To be properly isomorphic to the Newcomb's problem, the chance of the predictor being wrong should approximate to zero.
If I thought that the chance of my friend's mother being wrong approximated to zero, I would of course choose to one-box. If I expected her to be an imperfect predictor who assumed I would behave as if I were in the real Newcomb's problem with a perfect predictor, then I would choose to two-box.
Hm, I think I still don't understand the one-box perspective, then. Are you saying that if the predictor is wrong with probability p, you would take two-boxes for high p and one box for a sufficiently small p (or just for p=0)? What changes as p shrinks?
Or what if Omega/Ann's mom is a perfect predictor, but for a random 1% of the time decides to fill the boxes as if it made the opposite prediction, just to mess with you? If you one-box for p=0, you should believe that taking one box is correct (and generates $1 million more) in 99% of cases and that two boxes is correct (and generates $1000 more) in 1% of cases. So taking one box should still have a far higher expected value. But the perfect predictor who sometimes pretends to be wrong behaves exactly the same as an imperfect predictor who is wrong 1% of the time.
You choose the boxes according to the expected value of each box choice. For a 99% accurate predictor, the expected value of one-boxing is $990,000,000 (you get a billion 99% of the time, and nothing 1% of the time,) while the expected value of two-boxing is $10,001,000 (you get a thousand 99% of the time, and one billion and one thousand 1% of the time.)
The difference between this scenario and the one you posited before, where Ann's mom makes her prediction by reading your philosophy essays, is that she's presumably predicting on the basis of how she wou...
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