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

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

Date: Wed, 2 Jun 1999 22:51:09 -0700

On Wed, Jun 02, 1999 at 10:20:44PM -0700, hal.domain.name.hidden wrote:

Yes, exactly.

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

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