**From:** Vladimir Nesov (*robotact@gmail.com*)

**Date:** Mon Mar 17 2008 - 07:29:37 MDT

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On Mon, Mar 17, 2008 at 1:21 AM, Jeff L Jones <jeff@spoonless.net> wrote:

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*> So you agree with me in the case of conditionally destroying copies,
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*> but not in the case of conditionally creating copies? I don't
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*> understand how you could come to two different conclusions there.
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*> It's the same thing going on. The question is about what you will
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*> see, not about "what will happen". More specifically, I think the
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*> question is about what the best thing for you to anticipate seeing is.
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*>
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I'll try to restate the problem to eliminate explicit mention of

observers from it, until after the solution is found. I assume your

main issue with it will revolve around what "anticipate" actually

means. If you still have problem with it, please explain what

anticipation is, why is it different from probability and what it is

for.

Say, there is a box with a sphere in it. We now will do one of two

possible things, with 50% chance of doing either:

- in first case, we toss a red cube in it;

- in second case, we toss in it a green cube and 99 spheres identical

to the one that is already in the box.

What is the probability of finding a green cube in a box, if I tell

you that in this particular outcome there is also at least one sphere

in it? The same as just finding a green cube, 50%. And I know that

there is at least one sphere in the box, thank you very much, there is

at least one in each case.

Now, in the second experiment, we start with the same original

setting, one sphere in the box, and with 50-50 chance we now do one of

two following things:

- in first case, we toss a red cube in it;

- in second case, we toss in it a green cube, and additionally with

90% probability we remove a sphere from the box.

What is the probability of finding a green cube in a box, if I tell

you that in this particular outcome there is also at least one sphere

in it? Now this information about the presence of sphere is useful,

because in some outcomes in second cases there isn't one, it gets

removed.

Probability of finding a green cube is 50%, and of these 50% only in

0.5*0.1=0.05 of cases have a sphere, of total 0.5+0.05 of cases having

a sphere, so probability of finding a green cube is 1/11.

And, by the way, you are a remotely controlled robot whose mind is

implemented by a computer running inside one of those spheres, I won't

tell you which one. Lucky you there was at least one sphere after the

experiment.

-- Vladimir Nesov robotact@gmail.com

**Next message:**Peter de Blanc: "Re: Friendliness SOLVED!"**Previous message:**Peter de Blanc: "Re: Friendliness SOLVED!"**In reply to:**Jeff L Jones: "Re: Is a Person One or Many?"**Next in thread:**Stathis Papaioannou: "Re: Is a Person One or Many?"**Reply:**Stathis Papaioannou: "Re: Is a Person One or Many?"**Reply:**Jeff L Jones: "Re: Is a Person One or Many?"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

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