On 13/07/2016 11:36 pm, Bruno Marchal wrote:
On 11 Jul 2016, at 13:49, Bruce Kellett wrote:
On 11/07/2016 9:31 pm, Bruno Marchal wrote:
*Holiday Exercise:*
A guy undergoes the Washington Moscow duplication, starting again
from Helsinki.
Then in Moscow, but not in Washington, he (the one in Moscow of
course) undergoes a similar Sidney-Beijing duplication.
I write P(H->M) the probability in H to get M.
In Helsinki, he tries to evaluate his chance to get Sidney.
With one reasoning, he (the H-guy) thinks that P(H-M) = 1/2, and
that P(M-S) = 1/2, and so conclude (multiplication of independent
probability) that P(H-S) = 1/2 * 1/2 = 1/4.
But with another reasoning, he thinks that the duplications give
globally a triplication, leading eventually to a copy in W, a copy
in S and a copy in B, and so, directly conclude P(H-S) = 1/3.
So, is it 1/4 or 1/3 ?
Neither. The probability that the guy starting from Helsinki gets to
Sydney is unity.
Try to convince the guy who gets to Beijing, or the one who stayed in
Washington. He knows that the probability evaluated in Helsinki was
not P(Sidney) = 1.
We start with John Clark in Helsinki, so P(JC ~ H) = 1. By construction,
after the duplication and so on, P(JC ~ W) = P(JC ~ S) = P(JC ~ B) = 1.
(I use '~' as a shorthand for 'in' or 'sees'.) JC in Helsinki knows the
protocol, so he can easily see that these are the correct probabilities.
So, as I said, the probability that the guy starting from Helsinki gets
to Sydney is unity. Any other interpretation of this scenario involves
an implicit appeal to dualism -- there is "one true JC" that goes
through these duplications, and he can only ever end up in just one place.
As John Clark has correctly pointed out, your intuition and formalism
simply does not work in the presence of person-duplicating machines.
There is no single 1p view -- there are three possible 1p views in the
triplication scenario. So, again, John Clark is right when he says that
JC ~ H will see three cities (W, S, and B) after the experiment is
completed. If, as you claim, he will see only one city, you have to have
some dualist 'nut or core' that survives in only one of your copies.
Of course, as I said some time ago, the easiest resolution of you
logical conundrums is that JC ~ H does not survive, and that there are
three new persons, one in each city, so the probability that JC in H
will see Sydney is exactly zero.
Looking at the more realistic quantum realization of this triplication
scenario, we can formulate that as follows. We prepare a spin-half atom
with spin along the x-axis, then pass it through an S-G magnet oriented
along the y-axis, getting two possibilities, which we can call up and
down. We then take the up channel and pass that through a further S-G in
the x-direction, getting two further possibilities of left or right.
Let us perform this experiment many times and count the number of
particles in each of the three possible final states (down, left, and
right). If this is a real laboratory experiment, in which detection of a
particle in any channel leads to irreversible decoherence and the
formation of a separate world containing just that result, we will find
approximately half the particles end up in the down state, and
approximately a quarter in each of the left and right states. This gives
the most reliable estimate of the real probabilities for the outcome
from the given initial state.
If you take the MWI view, then you get one down, one left, and one right
in every run of the experiment, so the probability for each outcome is
unity. In order to get probability of 1/4 for left, say, you have to
detect the absence of a particle in the down state (so that the particle
is certainly in the up state) for which the probability is 1/2.
Actually, your preferred answer -- that the probability P(H->S) = 1/3,
is possible only in a fully dualist model. You are essentially claiming
that as the scenario puts John Clark's in all three cities, it is purely
a random chance that selects one of them to be the "true" John Clark --
a dualist "core" is assigned to one of these copies purely by chance.
This is the problem with probabilities in the MWI -- how do you
interpret probabilities when all possible outcomes occur with
probability one?
The probabilities concern the relative first person experiences.
Computationalism guaranties that there will be only one outcome.
You are simply talking nonsense, here. "Relative first person
experiences"? Relative to what? The scenario, and computationalism
guarantees that there will be all three outcomes. If there is only one
"1p" experience, then that can only be chosen dualistically. If there is
duplication (triplication) then your intuitions break down. I have to
say it -- John Clark has been right all along.
Bruce
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