(unknown charset) RE: normalization

From: (unknown charset) Jacques M. Mallah <jqm1584.domain.name.hidden>
Date: Wed, 12 Jan 2000 18:22:35 -0500 (EST)

On 7 xxx -1, Marchal wrote:
> Jacques M. Mallah wrote:
> >Marchal wrote:
> >>() I agree with Jacques Mallah that the measure depends on the number of
> >>implementations
> >
> > Please explain that remark.
> Well, it seems you say exactly that in some of your posts. I will try
> to find them in the archive. I think you were talking on "actual" "real"
> "physical" implementations where I speak about the computations
> occuring in the running UD, where "running" does not need to be actual
> running.

        Actually that doesn't matter. For me a physical system is just a
mathematical structure which "exists". So let all computations, for
example, exist.
        But for me the term "implementation" has a special definition
which I am still trying to formulate precisely, due to the implementation
problem, so I'm still not sure how you distinguish between a computation
and an implementation of a computation.

> It is linked to the following question: you are read and destroyed at
> Brussels and reconstituted in one exemplar at Washington and 9 identical
> exemplars in 9 thoroughly identical virtual simulation of Moscow.
> What is, at Brussels, your first person expectation to find yourself
> in Washington and Moscow ?

        Well, first there is the question of what you mean by a 1st person
expectation and by "yourself". There is no 1 of the 10 copies which is
linked more than the others to the original copy. So depending on the
definition I would either say that you are only at Brussels, or that you
are at all three places.
        But for practical purposes (such as, if you knew only German,
deciding which language to learn before being scanned, English or
Russian), I would of course say that, after the reconstitution, your
effective probability of seeing Moscow is 90%, so Russian is more
"likely" to come in handy.

> If you say that your expectation to find yourself at Maoscow is bigger
> than the one to find yourself at Washington, then, by definition, I will
> say that you link the measure with the number of (actual, here) running
> computations.

        OK, not just link it, but say it is directly proportional
(but see above). But my point is that if you do that you reject the
RSSA. For example, suppose that a year later (on 1/1/2001), the copies at
Moscow are destroyed except for one, while at Washington 8 new copies of
the version of you that has been there are made. During 2000, what is the
effective probability that you would be at Moscow? It is 90%; next year,
it is 10%.
        But according to the RSSA that would not be the case. If you
are 90% likely to be in Moscow in 2000, each type of version of you could
only "flow into" its own continuation, so the eff. pr. of being at Moscow
woulkd remain at 90% for 2001. (Chris Maloney would claim that the
eff. pr. of being at Moscow would be 10% for both years).
        The ASSA, on the other hand, is naturally compatible with the
simple rule that the effective probability is proportional to the number
of copies. So do you, in fact, agree with that rule?

                         - - - - - - -
               Jacques Mallah (jqm1584.domain.name.hidden)
         Physicist / Many Worlder / Devil's Advocate
"I know what no one else knows" - 'Runaway Train', Soul Asylum
             My URL: http://pages.nyu.edu/~jqm1584/
Received on Wed Jan 12 2000 - 15:26:42 PST

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