# Re: normalization

From: <GSLevy.domain.name.hidden>
Date: Tue, 18 Jan 2000 16:27:49 EST

In a message dated 01/18/2000 1:09:02 PM Pacific Standard Time,
jqm1584.domain.name.hidden writes:

> On Tue, 18 Jan 2000 GSLevy.domain.name.hidden wrote:
> > jqm1584.domain.name.hidden writes:
> > > The RSSA is not another way of viewing the world; it is a
> > > category error.
> >
> > I use the RSSA as the basis for calculating what I call the relative
> > probability, in this group the first person probability, or,
equivalently,
>
> > the probability conditional on the life of the observer. The ASSA is by
> > extension, the assumption for calculating the 3rd person probability.
> >
> > Let us perform a thought experiment.
> > Imagine that you are the scientist in the Schroedinger cat experiment.
>
> Scratch that. Right now let's stick to the example with Bruno and
> the 3 cities, because it's better for the current point.
> Suppose Bruno, in 1999, wants to know if he is more likely to be
> in Washington or in Moscow during 2001.
> First of all, that is not a well defined question, because
> "Bruno" must be defined. Suppose we define it to mean the set of all
> Bruno-like observations, where by "Bruno-like" we can assume we know what
> qualifies.
> But then the question becomes meaningless, because it is 100%
> certain that he will be in *both* cities. A 3rd person would have to
> agree with that, he is in *both* cities.
> So let's ask a meaningful question. Among the set of Bruno-like
> observations in 2001, what is the effective probability of such an
> observation being in Moscow?
> This is just a conditional effective probability so we use the
> same rule we always use:
> p(Moscow|Bruno in 2001) =
> M(Moscow, Bru. 2001) / [M(Moscow, Bru. 2001) + M(Washington, Bru. 2001)]
> where M is the measure.
> So in this case the conditional effective probability of him
> seeing Moscow at that time is 10%, and in *1999* he knows he should brush
> up on his English because his future 'selves' will be affected by that.
>

Fine, you have computed the third person probability. Unfortunately, your
example does not have the option of having an independent observer, and
therefore does not illustrate the concept I am trying to communicate.