I’ve been fiddling around with this in my head. I arrived at this argument for one-boxing:
Let us suppose a Rule, that we shall call W:
FAITHFULLY FOLLOW THE RULE THAT, IF FOLLOWED FAITHFULLY, WILL ALWAYS OFFER THE GREATEST CHANCE OF THE GREATEST UTILITY
To prove W one boxes, let us list all logical possibilities, which we’ll call W1 W2 and W3: W1 always one-boxing W2 always two boxing, and W3 sometimes one-boxing and sometimes two boxing. Otherwise, all of these rules are identical in every way, and identical to W in every way. Imagining that we’re Omega, we’d obviously place nothing in the box of the agent which follows W2, since it knows that agent would two-box.. Since this limits the utility gained, W2 is not W. W3 is a bit trickier, but a variant of W3 which two-boxes most of the time will probably not be favoured by Omega, since this would reduce his chance of being correct in his prediction. This reduces the chance of getting the greatest utility by however much, and thus, disqualifies all close to W2 variants of W3. A perfect W1 would guarantee that the box would contain 1,000,000 dollars, since Omega would get it’s prediction wrong in not rewarding an agent who one-boxes. However, this rule GUARANTEES not getting the 1,001,000 dollars, and therefore is sub -optimal. Because of Omega’s optimization, there is no such rule in which that is the most likely option, but if there is such a rule in which this is second-most-likely, that would probably be W. In any case, W favours B over A.
I was going to argue that W is more rational than a hypothetical rule Z which I think is what makes most two-boxers two-box, but maybe I’ll do that later, when I’m more sure I have time.
I’ve been fiddling around with this in my head. I arrived at this argument for one-boxing: Let us suppose a Rule, that we shall call W: FAITHFULLY FOLLOW THE RULE THAT, IF FOLLOWED FAITHFULLY, WILL ALWAYS OFFER THE GREATEST CHANCE OF THE GREATEST UTILITY To prove W one boxes, let us list all logical possibilities, which we’ll call W1 W2 and W3: W1 always one-boxing W2 always two boxing, and W3 sometimes one-boxing and sometimes two boxing. Otherwise, all of these rules are identical in every way, and identical to W in every way. Imagining that we’re Omega, we’d obviously place nothing in the box of the agent which follows W2, since it knows that agent would two-box.. Since this limits the utility gained, W2 is not W. W3 is a bit trickier, but a variant of W3 which two-boxes most of the time will probably not be favoured by Omega, since this would reduce his chance of being correct in his prediction. This reduces the chance of getting the greatest utility by however much, and thus, disqualifies all close to W2 variants of W3. A perfect W1 would guarantee that the box would contain 1,000,000 dollars, since Omega would get it’s prediction wrong in not rewarding an agent who one-boxes. However, this rule GUARANTEES not getting the 1,001,000 dollars, and therefore is sub -optimal. Because of Omega’s optimization, there is no such rule in which that is the most likely option, but if there is such a rule in which this is second-most-likely, that would probably be W. In any case, W favours B over A. I was going to argue that W is more rational than a hypothetical rule Z which I think is what makes most two-boxers two-box, but maybe I’ll do that later, when I’m more sure I have time.