On Mon, Feb 17, 2020 at 4:13 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:
> On 2/16/2020 2:17 PM, Bruce Kellett wrote:
>
> That is where the proof given by Kent comes into play. If in the N trials
> you observe pN zeros and (1-p)N ones, you estimate the probability for zero
> to be p, within certain confidence limits that depend on the number of
> trials. Note that this is precisely the 1p perspective, one person taking
> his actual data and making some estimates. This person then considers that
> some other person might have obtained r zeros, rather than the pN that he
> obtained. Applying the binomial theorem, he estimates the probability for
> this to occur as p^r(1-p)^{N-r}. This goes to zero in the limit as N
> becomes very large, so our original observer believes that he has the
> correct probability, since the probability of results significantly deviant
> from his goes to zero as N becomes large.
>
>
> The problem, of course, is that this reasoning applies equally well for
> all the inhabitants (from their individual first-person perspectives),
> whatever relative frequency p they see on their branch. All of them
> conclude that their relative frequencies represent (to a very good
> approximation) the branch weights. They clearly can't all be right, so
> either there is no actual probability underlying the events and their
> calculations are misguided, or the theory itself is incoherent.
>
>
> But exactly the same reasoning applies for any given true value of p.
> There will be different estimates by different experimenters and they can't
> all be right. Each will infer that any proportion other than the one he
> observed will have zero measure in the limit N->oo.
>
Exactly right. That is what my example of spin measurements on an ensemble
of equally prepared spin states comes into play. If all 2^N bit strings are
realized for one orientation of the S-G magnet, then exactly the same 2^N
bit strings are realized for every other orientation. Consequently, the
coefficients in the expansion play no role in determining the data, and it
makes no sense to talk of "the true value of p". There is no such true
value if all values are realized.
In Kent's thought experiment, if you consider the self-location as
> probabilistic then it's exactly the same as taking a sample of N from an
> ensemble for which p=0.5 is the true proportion. I think you prove too
> much by saying the estimate of any proportion of the other observers has
> zero measure in the limit, therefore everybody is wrong.
>
That is a strange thing to say -- I prove too much by showing that the
whole thing makes no sense? If you take a sample of N from an ensemble with
true proportion p=0.5????? The trouble is that you get the same ensemble
even if the true proportion is 0.99, or 0.01. or any other value.
If instead you estimate how many other experimenters will get estimates
> which are consistent with yours by being of high probability in your
> posterior Bayesian distribution, with high probability you will find that
> most of them will.
>
Exactly. Even if you estimate p=0.01, you will dismiss branches with
approximately equal numbers of zeros and ones as highly unlikely, and you
expect other experimenters to verify your results. If the number of trials
N is large, there are N(N-1)/2 branches with exactly 2 zeros and N-2 ones.
The probability for N/2 zeros and N/2 ones is (2/N)^N/2*(1-2/N)^N/2 ~
N^{-N/2}, which goes to zero very rapidly for large N.
There is no "intrinsic probability" in your scenario.
>>
>>
>> If there is no probability, what do you expect when you are still in
>> Helsinki. If you predict that you die, then you reject Mechanism (assumed
>> here). If you predict P(W) = 1, the city in Moscow will understand that the
>> prediction was wrong. If you predict that your history is the development
>> of PI, then only 1/2^N will be be confirmed, etc.
>>
>
>
> I turn the tables on you here, Bruno. You are confusing the 1p and 3p
> pictures. From each individual's personal perspective, he concludes,
> according to above argument, that his are the correct probabilities. It is
> only from the outside, third-person perspective, that we can see that he
> represents only a small fraction of the total population of 2^N branches.
>
> What is you prediction, if there is no probability. Keep in mind that “W”
>> and “M” does not refer to self-localisation, but to the first person
>> experience. Do you agree that in this case W and M are incompatible.
>> I just try to understand.
>>
>
> As I said, I make no prediction, since I do not think that the concept of
> probability can be meaningfully applied in cases of person duplication,
> such as the WM scenario, or, for that matter, Everettian quantum mechanics.
>
> This is also Adrian Kent's objection to MWI, and it will also nullify any
>> benefit you might seek to gain from the "frequency operator" -- every
>> "first person" will get a different eigenvalue in the limit of infinite
>> trials..
>>
>>
>> That is not correct. If it is the frequency operator which is measure, it
>> gives the Born Probabilities, at least if the “simple” derivation is
>> correct.
>>
>
> No,that argument is mistaken, as Kent's general argument in terms of the
> binomial expansion shows. All 2^N persons will use the frequency operator
> to conclude that their probabilities are the correct ones. Some will be
> seriously wrong,
>
>
> But almost all will intersubjectively agree that p is near 0.5. Science
> theories are based on intersubjective agreement...not personal expriences.
>
But the true probability, as given by the amplitudes, might be very far
from 0.5. This shows that the data are independent of the Born amplitudes
-- the Born rule does not apply, and cannot be simply grafted on.
> so the frequency operator is not a reliable indicator of probability.
>
> Incidentally, the fact that there are more bit strings in the set of all
> 2^N bit strings with approximately equal numbers of 0 and 1 results is a
> consequence of the binomial expansion when there are only two possible
> outcomes, as in the cases we have considered -- it is no more fundamental
> than that, and does not reflect some 3p-preferred probability.
>
> But my question is independent of Everett, so even if Kent is correct for
>> QM, it remains false for Mechanism. Let us agree first on the simple
>> Mechanist case, and then come back to Everett.
>>
>
>
> Kent posed his argument in terms of completely classical simulations, so
> it is precisely parallel to your WM-duplication scenario. I have applied
> the argument to Everettian QM because of the parallels between the two:
> Everett is just like the classical duplication case since it is completely
> deterministic and every possibility occurs on every trial. The only real
> difference is that the different outcomes in QM occur on different branches
> which, by decoherence, cannot interact or be aware of each other. So there
> is no effective 3p perspective in QM as there is in the WM-duplication.
> Arguments about the proportion of individuals who see particular sets of
> outcomes in QM are arguments from the 3p perspective, and it can be argued
> that in the absence of any possible 3p observer, such arguments are invalid.
>
>
> Then the arguments about every estimate being contrary to other estimates
> is also invalid.
>
Why should you think that? The estimates of probability that the individual
observer makes are all strictly first person estimates -- they use their
own data for their estimates, they do not take any third person view of the
situation. It is the argument that most observers will lie towards the
centre of the binomial distribution for two results that is an invalid
appeal to the third person view, and an appeal which leads to manifestly
wrong results if the true probability is far from 0.5.
Bruce
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