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Suppose that I have access to a machine that allows me to input a positive integer (perhaps up to ten decimal digits) and the machine will -- depending only on the input -- output a statement.

If the input is odd, then the output will be either "the input number is a raven number" or the statement "the input number isn't a raven number."

If the input is even, then the output will be either "the input number is black" or "the input number isn't black."

Suppose that I claim to be looking for evidence to support the claim that for every positive, odd integer n, if n is a raven number, then (n + 1) is a black number.

However, suppose that I use the following procedure. I generate a random ten-digit even number n, and input it into the machine. If the machine says that the input isn't black, then I put (n - 1) into my list of numbers that I will not allow anybody to enter into the machine.

If the machine says that the even number n is black, then I enter (n - 1) into the machine. If the machine indicates that (n - 1) is a raven number, then I claim to have additional evidence to support the claim: "for every positive, odd integer n, if the machine classifies n as a raven number, then the machine classifies (n + 1) as a black number."

Now, there could be many odd numbers n, such that n is a raven number and (n + 1) is a black number, and there could also be many odd numbers n, such that n is a raven number, and (n + 1) isn't a black number. However, using the above procedure, I won't find any examples that disconfirm the claim.

Given that the procedure was designed to ensure that no examples that disconfirm the claim will be discovered using the procedure, how can anybody assert that the details discovered using the procedure tend to confirm the claim?

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    I think the question Title is wrong. Rather than "… both a raven and black …" should be "… neither a raven nor black …". See Raven paradox - Wikipedia – Ray Butterworth Oct 8 at 16:27
  • I assume you are talking about the infinite set of positive integers and are leaving the algorithms for determining "raven" and "black" unspecified so we cannot attempt to construct an induction step and use mathematical induction. – Frank Hubeny Oct 8 at 16:59
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    As designed, the procedure seems to assume what it sets out to confirm. The procedure is arranged to create only data that justifies a preordained conclusion. – Mark Andrews Oct 8 at 17:00
  • Does anybody actually assert something based on using your procedure? A reference might help. But real life ravens are unlikely to be modeled by it. So whatever Bayesian justifications can be offered for conclusions about ravens, they would not apply to what is described in the post. So it is unclear what you are asking. – Conifold Oct 9 at 0:06
  • @Conifold: I haven't seen it asserted with reference to the particular procedure that I described, but I have seen the claim that collecting more examples of objects that have both the raven quality and the blackness quality tends to confirm the hypothesis that all ravens are black, with no restriction specified on the procedure that is used. With no restriction specified, my procedure is available. – Ren Eh Daycart Oct 9 at 0:44
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Your setup and experiment are analogous to the following more general scenario:

Suppose you want to provide evidence for the claim that all As are Bs. To do so, you design an experiment that only ever looks at Bs, and willfully ignores anything that isn't a B. If you find a B that isn't an A, no big deal; this doesn't contradict your hypothesis that all As are Bs. If you find a B that is an A, you have confirming instance of your hypothesis that all As are Bs. This is a poorly designed experiment. To try and test the hypothesis that all As are Bs, you should start by looking at the set of available As, not the set of available Bs.

Here's a more concrete example. Suppose your hypothesis is that all university students will vote for candidate X. Instead of doing the natural thing and polling university students at random, you only poll people who are known supporters of candidate X. Among them will be university students, so that all university students you've polled will therefore be voting for candidate X. You've rigged your experiment to get the desired result.

Or, in your raven experiment, it's like only ever looking at black numbers. (Though your machine may find numbers that aren't black, by your own design you don't follow through and check the predecessor of that number to see if it's a raven number.) Once you only check the predecessors of black numbers, you'll only ever find a raven number whose successor is a black number.

Given that the procedure was designed to ensure that no examples that disconfirm the claim will be discovered using the procedure, how can anybody assert that the details discovered using the procedure tend to confirm the claim?

It would be irrational for them to assert this for the above reasons. They've designed their procedure poorly, by methodologically excluding anything that has the possibility of being a counterexample. The only way to find a counterexample to "all As are Bs" is to find an A which isn't a B, and if you're looking only at Bs, you're looking in the wrong place.

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What is the justification for the claim that observing something that is neither a raven nor black increases the likelihood that all ravens are black?

This isn't a formal answer, but it might help understand the reasoning behind the apparent paradox.

Suppose you have a box that contains N birds, R of which are ravens, and B of which are black. You don't know the values of any of those three numbers. If someone removes a white dove from the box, will that increase the probability that all the ravens are black?


Consider the trivial case of only two birds, knowing that one is black and one is a raven. There's a 50% chance that the raven is black, but if someone removes a white dove, there is a 100% chance that the remaining bird is a black raven.

As a small example, suppose there are N=5 birds, R=2 of which are ravens, and B=3 of which are black.

  • If someone removes a raven from the box, the probability that it is black is 3 / 5 (3 black birds out of 5).

  • If someone then removes the second raven from the box, the probability that it is black is 2 / 4 (2 black birds out of 4).

The probability that both ravens were black is thus 3/5 × 2/4, or 30%.

Now suppose someone removes a white dove from the box first, and then we repeat the experiment with the remaining 4 birds.

  • If someone removes a raven from the box, the probability that it is black is 3 / 4 (3 black birds out of 4).

  • If someone then removes the second raven from the box, the probability that it is black is 2 / 3 (2 black birds out of 3).

The probability that both ravens were black is thus 3/4 × 2/3, or 50%.

That is, removing the white dove did increase the probability that all the ravens were black.

If you look at what actually happened, that really isn't so surprising.


In the general case, if someone removes the ravens one at a time, the chances that the first one is black are R / N. The chances that the second one is black are (R-1) / (N-1). And so on.

If all R ravens are removed, and all of them turn out to be black, the chances of this happening are:

(B / N)  ×  ((B-1) / (N-1))  ×  …  ×  ((1+B-R) / (1+N-R)).

Now if instead, someone first removes a bird that isn't a raven and isn't black, say a white dove, and then we repeat this same experiment starting with N-1 birds, the chances that all the ravens are black will be (replacing N by N-1):

(B / (N-1))  ×  ((B-1) / (N-2))  ×  …  ×  ((1+B-R) / (N-R)).

Compare these two probability calculations:

  • There are R terms in each (one for each raven).
  • For each term the numerators are the same.
  • For each term, the second numerator is smaller than the first numerator.

That means that for each term the one in the second probability is larger than the first.

That means that the value of the final probability is greater for the first expression than for the second.

That is, by removing a white dove from the box, the probability that all ravens are black was increased.


The apparent paradox happens when you consider a much larger example, such as all the birds on Earth.

You wonder, "Are all ravens black?". A few minutes later, you see a white dove fly by and think, "It's looking more likely that all ravens are black.".

  • “If someone removes a raven from the box, the probability that it is black is 3 / 5 (3 black birds out of 5).” That probability calculation might work if you replace “birds” with “coins”, “ravens” with “dimes”, and “black” with “heads up”, provided that we have repeated trials, and our method of flipping coins produces a uniform distribution of random outcomes. – Ren Eh Daycart Oct 9 at 1:56
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The question is given a positive, odd integer n can we check two things

  1. n is a raven number based on some effective method run by a machine
  2. n + 1 is a black number based on another effective method run by a machine

and then claim, based on empirical testing of positive, odd integers up to 10 decimal digits, that the following is true

for every positive, odd integer n, if the machine classifies n as a raven number, then the machine classifies (n + 1) as a black number.

For a finite set of integers each with no more than 10 decimal digits there should be an answer, true or false. Furthermore, if one did find a raven m such that m + 1 is not black that would mean for all integers n, an infinite set, that the statement is false. However, that is not the case. By assumption we always find the raven-black pair.

Note that this problem is set up as a thought experiment so that there is no opportunity to use mathematical induction because we are deliberately not given enough information to create an induction step.

Intuitively we may not expect to find a counterexample, however, we have no other basis on which to assign a likelihood that we will never find a counterexample. We have to perform an infinite number of tests to find out. Since we can never check an infinite set even with an effective method for each test no likelihood can be assigned.

This problem is similar to whether there exist infinitely many prime pairs that Michael Dummett brings up in the context of God's omniscience and realism: Does God know whether there are infinitely many prime pairs or not?

Although there is an effective method for checking whether any particular natural number is prime and in principle God might know this, the number of primes is infinite. There is no effective method (so far) of checking whether there are infinitely many prime pairs without performing an infinite number of tests even for God.

Because of this Dummett writes: (page 350)

The constructivist allows that it is determinate, for every natural number, whether it is prime or composite; he denies that it follows that the proposition that there are infinitely many prime pairs is determinately either true or false.

Based on this analogous situation, and taking a constructivist position, the possibility of there being a raven-black pair is not determinately either true or false. It would be meaningless then to assess the likelihood that the proposition has either one or the other of these truth values.


Dummett, M. (1991). The logical basis of metaphysics. Harvard university press. Retrieved from Internet Archive at https://archive.org/details/logicalbasisofme0000dumm/page/349

  • The question mentioned "(perhaps up to ten decimal digits)", but you increased that first to "10 million decimal digits" and then to infinitely many possible integers (and therefore to representations via arbitrarily long sequences of Hindu-Arabic numerals). That seems to be a change of question. Are you familiar with the historical claim by constructive mathematicians that it is "meaningless" to assert that the digit sequence "0123456789" occurs in the decimal expansion of pi? It is now known that it does occur. It's not merely meaningful, but also true, and known to be true. – Ren Eh Daycart Oct 9 at 2:06
  • @RenEhDaycart I changed it to 10 decimal digits. – Frank Hubeny Oct 9 at 3:18

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