On Sat, Sep 5, 2020 at 2:42 PM 'Brent Meeker' via Everything List < everything-list@googlegroups.com> wrote:
> On 9/4/2020 7:02 PM, Bruce Kellett wrote: > > On Sat, Sep 5, 2020 at 11:29 AM 'Brent Meeker' via Everything List < > <everything-list@googlegroups.com>everything-list@googlegroups.com> wrote: > >> >> But the theory isn't about the probability of a specific sequence, it's >> about the probability of |up> vs |down> in the sequence without regard for >> order. So there will, if the theory is correct, be many more sequences >> with a frequency of |up> near some theoretically computed proportion |a|^2 >> than sequences not near this proportion. >> > > > The theory is about the probabilitiies of observations. The observation in > question here is a sequence of |up> / |down> results, given that the > probability for each individual outcome is 0.5. If the theory cannot give a > probability for the sequence, > > > It can. But QM only predicts the p=0.5. To have a prediction for a > specific sequence HHTTHHHTTHTHTH... you need extra assumptions about > indenpendence. > Sure. And independence of the sequential observations is clearly implied by the set up. > And given those assumptions your theory will be contradicted with near > certainty. > Why? > Which is why I say the test of QM is whether p=0.5 is consistent with the > observed sequence in the sense of predicting the relative frequency of H > and T, not in the sense of predicting HHTTHHHTTHTHTH... > I am not attempting to predict a particular sequence. All that I have said is that the probability of any such sequence in N independent trials is 1/2^N. And that is simple probability theory, which cannot be denied. then multiply the probabilities for each particular result in your sequence > of measurements. The number of sequences with particular proportions of up > or down results is irrelevant for this calculation. > > Again, you are just attempting to divert attention from the obvious result > that the Born rule calculation gives a different probability than expected > when every outcome occurs for each measurement. In the Everett case, every > possible sequence necessarily occurs. This does not happen in the genuine > stochastic case, where only one (random) sequence is produced. > > > In the Everett theory a measurement of spin up for a particle prepared in > spin x results in two outcomes...only one is observed. If that is enough to > dismiss Everett then all the this discussion of probability and the Born > rule is irrelevant. > I have no idea what you are talking about! Nothing like that was ever suggested. Everett predicts that in such a measurement, both outcomes obtain -- in separate branches. But the probability of this is one. Repeat N times. N time one is still just one. There is nothing more to it than that. I think you are being desperate in your attempts to play 'advocatus diaboli'. The point is that the Born rule is inconsistent with Everett. 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. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLRcLwLQa2SLzViEqkZX6nU1%3Dp6SXUd7WA0nuzaiyOw0yg%40mail.gmail.com.
