>
> On Tue, 26 Oct 1999, Russell Standish wrote:
> [JM wrote] [&BTW I am getting tired of RS omitting the attribution]
^^^ Blame my email software. I almost always leave the .signatures in
to make it obvious who I'm responding to.
> > > That's total BS.
> > > I'll review, although I've said it so many times, how effective
> > > probabilities work in the ASSA. You can take this as a definition of
> > > ASSA, so you can NOT deny that this is how things would work if the ASSA
> > > is true. The only thing you could try, is to claim that the ASSA is
> > > false.
> > > The effective probability of an observation with characteristic
> > > 'X' is (measure of observations with 'X') / (total measure).
> > > The conditional effective probability that an observation has
> > > characteristic Y, given that it has characteristic X, is
> > > p(Y|X) = (measure of observations with X and with Y) / (measure with X).
> > > OK, these definitions are true in general. Let's apply them to
> > > the situation in question.
> > > 'X' = being Jack Mallah and seeing an age for Joe Shmoe and for
> > > Jack Mallah, and seeing that Joe also sees both ages and sees that Jack
> > > sees both ages.
> >
> > I shall take X = being Jack Mallah. The rest is irrelevant.
> >
> > > Suppose that objectively (e.g. to a 3rd party) Jack and Joe have
> > > their ages drawn from the same type of distribution. (i.e. they are the
> > > same species).
> > > Case 1: 'Y1' = the age seen for Joe is large.
> > > Case 2: 'Y2' = the age seen for Jack is large.
> > > Clearly P(Y1|X) = P(Y2|X).
> >
> > Sorry, not so clear. It is true by symmetry that p(Y1)=p(Y2).
> >
> > p(Y1|X) = p(Y1&X)/p(X)
> > p(Y2|X) = p(Y2&X)/p(X)
> >
> > Why do you assume p(Y1&X) = p(Y2&X)? I can see no reason. They
> > certainly aren't symmetrical. About all one can say from symmetry is
> > p(Y1&X) = p(Y2&Z), where Z = being Joe Schmoe.
>
> I must disrespectfully disagree.
> It is obvious that p(Y1&X) = p(Y1&Z), because in all instances
> in
It is not obvious, for the same reason that p(Y1&X) = p(Y2&X) is not obvious.
If QTI is true, then it is clearly not true. Don't assume what you're
trying to prove.
> which there is an observation with Y1 & X, there is observation by Joe
> Shmoe with Y1 & Z, of equal measure. That's why I added the extra
> conditions, to make it real obvious.
> (Since there are no near-zombies in the ASSA. They are both there
> (in that branch/universe), both with human brains, so they get the same
> measure.)
> So p(Y1&X) = p(Y2&Z) = p(Y1&Z). OK we have shown it for Joe,
> Jack's case works the same way: p(Y1&X) = p(Y2&X).
>
> - - - - - - -
> Jacques Mallah (jqm1584.domain.name.hidden)
> Graduate Student / Many Worlder / Devil's Advocate
> "I know what no one else knows" - 'Runaway Train', Soul Asylum
> My URL: http://pages.nyu.edu/~jqm1584/
>
>
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Dr. Russell Standish Director
High Performance Computing Support Unit,
University of NSW Phone 9385 6967
Sydney 2052 Fax 9385 6965
Australia R.Standish.domain.name.hidden
Room 2075, Red Centre
http://parallel.hpc.unsw.edu.au/rks
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Received on Wed Nov 03 1999 - 22:49:55 PST