On Fri, Mar 6, 2020 at 11:33 AM Bruce Kellett <[email protected]> wrote:
> On Fri, Mar 6, 2020 at 11:08 AM 'Brent Meeker' via Everything List < > [email protected]> wrote: > >> On 3/5/2020 3:33 PM, Bruce Kellett wrote: >> >> No, it doesn't. Just think about what each observer sees from within his >> branch. >> >> >> It's what an observer has seen when he calculates the statistics after N >> trials. If a>b there will be proportionately more observers who saw more >> 0s than those who saw more 1s. Suppose that a2=2/3 and b2=1/3. Then at >> each measurement split there will be two observers who see 0 and one who >> sees 1. So after N trials there will be N^3 observers and most of them >> will have seen approximately twice as many 0s as 1s. >> > > > From within any branch the observer is unaware of other branches, so he > cannot see these weights. His statistics will depend only on the results on > his branch. In order for multiple branches to count as probabilities, you > have to appeal to some Self-Selecting-Assumption (SSA) in the 3p sense: you > have to consider that the observer self-selects at randem from the set of > all observers. Then, since there are more branches according to the > weights, the probability that the randomly selected observer will see a > branch that is multiplied over the ensemble will depend on the number of > branches with that exact sequence. But this is not how it works in > practise, because each observer can ever only see data within his branch, > even if that observer is selected at random from among all observers, he > will calculate statistics that are independent of any branch weights. > > Bruce > To put this another way., if a=sqrt(2/3) and b=sqrt(1/3), then if an observer is to conclude, from his data, that 0 is twice as likely as 1, he must see approximately twice as many zeros as ones. This cannot be achieved by simply multiplying the number of branches on a zero result. Multiplying the number of branches does not change the data within each branch, so observers will obtain exactly the same statistics as they would for a=b=1/sqrt(2). As I have repeatedly said, the data on each (and every) branch is independent of the weights or coefficients. This is a trivial consequence of having every result occur on every trial. Even if zero has weight 0.99, and one has weight 0.01, at each fork there is still one branch corresponding to zero, and one branch corresponding to one. Multiplying the number of zero branches at each fork does not change the statistics within individual branches. And it is the data from within his branch that the physicist must use to test the theory. Even if he is selected at random from some population where the number of branches is proportional to the weights, he still has only the data from within a single branch against which to test the theory. Multiplying branches is as irrelevant as imposing branch weights. That is where I think the attempt to force the Born rule on to Everett must inevitable fail -- there is no way that one can arrange fork dynamics so that there will always be twice as many zeros as ones along each branch (for the case a^2=2/3, b^2=1/3). In the full set of all 2^N branches there will, of course, be branches in which this is the case. But that is just because when every possible bit string is included, that possibility will also occur. The problem is that the proportion of branches for which this is the case becomes small as N increases. Consequently, the majority of observers will conclude that the Born rule is disconfirmed. This is not in accordance with observation, so Everett fails as a scientific theory -- it cannot account for our observation of probabilistic results. 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 [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLTgbtOztrq%3DWSqnTCsaLOm8gxmgJSMwYGcGYcho5uv8sw%40mail.gmail.com.
