The odds form of Bayes' rule works for any two hypotheses and $H_{j},$ and looks like this:

$\begin{matrix} \frac{P (H_{i} ∣ e)}{P (H_{j} ∣ e)} = \frac{P (H_{i})}{P (H_{j})} \times \frac{P (e ∣ H_{i})}{P (e ∣ H_{j})} . \\ (1) \end{matrix}$

The probabilistic form of Bayes' rule requires a hypothesis set $H_{1}, H_{2}, H_{3}, \dots$ that is mutually exclusive and exhaustive, and looks like this:

$\begin{matrix} P (H_{i} ∣ e) = \frac{P (e ∣ H_{i}) \cdot P (H_{i})}{\sum_{k} P (e ∣ H_{k}) \cdot P (H_{k})} . \\ (2) \end{matrix}$

We will now show that equation (2) follows from equation (1). Given a collection $H_{1}, H_{2}, H_{3}, \dots$ of mutually exclusive and exhaustive hypotheses and a hypothesis $H_{i}$ from that collection, we can form another hypothesis $\neg H_{i}$ consisting of all the hypotheses $H_{1}, H_{2}, H_{3}, \dots$ except $H_{i} .$ Then, using $\neg H_{i}$ as $H_{j}$ and multiplying the fractions on the right-hand side of equation (1), we see that

$\frac{P (H_{i} ∣ e)}{P (\neg H_{i} ∣ e)} = \frac{P (H_{i}) \cdot P (e ∣ H_{i})}{P (\neg H_{i}) \cdot P (e ∣ \neg H_{i})} .$

$P (\neg H_{i}) \cdot P (e ∣ \neg H_{i})$ is the prior probability of $\neg H_{i}$ times the degree to which $\neg H_{i}$ predicted $e .$ Because $\neg H_{i}$ is made of a bunch of mutually exclusive hypotheses, this term can be calculated by summing $P (H_{k}) \cdot P (e ∣ H_{k})$ for every $H_{k}$ in the collection except $H_{i} .$ Performing that replacement, and swapping the order of multiplication, we get:

$\frac{P (H_{i} ∣ e)}{P (\neg H_{i} ∣ e)} = \frac{P (e ∣ H_{i}) \cdot P (H_{i})}{\sum_{k \neq i} P (e ∣ H_{k}) \cdot P (H_{k})} .$

These are the posterior odds for $H_{i}$ versus $\neg H_{i} .$ Because $H_{i}$ and $\neg H_{i}$ are mutually exclusive and exhaustive, we can convert these odds into a probability for $H_{i},$ by calculating numerator / (numerator + denominator), in the same way that $3 : 4$ odds become a 3 / (3 + 4) probability. When we do so, equation (2) drops out:

$P (H_{i} ∣ e) = \frac{P (e ∣ H_{i}) \cdot P (H_{i})}{\sum_{k} P (e ∣ H_{k}) \cdot P (H_{k})} .$

Thus, we see that the probabilistic formulation of Bayes' rule follows from the odds form, but is less general, in that it only works when the set of hypotheses being considered are mutually exclusive and exhaustive.

We also see that the probabilistic formulation converts the posterior odds into a posterior probability. When computing multiple updates in a row, you actually only need to perform this "normalization" step once at the very end of your calculations — which means that the odds form of Bayes' rule is also more efficient in practice.