Re: normalization

From: Russell Standish <R.Standish.domain.name.hidden>
Date: Thu, 13 Jan 100 15:37:53 +1100 (EST)

> > 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).

Interesting way of putting it. I quite agree, the RSSA says that in
1999, the probability of being in Moscow in 2001 is 90%, and 10% of
being in Washington. But then effective probability is not related to
absolute measure, rather the conditional measure - so no problems here.

> 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/
>
>



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Dr. Russell Standish Director
High Performance Computing Support Unit,
University of NSW Phone 9385 6967
Sydney 2052 Fax 9385 6965
Australia R.Standish.domain.name.hidden
Room 2075, Red Centre http://parallel.hpc.unsw.edu.au/rks
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Received on Wed Jan 12 2000 - 20:35:18 PST

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