 Provided \(p(E)\) is less than 1, this amounts to multiplying \(p(H)\) by a ratio greater than 1, which means \(p(H\mid E)\) comes out larger than \(p(H)\). Moreover, since 1 is the greatest quantity that can appear in the numerator, the case where \(H\) entails \(E\) and thus \(p(E\mid H)=1\) gives the greatest possible boost to the probability of \(H\). In other words, confirmation is greatest when the theory fits the evidence as well as possible.