On Thursday, 14 July 2016, Bruce Kellett <bhkell...@optusnet.com.au> wrote:
> On 13/07/2016 11:36 pm, Bruno Marchal wrote: > > On 11 Jul 2016, at 13:49, Bruce Kellett wrote: > > On 11/07/2016 9:31 pm, Bruno Marchal wrote: > > > *Holiday Exercise:* > > A guy undergoes the Washington Moscow duplication, starting again from > Helsinki. > Then in Moscow, but not in Washington, he (the one in Moscow of course) > undergoes a similar Sidney-Beijing duplication. > > I write P(H->M) the probability in H to get M. > > In Helsinki, he tries to evaluate his chance to get Sidney. > > With one reasoning, he (the H-guy) thinks that P(H-M) = 1/2, and that > P(M-S) = 1/2, and so conclude (multiplication of independent probability) > that P(H-S) = 1/2 * 1/2 = 1/4. > > But with another reasoning, he thinks that the duplications give globally > a triplication, leading eventually to a copy in W, a copy in S and a copy > in B, and so, directly conclude P(H-S) = 1/3. > > So, is it 1/4 or 1/3 ? > > > Neither. The probability that the guy starting from Helsinki gets to > Sydney is unity. > > > Try to convince the guy who gets to Beijing, or the one who stayed in > Washington. He knows that the probability evaluated in Helsinki was not > P(Sidney) = 1. > > > We start with John Clark in Helsinki, so P(JC ~ H) = 1. By construction, > after the duplication and so on, P(JC ~ W) = P(JC ~ S) = P(JC ~ B) = 1. (I > use '~' as a shorthand for 'in' or 'sees'.) JC in Helsinki knows the > protocol, so he can easily see that these are the correct probabilities. > So, as I said, the probability that the guy starting from Helsinki gets to > Sydney is unity. Any other interpretation of this scenario involves an > implicit appeal to dualism -- there is "one true JC" that goes through > these duplications, and he can only ever end up in just one place. > That is the source of the effect being discussed: JC and all of us feel that personal identity only transfers to the one true JC, despite the evidence of all the other clearly visible JC's. This is because "I" am the person who has my experiences, and I am not telepathically linked to my copies. > As John Clark has correctly pointed out, your intuition and formalism > simply does not work in the presence of person-duplicating machines. There > is no single 1p view -- there are three possible 1p views in the > triplication scenario. So, again, John Clark is right when he says that JC > ~ H will see three cities (W, S, and B) after the experiment is completed. > If, as you claim, he will see only one city, you have to have some dualist > 'nut or core' that survives in only one of your copies. > > Of course, as I said some time ago, the easiest resolution of you logical > conundrums is that JC ~ H does not survive, and that there are three new > persons, one in each city, so the probability that JC in H will see Sydney > is exactly zero. > > Looking at the more realistic quantum realization of this triplication > scenario, we can formulate that as follows. We prepare a spin-half atom > with spin along the x-axis, then pass it through an S-G magnet oriented > along the y-axis, getting two possibilities, which we can call up and down. > We then take the up channel and pass that through a further S-G in the > x-direction, getting two further possibilities of left or right. > > Let us perform this experiment many times and count the number of > particles in each of the three possible final states (down, left, and > right). If this is a real laboratory experiment, in which detection of a > particle in any channel leads to irreversible decoherence and the formation > of a separate world containing just that result, we will find approximately > half the particles end up in the down state, and approximately a quarter in > each of the left and right states. This gives the most reliable estimate of > the real probabilities for the outcome from the given initial state. > > If you take the MWI view, then you get one down, one left, and one right > in every run of the experiment, so the probability for each outcome is > unity. In order to get probability of 1/4 for left, say, you have to detect > the absence of a particle in the down state (so that the particle is > certainly in the up state) for which the probability is 1/2. > > Actually, your preferred answer -- that the probability P(H->S) = 1/3, is > possible only in a fully dualist model. You are essentially claiming that > as the scenario puts John Clark's in all three cities, it is purely a > random chance that selects one of them to be the "true" John Clark -- a > dualist "core" is assigned to one of these copies purely by chance. > > > This is the problem with probabilities in the MWI -- how do you interpret > probabilities when all possible outcomes occur with probability one? > > > The probabilities concern the relative first person experiences. > Computationalism guaranties that there will be only one outcome. > > > You are simply talking nonsense, here. "Relative first person > experiences"? Relative to what? The scenario, and computationalism > guarantees that there will be all three outcomes. If there is only one "1p" > experience, then that can only be chosen dualistically. If there is > duplication (triplication) then your intuitions break down. I have to say > it -- John Clark has been right all along. > > Bruce > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to everything-list+unsubscr...@googlegroups.com > <javascript:_e(%7B%7D,'cvml','everything-list%2bunsubscr...@googlegroups.com');> > . > To post to this group, send email to everything-list@googlegroups.com > <javascript:_e(%7B%7D,'cvml','everything-list@googlegroups.com');>. > Visit this group at https://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/d/optout. > -- Stathis Papaioannou -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
