conditional probability - conditioned twice Let $p(\cdot)$ be a discreet probability function and $A,B,C$ be events. What does $p(A|B|C)$ mean? Is this the same as $p(A|B,C)$ or is it:
if we treat $D=A|B$ as another event, $p(D|C) = \frac{p(D\cap C)}{p(C)}$?
How would $p(D \cap C)$ be computed then with $D$ defined earlier?
 A: I think $$\frac{P\left(A\cap B\mid C\right)}{P\left(B\mid C\right)}=\frac{P\left(A\cap B\cap C\right)}{P\left(C\right)}/\frac{P\left(B\cap C\right)}{P\left(C\right)}=\frac{P\left(A\cap B\cap C\right)}{P\left(B\cap C\right)}=P\left(A\mid B\cap C\right)$$
has the best 'chances' here :).
A: Yes $\mathbb{P}(A|B|C) = \mathbb{P}(A|B,C)$
Since
$$\mathbb{P}(A,B)=\mathbb{P}(A|B)\cdot\mathbb{P}(B)$$
Remember that $\mathbb{P}(\cdot |C)$ also is a probability in the same space, 
$$\mathbb{P}(A,B|C)=\mathbb{P}(A|B|C)\cdot\mathbb{P}(B|C)$$
Then
$$\mathbb{P}(A|B|C) = \frac{\mathbb{P}(A,B|C)}{\mathbb{P}(B|C)}=\frac{\mathbb{P}(A,B,C)}{\mathbb{P}(B,C)}=\mathbb{P}(A|B,C)$$
A: Sum Rule:
$$p(a) = \sum_{b}p(a,b)$$
As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:
$$p(a|c) = \sum_{b}p(b, a|c)$$
For the product rule:
$$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?
Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?
Yes: Aim to prove:  $$p(a|b|c) = p(a|b,c)$$
$$p(a,b|c) = p(a|b|c)p(b|c)$$
$$p(a|b|c) = \dfrac{p(a,b|c)}{p(b|c)} = \dfrac{ \dfrac{p(a,b,c)}{p(c)}} {\dfrac{p(b,c)}{p(c)}} = \dfrac{p(a,b,c)}{p(b,c)} = \dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$
