On Fri, Mar 6, 2020 at 11:14 AM 'Brent Meeker' via Everything List < [email protected]> wrote:
> On 3/5/2020 3:44 PM, Bruce Kellett wrote: > > OR postulate that the splits are into many copies so that the branch count >> gives the Born statistics. >> > > > That has possibilities, but I think it cannot work either. After all, each > observer just sees a sequence of results -- he is unaware of other branches > or sequences, so does not know how many branches are the same as his. The > 1p/3p distinction comes into play again. Any attempt to make multiple > branches reproduce probabilities necessarily confuses this distinction. You > have to think in terms of what data an observer actually obtains. Thinking > about what happens in the "other worlds" is illegitimate. > > > Consider the many copies case as an ensemble and it will reproduce the >> Born statistics even though it is deterministic. This is easy to see >> because every sequence a single observer has seen is the result of a random >> choice at the split of which path you call "that observer". >> > > > But the weights do not influence that split, so the observer cannot see > the weights. > > > Not weights, multiple branches. > The observer cannot see multiple branches from within a branch, either. 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/CAFxXSLT%2BL5kG%2BTWJu3p1qEs_dpD0EphWH_q-siGyztbXiedpdw%40mail.gmail.com.
