Re: Doomsday Argument (was: a baysian solution)

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Date: Wed, 22 Apr 1998 20:59:07 -0700

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On Wed, Apr 22, 1998 at 10:41:10PM +0000, Nick Bostrom wrote:
> Wei Dai <weidai.domain.name.hidden> wrote:
> > 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.
>
> In the hypothetical reasoning I gave, D_1 is the negation of D_2.
> (Either there exists someone called Mr. X, or it is not the case that
> there exists someone aclled Mr. X.". This implies P(A and D_1)+P(A
> and D_2) = P(A). This condition isn't satisfied for the numbers you
> give, so your counterexample fails.

I don't understand your reasoning then. I thought you were saying that
P(universe is type A | There exist someone called Mr. X) > 1/2 for all X
and "There exist someone called Mr. X" is true for some X, therefore
P(universe is type A) > 1/2. This is analogous to saying P(A|D_i) > 1/2
for all i and D_i is true for some i, therefore, P(A) > 1/2. My example
shows this kind of reasoning is not valid. If this is not your reasoning,
then please explain how you derived P(A) > 1/2.

Something strange is going on here, but I don't know what. If I'm right,
everyone in this scenario should believe the universe is equally likely to
be type A as type B and then change his mind after learning his name. A
simple Dutch book argument suggests this is irrational, but as far as I
can see the probability computations are correct. May I'm missing
something obvious.
Received on Wed Apr 22 1998 - 20:59:09 PDT