What is $P(A | B,C)$ when $B$, $C$ are both independent Why is it true that $P(A|B,C) \neq P(A|B) P(A|C)$ when $B$ and $C$ are independent?
Intuitively I can understand that the LHS probability should be greater than the RHS and the above equation doesn't hold that. But is there any proof to convince me.
Another question: say if $A$ and $C$ are independent then can I tell
$P(A|BC) = P(A|B)$?
 A: I.
$P(A|B,C) = \frac{P(A,B,C)}{P(B,C)} = \frac{P(A,B,C)}{P(B)P(C)}$ (1)
$P(A|B)P(A|C) = \frac{P(A,B)P(A,C)}{P(B)P(C)}$ (2)
Equality depends on $P(A,B,C) = P(A,B)P(A,C)$, which usually does not hold.
II. still not true
Example: throw 4-sided dice
A = {get 1 or 2}
C = {get 1 or 3}
B = {get 1, 2 or 3}
you can check by definition that A,C are independent, but
P(A|B,C) = 1/2
P(A|B) = 2/3
A general comment on such statements of conditional probability: sometimes they may seem true, but if you cannot prove it from definition, they are usually not.
Edit: Minor fix replacing $p(C)$ with $P(C)$ in the first equation.
A: The whole concept of independence gets complicated once you move beyond 2 cases.  The best example I have seen of this came from a textbook (I think it was by Feller):
Consider that your population is the following $(A,B,C)$ triplets:
$$(1,2,3); (1,3,2);
(2,1,3); (2,3,1);
(3,1,2); (3,2,1);
(1,1,1); (2,2,2); (3,3,3)
$$
So if you select the first triplet, then $A=1, B=2, C=3$.
It is simple to verify that $A$ and $B$ are independent (pairwise) since $p(A=a|B) = 1/3 = p(A=a)$, etc. Also $A$ and $C$ are pairwise independent and $B$ and $C$ are pairwise independent, but given any 2, you know exactly what the 3rd one will be (complete dependence).
