First, I’m gonna clarify some terms to make this more precise. Let Y be a person psychologically continuous with your present self. P(there is some Y that observes surviving a suicide attempt|Quantum immortality) = 1. Note MWI != QI. But QI entails MWI. P(there is some Y that observes surviving a suicide attempt| ~QI) = p.
It follows from this that P(~(there is some Y that observes surviving a suicide attempt)|~QI) = 1-p.
I don’t see a confusion of levels (whatever that means).
I still see a problem here. Substitute quantum suicide → quantum coinflip, and surviving a suicide attempt → observing the coin turning up heads.
Now we have P(there is some Y that observes coin falling heads|MWI) = 1, and P(there is some Y that observes coin falling heads|Copenhagen) = p.
So any specific outcome of a quantum event would be evidence in favor of MWI.
I think that works actually. If you observe 30 quantum heads in a row you have strong evidence in favor of MWI. The quantum suicide thing is just a way of increasing the proportion of future you’s that have this information.
If you observe 30 quantum heads in a row you have strong evidence in favor of MWI.
But then if I observed any string of 30 outcomes I would have strong evidence for MWI (if the coin is fair, “p” for any specific string would be 2^-30).
Sorry, now I have no idea what we’re talking about. If your experiment involves killing yourself after seeing the wrong string, this is close to the standard quantum suicide.
If not, I would have to see the probabilities to understand. My analysis is like this: P(I observe string S | MWI) = P(I observe string S | Copenhagen) = 2^-30, regardless of whether the string S is specified beforehand or not. MWI doesn’t mean that my next Everett branch must be S because I say so.
The reason why this doesn’t work (for coins) is that (when MWI is true) A=”my observation is heads” implies B=”some Y observes heads”, but not the other way around. So P(B|A)=1, but P(A|B) = p, and after plugging that into the Bayes formula we have P(MWI|A) = P(Copenhagen|A).
Can you translate that to the quantum suicide case?
I still see a problem here. Substitute quantum suicide → quantum coinflip, and surviving a suicide attempt → observing the coin turning up heads.
Now we have P(there is some Y that observes coin falling heads|MWI) = 1, and P(there is some Y that observes coin falling heads|Copenhagen) = p.
So any specific outcome of a quantum event would be evidence in favor of MWI.
I think that works actually. If you observe 30 quantum heads in a row you have strong evidence in favor of MWI. The quantum suicide thing is just a way of increasing the proportion of future you’s that have this information.
But then if I observed any string of 30 outcomes I would have strong evidence for MWI (if the coin is fair, “p” for any specific string would be 2^-30).
You have to specify a particular string to look for before you do the experiment.
Sorry, now I have no idea what we’re talking about. If your experiment involves killing yourself after seeing the wrong string, this is close to the standard quantum suicide.
If not, I would have to see the probabilities to understand. My analysis is like this: P(I observe string S | MWI) = P(I observe string S | Copenhagen) = 2^-30, regardless of whether the string S is specified beforehand or not. MWI doesn’t mean that my next Everett branch must be S because I say so.
The reason why this doesn’t work (for coins) is that (when MWI is true) A=”my observation is heads” implies B=”some Y observes heads”, but not the other way around. So P(B|A)=1, but P(A|B) = p, and after plugging that into the Bayes formula we have P(MWI|A) = P(Copenhagen|A).
Can you translate that to the quantum suicide case?