Technically, when viewed on the log-odds scale, it is exactly the same amount of evidence for either hypothesis.
On the (0;1) scale of probabilities it is, however, stronger evidence for flawed experiment. E.g. if we had a null hypothesis H0 at a prior P of 0.25 and two hypotheses H1 and H2 at 0.25 and 0.5, respectively, and we see evidence that has a likelihood of 5:1 for each of these hypotheses over H0. Then we have posterior P(H0|D)/P(H1|D)=1/5 and P(H0|D)/P(H2|D)=1/10, and after normalization, P(H0)=1/16, P(H1)=5/16, P(H2)=10/16. So, in this situation, P(H2) has increased by 2⁄16 and P(H1) only by 1⁄16. Which is what I meant in my comment: the probability of the more likely hypothesis increases faster, so H2 becomes more more probable than H1 :) Although on the log-odds scale, of course, both H1 and H2 received the same amount of evidence.
Technically, when viewed on the log-odds scale, it is exactly the same amount of evidence for either hypothesis.
On the (0;1) scale of probabilities it is, however, stronger evidence for flawed experiment. E.g. if we had a null hypothesis H0 at a prior P of 0.25 and two hypotheses H1 and H2 at 0.25 and 0.5, respectively, and we see evidence that has a likelihood of 5:1 for each of these hypotheses over H0. Then we have posterior P(H0|D)/P(H1|D)=1/5 and P(H0|D)/P(H2|D)=1/10, and after normalization, P(H0)=1/16, P(H1)=5/16, P(H2)=10/16. So, in this situation, P(H2) has increased by 2⁄16 and P(H1) only by 1⁄16. Which is what I meant in my comment: the probability of the more likely hypothesis increases faster, so H2 becomes more more probable than H1 :) Although on the log-odds scale, of course, both H1 and H2 received the same amount of evidence.