From: Marcello Mathias Herreshoff (m@marcello.gotdns.com)
Date: Thu Sep 15 2005 - 00:50:37 MDT
On Wed, Sep 14, 2005 at 10:36:32PM -0700, Eliezer S. Yudkowsky wrote:
> Ben Goertzel wrote:
> >Anyway, if anyone feels like checking some interesting elementary
> >probability theory calculations, see:
> >
> >http://www.goertzel.org/new_essays/hempel.htm
> >
> >What I seem to have done there is to construct a simple case where there
> >seems to be no Hempel paradox according to elementary probability theory.
> >That is, I define a population of N entities containing at least one raven,
> >but where drawing a random nonblack entity from the population and
> >observing
> >it to be a nonraven does not change one's rational estimate of
> >P(black|raven). Unless I made a calculational error, in which case I would
> >be curious to discover what it is ;-)
> I find your lack of faith disturbing.
>
> I've got to sleep now, so this is a fine opportunity for aspiring
> probability theorists on SL4 to test their wings. Props for anyone who
> points out flaws in Ben's reasoning before I write a response tomorrow.
> These props are also available to Ben, but if he doesn't get it on his
> own, and I have to write up the flaws myself, I want Ben to never again
> claim that Bayesian probability is quote wrong unquote.
This from http://www.goertzel.org/new_essays/hempel.htm :
>> Given only this knowledge, the probability that the bag is in State 2 would
>> seem to be
>> b_2 = a_2 / (a_2 + a_4)
>> and the probability that the bag is in State 4 would seem to be
>> 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
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