> On 16 Feb 2020, at 23:17, Bruce Kellett <[email protected]> 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.
I guess by W and M you mean the W-guy and the M-guy, in some third person
description. Yet, the question is asked to the H-guy, before he presses the
button, and concerns what he is executed to feel after having push on the
button.
Do you accept that what you said above entails that the H-guy can predict that
he (whoever he will become) will drink some coffee with probability 1?
>> 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.
OK. Of course the M-man is an honorable continuer of the H-guy. Mechanism gives
him the right today that he is the H-guy, like he gives that right to the
M-guy. (Personal identity, as we have define it is not transitive: with the M
and W guys are the H-guy, but the W-guy has diverged from the M-guys, and can
be reasonably said have become two different person.
> So what? We are talking about estimating the probability from repeated trials
> -- there is no other sensible way to estimate empirical probabilities.
There are, but I appreciate the frequency approach, and it works for both the
first person singular and the first person plural. Bayes and the Dutch argument
works only for the first person plural, where population of individuals are
duplicated, and the statsitical reasoning is made internally in each copy.
> 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,
I am nt sure why you says this.
> so we must estimate probabilities from the relative frequencies in a number
> of trials.
But we can do that, and its shows that the probability of having wake up in M k
times, in a n-iteration, is given by the binomial k, n times 1/2^n.
>
>> 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.
Good!
> 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.
That is the error. And as they are aware of the protocol, they can fix this.
> 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.
The prediction must be valid for all the copies. Are you OK that the H-guy
predicts “W or M”, instead of “W”, “M” and “W and M”, for one trial? Are you OK
that the H-guy says (assuming mechanism of course) that he is sure to get some
coffee, but that he cannot be sure if it will taste like American or Russian
coffee?
>
>
>> 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.
That has to be accounted in the theory. I don’t see any problem. If they bet
P(W) = P(M) = 1/2, the proportion of people OK with this, divided by 2^n get
close to one. Yes, some will have wrong probabilities, and we can imagine that
the guy who get the 100 firs decimal of PI in binary correct might be compelled
to bet that this will continue for him, but we know that only one among the 2
following copies will assess this. The error is akin to the joke of the guy
taking a bomb with him when travelling by planes, because he heard that the
probability that there are two bombs in a plane is null (lol).
It might seem counter-intuive, but even if you get the 1000 first binary-digits
of PI, you better bet on 1/2, as all continuers will assess, except one.
>
> 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.
I do have myself some reason to criticise the use of probability, as I
explained 20 years ago here. It is more a credibility, and actually it is a
notion of quantum credibility. The point is that the H-guy can predict he will
the coffee with certainty, but cannot predict its taste, except by an OR (it
will taste American or Russian). Then, the numeral identity favours, in this
simple and non realist scenario, that P = 1/2, as almost all copies confirmed
in the long run.
>
>> 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, so the frequency operator is not a reliable indicator of
> probability.
Some will be wrong, but the majority will be correct. As the number of
iteration increase, we get near one. And in the “real scenario”, that is
confronted to the arithmetical reality which emulates all 2^aleph_0 consistent
continuations, those having the wrong probability constitue a set of real of
measure 0.
>
> 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.
The point is that those proportion are given by the binôme coefficient, or the
Gauss curves with iterated a lot.
>
> 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.
OK, but only up to that transfinite multiplication.
> 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.
The 3p are always theoretical insight, and in this case, it is wave itself. We
cannot see it, but we cannot see the theory, only events and objects that the
theory is about.
> 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.
But in the WM-duplication scenario, this cannot be applied, and that is what we
have to discuss first, I think. Now I am glad you see that Everett uses
(instinctively we could say) the (simple) self-duplication idea in the
universal wave, that will help to see the “many-histories” interpretation
developed by the universal machines in arithmetic.
Bruno
>
> Bruce
>
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