OK. The way I’ve understood the problem with Omega is that Omega is a perfect predictor so you have 2 options and 2 outcomes:
you two box --> you get $2,000 ($1000 in each box)
you one box --> you get 1M ($1M in one box, $1000 in the second box)
If Omega is not a perfect predictor, it’s possible that you two box and you get 1,001,000. (Omega incorrectly predicted you’d one box.)
However, if you are likely to 2box using this reasoning, Omega will adjust his prediction accordingly (and will even reduce your winnings when you do 1box—so that you can’t beat him).
My solution was to 1box almost always—so that Omega predicts you will one box, but then ‘cheat’ and 2-box almost never (but sometimes). According to Greg, your ‘sometimes’ has to be over a set of measure 0, any larger than that and you’ll be penalized due to Omega’s arithmetic.
What really important thing is at stake?
Nothing—if only an extra thousand is at stake, I probably wouldn’t even bother with my quantum caveat. One million dollars would be great anyway. But I can imagine an unfriendly Omega giving me choices where I would really want to have both boxes maximally filled … and then I’ll have to realize (rationally) that I must almost always 1 box, but I can get away with 2-boxing a handful of times. The problem with a handful, is that how does a subjective observer choose something so rarely? They must identify an appropriately rare quantum event.
So this job could even be accomplished by flipping a quantum coin 10000 times and only two-boxing when they come up tails each time. You’re just looking for a decision mechanism that only applies in a handful of branches.
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.
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.
OK. The way I’ve understood the problem with Omega is that Omega is a perfect predictor so you have 2 options and 2 outcomes:
you two box --> you get $2,000 ($1000 in each box)
you one box --> you get 1M ($1M in one box, $1000 in the second box)
If Omega is not a perfect predictor, it’s possible that you two box and you get 1,001,000. (Omega incorrectly predicted you’d one box.)
However, if you are likely to 2box using this reasoning, Omega will adjust his prediction accordingly (and will even reduce your winnings when you do 1box—so that you can’t beat him).
My solution was to 1box almost always—so that Omega predicts you will one box, but then ‘cheat’ and 2-box almost never (but sometimes). According to Greg, your ‘sometimes’ has to be over a set of measure 0, any larger than that and you’ll be penalized due to Omega’s arithmetic.
Nothing—if only an extra thousand is at stake, I probably wouldn’t even bother with my quantum caveat. One million dollars would be great anyway. But I can imagine an unfriendly Omega giving me choices where I would really want to have both boxes maximally filled … and then I’ll have to realize (rationally) that I must almost always 1 box, but I can get away with 2-boxing a handful of times. The problem with a handful, is that how does a subjective observer choose something so rarely? They must identify an appropriately rare quantum event.
So this job could even be accomplished by flipping a quantum coin 10000 times and only two-boxing when they come up tails each time. You’re just looking for a decision mechanism that only applies in a handful of branches.
Yes, exactly.
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.