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Eliezer_Yudkowsky comments on Open Thread: February 2010 - Less Wrong
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So long as you make your Newcomb's choice for what seem like good reasons rather than by flipping a quantum coin, it is likely that very many of you will pick the same good reasons, and that Omega can easily achieve 99% or higher accuracy. I would expect almost no Eliezer Yudkowskys to two-box - if Robin Hanson is right about mangled worlds and there's a cutoff for worlds of very small amplitude, possibly none of me. Remember, quantum branching does not correspond to high-level decisionmaking.
Yes, most Eliezer Yudkowskys will 1-box. And most byrnemas too. But the new twist (new for me, anyway) is that the Eliezer's that two-box are the ones that really win, as rare as they are.
The one who wins or loses is the one who makes the decision. You might as well say that if someone buys a quantum lottery ticket, the one who really wins is the future self who wins the lottery a few days later; but actually, the one who buys the lottery ticket loses.
The slight quantum chance that EY will 2-box causes the sum of EYs to lose, relative to a perfect 1-boxer, assuming Omega correctly predicts that chance and randomly fills boxes accordingly. The precise Everett branches where EY 2-boxes and where EY loses are generally different, but the higher the probability that he 1-boxes, the higher his expected value is.
And, also, we define winning as winning on average. A person can get lucky and win the lottery -- doesn't mean that person was rational to play the lottery.
Interestingly, I worked through the math once to see if you could improve on committed 1-boxing by using a strategy of quantum randomness. Assuming Omega fills the boxes such that P(box A has $)=P(1-box), P(1-box)=1 is the optimal solution.
Interesting. I was idly wondering about that. Along somewhat different lines:
I've decided that I am a one-boxer,and I will one box. With the following caveat: at the moment of decision, I will look for an anomaly with virtual zero probability. A star streaks across the sky and fuses with another one. Someone spills a glass of milk and halfway towards the ground, the milk rises up and fills itself back into the glass. If this happens, I will 2-box.
Winning the extra amount in this way in a handful of worlds won't do anything to my average winnings-- it won't even increase it by epsilon. However, it could make a difference if something really important is at stake, where I would want to secure the chance that it happens one time in the whole universe.
Let p be the probability that you 2-box, and suppose (as Greg said) that Omega lets P(box A empty) = p with its decision being independent of yours. It sounds like you're saying you only care about the frequency with which you get the maximal reward. This is P(you 2-box)*P(box A full) = p(1-p) which is maximized by p=0.5, not by p infinitesimally small.
Why is this comment being down-voted? I thought it was rather clever to use Omega's one weak spot -- quantum uncertainty -- to optimize your winnings even over a set with measure zero.
Because Omega is going to know what triggers you would use for anomalies. A star streaking across the sky is easy to see coming if you know the current state of the universe. As such, Omega would know you are about to two-box even though you are currently planning to one-box.
When the star streaks across the sky, you think, "Ohmigosh! It happened! I'm about to get rich!" Then you open the boxes and get $1000.
Essentially, it boils down to this: If you can predict a scenario where you will two-box instead of one-box than Omega can as well.
The idea of flipping quantum coins is more fool proof. The idea of stars streaking or milk unspilling is only hard for us to see coming. Not to mention it will probably trigger all sorts of biases when you start looking for ways to cheat the system.
Note: I am not up to speed on quantum mechanics. I could be off on a few things here.
OK, right: looking for a merging of stars would be a terrible anomaly to use because that's probably classical mechanics and Omega-predictable. The milk unspilling would still be a good example, because Omega can't see it coming either. (He can accurately predict that I will two-box in this case, but he can't predict that the milk will unspill.)
I would have to be very careful that the anomaly I use is really not predictable. For example, I screwed up with the streaking star. I was already reluctant to trust flipping quantum coins, whatever those are. They would need to be flipped or simulated by some mechanical device and may have all kinds of systematic biases and impracticalities if you are actually trying to flip 10^23^23 coins.
Without having plenty of time to think about it, and say, some physicists advising me, it would probably be wise for me to just one-box.
I didn't down vote but I confess I don't really know what you're talking about in that comment. Why would you two box in that case? What really important thing is at stake? I don't get it.
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.