On 2/16/2020 2:17 PM, Bruce Kellett wrote:
On Mon, Feb 17, 2020 at 1:27 AM Bruno Marchal <[email protected]
<mailto:[email protected]>> wrote:
On 14 Feb 2020, at 22:48, Bruce Kellett <[email protected]
<mailto:[email protected]>> wrote:
On Sat, Feb 15, 2020 at 1:35 AM Bruno Marchal <[email protected]
<mailto:[email protected]>> wrote:
Just to be clear, are you OK with P(W) = 1/2 in the
WM-duplicatipon, when “W” refers to the first person experience?
No. As I have said before, the H-man has no basis on which to
assign any probability at all to the possibility that he will see
W (or M) tomorrow,
Do you accept the idea that if we offer him (to the two copies,
thus) a cup of coffee after reconstitution, in both M and W, that
he can say in Helsinki that if mechanism is correct, he will drink
coffee with probability one? What would you say if you were the H-guy?
If all copies are given a cup of coffee, then it is certain that W and
M will drink coffee--by hypothesis.
The trouble is that probabilities tend to be defined by the limit
of relative frequencies over a large number of trials.
But one trial is enough to refute P(W) = 1 and P(M) = 1. Or to
refute P(W & M) = 1, given that W and M are incompatible first
person experience (none of the copies will feel to be in two
cities at once).
One trial is enough to refute P(W)=1 if you take the view of the
M-man. So what? We are talking about estimating the probability from
repeated trials -- there is no other sensible way to estimate
empirical probabilities. You can estimate probabilities in the
single-world case, say for coin tosses: if you assume that the coin
and the tossing method are fair, then the probabilty for "heads"
equals the probability for "tails"; again, by hypothesis. But in the
duplication case we do not have this possibility available, so we must
estimate probabilities from the relative frequencies in a number of
trials.
If you perform the WM-duplication N times, there will be 2^N
"first person experiences”
OK.
and many of them will assign probabilities greatly different from
0.5.
Not at all. In the limit most will say that it looks like white
noise: arbitrary sequence. We can show that most histories
(sequence of W and M) will be algorithmically incompressible, and
if the copies met, they can see that their population is well
described by the Pascal triangle (or Newton’s binomial).
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.
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.
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.
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 exepriences.
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.
Brent
Bruce
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