Let

A= John will buy a burglar alarm by next year

and

B= John's home will be burglarized by next year.

Most people would say that

Pr(A|B) > Pr(A|-B) and Pr(B|-A) > Pr(B|A).

Note that in (A|B) the condition B is naturally taken to be the cause of A: John bought the alarm because his house was broken into. By contrast, in (A|-B), the condition -B is naturally taken to be diagnostic of A: the fact that John's house was not burglarized is taken as evidence for the claim that he has bought an alarm system. The same applies to the other two conitionals, where the first is causal and the second diagnostic. There is ample psychological evidence that we tend to favor causal over diagnostic inferences, and that's why most of us would say that Pr(A|B) > Pr(A|-B) and Pr(B|-A) > Pr(B|A).

However, this is wrong, that is, demonstrably in conflict with the rules of probability. Here is why.

 

We are going to prove that

Pr(A|B) > Pr(A|-B) if and only if Pr(B|A) > PR(B|-A)                                         (1)

To prove it, we need, first, to prove a lemma.

Lemma to be proved:

Pr(A|B) > Pr(A|-B) if and only if Pr(A|B) > Pr(A).                                               (2)          

First we prove that

If Pr(A|B) > Pr(A|-B), then Pr(A|B) > Pr(A).                                                       (3)

Proof:

i.                      Pr(A) =Pr(A&B) + Pr(A&-B)

This is because A is logically equivalent to (A&B) v (A& -B), and two logically equivalent statements have the same probability.

ii.                    Pr(A) = Pr(B) Pr(A|B) + Pr(-B) Pr(A|-B)

iii.                 Pr(A) < Pr(B) Pr(A|B) + Pr(-B) Pr (A|B)

We go from (ii) to (iii) simply by substituting Pr(A|B) in place of Pr(A|-B) in (ii).  Given (3), we change the "=" with "<".

iv.                  Pr(A) < Pr(A|B) {Pr(B) + Pr(-B)}

v.                    Pr(A) < Pr(A|B).

We go from (iv) to (v) because Pr(B) + Pr(-B)=1.

Now we prove that

If Pr(A|B) > Pr(A), then Pr(A|B) > Pr(A|-B).                                                      (4)

Proof:

i.                      Pr(A) < Pr(A|B)

ii.                    Pr(A) < Pr(A|B) {Pr(B) + Pr(-B)}

iii.                 Pr(A) < Pr(B) Pr(A|B) + Pr(-B) Pr(A|B)

iv.                  Pr(A) = Pr(B) Pr(A|B) + Pr(-B) Pr(A|-B)

v.                    Pr(A|B) > Pr(A|-B).

We go from (iii)-(iv) to (v) because the right sides of (iii) and (iv) differ only in that (iii) has Pr(A|B) at the end and (iv) has Pr(A|-B).  

Since (3) and (4) together amount to (2), this concludes the proof the lemma.

Now we use the lemma to prove (1).

i.                      Pr(A|B) > Pr(A|-B)

ii.                    Pr(A|B) > Pr(A)                                        (By the lemma just proved)

iii.                 Pr(A&B)/Pr(B) > Pr(A)                        (Because Pr(A&B) = Pr(B) Pr(A|B)).

iv.                  Pr(A&B)/Pr(A) > Pr(B)                       (Algebra on (iii))

v.                    Pr(B|A) > Pr(B), QED.

Hence, it is wrong (contrary to the rules of probability) to believe both that Pr(A|B) > Pr(A|-B) and Pr(B|A) < Pr(B|-A). As usual, our intuitions about conditional probability lead us astray.