 Whether we prefer the subjectivist’s response to Hume’s problem or the objectivist’s, a crucial element is still missing. Earlier we noted that justifying a Carnapian assignment of prior probabilities only gets us half way to a solution. We still have to turn these prior probabilities into posterior probabilities: initially, the probability of tails on the tenth toss was \(1/2\), but after observing the first 9 tosses come out tails, it’s supposed to be \(10/11\). Having justified our initial assignment of probabilities—whether the subjectivist way or the objectivist way—we can prove that \(p(T_{10}\mid T_{1\ldots9})=10/11\) compared to \(p(T_{10})=1/2\). But that doesn’t mean the new probability of \(T_{10}\) is \(10/11\). Remember, the symbolism \(p(T_{10}\mid T_{1\ldots9})\) is just shorthand for the fraction \(p(T_{10} \wedge T_{1\ldots9})/p(T_{1\ldots9})\). So the fact that \(p(T_{10}\mid T_{1\ldots9})=10/11\) just means that this ratio is \(10/11\), which is still just a fact about the initial, prior probabilities.