On Tue, 26 Oct 1999, Russell Standish wrote: [JM wrote] [&BTW I am getting tired of RS omitting the attribution] > > 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 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 ([EMAIL PROTECTED]) 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/