On Thu, Mar 05, 2020 at 09:45:38PM +1100, Bruce Kellett wrote: > On Thu, Mar 5, 2020 at 5:26 PM Russell Standish <[email protected]> wrote: > > > But a very large proportion of them (→1 as N→∞) will report being > within ε (called a confidence interval) of 50% for any given ε>0 > chosen at the outset of the experiment. This is simply the law of > large numbers theorem. You can't focus on the vanishingly small > population that lie outside the confidence interval. > > > This is wrong.
Them's fighting words. Prove it! > In the binary situation where both outcomes occur for every > trial, there are 2^N binary sequences for N repetitions of the experiment. > This > set of binary sequences exhausts the possibilities, so the same sequence is > obtained for any two-component initial state -- regardless of the amplitudes. > You appear to assume that the natural probability in this situation is p = 0.5 > and, what is more, your appeal to the law of large numbers applies only for > single-world probabilities, in which there is only one outcome on each trial. I didn't mention proability once in the above paragraph, not even implicitly. I used the term "proportion". That the proportion will be equal to the probability in a single universe case is a frequentist assumption, and should be uncontroversial, but goes beyond what I stated above. > > In order to infer a probability of p = 0.5, your branch data must have > approximately equal numbers of zeros and ones. The number of branches with > equal numbers of zeros and ones is given by the binomial coefficient. For > large > even N = 2M trials, this coefficient is N!/M!*M!. Using the Stirling > approximation to the factorial for large N, this goes as 2^N/sqrt(N) (within > factors of order one). Since there are 2^N sequences, the proportion with n_0 > = > n_1 vanishes as 1/sqrt(N) for N large. I wasn't talking about that. I was talking about the proportion of sequences whose ratio of 0 bits to 1 bits lie within ε of 0.5, rather than the proportion of sequences that have exactly equal 0 or 1 bits. That proportion grows as sqrt N. > > 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. The confidence interval ε is fixed. 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. > > > > > The crux of the matter is that all branches are equivalent when both > outcomes > > occur on every trial, so all observers will infer that their observed > relative > > frequencies reflect the actual probabilities. Since there are observers > for all > > possibilities for p in the range [0,1], and not all can be correct, no > sensible > > probability value can be assigned to such duplication experiments. > > I don't see why not. Faced with a coin flip toss, I would assume a > 50/50 chance of seeing heads or tails. Faced with a history of 100 > heads, I might start to investigate the coin for bias, and perhaps by > Bayesian arguments give the biased coin theory greater weight than the > theory that I've just experience a 1 in 2^100 event, but in any case > it is just statistics, and it is the same whether all oputcomes have > been realised or not. > > > The trouble with this analogy is that coin tosses are single-world events -- > there is only one outcome for each toss. Consequently, any intuitions about > probabilities based on such comparisons are not relevant to the Everettian > case > in which every outcome occurs for every toss. Your intuition that it is the > same whether all outcomes are realised or not is simply mistaken. > > > > The problem is even worse in quantum mechanics, where you measure a > state > such > > as > > > > |psi> = a|0> + b|1>. > > > > When both outcomes occur on every trial, the result of a sequence of N > trials > > is all possible binary strings of length N, (all 2^N of them). You then > notice > > that this set of all possible strings is obtained whatever non-zero > values of a > > and b you assume. The assignment of some propbability relation to the > > coefficients is thus seen to be meaningless -- all probabilities occur > equal > > for any non-zero choices of a and b. > > > > For the outcome of any particular binary string, sure. But if we > classify the outcome strings - say ones with a recognisable pattern, > or when replayed through a CD player reproduce the sounds of > Beethoven's ninth, we find that the overwhelming majority are simply > gobbledegook, random data. > > > Sure. Out of all possible binary strings of length N, most will resemble > random > noise. Though if N is large enough, all the works of Shakespeare will be > encoded, in order. And an increasingly large number of times as N -> oo. I do > not see that this is in any way relevant to the issues at hand. > > > And the overwhelming majority of those will > have a roughly equal number of 0s and 1s. > > > Now that is simply false, as shown above. > > > For each of these > categories, there will be a definite probability value, and not all > will be 2^-N. For instance, with Beethoven's ninth, that the tenor has > a cold in the 4th movement doesn't render the music not the ninth. So > there will be set of bitstrings that are recognisably the ninth > symphony, and a quite definite probability value. > > > > There will be a definite number of such strings encoding something close to > Beethoven's ninth. And they will also all have similar proportions of zeros > and > ones, and thus represent similar probabilities. But again, this is not > relevant > to the underlying issue. > > > > > You may counter that the assumption that an observer cannot see all > > outcomes is an extra thing "put in by hand", and you would be right, > > of course. It is not part of the Schroedinger equation. But I would > > strongly suspect that this assumption will be a natural outcome of a > > proper theory of consciousness, if/when we have one. Indeed, I > > highlight it in my book with the name "PROJECTION postulate". > > > > This is, of course, at the heart of the 1p/3p distinction - and of > > course the classic taunts and misunderstandings between BM and JC > > (1p-3p confusion). > > > > > > I know that it is a factor of the 1p/3p distinction. My complaint has > > frequently been that advocates of the "p = 0.5 is obvious" school are > often > > guilty of this confusion. > > > > > > Incidently, I've started reading Colin Hales's "Revolution of > > Scientific Structure", a fellow Melburnian and member of this > > list. The interesting proposition about this is Colin is proposing > > we're on the verge of a Kuhnian paradigm shift in relation to the > role > > of the observer in science, and the that this sort of > misunderstanding > > is a classic symptom of such a shift. > > > > > > > > Elimination of the observer from physics was one of the prime > motivations > for > > Everett's 'relative state' idea. Given that 'measurement' and 'the > observer' > > play central roles in variants of the 'Copenhagen' interpretation. > > > > Yes - but not everyone is pure Everett, even if they're many worlds. I > have often argued publicly that the observer needs to be front and > centre in ensemble theories. It is also true of Bruno's > computationalism - the observer is front and centre, and characterised > by being a computation. Maybe it's so, maybe it ain't, but at least the > idea gets us out of the morass that science of conscioussness is in. > > > This may well be the case. But I have been concerned primarily with the > possibility of developing some useful notion of probability in Everettian > quantum mechanics, when every possible outcome occurs (in different branches) > on every trial. This is relevant to Bruno's WM-duplication scenario, but > probably not for your plenum consisting of every possible bit string -- I only > consider all possible bit strings of length N in N repetitions of the > experiment, which is far fewer that all possible bit strings of any length. > Consciousness studies are outside my brief, and I follow standard physics > practice in eliminating consideration of the role of the observer -- > everything > is just quantum mechanics in this approach. > > 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/ > CAFxXSLSJvyZ1ud1KtD6%3DFJm9we%3Dx38dnRZRr4o-Mpmse%2BpZg7g%40mail.gmail.com. -- ---------------------------------------------------------------------------- Dr Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders [email protected] http://www.hpcoders.com.au ---------------------------------------------------------------------------- -- 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/20200308071420.GA2903%40zen.
