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From: Wei Dai <weidai.domain.name.hidden>

Date: Tue, 21 Apr 1998 11:21:49 -0700

On Tue, Apr 21, 1998 at 03:03:47AM +0000, Nick Bostrom wrote:

*> Both are randomly choosen (according to the doomsdayer) but only from
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*> the set of *actual* names and birth ranks.
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Let's come back to this after we resolve the following technical issue.

*> > Since "There is at least one person called Nick" and "There is at least
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*> > one person called Sam" are not mutually exclusive, you can't sum over all
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*> > possible names and conclude that the real universe is twice as likely to
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*> > be type B before you learn your name.
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*>
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*> But "There is at least one person called Nick." *is* incompatible
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*> with "There is nobody called Nick.". So if the above reasoning were
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*> correct, then you would say: "Suppose initially the A-universe
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*> hypothesis is equally probable as the B-universe hypothesis.
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*> Suppose my name is Mr. X. Since P(There exists someone called Mr.
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*> X|A-universe hpothesis) is less than P(There exists someone called
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*> Mr. X| B-universe hypothesis). Therefore the B-hypthesis is more
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*> likely (even before you get to know what X stands for).
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I don't think probability theory allows you to reason this way. Suppose

you have four hypotheses A, B, D_1, D_2, such that P(A and B)=0, P(D_1 or

D_2)=1, P(D_1|A)=1, P(D_2|A)=1, P(D_1|B)=1/2, P(D_2|B)=1/2. You are not

allowed to then conclude P(A)>P(B). For example you could have P(A)=1/2,

P(B)=1/2, P(A and D_1)=1/2, P(A and D_2)=1/2, P(B and D_1)=1/4, P(B and

D_2)=1/4.

Received on Tue Apr 21 1998 - 12:50:46 PDT

Date: Tue, 21 Apr 1998 11:21:49 -0700

On Tue, Apr 21, 1998 at 03:03:47AM +0000, Nick Bostrom wrote:

Let's come back to this after we resolve the following technical issue.

I don't think probability theory allows you to reason this way. Suppose

you have four hypotheses A, B, D_1, D_2, such that P(A and B)=0, P(D_1 or

D_2)=1, P(D_1|A)=1, P(D_2|A)=1, P(D_1|B)=1/2, P(D_2|B)=1/2. You are not

allowed to then conclude P(A)>P(B). For example you could have P(A)=1/2,

P(B)=1/2, P(A and D_1)=1/2, P(A and D_2)=1/2, P(B and D_1)=1/4, P(B and

D_2)=1/4.

Received on Tue Apr 21 1998 - 12:50:46 PDT

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