On Thu, Mar 5, 2020 at 5:26 PM Russell Standish <li...@hpcoders.com.au> wrote:
> On Thu, Mar 05, 2020 at 11:34:55AM +1100, Bruce Kellett wrote: > > On Thu, Mar 5, 2020 at 10:39 AM Russell Standish <li...@hpcoders.com.au> > wrote: > > > > ISTM - probability is all about what an observer observes. Since the > > observer cannot see all outcomes, an objection based on all outcomes > > occurring seems moot to me. > > > > > > The fact that the observer cannot see all outcomes is actually central > to the > > argument. If, in the person-duplication scenario, the participant naively > > assumes a probability p = 0.5 for each outcome, such an intuition can > only be > > tested by repeating the duplication a number of times and inferring a > > probability value from the observed outcomes. Since each observer can > see only > > the outcomes along his or her particular branch (and, ipso facto, is > unaware of > > the outcomes on other branches), as the number of trials N becomes very > large, > > only a vanishingly small proportion of observers will confirm their 50/50 > > prediction . This is a trivial calculation involving only the binomial > > coefficient -- Brent and I discussed this a while ago, and Brent could > not > > fault the maths. > > 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. 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. 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. 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. > > 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 everything-list+unsubscr...@googlegroups.com. 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