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Old evidence challenges a simple account of confirmation as actual conditional updating, but does not refute every probabilistic treatment of the raven paradox. The strongest response distinguishes support from newly acquired information and makes the sampling procedure explicit.

Let $H$ say that every raven is black. Its contrapositive says that every nonblack thing is a nonraven. Since these statements are logically equivalent, evidence should support them equally. Combining this equivalence requirement with the unrestricted rule that every positive instance confirms a universal generalization appears to make discovering a white shoe support $H$. The tension is between that result and the judgement that such a discovery is irrelevant.^[ref]

Bayesian confirmation replaces an automatic instance rule with a comparison of likelihoods. For background $K$, positive prior probability for both $H$ and its alternative, and evidence $E$ of positive probability,The equivalence follows directly from Bayes' theorem. A statement's grammatical appearance as a positive instance does not determine either likelihood.

Consider an intentionally simple finite model. There are $r$ ravens and $M$ nonblack objects, with these counts fixed under both hypotheses. Under $H$ none of the nonblack objects is a raven; under the sole alternative $H_1$, exactly one is. Selecting uniformly among the nonblack objects and discovering a nonraven has likelihoods $1$ and $1-1/M$. Selecting uniformly among the ravens and discovering a black one has likelihoods $1$ and $1-1/r$. With equal prior probabilities, the respective posterior probabilities of $H$ areIf $M\gg r$, the first discovery supplies far less support. For $M=10^6$ and $r=100$, the increases over $1/2$ are approximately $2.5\times 10^{-7}$ and $0.00251$. This illustrates a possible resolution of the intuitive disparity, not a theorem that every nonblack nonraven always confirms $H$. More complex alternatives can reverse the comparison, and examining something already known to be a shoe need supply no relevant new information.^[ref]

Now suppose $E$ is already in $K$. Then $P(E\mid K)=1$, and ordinary conditioning givesThere is no further increase from learning the same fact again. Yet an already established observation can become an important reason to accept a subsequently developed theory. Clark Glymour's problem of old evidence exposes this mismatch between actual updating and scientific assessments of evidential support.^[ref]

One response evaluates the observation relative to an earlier or counterfactual background that does not contain it. That can represent how strongly a theory accounts for the observation compared with alternatives. It must specify that background rather than pretend to remove knowledge uniquely from a person's present beliefs. Another response concerns learning that a theory successfully accounts for the observation. That discovery can be new even when the observation is old. Modelling it requires attention to limited reasoning and changing hypotheses; an ideal agent already knowing every logical consequence cannot simply learn an entailment it previously overlooked.

Neither manoeuvre licenses counting the same evidence twice, or ignoring that parameters were selected to fit it. Neither establishes a unique complete theory of confirmation. Nevertheless, the old-evidence objection concerns the interpretation and completeness of a probabilistic framework; it does not invalidate the likelihood calculation for genuinely new sampling information.

The preferable solution combines a background-sensitive account of support with rejection of unrestricted instance confirmation. It preserves logical equivalence while allowing a white shoe to provide tiny support, no support or contrary support under different justified models. This is more explanatory than assigning confirmation solely from logical form, although assessing old evidence requires additional work beyond routine conditioning.