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?
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?