> -----Original Message-----
> From: Jacques Mallah [mailto:[EMAIL PROTECTED]]
>     Nope!  It's just that with FIN, your expected age
> diverges.  If you want
> to say that's impossible, fine with me.  FIN is logically
> impossible for a
> sane person to believe!
>     But there's one exception: your brain can only hold a
> limited amount of
> information.  So it's possible to be too old to remember how
> old you are.
> *Only if you are that old, do you have a right to not reject
> FIN on these
> grounds.*  Are you that old?
>     (Of course, you must still reject it on other grounds!)

Yeah, that's one of my objections to QTI. Although perhaps add-on memory chips will 
become available one day :-)


> >That's OK so far. And it turns out correctly for most cases (i.e.
> >99.99999999(etc)% of observers WILL turn out to have ages of
> infinity (if
> >QTI etc)). But an infinitesimal fraction won't - including
> everyone you
> >observe around you (the multiverse is very very very (keep
> typing "very"
> >til doomsday) big! (assuming MWI)).
>     Right.  Do you think you are in an infinitesimal
> fraction, or in a
> typical fraction?

Infinitesimal, if QTI is correct, otherwise fairly typical. Assuming QTI is correct 
and ignoring any other objections to it, it's
*possible* for me to be in an infinitesimal fraction - in fact it's necessary.

> > > In the same way, the SSA helps you guess things.  It's
> just a procedure
> >to follow which usually helps the people that use it to make correct
> >guesses.
> >
> >It doesn't seem to help in this case though. I don't need to
> guess my age,
> >it's a given.
>     Maybe the following example will help.
>     Suppose there are two possibilities:
> 1.  90% of people see A, 10% see B
> 2.  10% of people see A, 90% see B
>     You see A.  But you want to know whether #1 or #2 is
> true.  A priori,
> you feel that they are equally likely to be true.  Should you
> throw up your
> hands simply because both #1 and #2 are both consistent with your
> observation?  No.  So use Bayes' theorem as follows:
> p(1|A) = [p(A|1) p_0(1)] / [p(A|1) p_0(1) + p(A|2) p_0(2)]
>        = [  (.9)  (.5) ] / [  (.9)  (.5)  +   (.1) (.5)  ] = .9
>     So you now think #1 is 90% likely to be true, if you use
> this procedure.
>   So you will guess #1.  OK, lets try and check to see if
> this procedure is
> good.
> If #1 is true then 90% of people who use the procedure guess
> #1 (right).
> If #2 is true then 10% of people who use the procedure guess
> #1 (wrong).
>     Well I'd say that's pretty good, and also the best you can do.
>     I gotta go.

I agree, but according to QTI I *must* pass through a phase when I see the unlikely 
bits, no matter how unlikely it is that a
typical moment will fall into that phase. Even if I later spend 
99.9999999999999999999....% of my observer moments seeing the stars
going out one by one, there still has to be that starting point!

Maybe we should resume this disussion in 1,000,000 years - or not, as the case may be.


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