Jerry Clark wrote:
> > Let R be the ratio of blue to green SAS's, R = b/g. I want to compute
> > the probability distribution as a function of R on the domain [0,1]:
> I'm assuming you mean 'R = b/(b+g)'...
> > P(R) P(1 blue | R)
> > P(R | 1 blue) = ----------------------
> > P(1 blue)
> > Now, P(R) is the prior, and I said above that the assumption is that
> > this is constant over [0,1], and P(1 blue) = 50%, so P(R)/P(1 blue) is
> > a constant, and is just a normalizing factor.
> > Clearly, P(1 blue | R) = R. Therefore, with the normalizing condition
> > that the integral of P from 0 to 1 = 1, we get
> > P(R | 1 blue) = 2R
> > Which has an expection value for R of 2/3. That is, given a sample of
> > one blue SAS, and the prior mentioned above, I would compute that the
> > probablity of any SAS being blue is 2/3.
> > >
> > > >
> > > > So if the set of life-SAS's is not isomorphic to the set of (3+1
> > > > dim. pseudo-Riemannian manifold quantum field)-SAS's, then we'd
> > > > have no a priori reason to assume that the measures of these sets
> > > > are the same. If the measures are different, then one is larger
> > > > than the other. My money will be on our set having the larger
> > > > measure. If the measures
> > > > are transfinite but of different orders, then I conclude that the
> > > > probability of finding oneself to be a life-SAS is zero.
> > I stand by this. Note that I explicitly give a new prior: *if* the
> > measures of each set are of different orders of transfinite numbers.
> > Using blue and green again, that would mean that either R=0 or R=1.
> > We still have the ratio of prior probabilities, P(R)/P(1 blue), is a
> > constant. But now P(1 blue|R=0) = 0, and P(1 blue|R=1) = 1. The
> > normalization integral becomes a sum, and we get, simply,
> > P(R=0 | 1 blue) = 0
> > P(R=1 | 1 blue) = 1
> > Which is pretty easy to interpret.
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