Conditional Probabilities and the Probabilities of Conditionals

A conditional probability is represented P(A | B), read "the probability of A given B", and is (by definition) equivalent to P(A & B) / P(B). To see how this works, consider a simple example. Let's take a fair die and roll it. What is the probability that we will role and even number? Clearly, it's 1/2. And what is the probability that we will role a 6? Clearly, it's 1/6. Now, what is the probability that we'll roll a 6 given that we roll an even number? In other words, what is P(roll a 6 | roll an even number)? By definition of conditional probability this is equivalent to P(roll a 6 and roll an even number) / P(roll an even number). This equals (1/6)/(1/2), which equals 1/3.

Some philosophers (including myself a few years back) have thought that conditional probabilities could be used to represent the probabilities of conditionals. This view is commonly referred to as Stalnaker's hypothesis (SH), after Robert Stalnaker, a prominent philosopher who explicitly proposed the idea. In other words,

(SH) P(If p then q) = P(q | p).

Unfortunately, while Stalnaker's hypothesis is prima facie plausible, it is demonstrably false, as famously shown by David Lewis (an achievement that went a long way toward establishing Lewis's reputation as one of the smartest philosophers on the planet until his death in 2001). I won't try to run through Lewis's proof here, but suffice to say that it is now universally agreed upon by scholars working on conditionals that Stalnaker's hypothesis fails.

But if the probabilities of conditionals cannot be equated with conditional probabilities, what can they be equated with? Intuitively, it seems like the notion of the probability of a conditional ought to make sense and that there should, in principle at least, be some way of estimating a value for large classes of conditionals, if not all. I think this is right, though I should mention that not everyone agrees. Philosopher Ernest Adams has famously defended the view that there are no probabilities of conditionals. In other words, he holds that there is nothing that can be called the value of P(If p then q). He goes on to propose that the conditional probability P(q | p) measures the assertibility (but not the probability) of the conditional if p then q.

I think Adams is wrong and that there's a very straightforward way of thinking about the probabilities of conditionals. Consider if p then q. If p entails q, then it seems obvious that P(If p then q) ought to equal one. Similarly, if p and q are mutually incompatible (p entails ~q), then it seems obvious that P(If p then q) ought to equal zero. But if those two cases clearly have probabilities, then it's hard to see why cases in which p is compatible with but does not entail q should not have probabilities. Suppose we think of this like an argument with p as a premise and q as the conclusion (a natural model because every argument can be written as a conditional). Since p does not entail q, we have an enthymeme:

p
(unstated premise)
∴ q

Now, what proposition do we need for an unstated premise to make this argument valid? Let's call this enthymematic premise the deductive complement of p in relation to q and represent it by X. X's job is to supply any information in q that is not already in p, such that (p+X) entails q. My proposal, then, is this (I'll call it Rhoda's hypothesis, RH):

(RH) (a) If p entails q then P(If p then q)=1. (b) If p entails ~q, then P(If p then q)=0. (c) If p entails neither q nor ~q, then P(If p then q) = P(X), where X is whatever information is contained in q that is not already contained in p.

The notion of information 'contained in' a proposition may be explicated via the notion of logical entailment. Propositions p and q contain the same information if and only if they have exactly the same entailments. If p entails q, then anything entailed by q is also entailed by p. X is equivalent to the conjunction of all propositions entailed by q that are not entailed by p.

In some cases, X will be logically equivalent to 'If p then q', but it will never be logically stronger than that, and often it will be logically weaker. I'll leave it as an exercise for the reader to explain why.