The above formula is usually called “odds form of Bayes formula”. We get the standard form P(x|D)=P(x)×P(D|x)P(D) by letting y=D in the odds form, and we get the odds form from the standard form by dividing it by itself for two hypotheses (P(D) cancels out).
The serious problem with the standard form of Bayes is the P(D) term, which is usually hard to estimate (as we don’t get to choose what D is). We can try to get rid of it by expanding P(D)=P(D|x)P(x)+P(D|¬x)P(¬x), but that’s also no good, because now we need to know P(D|¬x). One way to state the problem with this is to say that a hypothesis for given observations is a description of a situation that makes it possible to estimate the probability of those observations. That is, x is a hypothesis for D if it’s possible to get a good estimate of P(D|x). To evaluate an observation, we should look for hypotheses that let us estimate that conditional probability; we do get to choose what to use as hypotheses. So the problem here is that if x is a hypothesis for D, it doesn’t follow that ¬x is a hypothesis for D or for anything else of interest. The negation of a hypothesis is not necessarily a hypothesis. That is why it defeats some of the purpose of moving over to using the odds form of Bayes if we let y=¬x, as it’s sometimes written.
The above formula is usually called “odds form of Bayes formula”. We get the standard form P(x|D)=P(x)×P(D|x)P(D) by letting y=D in the odds form, and we get the odds form from the standard form by dividing it by itself for two hypotheses (P(D) cancels out).
The serious problem with the standard form of Bayes is the P(D) term, which is usually hard to estimate (as we don’t get to choose what D is). We can try to get rid of it by expanding P(D)=P(D|x)P(x)+P(D|¬x)P(¬x), but that’s also no good, because now we need to know P(D|¬x). One way to state the problem with this is to say that a hypothesis for given observations is a description of a situation that makes it possible to estimate the probability of those observations. That is, x is a hypothesis for D if it’s possible to get a good estimate of P(D|x). To evaluate an observation, we should look for hypotheses that let us estimate that conditional probability; we do get to choose what to use as hypotheses. So the problem here is that if x is a hypothesis for D, it doesn’t follow that ¬x is a hypothesis for D or for anything else of interest. The negation of a hypothesis is not necessarily a hypothesis. That is why it defeats some of the purpose of moving over to using the odds form of Bayes if we let y=¬x, as it’s sometimes written.