Yvain comments on Quantum Russian Roulette - Less Wrong
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I vaguely remember this being discussed here before, and people deciding it wouldn't work. Before the coin-flipper is run, you have a 1/2^20 chance of seeing your number come up, whether many worlds is true or false. That means that seeing the number come up doesn't tell you anything about whether MW is true or not. It just tells you you're extremely lucky: either lucky enough that the coin-flipper got a very specific number, or lucky enough to have ended up in the very specific universe where the flipper got that number.
I don't really buy that argument. It would apply to any measurement scenario. You could say in the two-mirror experiment: "These dots on the screen don't mean a thing, we just got extremely lucky." Which is of course always a theoretical possibility.
Of course you can derive that you were extremely lucky, but also that "someone got extremely lucky" [SGEL]. If you start with some arbitrary estimates e.g. P(SWI)=0.5 and P(MWI)=0.5 and try to update P(MWI) by using Bayesian inference, you get:
By P(SGEL|SWI)=1/2^20
P(SGEL|MWI)=1
You get:
P(MWI|SGEL)=P(SGEL|MWI)P(MWI)/(P(SGEL|SWI)P(SWI)+P(SGEL|MWI)P(MWI))=
0.5/((1/2^20)0.5 + 0.5)=1/(1+1.2^20) ~ 1-1/2^20
Well, yes, but we can't peek into other Everett branches to check them for lucky people.
I don't see why you wanted to. You could only increase P(MWI) by finding there any.