On 13 Aug 2017, at 00:58, Bruce Kellett wrote:
On 12/08/2017 5:56 pm, Bruno Marchal wrote:
On 12 Aug 2017, at 04:12, Bruce Kellett wrote:
On 12/08/2017 3:22 am, Bruno Marchal wrote:
On 11 Aug 2017, at 13:40, Bruce Kellett wrote:
Are you telling us that P(W) ≠ P(M) ≠ 1/2. What do *you*
expect when pushing the button in Helsinki?
I expect to die, to be 'cut', according to the protocol. The
guys in W and M are two new persons, and neither was around in H
to make any prediction whatsoever.
Fair enough.
You think the digital mechanism thesis is wrong.
Correct.
There is a fundamental problem with your person-duplication
thought experiments. This is that the way in which you interpret
the scenario inherently involves an irreducible 1p-3p confusion.
The first person (1p) concerns only things that the person can
experience directly for himself. It cannot, therefore, involve
things that he is told by other people, because such things are
necessarily third person (3p) knowledge -- knowledge which he does
not have by direct personal experience. So our subject does not
know the protocol of the thought experiment from direct experience
(he has only been told about it, 3p). When he presses the button
in the machine, he can have no 1p expectations about what will
happen (because he has not yet experienced it). He presses the
button in the spirit of pure experimental enquiry -- "what will
happen if I do this?" His prior probability for any particular
outcome is zero.
That is just plain false. The guy in Helsinki knows the protocol,
and he assumes Mechanism. So he knows that P(W) = P(M) , and that
P(W) ≠ 0, and P(M) ≠ 0, and P(X) for any X different from W and
M is equal to 0.
He only knows the protocol because he has been told about it. How
does he know that he isn't being lied to? Knowledge of the protocol
is 3p knowledge.
If he is lied to, it is a different thought experiment. Obviously, if
he opens the door, and see Vienna, his prediction "W v M" was false,
but that change noting to the reasoning.
So when he presses the button in Helsinki, and opens the door to
find himself in Moscow, he will say, "WTF!". In particular, he
will not have gained any 1p knowledge of duplication. In fact, he
is for ever barred from any such knowledge.
Yes, that is the 3p/1p confusion that John Clark is doing. He told
me that the guy in Moscow says "I knew it", or I predicted it, by
saying that P(W & M) = 1.
You try to help John C., but you contradict his "theory" (which is
indeed based on the 1p/3p confusion).
I suggest that the whole of step 3 is based on a 1p/3p confusion. If
the duplicated subject does not have 3p knowledge of the protocol,
he will never be aware of being duplicated.
Explain this to John. We agree.
In fact, he can never get first person knowledge of that
duplication, even if he is, in fact, duplicated.
Which I sum up often by: the candidate does not feel the split, which
is of course an allusion to the similar phenomenon happening with
Everett. You make my point, thanks.
If he repeats the experiment many times, he will simply see his
experiences as irreducibly random between M and W, with some
probability that he can estimate by keeping records over a period
of time. If you take the strict 1p view of the thought experiment,
the parallel with the early development of QM is more evident. In
QM, no-one has the 3p knowledge that all possible outcomes are
realized (in different worlds).
So, before pressing the button in H, his prior probabilities are
p(M) = p(W) = 0, with probably, p(H) = 1.
What?
(I recall that in H the person is annihilated).
He doesn't know this because he doesn't know the protocol.
?
The question is : do you agree with step 3, step 4, step 5, step 6,
step 7?
In all case, he knows the protocol. I am OK with different thought
experience, but then propose them in another thread, and explain what
you want to prove. You can't change the premise of the reasoning in
the middle without explaining why?
On the other hand, if you allow 3p knowledge of the protocol to
influence his estimation of probabilities before the experiment,
you can't rule out 3p knowledge that he can gain at any time after
pressing the button. In which case, the 1p-3p confusion is
complete, p(M) = p(W) = 1, and he can expect to see both cities.
In that case, the pure 1p view becomes irrelevant.
W and M, as they have defined (they concern the experience of
opening a door and describing which city is seen) are incompatible
experience, and the protocol entails that P(W) = P(M). If P(M) =
P(W) = 1, you get a probability equal to 2.
Rubbish. You have no basis for adding these probabilities because
they are not independent events.
The third axiom of Kolmogorov asks only that the events are
incompatible, or disjoint (in the algebra of events).
In this case the probability is on the algebra of first person
distinguishible events, and the events "seeing M" is first person
incompatible with the event 'seeing M".
In the "simple" self-duplication experience, the first person
indeterminacy is 3p equivalent with the perfect coin throw.
What would be wrong is to infer from this that in front of the
universal dovetailing, or "in front of the sigma_1 arithmetical
reality", the probabilities remains simple to evaluate. We get two
difficulties: an infinite number of "reconstitutions", and infinitely
many are not distinguishable (with more complex nets of bifurcations
and fusions, ...but then guess what: computer science/mathematical
logic excels on pronouns and indexicals, and so it is relatively easy
to extract the logical structure of the subject and of its observable.
Up to now, it works, and, unlike the physical sciences, it clearly
explains the fundamental nuance between the qualia and the quanta. Of
course, there are many open problems.
And of course the 1500 years of brainwashing in theology does not help.
With the UD, eventually we will have a notion of credibility in
place of probability, but this does not happen in the self-
duplication protocol. You just cannot have P(M), or P(W) = 1,
because that is refuted directly by both copies, when asked about
their first person experience.
Before the procedure, neither has any probability estimate.
False. They have, P(W) = P(M) = 1/2.
Much later, you could understand that this is just impossible, but
this means that a perfect self-duplication is impossible. The reason
is that you cannot garantie a total annihilation in Helsinki. The
correct prediction is more 1/2 minus epsilon, with epsilon measuring
the proportion of computation where you survive the annihilation in
Helsinki.
But, at this stage, we suppose the elimination in Helsinki to be
absolute, and that is part of the theoretical protocol, just to
understand enough to move on step 4.
After the procedure, p(M) = p(W) = 1
?
Ah! I see, I guess you come back on the third person view on the two
first person experiences. But by definition, we are talking abaout the
first person expectations. It is the 1p ===> 3-1p moves. In that
sense p(M) = p(W) = 1, but that has no relevance for the consequences
of the reasoning.
because they know with certainty where they are, and the results are
not independent.
They are first person disjoint, and that is enough for additivity here.
It really looks like you are doing the usual Clark's trick consisting
in taking the outside 3p view, when we are actually and precisely
discussing about the 1p view expectations. That is equivalent with a
first person elimination, a common error among some religious-
materialists.
Bruno
Bruce
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