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.
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. In fact, he can never get first
person knowledge of that duplication, even if he is, in fact, duplicated.
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.
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.
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. After the
procedure, p(M) = p(W) = 1 because they know with certainty where they
are, and the results are not independent.
Bruce
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