TITLE: If $P(A) \neq 0$ and $P(B) \neq 0$, then $P(B|A) \geq P(B)$ is equivalent to $P(A|B) \geq P(A)$
QUESTION [9 upvotes]: I am puzzled by the intuition behind the following fact:
If $P(A) \neq 0$ and $P(B) \neq 0$, then $P(B|A) \geq P(B)$ is equivalent to $P(A|B) \geq P(A)$. 
This is easy enough to show by definition of conditional probability, but I would like to have some sort of geometric intuition behind this. 
I can create positive example pictures, but not one that shows me why this statement must hold.
I would greatly appreciate some help. Thanks

REPLY [6 votes]: First, note that this theorem is equivalent to

If $P(A) \neq 0$ and $P(B) \neq 0$ and 
  $$P(B|A) \geq P(B), \tag{1}$$ then $$P(A|B) \geq P(A).\tag{2}$$

because you can just swap $A$ and $B$ in the theorem to get the direction (2) $\Rightarrow$ (1).

Real life intuition (aka Bayesian inference): 
Consider a thing $B$ that's more likely when you know $A$ happens than it is when you don't know whether $A$ happens. Mathematically, this is condtion (1). Now if you observe $B$, it is actually evidence for $A$. Which mathematically means (2).
Example:
You know that rigged dice tend to give $6$s more often than fair dice.  You pick up a die you find somewhere. As you really haven't encountered that many significantly rigged dice in your life, but have seen quite many almost fair dice, you initially think that your die is likely close to fair. You throw the die 20 times and observe 20 sixes. What do you deduce? You start thinking that the die is very probably rigged.
Mathematically: Let $A$ be the event that the die is significantly rigged, and $B$ be the event that you get 20 sixes in a row. The fact that rigged dice give more sixes than fair dice means $P(B|A)>P(B)$. Before observing the throws, you think the probability that the die is rigged is $P(A)$, which is very small. But once you see the throws, you think that the probability is $P(A|B)$. The new probability is a lot larger, which means that $P(A|B)>P(A)$. The fact that the new probability is a lot larger was based on your knowledge that $P(B|A)>P(B)$.