If P(I survive|MWI) = 1, and P(I survive|Copenhagen) = p, then what is the rest of that probability mass in Copenhagen interpretation?
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).
ETA: And, of course, the problem with “anthropic probabilities” gets even harder when you consider copies and merging, simulations, Tegmark level 4, and Boltzmann brains (The Anthropic Trilemma). I’m not sure if there even is a general solution. But I strongly suspect that “you can prove MWI by quantum suicide” is an incorrect usage of probabilities.
I don’t know if this is the point you meant to make but the existence of these other hypotheses that could imply anthropic immortality definitely does get in the way of providing evidence in favor of Many Worlds through suicide. Surviving increases the probability of all of those hypotheses (to different extents but not really enough to distinguish them).
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?
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 don’t know if this is the point you meant to make but the existence of these other hypotheses that could imply anthropic immortality definitely does get in the way of providing evidence in favor of Many Worlds through suicide. Surviving increases the probability of all of those hypotheses (to different extents but not really enough to distinguish them).
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?