On 11 Jul 2016, at 18:56, smitra wrote:
On 11-07-2016 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. This is the problem with probabilities in the MWI --
how do you interpret probabilities when all possible outcomes occur
with probability one?
Bruce
In duplication experiments the prior probability to exist at all in
any of the possible states increases after the duplication, while in
unitary QM this is conserved (except if one or more of the possible
outcomes is death).
It is plausibly conserved with Mechanism too, if the arithmetical
quantum logics (the one extracted from Z1*, or X1*, or S4Grz1)
justifies the linear rule Y = II. And normally, if QM is empirically
correct, and computationalism is correct, they should match.
The correct way to analyze Bruno style duplication arguments is to
start with assigning some measure m to the observer before any
duplication is carried out, in this case the observer at H.
The probabilities are relative, and conditioned implicitly by P(H) = 1.
Then H gives rises to two copies in states W and M (we can call them
copies, but they are actually different observers as they have
different memories stored in their brains, so they are different
algorithms).
They are the same programs, but with different input. By the SMN
theorem, you *can* conceive them as different programs. OK.
The measures will be m for each of these observers. Then W is not
going to be copied, while M gives rise to S and B, so we end up with
3 observers each with measure m. The probability is thus 1/3.
That is the correct answer .... if the guy remains unconscious in
Moscow. But I doubt it is 1/3 in case he does wake up. In that case I
would say it is 1/4. (To be sure, I use Gödel-Löb and self-reference,
which works only for the case P=1, already non trivial, and quantum-
like, to avoid such probability question which are premature, but we
can also speculate a bit).
In an analogue MWI setting, the outcome is different, at each
duplication the measure for a particular outcome is halved. W thus
has a measure of m/2, while S and B each have a measure of m/4, the
probability is thus 1/4.
Hmm... I would like to see the analogue. I think the devil is in the
details here. The analogue of letting the Moscow guy unconscious is
the interference when we don't make a measurement at some intermediate
state of a quantum computation (say).
My opinion is P(H->S) = 1/4, if the guy in Moscow remains awake, and
P(H->S) = 1/3 if he remains unconscious there.
Bruno
Saibal
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