# Exploring Hempel’s Paradox of Confirmation

Ben Goertzel

September 14, 2005

I'm curious if anyone can shoot any holes in this line of argument...

To try to understand the Hempel confirmation paradox better, I decided to work out a very simple example in detail.  The simplest possible case involves a population size of 2.  In this case, so far as I can tell, the Hempel paradox doesn't arise according to a probabilistic analysis.  (But maybe I made a calculation error, I didn't spend that much time on it ;-)

Anyway, here goes....

Consider a bag containing 2 objects, one of which is known to be a raven.  For simplicity of discussion, suppose all the objects in the universe (including ravens) are either black or white.

Suppose there is also a midget in the bag who, when you ask him, will obey the command: "Show me a random non-black object from the bag."

There is a possibility that both objects in the bag are ravens.  In this case, there is no way to draw a random non-black object and get a non-raven, so the midget has to apologize and not give you anything.

The question is, suppose you see what the midget shows you and it's a non-raven.  Then does this observation affect your rational estimate of the probability

P(black|raven)

(where it is assumed that this probability is evaluated relative purely to the universe consisting of the two objects in the bag)?

There are seven distinguishable states for the interior of the bag, each of which one may assign a certain prior probability.

Raven     Non-raven

State 1:     B        B

State 2:     B          W

State 3:   W         B

State 4:     W          W

Raven     Raven

State 5:     B        B

State 6:     B          W

State 7:   W         W

(B=black, W=white)

(Note that we don't need a state of the form

Raven     Raven

W        B

because this would be identical to State 6: the definition of State 6 is simply that there are two ravens, one black and one white.)

Let the prior probability of each of these four states be called a_1, a_2, a_3, a_4, a_5, a_6, a_7.

If the midget gives you a nonblack entity when asked, that means the bag must be in states 1, 2, 3 or 4.

Now, suppose the midget has drawn a random nonblack entity from the bag and it's a nonraven.

This means the bag must be in States 2 or 4.

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)

So, after drawing a random nonblack entity and seeing it's a nonraven, the estimated odds that ravens are black in the population inside the bag should be b_2

Now suppose our initial assumption was that

a_2 = c * a_4

Then we find that

b_2 = a_2 / (a_2 + a_2/c) = c /(1+c)

b_4 = a_4 / (ca_4 + a_4) =  1/(1+c)

So,

b_2 = c * b_4

Now,

r = P(black|raven) / P(white|raven)

uniquely determines P(black|raven) via

P(black|raven) = r/(r+1)

So if the ratio is unchanged via the observation of a white nonraven, i.e. if

P_prior(black|raven) / P_prior(white|raven) =

P(black|raven & I have chosen a random white entity and found it to be a nonraven) /

P(white|raven & I have chosen a random white entity and found it to be a nonraven)

(as we see from the fact that b_2/b_4 = a_2/a_4) then

P_prior(black|raven) =

P(black|raven & I have chosen a random white entity and found it to be a nonraven)

which means that

P(black|raven)

is unchanged via the process of observing a random white entity and finding it to be a raven.

So there is no Hempel paradox in this case.  Here we are able to observe evidence increasing our estimate of

P(non-raven|non-black)

without affecting our estimate of

P(black|raven)

To see why we have increased our estimate of P(non-raven|non-black), we need to look at the three other possible states of the universe, ignored above:

Before observing any entities, the estimate one has is

P(non-raven|white) =

P(non-raven & white) / P(white) =

(.5 a_2 + .5 a_4) / (a_7 + .5 a_6 + .5 a_2 + .5 a_3 + a_4)

After observing a random white entity and observing it to be a non-raven, we know we are in states 2 or 4, and so our correct estimate is

P(non-raven|white) =

P(non-raven & white) / P(white) =

(.5 a_2 + .5 a_4) / (.5 a_2 +  a_4)

which is larger (since the possible states with no non-ravens have been removed from consideration).