On Fri, Mar 6, 2020 at 10:15 AM 'Brent Meeker' via Everything List < [email protected]> wrote:
> On 3/5/2020 1:57 PM, Bruce Kellett wrote: > > On Fri, Mar 6, 2020 at 8:08 AM 'Brent Meeker' via Everything List < > [email protected]> wrote: > >> On 3/5/2020 2:45 AM, Bruce Kellett wrote: >> >> >> Now sequences with small departures from equal numbers will still give >> probabilities within the confidence interval of p = 0.5. But this >> confidence interval also shrinks as 1/sqrt(N) as N increases, so these >> additional sequences do not contribute a growing number of cases giving p ~ >> 0.5 as N increases. So, again within factors of order unity, the proportion >> of sequences consistent with p = 0.5 decreases without limit as N >> increases. So it is not the case that a very large proportion of the binary >> strings will report p = 0.5. The proportion lying outside the confidence >> interval of p = 0.5 is not vanishingly small -- it grows with N. >> >> >> I agree with you argument about unequal probabilities, in which all the >> binomial sequences occur anyway leading to inference of p=0.5. But in the >> above paragraph you are wrong about the how the probability density >> function of the observed value changes as N->oo. For any given interval >> around the true value, p=0.5, the fraction of observed values within that >> interval increases as N->oo. For example in N=100 trials, the proportion >> of observers who calculate an estimate of p in the interval (0.45 0.55) is >> 0.68. For N=500 it's 0.975. For N=1000 it's 0.998. >> >> Confidence intervals are constructed to include the true value with some >> fixed probability. But that interval becomes narrower as 1/sqrt(N). >> So the proportion lying inside and outside the interval is relatively >> constant, but the interval gets narrower. >> > > > I think I am beginning to see why we are disagreeing on this. You are > using the normal approximation to the binomial distribution for a large > sequence of trials with some fixed probability of success on each trial. In > other words, it is as though you consider the 2^N binary strings of length > N to have been generated by some random process, such as coin tosses or the > like, with some prior fixed probability value. Each string is then > constructed as though the random process takes place in a single word, so > that there is only one outcome for each toss. > > Given such an ensemble, the statistics you cite are undoubtedly correct: > as the length of the string increases, the proportion of each string within > some interval of the given probability increases -- that is what the normal > approximation to the binomial gives you. And as N increases, the confidence > interval shrinks, so the proportion within a confidence interval is > approximately constant. But note these are the proportions within each > string as generated with some fixed probability value. If you take an > ensemble of such strings, the the result is even more apparent, and the > proportion of strings in which the probability deviates significantly from > the prior fixed value decreases without limit. > > That is all very fine. The problem is that this is not the ensemble of > strings that I am considering! > > The set of all possible bit strings of length N is not generated by some > random process with some fixed probability. The set is generated entirely > deterministically, with no mention whatsoever of any probability. Just > think about where these strings come from. You measure the spin of a > spin-half particle. The result is 0 in one branch and 1 in the other. Then > the process is repeated, independently in each branch, so the 1-branch > splits into a 11-branch and a 10-branch; and the 0-branch splits into a > 01-branch and a 00-branch. This process goes on for N repetitions, > generating all possible bit strings of length N in an entirely > deterministic fashion. The process is illustrated by Sean Carroll on page > 134 of his book. > > Given the nature of the ensemble of bit strings that I am considering, the > statistical results I quote are correct, and your statistics are completely > inappropriate. This may be why we have been talking at cross purposes. I > suspect that Russell has a similar misconception about the nature of the > bit strings under consideration, since he talked about statistical results > that could only have been obtained from an ensemble of randomly generated > strings. > > > Yes, I understand that. And I understand that you have been talking about > Everett's original idea in which at each split both results obtain, one in > each branch...with no attribute of weight or probability or other measure. > It's just 0 and 1. Which generates all strings of zeros and ones. This > ensemble of sequences has the same statistics as random coin flipping > sequences, even though it's deterministic. > That is, in fact, false. It does not generate the same strings as flipping a coin in single world. Sure, each of the strings in Everett could have been obtained from coin flips -- but then the probability of a sequence of 10,000 heads is very low, whereas in many-worlds you are guaranteed that one observer will obtain this sequence. There is a profound difference between the two cases. But it doesn't have the same statistics as flipping an unfair coin, i.e. > when a=/=b. > Actually, that is incorrect too. The sequence of bit strings is independent of the superposition weights. So you get the same strings whatever the supposed bias of your coin -- in the many-worlds case, of course. The statistics are different in the single-world case, but I am not considering that. So to have a multiple world interpretation that produces statistics > agreeing with the Born rule one has to either assign weights to each of the > two worlds at a split and then interpret those as probability amplitudes > One can attempt this, but such weights are still invisible in the data that each observer obtains, so it doesn't work in practice. 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. 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/CAFxXSLSEQz9EEfGmWK-TZCGEbf5%3DKtywuCDspSH2P7XfWJODXg%40mail.gmail.com.
