= Solution
The <raven paradox> combines a plausible equivalence principle with an overly unrestricted account of positive-instance <scientific confirmation>. Let $R(x)$ and $B(x)$ mean that $x$ is a raven and black respectively. Then
$$
H:\ \forall x\,(R(x)\Rightarrow B(x))
\quad\Longleftrightarrow\quad
\forall x\,(\neg B(x)\Rightarrow\neg R(x)).
$$
A white shoe is a positive instance of the right-hand generalization. If every positive instance confirms its universal generalization, and <logically equivalent> statements receive the same evidential support, inspecting the shoe seems to confirm the original raven claim. There is no contradiction in the equivalence itself. The tension concerns the evidential rule and the intuition that an arbitrary shoe teaches us nothing about birds.<Studies in the Logic of Confirmation I by Carl Hempel>[Section 5: instance confirmation, equivalent hypotheses and nonblack nonravens]{ref}
The right response is to make background knowledge and sampling explicit. Under <Bayesian confirmation>, evidence increases support for $H$ precisely when its <likelihood> is greater under $H$ than under the alternative, assuming nondegenerate <prior probabilities>. A fact can therefore be compatible with $H$ yet provide no new support. The mere existence of a white shoe does not determine the needed <likelihood ratio>.<How Bayesian Confirmation Theory Handles the Paradox of the Ravens by Branden Fitelson and James Hawthorne>[Sections 1–3: background information, competing confirmation conditions and quantitative comparisons]{ref}
Consider a deliberately simple finite model. There are $N$ objects, exactly $r$ ravens, and $M=N-r$ nonravens, with $M>r>1$. All nonravens are white. Under $H_0$ all ravens are black; under $H_1$ exactly one raven is white. Draw uniformly from the entire population. The event of drawing a white nonraven has <probability> $M/N$ under either <hypothesis>. Its <Bayes factor> is one and it leaves the <prior odds> unchanged. A black raven instead has <likelihoods> $r/N$ and $(r-1)/N$, so gives <Bayes factor> $r/(r-1)$ for $H_0$.
Now change the protocol: draw uniformly from the \i[white objects], then discover that the selected object is a nonraven. This result has <likelihoods> $1$ and $M/(M+1)$. It gives the small positive <Bayes factor>
$$
\boxed{\mathrm{BF}_{01}=\frac{M+1}{M}.}
$$
For $N=1000$ and $r=10$, this is $991/990$, whereas sampling a black raven from the full population gives $10/9$. The white nonraven can now help because the search could have revealed a white raven. The two experiments have different selection information, so their differing evidential values do not violate <logical equivalence>.
This model establishes possibilities, not a universal numerical law about shoes. Different alternatives, population information or selection procedures change the result. If the shoe's being a nonraven was already known, discovering that it is white need not add anything. Conversely, ruling out a potential counterexample can supply a small amount of information. \b[Retain logical equivalence, but reject the unconditional instance-confirmation rule.] Once the complete evidential situation is specified, neither zero support nor weak positive support is paradoxical, and neither amounts to proof of a universal law.
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