Definition of Conditional Probability with multiple conditions Specifically, say I have two events, A and B, and some distribution parameters $ \theta $, and I'd like to look at $P(A | B,\theta)$.
So, the simplest definition of conditional probability is, given some events A and B, then $P(A|B) = \frac{P(A \cap B)}{P(B)}$.  So if there are multiple events to condition on, like I have above, could I say that $P(A | B,\theta) \stackrel{?}{=} \frac{P((A | \theta)\cap(B | \theta))}{P(B|\theta)}$  or am I looking at the in totally the wrong way?  I tend to psych myself out when I deal with probability sometimes, I'm not really sure why.
 A: You can do a little trick. Let $(B \cap \theta) = C$. Now you can write
$$P(A|B, \theta) = P(A|C).$$
The problem reduces to that of a conditional probability with only one condition:
$$P(A|C) = \frac{P(A \cap C)}{P(C)}$$
Now fill in $(B \cap \theta)$ for $C$ again and you have it:
$$\frac{P(A \cap C)}{P(C)} = \frac{P(A \cap (B \cap \theta))}{P(B \cap \theta)}$$
And this is the result that you wanted to get to. Let's write this in exactly the form you had when you originally stated the question:
$$P(A|B , \theta) = \frac{ P(A \cap B \cap \theta) }{ P(B \cap \theta) }$$
As to your second question, why it is that probability freaks you out: it is one of the findings from psychological research that humans are not very good at probabilistic reasoning ;-). It was a bit hard for me to find a reference that I can point you to. But the work of Daniel Kahneman is certainly very important in this regard. 
A: I think you probably want this:
$$\rm{P}(A|B,\theta) = \frac{\rm{P}(A\cap B|\theta)}{\rm{P}(B|\theta)}$$
I often find it confusing thinking about how to manipulate probabilities. With multiple conditions, I find it easiest to think about it this way:


*

*temporarily remove the condition(s) that you want to remain as conditions in your result. In this case write $\rm{P}(A|B)$, taking out $\theta$.

*apply the normal rules. In this case $\rm{P}(A|B) = \rm{P}(A\cap B)/\rm{P}(B)$.

*restore the condition(s) that were removed. In this case, restore $\theta$, to get the result $\rm{P}(A|B,\theta) = \rm{P}(A\cap B|\theta)/\rm{P}(B|\theta)$.

