it's composite 99.9% of the time, and you get $2M.
The one thing we know about Omega is that it picks composites iff it predicts you will two-box. In the Meanmega example, it picks the numbers so that you two-box whenever it can, that just means whenever the lottery number is composite. So in all those cases, you get $2M. That you would have gotten anyway. Huh. And $1M from one-boxing if the lottery number is prime. Whereas, if you one-box, you get $1M 99.9% of the time, plus a lot of money from the lottery anyway. OK, so you're completely right. I might have to think about this more.
Assuming Manfred is completely right, how many non-identical numbers should it take before you decide you're not dealing with Meanmega and can start two-boxing when they're the same?