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gregconen comments on Open Thread: February 2010 - Less Wrong
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The math is actually quite straight-forward, if anyone cares to see it. Consider a generalized Newcomb's problem. Box A either contains $A or nothing, while box B contains $B (obviously A>B, or there is no actual problem). Let Pb the probability that you 1-box. Let Po be the probability that Omega fills box A (note that only quantum randomness counts, here. If you decide by a "random" but deterministic process, Omega knows how it turns out, even if you don't, so Pb=0 or 1). Let F be your expected return.
Regardless of what Omega does, you collect the contents of box A, and have a (1-Pb) probability of collecting the contents of box B. F(Po=1)= A + (1-Pb)B
F(Po=0)=(1-Pb)B
For the non-degenerate cases, these add together as expected. F(Po, Pb) = Po(A + (1-Pb)B) + (1-Po)[(1-Pb)B]
Suppose Po = Pb := P
F(P) = P(A + (1-P)B) + [(1-P)^2] B
=P(A + B - PB) + (1-2P+P^2) B
=PA + PB - (P^2)B + B - 2PB + (P^2)B
=PA + PB + B - 2PB
=B + P(A-B)
If A > B, F(P) is monotonically increasing, so P = 1 is the gives maximum return. If A<B, P=0 is the maximum (I hope it's obvious to everyone that if box B has MORE money than a full box A, 2-boxing is ideal).
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.