I think the agent should take the bet, and the double counting is actually justified. Epistemic status: Sleep deprived.
The number of clones that end up betting along with the agent is an additional effect of its decision that EDT-with update is correctly accounting for. Since “calculator says X” is evidence that “X = true”, selecting only clones that saw “calc says X” gives you better odds. What seems like a superfluous second update is really an essential step—computing the number of clones in each branch.
Consider this modification: All N clones bet iff you do, using their own calculator to decide whether to bet on X or ¬X.
This reformulation is just the basic 0-clones problem repeated, and it recommends no bet.
if X, EVT = ¯100 = 0.99 × N winners × $10 − 0.01 × N losers × $1000 if ¬X, EVT = ¯100 = 0.99 × N winners × $10 − 0.01 × N losers × $1000
Now recall the “double count” calculation for the original problem.
if X, EVT = 9900 = 0.99 × N winners × $10 if ¬X, EVT = ¯10 = −0.01 × N losers × $1000
Notice what’s missing: The winners when ¬X and, crucially, the losers when X. This is a real improvement in value—if you’re one of the clones when X is true, there’s no longer any risk of losing money.
Yeah, I think this is right. It seems like the whole problem arises from ignoring the copies of you who see “X is false.” If your prior on X is 0.5, then really the behavior of the clones that see “X is false” should be exactly analogous to yours, and if you’re going to be a clone-altruist you should care about all the clones of you whose behavior and outcomes you can easily predict.
I should also point out that this whole setup assumes that there are 0.99N clones who see one calculator output and 0.01N clones who see the opposite, but that’s really going to depend on what exact type of multiverse you’re considering (quantum vs. inflationary vs. something else) and what type of randomness is injected into the calculator (classical or quantum). But if you include both the “X is true” and “X is false” copies then I think it ends up not mattering.
I think the agent should take the bet, and the double counting is actually justified. Epistemic status: Sleep deprived.
The number of clones that end up betting along with the agent is an additional effect of its decision that EDT-with update is correctly accounting for. Since “calculator says X” is evidence that “X = true”, selecting only clones that saw “calc says X” gives you better odds. What seems like a superfluous second update is really an essential step—computing the number of clones in each branch.
Consider this modification: All N clones bet iff you do, using their own calculator to decide whether to bet on X or ¬X.
This reformulation is just the basic 0-clones problem repeated, and it recommends no bet.
Now recall the “double count” calculation for the original problem.
Notice what’s missing: The winners when ¬X and, crucially, the losers when X. This is a real improvement in value—if you’re one of the clones when X is true, there’s no longer any risk of losing money.
Yeah, I think this is right. It seems like the whole problem arises from ignoring the copies of you who see “X is false.” If your prior on X is 0.5, then really the behavior of the clones that see “X is false” should be exactly analogous to yours, and if you’re going to be a clone-altruist you should care about all the clones of you whose behavior and outcomes you can easily predict.
I should also point out that this whole setup assumes that there are 0.99N clones who see one calculator output and 0.01N clones who see the opposite, but that’s really going to depend on what exact type of multiverse you’re considering (quantum vs. inflationary vs. something else) and what type of randomness is injected into the calculator (classical or quantum). But if you include both the “X is true” and “X is false” copies then I think it ends up not mattering.