Does it always hold that $\mathbb {P}(A | B) = \mathbb {P}(B | A)$?

Question

I read that if $A$ and $B$ are independent then it holds that $\mathbb {P}(A \cap B) = \mathbb {P}(A)\mathbb {P}(B)$.

Thus, $\mathbb {P}(A | B) = \frac {\mathbb {P}(A)\mathbb {P}(B)}{\mathbb {P}(B)} = \mathbb {P}(A)$ and also $\mathbb {P}(B | A) = \frac {\mathbb {P}(B)\mathbb {P}(A)}{\mathbb {P}(A)} = \mathbb {P}(B)$.

Does this mean that $\mathbb {P}(A | B) = \mathbb {P}(B | A)$ is not always true?

Answer 1

Assume that $P(A)>0$ and $P(B) >0$ so that the conditional probabilities are defined. $P(B|A)=\frac {P(A\cap B)} {P(A)}$ and $P(A|B)=\frac {P(A\cap B)} {P(B)}$. These numbers are equal iff $P(A\cap B)=0$ or $P(A)=P(B)$. There are planty of cases where this is not satisfied. One simple example is when $B=\Omega$ (the sample space) and $A$ is any event with $0 <P(A) <1$.

Answer 2

Of course $\mathbb {P}(A | B) = \mathbb {P}(B | A)$ is not always true. A coin coming heads and a die throwing 6 are independent events but the chance of heads is not the same as that of throwing a 6. (Note that $P(6) = P(6|\text{heads}) = \frac{1}{6}$ while $P(\text{heads}|6) = P(\text{heads}) =\frac12$)