 What then if the sampling in not constrained \((k = \top)\) and the evidence now amounts to the finding of a black raven, \(e = raven(a) \wedge black(a)\), versus a non-black non-raven, \(e^* =\neg black(a) \wedge \neg raven(a)\)? We’ve already seen that, for either Hempelian or HD-confirmation, \(e\) and \(e^*\) are on a par: both Hempel-confirm \(h\), none HD-confirms it. In the former case, the original Hempelian version of the ravens paradox immediately arises; in the latter, it is avoided, but at a cost: \(e\) is declared flatly irrelevant for \(h\)—a bit of a radical move. Can the Bayesian do any better? Quite so. Consider the following conditions: