Is conditional probability transitive? Is conditional probability "transitive" (or how else this is called - please, explicate) in the meaning that $P(a \mid c)=P(a \mid b)P(b \mid c)$ ? Intuitively this seems so, but could you comment/proove on how to understand this (or refute)?
 A: From the definition of conditional probability, we have
$$
P(A\mid B)P(B\mid C) = \frac{P(A\cap B)}{P(B)} \frac{P(B\cap C)}{P(C)}
$$
and
$$
P(A\mid C) = \frac{P(A\cap C)}{P(C)}.
$$
It's not clear to me why these should be equal. Let's find a counterexample.
Suppose we roll a fair die. Let $A$ be the event that $1$ shows, let $B$ be the event that an even number appears, and let $C$ be the event that either $1$, $2$, or $3$ shows.
Then
$$
P(A\mid C) = 1/3
\quad\text{and}\quad
P(A\mid B) = 0
\quad\text{and}\quad
P(B\mid C) = 1/3.
$$
This example shows that the desired formula does not hold in general.
A: Something of which that proposition is reminiscent is true:
$$
\Pr(A\ \&\ B \mid C) = \Pr(A\mid B\ \&\ C)\Pr(B\mid C). \tag 1
$$
Now suppose $A\subseteq B\subseteq C.$ In that case we have
$$
\Pr(A \mid C) = \Pr(A\mid B)\Pr(B\mid C) \tag 2
$$
since in the case $A\subseteq B\subseteq C,$ the proposition $(2)$ follows from $(1).$
A: It is not so. 
Maybe your intuition is that since with $A$, $B$, and $C$ being numbers we have:
$$\frac{A}{C}=\frac{A}{B} \cdot \frac{B}{C}$$
you may think something similar works for events as well.... but that is not true.
In fact, we don't even have:
$$P(A|B) \cdot P(B) = P(A)$$
for what we do have is:
$$P(A|B) \cdot P(B) = P(A \cap B)$$
So, if you are thinkng of treating the '$|$' as a kind of division ... don't!
In fact, $|$ is not an operator, and $A|B$ is not an event. It is a common mistake to see them that way, and I suspect you are doing something similar.
