> -----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 :-) (SNIP) > >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. Charles
