**From:** Marcello Mathias Herreshoff (*m@marcello.gotdns.com*)

**Date:** Thu Sep 15 2005 - 00:50:37 MDT

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On Wed, Sep 14, 2005 at 10:36:32PM -0700, Eliezer S. Yudkowsky wrote:

*> Ben Goertzel wrote:
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*> >Anyway, if anyone feels like checking some interesting elementary
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*> >probability theory calculations, see:
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*> >
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*> >http://www.goertzel.org/new_essays/hempel.htm
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*> >
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*> >What I seem to have done there is to construct a simple case where there
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*> >seems to be no Hempel paradox according to elementary probability theory.
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*> >That is, I define a population of N entities containing at least one raven,
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*> >but where drawing a random nonblack entity from the population and
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*> >observing
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*> >it to be a nonraven does not change one's rational estimate of
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*> >P(black|raven). Unless I made a calculational error, in which case I would
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*> >be curious to discover what it is ;-)
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*> I find your lack of faith disturbing.
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*>
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*> I've got to sleep now, so this is a fine opportunity for aspiring
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*> probability theorists on SL4 to test their wings. Props for anyone who
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*> points out flaws in Ben's reasoning before I write a response tomorrow.
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*> These props are also available to Ben, but if he doesn't get it on his
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*> own, and I have to write up the flaws myself, I want Ben to never again
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*> claim that Bayesian probability is quote wrong unquote.
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This from http://www.goertzel.org/new_essays/hempel.htm :

*>> Given only this knowledge, the probability that the bag is in State 2 would
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*>> seem to be
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*>> b_2 = a_2 / (a_2 + a_4)
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*>> and the probability that the bag is in State 4 would seem to be
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*>> b_4 = a_4 / (a_2+ a_4)
*

It might at first sight seem to be true, but it isn't.

As a simplification let us only consider states two and four.

Raven Non-Raven

State 2: B W

State 4: W W

Let H be a random variable with value 2 or 4 for which state we are in.

Let E be the random variable with value r (the midget retrieves a Raven) or n

(the midget retrieves a non-raven.)

That is:

a_2 = P(H=2)

a_4 = P(H=4)

b_2 = P(H=2|E=n)

b_4 = P(H=4|E=n)

We want to calculate:

p(H=2|E=r) and p(H=4|E=n)

Now apply Bayes law:

p(H=2|E=n) = P(H=2)*P(E=n|H=2)/P(E=n)

p(H=4|E=n) = P(H=4)*P(E=n|H=4)/P(E=n)

Now we can divide to get the odds ratio:

b_2 / b_4 = (a_2*P(E=n|H=2))/(a_4*P(E=n|H=4))

Now Ben would be completely right about his assumption if:

P(E=n|H=2) = P(E=n|H=4)

However, this is not true:

P(E=n|H=2) = 1

P(E=n|H=4) = 1/2 (because there is another non-black object around for the

midget to stumble on, namely the raven)

Instead we get:

b_2/b_4 = 2*a_2/a_4

In short, the odds ratio for all ravens being black has been increased by a

factor of two when our midget procures a non-black object and it turns out to

not be a raven. This most definitely is evidence that all ravens are black!

-=+Marcello Mathias Herreshoff

**Next message:**Ben Goertzel: "RE: Hempel's paradox redux"**Previous message:**Eliezer S. Yudkowsky: "Re: Hempel's paradox redux"**In reply to:**Eliezer S. Yudkowsky: "Re: Hempel's paradox redux"**Next in thread:**Ben Goertzel: "RE: Hempel's paradox redux"**Reply:**Ben Goertzel: "RE: Hempel's paradox redux"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

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