TITLE: Multiple Conditioning on Event Probabilities
QUESTION [7 upvotes]: I am trying to understand what's wrong with the following logic related to "multiple conditioning." Why is the probability of [(A given B) given C] not the same as the probability of [A given (B and C)] ? I know it's not true, but only because numbers disagree. I am having a hard time parsing what's wrong with the logic.

REPLY [2 votes]: You cannot apply standard probability rules to conditional events of the form (A | B) except when A and B are ordinary two-valued events. Conditional events may be true, false or null (so the logic of conditional events is a three-valued logic). The calculus of probability for conditional events requires generalised formulas. The matter becomes very simple if two unary truth-functional connectives are introduced: T(p) and H(p). The first returns true if p is true and false if p is either false or null. The second returns true if p is either true or false and false if p is null. Every conditional event p may be written as (T(p) | H(p)) and its probability is given by the following general equation: Pr(p) = Pr(T(p))/Pr(H(p)). If p = B|C where B and C are ordinary events, the formula yields Pr(T(B|C))/Pr(H(B|C)). But, in this special case T(B|C) = (B & C) and H(B|C) = C, so that if B and C are ordinary events it holds:
Pr(B | C) = Pr(B & C)/Pr(C).