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Jordan comments on Open Thread: February 2010 - Less Wrong
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I'm not sure why you take Po = Pb. If Omega is trying to maximize his chance of predicting correctly then he'll take Po = 1 if Pb > 1/2 and Pb = 0 if Pb < 1/2. Then, assuming A > B / 2, the optimal choice is Po = 1/2.
Actually, if Omega behaves this way there is a jump discontinuity in expected value at Po=1/2. We can move the optimum away from the discontinuity by postulating there is some degree of imprecision in our ability to choose a quantum coin with the desired characteristic. Maybe when we try to pick a coin with bias Po we end up with a coin with bias Po+e, where e is an error chosen from a uniform distribution over [Po-E, Po+E]. The optimal choice of Po is now 1/2 + 2E, assuming A > 2EB, which is the case for sufficiently small E (E < 1/4 suffices). The expected payoff is now robust (continuous) to small perturbations in our choice of Po.
A good point.
Your solution does have Omega maximize right answers. My solution works if Omega wants the "correct" result summed over all Everett branches: for every you that 2-boxes, there exists an empty box A, even if it doesn't usually go to the 2-boxer.
Both answers are correct, but for different problems. The "classical" Newcomb's problem is unphysical, just as byrnema initially described. A "Quantum Newcomb's problem" requires specifying how Omega deals with quantum uncertainty.
Interesting. Since the spirit of Newcomb's problem depends on 1-boxing have a higher payoff, I think it makes sense to additionally postulate your solution to quantum uncertainty, as it maintains the same maximizer. That's so even if the Everett interpretation of QM is wrong.