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The Ravens Paradox

 

Hempel first discovered the ravens paradox in 1965. It consists of five statements:

  1. All ravens are black is logically equivalent to all non-black things are non-raven (assuming a finite number of things in the world).

  2. Providing all the ravens that I find are black, the more I find the more likely it is that all ravens are black.

  3. Mimicking statement [2], the more non-black non-ravens I find the more likely it is that all non-black things are non-raven.

  4. Using statement [1] we can conclude that the more non-black non-ravens I find the more likely it is that all ravens are black (statements [3] and [4] are logically equivalent).

  5. Common sense dictates that we can learn nothing about the colour of ravens by looking at non-ravens.

 

Black Raven                                                                 Non-Black Non-Raven

Black Raven                                                    Non-Black Non-Raven

 

 

Statements [4] and [5] contradict each other and this is the root of the paradox.

One of the five statements must be incorrect. Let us examine them in turn.

 

[1] This statement is one of pure logic and is correct (in the world of logic).

 

 

[2] If this statement is true we should be able to mathematically calculate the probabilities. We can, and we can use them to prove that for each new raven (which must be black) that is found the probability that all ravens are black is increased.

For the mathematical proof of statement [2]; click here

 

[3] and [4] Similarly, if these statements are true, we should be able to find the probabilities. We can, and they lead us to two surprising results:

  1. That the more non-black non-ravens one finds the higher the probability that all ravens are black.
  2. That more black non-ravens one finds the lower the probability that all ravens are black.

For the mathematical proof of statements [3] and [4]; click here

 

We have now proved statements [1], [2], [3] and [4] but we are still left with the contradictory statement [5]

 

 

Conclusion

We have proved that:

  1. In a finite thing problem it is possible to calculate the probability that all ravens are black if the number of black and non-black things are known and the number of ravens and non-ravens.

  2. This probability increases when either black ravens and non-black non-ravens are found but that this probability decreases when black non-ravens are found. (Clearly, finding a non-black raven reduces this probability to zero.)

 

However this does not make life any easier for our ornithologist (and we can therefoe resolve the apparently contradictory statement [5]). Although he can increase the probability that all ravens are black that is all he can do. He can never learn anything useful. Neither the number of ravens / non-ravens or black / non-black things is or can ever be known in the real world. This means our ornithologist can never generate the probability that all ravens are black. He may know it is increasing but it will always lie between 0 and 1. Where it is on that line he will never know and can therefore learn nothing without studying ravens.

 

 

Further Points

  1. As the number of ravens in the world increases (if we know the elusive total figures) it becomes more important to find non-ravens. Indeed evidence of non-black non-ravens lends greater confirmation to the hypothesis that “All ravens are black” than evidence of black ravens when there are more ravens than non-ravens.

  2. The confusion of statement [5] (that nothing can be learnt about ravens from non-ravens) ocurrs because there are so many more non-ravens in the world than ravens. Referring to point (1) above, we can see that because of this the evidence of non-black non-ravens offers far less confirmation than evidence of black ravens, so much so that common sense takes it as zero.

  3. In the same way that finding a white shoe (non-black non-raven) increase the probability that all ravens are black it also increases the probability that all ravens are yellow. One piece of evidence can be used to support two contradictory hypotheses. A clear example of this fact, that is so easy to abuse in science, can be seen if you imagine somebody who doesn't know what day of the week it is. If you tell them it isn't Monday, you increase the probability in their minds that it is Tuesday, Wednesday or indeed any of the other days (except Monday). Your one comment is evidence to all of the six contradictory claims that it is any other particular day.

 


 

For the mathematical proof of statement [2]; click here

For the mathematical proof of statements [3] and [4]; click here

 


     
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Mathematical proof for statement [2]

 

Taking the:

Fraction of ravens that are black   = P

Total No. of Ravens looked at       = N

Total No. that are black                  = B

 

Assuming random picking of ravens, we can use the Binomial theorem for a two-result question (Is the raven black?). This gives us:

Probability of finding B black ravens out of N ravens = PB . (1-P)(N-B)

 

Finding and non-black ravens immediately disproves the hypothesis that all ravens are black, so let us assume that all the ravens we find are black. i.e. let us take the important case of B = N. Now we have:

Probability of finding B black ravens = PB

Which, fortunately is less than or equal to one as P < 1

 

Let us finally consider, the probability of finding one more black raven than we have already. i.e. B + 1 ravens.

Probability of finding B+1 black ravens = PB+1

We can clearly see that: PB+1 < PB

 

If we assume ourselves not to be infinitely lucky (i.e. we aren’t so lucky (or unlucky) that we can go on picking black ravens when in reality they aren’t all black) we must set a confidence interval. e.g. we cannot accept a less than 2% chance of picking a certain sample (of all black) ravens, we have to raise our estimate of P.

 

Once a confidence level has been set and reached, at a certain level of P.

(e.g. Confidence level of 2% (Probability = 0.02)

Estimate for P = 0.75 (i.e a three-quarters of all ravens are black)

Confidence level reached at only 6 ravens. (P6 < 0.02))

 

Then if we find another raven that is black, we are forced to raise our estimate of P (the chances of finding 7 ravens at P = 0.75 are just too low to reasonably accept).

 

So, finding a black raven raises our estimate for P, which is the same as raising our estimate for the proportion of ravens that are black, which is the same as raising the probability that in fact all ravens are black. We have therefore proved statement [2], which was “Providing all the ravens I find are black, the more I find the more likely it is that all ravens are black.”

Click here to return to the main argument

 

Mathematical proof for statements [3] and [4]

 

We are now considering the case where we are not looking at ravens (they move!), only non-ravens. Let us take:

Total No. of things in the world                     = T

Total No. of Ravens in the world                  = R

Total No. of non-black things in the world    = W

 

Using Combination theory we find:

Probability that all ravens are black =

 

If we then take Pand 0 as the probability that all ravens are black when T= T0 W = Wfound 0 and P1 as the probability that all ravens are black when we have one more non-black non-raven we have:

 

We can follow the maths through to show that as:         

so is P0 < P1

 

i.e. the surprising result that the more non-black non-ravens one finds the higher the probability that all ravens are black.

 

Note: By similar methods it can also be shown that more black non-ravens one finds the lower the probability that all ravens are black.

Click here to return to the main argument

 

     
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