Perhaps a Bayesian approach would be illuminating. There are four kinds of objects in the world: black ravens, nonblack ravens, black nonravens, and nonblack nonravens. Call these A, B, C, and D. Let the probability you assign to the next object that you encounter being in one of these classes be p, q, r, and s respectively. Rather than having two competing hypotheses about the blackness of ravens, there is a prior distribution of the parameters p, q, r, and s.
(Note that the way I've set this up removes any concept of blackness common to black ravens and black nonravens. The astute -- more astute than me, for whom this is the last paragraph written -- may guess at once that P naq Q ner tbvat gb or rkpunatrnoyr va guvf sbezhyngvba, naq gurersber arvgure zber guna gur bgure pna or rivqrapr eryngvat gb gur inyhrf bs c naq d. I come back to this at the end.)
In a state of total ignorance, a reasonable prior for the distribution of (p,q,r,s) is that they are uniformly distributed over the tetrahedron in four-dimensional space defined by these numbers being in the range 0 to 1 and their sum being 1.
After observing numbers a, b, c, and d of the four categories, the posterior is (after a bit of mathematics) p^a q^b r^c s^d/K(a,b,c,d), where K(a,b,c,d) = a!b!c!d!/(N+3)!, where N = a+b+c+d. (The formula generalises to any number of categories, replacing 3 by the number of categories minus 1.)
The expectation value of p is K(a+1,b,c,d)/K(a,b,c,d) = (a+1)/(N+4), and similarly for q, r, and s. (Check: these add up to 1, as they should.)
How does the expectation value of p change when you observe that the N+1'th object you draw is an A, B, C, or D?
If it's an A, the ratio of the new expectation value to the old is (a+2)(N+4)/(a+1)(N+5). For large N this is approximately 1 + 1/(a+1) - 1/(N+5) > 1.
If it's a B (and the cases of C and D are the same) then the ratio is (N+4)/(N+5) = 1 - 1/(N+5) < 1.
So observing an A increases your estimate of the proportion of the population that are A, and observing anything else decreases it, as one would expect. That was just another sanity check.
Now consider the ratio q/p, the ratio of non-black to black ravens. The expectation of this, assuming a>0 (you have seen at least one black raven), is K(a-1,b+1,c,d)/K(a,b,c,d) = (b+1)/a. This increases to (b+2)/a when you observe a nonblack raven, and decreases to (b+1)/(a+1) when you observe a black one. (I would have calculated the expectation of q/(q+p), the expected proportion of ravens that are nonblack, but that is more complicated.)
If you have seen a thousand black ravens and no nonblack ones, the increase is from 1/1000 to 2/1000, i.e. a doubling, but the decrease is from 1/1000 to 1/1001, a tiny amount. On the log-odds scale, the first is 1 bit, the second is about 0.0014 bits.
On this analysis, observations of nonravens, whether black or not, have no effect on the expectation of the proportion of nonblack ravens.
If we reformulate the original hypothesis that all ravens are black as "q/p < 0.000001", then observing the 1001th raven to be green will pretty much kill that hypothesis, until we see of the order of a million black ravens in a row without a nonblack one. But the nonraven objects will continue to be irrelevant: C and D are exchangeable in this formulation of the problem.
Now reconsider the original paradox on its own terms. I will draw a connection with the grue paradox.
Suppose we accept the paradoxical argument that "All ravens are black" and "all nonblack things are nonravens" are logically equivalent, and therefore everything that is evidence for one is evidence for the other.
Let "X is bnonb" mean "X is a black raven or a nonblack nonraven." Consider the hypothesis that all ravens are bnonb, and its contrapositive, that all non-bnonb things are nonravens. In effect, we have exchanged C and D, but not A and B. Every argument that nonblack nonravens are evidence for all ravens being black is also an argument than nonbnonb nonravens are evidence for all ravens being bnonb. But substituting the definition of bnonb in the latter, it claims that black nonravens are evidence for the blackness of ravens. Hence both black and nonblack nonravens support the blackness of ravens.
But there's more. Swapping black and nonblack in all of the above would imply that both black and nonblack nonravens are evidence for the nonblackness of ravens.
At this point we appear to have proved that all nonravens are evidence for every hypothesis about ravens. I don't think the original paradox can be saved by arguing that yes, nonblack nonravens are evidence, just an utterly insignificant amount, as some do.
A further elaboration then occurred to me. If non-ravens are, as the above argument claims, not evidential for the properties of ravens, then neither are non-European ravens evidential for the properties of European ravens, which does not seem plausible. This amount of confusion suggests that some essential idea is missing. I had thought causality or mechanism, but the Google search suggested by that turned up this paper: "Infinitely many resolutions of Hempel's paradox" by Kevin Korb, which takes a purely Bayesian approach, which I think has som...
The raven paradox, originated by Carl Gustav Hempel, is an apparent absurdity of inductive reasoning. Consider the hypothesis:
H1: All ravens are black.
Inductively, one might expect that seeing many black ravens and no non-black ones is evidence for this hypothesis. As you see more black ravens, you may even find it more and more likely.
Logically, a statement is equivalent to its contrapositive (where you negate both things and flip the order). Thus if "if it is a raven, it is black" is true, so is:
H1': If it is not black, it is not a raven.
Take a moment to double-check this.
Inductively, just like with H1, one would expect that seeing many non-black non-ravens is evidence for this hypothesis. As you see more and more examples, you may even find it more and more likely. Thus a yellow banana is evidence for the hypothesis "all ravens are black."
Since this is silly, there is an apparent problem with induction.
Resolution
Consider the following two possible states of the world:
Suppose that these are your two hypotheses, and you observe a yellow banana (drawing from some fixed distribution over things). Q: What does this tell you about one hypothesis versus another? A: It tells you bananas-all about the number of black ravens.
One might contrast this with a hypothesis where there is one less banana, and one more yellow raven, by some sort of spontaneous generation.
Observations of both black ravens and yellow bananas cause us to prefer 1 over 3, now!
The moral of the story is that the amount of evidence that an observation provides is not just about whether it whether it is consistent with the "active" hypothesis - it is about the difference in likelihood between when the hypothesis is true versus when it's false.
This is a pretty straightforward moral - it's a widely known pillar of statistical reasoning. But its absence in the raven paradox takes a bit of effort to see. This is because we're using an implicit model of the problem (driven by some combination of outside knowledge and framing effects) where nonblack ravens replace black ravens, but don't replace bananas. The logical statements H1 and H1' are not alone enough to tell how you should update upon seeing new evidence. Or to put it another way, the version of induction that drives the raven paradox is in fact wrong, but probability theory implies a bigger version.
(Technical note: In the hypotheses above, the exact number of yellow bananas does not have to be the same for observing a yellow banana to provide no evidence - what has to be the same is the measure of yellow bananas in the probability distribution we're drawing from. Talking about "99 ravens" is more understandable, but what differentiates our hypotheses are really the likelihoods of observing different events [there's our moral again]. This becomes particularly important when extending the argument to infinite numbers of ravens - infinities or no infinities, when you make an observation you're still drawing from some distribution.)