 On Hook’s account, to be close to certain that if \(A, B\) is to give a high value to \(\mathbf{p}(A \supset B)\). How does \(\mathbf{p}(A \supset B)\) compare with \(\mathbf{p}_A (B)\)? In two special cases, they are equal: first, if \(\mathbf{p}(A \amp{\sim}B) = 0\) (and \(\mathbf{p}(A)\) is not 0), \(\mathbf{p}(A \supset B) = \mathbf{p}_A (B) = 1\) (i.e. 100%). Second, if \(\mathbf{p}(A) = 100\)%, \(\mathbf{p}(A \supset B) = \mathbf{p}_A (B) = \mathbf{p}(B)\). In all other cases, \(\mathbf{p}(A \supset B)\) is greater than \(\mathbf{p}_A (B)\). To see this we need to compare \(\mathbf{p}(A \amp{\sim}B)\) and \(\mathbf{p}(A \amp{\sim}B)/\mathbf{p}(A)\). Consider again the partition \(\{A \amp B, A \amp{\sim}B, {\sim}A\}. \mathbf{p}(A \amp{\sim}B)\) is a smaller proportion of the whole space than it is of the \(A\)-part — the part of the space in which \(A\) is true — except in the special cases in which \(\mathbf{p}(A \amp{\sim}B) = 0\), or \(\mathbf{p}({\sim}A) = 0\). So, except in these special cases, \(\mathbf{p}_A ({\sim}B)\) is greater than \(\mathbf{p}(A \amp{\sim}B)\). Now \(\mathbf{p}(A \supset B) = \mathbf{p}({\sim}(A \amp{\sim}B))\); and \(\mathbf{p}(A \amp{\sim}B) + \mathbf{p}({\sim}(A \amp{\sim}B)) = 1\). Also \(\mathbf{p}_A (B) + \mathbf{p}_A ({\sim}B) = 1\). So from \(\mathbf{p}_A ({\sim}B) \gt \mathbf{p}(A \amp{\sim}B)\) it follows that \(\mathbf{p}(A \supset B) \gt \mathbf{p}_A (B)\).