Wei Dai <weidai.domain.name.hidden> wrote:
> > > Since "There is at least one person called Nick" and "There is at least
> > > one person called Sam" are not mutually exclusive, you can't sum over all
> > > possible names and conclude that the real universe is twice as likely to
> > > be type B before you learn your name.
> >
> > But "There is at least one person called Nick." *is* incompatible
> > with "There is nobody called Nick.". So if the above reasoning were
> > correct, then you would say: "Suppose initially the A-universe
> > hypothesis is equally probable as the B-universe hypothesis.
> > Suppose my name is Mr. X. Since P(There exists someone called Mr.
> > X|A-universe hpothesis) is less than P(There exists someone called
> > Mr. X| B-universe hypothesis). Therefore the B-hypthesis is more
> > likely (even before you get to know what X stands for).
>
> 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.
_____________________________________________________
Nick Bostrom
Department of Philosophy, Logic and Scientific Method
London School of Economics
n.bostrom.domain.name.hidden
http://www.hedweb.com/nickb
Received on Wed Apr 22 1998 - 14:47:57 PDT