RE: Copies Count

From: Stathis Papaioannou <>
Date: Mon, 20 Jun 2005 17:04:54 +1000

Hal Finney writes:

>Stathis Papaioannou writes:
> > Here is another way of explaining this situation. When there are
> > parallel copies of you, you have no way of knowing which copy you are,
> > although you definitely are one of the copies during any given moment,
> > no telepathic links with the others or anything like that. If a
> > of the copies are painlessly killed, you notice nothing, because your
> > successor OM will be provided by one of the copies still going (after
> > this is what happens in the case of teleportation). Similarly, if the
> > of copies increases, you notice nothing, because during any given moment
> > are definitely only one of the copies, even if you don't know which one.
> > However, if your quantum coin flip causes 90% of the copies to have bad
> > experiences, you *will* notice something: given that it is impossible to
> > know which particular copy you are at any moment, or which you will be
> > next moment, then there is a 90% chance that you will be one of those
> > has the bad experience. Similarly, if you multiply the number of copies
> > tenfold, and give all the "new" copies bad experiences, then even though
> > "old" copies are left alone, you will still have a 90% chance of a bad
> > experience, because it is impossible to know which copy will provide
> > next OM.
>I'm not sure I fully understand what you are saying, but it sounds like
>you agree at least to some extent that "copies count". The number of
>copies, even running in perfect synchrony, will affect the measure of
>what that observer experiences, or as you would say, his subjective
>So let me go back to Bruno's thought experiment and see if I understand
>you. You will walk into a Star Trek transporter and be vaporized and
>beamed to two places, Washington and Moscow, where you will have two
>(independent) copies wake up. Actually they are uploads and running on
>computers, but that doesn't matter (we'll assume). Bruno suggests that
>you would have a 50-50 expectation of waking up in Washington or Moscow,
>and I think you agree.
>But suppose it turns out that the Moscow computer is a parallel
>processor which, for safety, runs two copies of your program in perfect
>synchrony, in case one crashes. Two synchronized copies in Moscow,
>one in Washington.
>Would you say in this case that you have a 2/3 expectation of waking up
>in Moscow?
>And to put it more sharply, suppose instead that in Washington you will
>have 10 copies waking up, all independent and going on and living their
>lives (to the extent that uploads can do so), sharing only the memory
>of the moment you walked into the transporter. And in Moscow you will
>have only one instance, but it will be run on a super-parallel computer
>with 100 computing elements, all running that one copy in parallel and
>So you have 10 independent copies in Washingon, and 100 copies that
>are all kept in synchrony in Moscow. What do you expect then? A 90%
>chance of waking up in Washington, because 9/10 of the versions of you
>will be there? Or a 90% chance of waking up in Moscow, because 9/10 of
>the copies of you will be there?
>I think, based on what you wrote above, you will expect Moscow, and that
>"copies count" in this case.
>If you agree that copies count when it comes to spatial location, I
>wonder if you might reconsider whether they could count when it comes
>to temporal location. I still don't have a good understanding of this
>situation either, it is counter-intuitive, but if you accept that the
>number of copies, or as I would say, measure, does make a difference,
>then it seems like it should apply to changes in time as well as space.

I agree that you will have a 90% chance of waking up in Moscow, given that
that is the *relative* measure of your successor OM when you walk into the
teleporter. This is the only thing that really matters with the copies, from
a selfish viewpoint: the relative measure of the next moment:

(a) If you are copied 100 times and 99% of the copies tortured, you will
certainly know this, as there is a 99% subjective probability that you will
be tortured.

(b) If you are copied 100 times and the copies allowed to diverge, then 99%
of the copies painlessly killed, that means you have a 99% chance of being
killed, because in two steps, (i) there is a 100% chance your next OM will
be one of the 100 copies; and (ii) there will be a 99% chance that you will
have become one of the copies that will be killed, and if you are, then
there will be 0% chance that you will have any "next moment".

(c) If you are copied 100 times and all the copies are kept running in
parallel, then 99% of the copies painlessly killed, you can't possibly know
that anything odd has happened at all, because there is a 100% chance that
your next OM will come from the one remaining copy. Similarly if there were
10^100 copies *kept running in perfect sync* and all but one terminated.

--Stathis Papaioannou

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Received on Mon Jun 20 2005 - 03:06:43 PDT

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