From: Ben Goertzel (ben@goertzel.org)
Date: Thu Sep 15 2005 - 06:38:13 MDT
Yes, this is the same as the "two coins" analogy, it's just more fully
formalized...
ben
> 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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