When does $P(A|B) = P(B|A)$? If A and B are events, when does $P(A|B) = P(B|A)$?
If it is not always true, please provide a counter example as I cannot.
 A: Too long for a comment. They are almost never the same. In fact $$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{P(A)}{P(B)}\frac{P(A\cap B)}{P(A)}=P(B\mid A) \frac{P(A)}{P(B)}$$ so they will be equal if and only if $P(A) = P(B)$ or $P(A\cap B) = 0$.

I'm guessing this question has come from confusion over the definition of $P(A\mid B)$. This is the probability that an event $A$ occurs given that we know that another event $B$ occurs.
For example, if $A$ is the event that I roll a $4$ on a regular $6$-sided dice, and $B$ is the event that the number I roll is even, then $P(A\mid B)$ is the probability that I roll a $4$ given that I know that my roll was even; my roll can be one of $2,4$ and $6$ so the answer is $\frac 13$.
As such, in order for $P(A\mid B)$ and $P(B\mid A)$ to be equal there would have to be a strange relationship between $A$ and $B$ - either the probabilities of the events happening would have to be be equal, or we would need $P(A\mid B) = P(B\mid A) = 0$.
A: You have $$P(A|B) = {P(A\cap B)\over P(B)},$$
and 
$$P(B|A) = {P(A\cap B)\over P(A)}.$$
What happens when you set these equal?  
A: You have a general identity that $P(A|B)P(B)=P(A\cap B)=P(B|A)P(B)$.
Counter example to $P(A|B)=P(B|A)$: throw a regular dice once, let $A$ be the event of landing $2$ and $B$ the event of landing an even number. Then, $P(A|B)=\frac{1}{3}$ whereas $P(B|A)=1$.
A: $P(A|B)=\frac{P(A\cap B)}{P(B)}$.
Now do the same for $P(B|A)$ and set them equal.  What can you say?
