Combining conditional dependent probabilities I want to calculate $P(x\mid AB)$ where $AB$ is a condition in which both $A$ and $B$ occur.
I know the following data points:
$P(x\mid A)$
$P(x\mid B)$
$P(A\mid B) =$ I can estimate a number. It's not $P(A)$ since there is some dependence between $A$ and $B$.
$P(B\mid A) =$ I can estimate a number. It's not $P(B)$ since there is some dependence between $A$ and $B$.
If necessary, I can also estimate the following:
$P(A)$
$P(B)$
$P(x)$
Is there a way for me to calculate $P(x\mid AB)$?
In case it matters, $x$, $A$, and $B$ are all binary events, either something happens or it doesn't.
Separately, is there a way to generalize this since I am also trying to estimate $P(x\mid ABCD)$ where $C$ and $D$ follow similar rules as $A$ and $B$ as described above.
Thanks for the help!
 A: You note that $A$ and $B$ are not (unconditionally) independent. However, if they are independent conditional on $x$,
$$
p(A,B|x) = p(A|x)\,p(B|x),
$$ 
then you have enough information to compute $p(x|A,B)$. 
First factor the joint distribution two ways:
$$
p(x,A,B) = p(x|A,B)\,p(A,B) = p(A,B|x)\,p(x).
$$
Using these two factorizations, write Bayes' rule:
$$
p(x|A,B) = \frac{p(A,B|x)\,p(x)}{p(A,B)}.
$$
You know $p(x)$. You also know $p(A,B)$, since
$$
p(A,B) = p(A|B)\,p(B) = p(B|A)\,p(A),
$$
and you know $p(A)$, $p(B)$, $p(A|B)$, and $p(B|A)$. 
If $A$ and $B$ are conditionally independent you only need $p(A|x)$ and $p(B|x)$, but you know these as well, since (using Bayes' rule again)
$$
p(A|x) = \frac{p(x|A)\,p(A)}{p(x)}
\qquad\text{and}\qquad
p(B|x) = \frac{p(x|B)\,p(B)}{p(x)},
$$
and you know $p(x|A)$ and $p(x|B)$.
Putting this together, one way to write the answer is
$$
p(x|A,B) = \frac{p(x|A)\,p(x|B)\,p(A)}{p(x)\,p(A|B)} .
$$
Without the assumption of conditional independence (or its equivalent) I don't think you can get the answer with what you know. 
