> -----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
> >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
> 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.