Conditions under which "explaining away" does vs. does not occur My understanding of "explaining away" is as follows. If there is an effect, C, that can result from two independent causes, A and B, then observing only C makes A more likely than observing both C and B. That is, P(A|C) > P(A|B, C) because C is already explained by B, which makes A less likely.
The concrete example I have in mind is Pearl's classical earthquake vs. burglary setting of an an alarm. Earthquake and burglary are the two independent causes. Alarm going off is the effect. If the alarm goes off, then there's some probability that an earthquake occurred. But if the alarm goes off and I know theres a burglary, then an earthquake becomes less likely.
My question is: if the two causes, A and B, are both discrete and binary, meaning it either happens or it does not, and the effect C is also discrete and binary, then is it ALWAYS the case that P(A|C) > P(A|B, C)? Are there conditions under which this relationship inequality is false?
 A: Seems so if $0<P(A)<1$, $0<P(C)<1$,$0<P(B)<1$  . Let $A$, $B$ be independent events, both cause $C$ to occur with probability 1.
$P(A)P(B)=P(A,B)$
$P(C|A)=P(C|B)=1$
Then $P(A|C) = \frac{P(A)}{P(C)} >P(A)$ by Bayes Rule. 
Now condition instead on $B$ and $C$:
$P(A|B,C) = \frac{P(A,B)P(C|A,B)}{P(B)P(C|B)} = \frac{P(A,B)}{P(B)} =\frac{P(A)P(B)}{P(B)} =P(A)$
If $P(B)$ was zero we can't condition on it, if $P(C)=1$ its occurence does not make $A$ more probable, if $P(A)=1$ it is already certain and if $P(B)=1$ then in this setup, $P(C)=1$. But if all are bounded away from 0 and 1, it seems to hold.
Edit: Independence of $A$ and $B$ is crucial here, and I only noticed now, reading the other answer, that you didn't ask about that case. If independence does not hold, it need not be true. 
A: It all depends on the effect of B on A. Does event B occuring increase or decrease the probability of event A occurring?
Imagine $A = \mathbb{I}_{\text{get an A+ on a test}}$, $B = \mathbb{I}_{\text{got at least 8 hours of sleep}}$, $C = \mathbb{I}_{\text{studied at least 10 hours}}$. We expect that both $B$ makes $A$ more likely and $C$ also makes $A$ more likely. When both $B$ and $C$ occur, you are more likely to get an A+ then if simply $C$ occurs and whether or not $B$ occurs is unknown. Thus,  $P(A|C) < P(A|B, C)$.
Apologies if you wanted a proof, but I think this example will give you intuition on why this is true.
