Re: practical reasoning and strong SSA

From: Wei Dai <weidai.domain.name.hidden>
Date: Wed, 2 Jun 1999 22:51:09 -0700

On Wed, Jun 02, 1999 at 10:20:44PM -0700, hal.domain.name.hidden wrote:
> So, are you saying that P(I observe "N[Pi]=3.14159" | PI == 3.14159)
> should be considered the probability, given that PI has that value, that
> a randomly chosen observer-moment is of me, sitting at a computer, seeing
> this value display, out of all observer-moments in all possible universes?
>
> In that case the probability would be extremely low, since only a tiny
> fraction of all observer-moments would consist of me doing that. Maybe
> it would be something like 1E-100.
>
> However, the probability P(I observe "N[Pi]=3.14159" | PI != 3.14159) would
> be even lower, since not only would the randomly chosen observer moment
> have to be me running mathematica, but it would also be necessary that
> mathematica is wrong. Maybe this probability would be 1E-110.
>
> We would then run the Bayesian formula with these two extremely low
> conditional probabilities. From this we would conclude that
> P(PI == 3.14159 | I observe "N[Pi]=3.14159") would be high, as we expect.
>
> Is this how you would work this problem, based on the strong SSA?

Yes, exactly.

> I'm not sure I am on the right track here...

You are. :) Now the point I was trying to make was that without the strong
SSA, P(I observe "N[Pi]=3.14159" | PI == 3.14159) would be ill-defined, so
we could not conclude that P(PI == 3.14159 | I observe "N[Pi]=3.14159")
is high. Do you agree?
Received on Wed Jun 02 1999 - 22:54:31 PDT