Bayes's theorem states $P(A\mid B) = \dfrac{P(B\mid A)\cdot P(A)}{P(B)}$. The intuition behind this is simple: if $B$ is true, then the probability that $A$ is true is the number of cases where $A$ is true out of all cases where $B$ is true.

Now, here is another formulation of the rule, just rearranging fractions: $\dfrac{P(A\mid B)}{P(A)} = \dfrac{P(B\mid A)}{P(B)}$. To me, what this says is that "if upon learning $B$ is true, we think $A$ is $x$ times more likely to be true than we previously thought, then upon learning $A$, $B$ is $x$ times more likely to be true than we previously thought." But this sentence does not seem similarly obvious to me. Is there a natural interpretation of $\dfrac{P(A\mid B)}{P(A)} = \dfrac{P(B\mid A)}{P(B)}$?

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    $\begingroup$ The equation says that conditioning on B increases the probability of A by the same ratio that conditioning on A increases the probability of B. (You are correct that this is somewhat non-obvious). In a sense, this means that if the fractions exceed 1, then $A$ and $B$ are "positively correlated" (synergistic), and that this relationship is symmetric. The relationship ties in with the concept of lift in data mining: en.wikipedia.org/wiki/Lift_(data_mining). $\endgroup$
    – Marconius
    Aug 25, 2015 at 21:26

2 Answers 2


I agree that this is less obvious, but you can go some way towards making intuitive sense of it by noting that it's obvious in three important cases:

If $A$ and $B$ are identical, it's obviously true by symmetry.

If $A$ and $B$ are independent, it's obviously true because both ratios must then be $1$.

If $A$ and $B$ are mutually exclusive, it's obviously true because both ratios must then be $0$.

Given this, it would be surprising if such simple ratios would manage to coincide at three different points but not in general.


You may be confusing yourself because you are skipping a step. Consider the definition of conditional probability. $$P(A|B)=\frac{P(A\cap B)}{P(B)}\implies P(A\cap B)=P(A|B)P(B)$$

The intersection of two sets is commutative, i.e. $$P(A\cap B) = P(B\cap A)=P(B|A)P(A)$$

Therefore, $P(A|B)P(B)=P(B|A)P(A)$


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