When writing the conditional probability $P(A|B)$, which is interpreted to be that given $B$ what is the probability of observing $A$, can one also claim that this says the probability of observing $A$ being caused by $B$, i.e., is there a causal connection between $B$ and $A$?
Thanks!
On
No, not at all. Consider, $B$="the flagpole's shadow is very short", $A$="the sun is high in the sky".
$P(A \mid B)$ could be very high, but $B$ does not cause $A$.
Here's another example that I believe is due to R.A. Fisher (but I may be wrong about that). Let $A$="you get lung cancer" and $B$="you smoke heavily". We know that $P(A \mid B)$ is quite high and are quite confident that $B$ in fact causes $A$.
But the causal claim doesn't follow from the high conditional probability alone. For suppose that $A$ and $B$ are causally unrelated but that $C$ is a common cause of both $A$ and $B$. For example, $C$ could be "you have a gene that causes both a strong desire to smoke and lung cancer." Then, given $B$, it's highly likely (we suppose) that $C$ is operative, and hence that $A$ occurs. So $P(A \mid B)$ is high. But, by assumption, $B$ and $A$ do not influence each other casually.
On
Good news - there's a positive answer to a perhaps not totally unrelated question below.
We know the definition of $P(A|B)$. For this to imply anything about causation you need to first give a precise mathematical definition of "$B$ caused $A$"...
Now, material implication is not causation (in fact the fact that material implication says nothing about causation is why it sometimes seems somehow "wrong" to beginners). But it does have a nice simple precise definition.
One might plausibly define the event $B\implies A$ by $$B\implies A=\{\omega\in\Omega:\omega\in B\implies\omega\in A\}.$$
(In other words: That definition says that $B\implies A$ is the set of $\omega$ such that $\omega\in A$ or $\omega\notin B$, or $$(B\implies A)=A\cup(\Omega\setminus B).)$$
And then it turns out that if $P(A|B)$ is large it does follow that $P(B\implies A)$ is large.
Note that the complement of $B\implies A$ is just $B\setminus A$. So $$1-P(B\implies A)=(P(B)-P(A\cap B))=P(B)(1-P(A|B))\le1-P(A|B).$$Proving
Ullrich's Astounding Theorem If $P(B)>0$ then $P(B\implies A)\ge P(A|B)$.
Consider $B$ being "a person was breathing this morning" and $A$ being "that person is dead".
$P(A \mid B)$ is the probability that the person died having been breathing earlier in the day. This is likely to be a small but nonzero number, approximately equal to the probability that a random person dies on any given day.
Nonetheless, you'd probably not want to conclude that breathing leads to death in any meaningful way. (Indeed, a better conclusion is that not breathing leads to death!)