In an earlier post, I proposed the following hypothesis concerning the probabilities of conditionals:
(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.
And I suggested that the term X in last part (c) is to be understood as the conjunction of all propositions entailed by q that are not entailed by p.
Now, it’d be nice to be able to reduce conditions (a) and (b) to (c) by treating them as limit cases of the latter. I’m just not sure how to formulate a convincing argument for that reduction.
In case (a), X drops out entirely, since if p entails q then it also entails every proposition entailed by q. Perhaps we could think of X in this case as a proposition with no content, i.e., a tautology. As a tautology, P(X) = 1. So if we are entitled to construe X in this way, then (a) can be reduced to (c).
In case (b), we could take a different tack. If p entails ~q, then by (a) P(If p then ~q) = 1. Now, if we accept (as I do) conditional excluded middle (CEM), then “If p then either q or ~q” is a necessary truth. Its probability therefore equals 1. Since q and ~q are mutually exclusive and jointly exhaustive, however, it seems to follow that P(If p then either q or ~q) = P(If p then q) + P(If p then ~q), which means that P(If p then q) = 0.
So if that’s right, then case (b) reduces to (a), which reduces to (c). But that’s assuming that X is a tautology in case (a) and that CEM is true, which is controversial.
Here’s another issue: What sort(s) of conditional could (RH) apply to? Clearly, (RH) will not work for material conditionals, for then if we let p = (r & ~r), we generate a contradiction since r & ~r materially implies both q and ~q. Hence by (RH), P(If p then q) would come out to both 1 and 0, which is absurd. The same result obtains for strict conditionals, since necessary falsehoods strictly imply everything.
I’m not bothered by this, however, since I don’t think ordinary language conditionals are equivalent to either material or strict conditionals. I reject the ex falso quodlibet (a false proposition implies anything) assumption of material implication and the ex contradictione quodlibet (a necessarily false proposition implies anything) assumption of strict implication. Instead, it seems to me that ordinary language conditionals assume that the antecedent has got to be relevant to the consequent in terms of its semantic content (and not just in terms of its truth value in the actual world or its distribution of truth values across all possible worlds). Unfortunately, I don’t have a developed theory of semantic relevance to offer.
Finally, reader Phil (“Oudeis Oudamou”) asks a couple questions. I’ll take the easier first.
Phil: First of all, do you accept the equivalence : prob[if a, then b]= [if a, then prob b]?
Alan: No. The equation doesn’t make sense as it stands. The LHS is equivalent to a number between 0 and 1. The RHS is a proposition. Whatever else they are, numbers aren’t propositions. I suspect when you wrote that you were thinking of the LHS as the proposition “Probably, if a then b” and wondering if I’d agree that that proposition is equivalent to the proposition “If a then probably b”. Again, then answer is no. You aren’t trying to catch me in the probabilistic equivalent of the necessitas consequentiae / necessitas consequentis fallacy, are you? That a conditional is true in most possible worlds (assuming the notion of “most” possible worlds is well-defined) does not imply that if the antecedent is true in the actual world then the consequent is true in most possible worlds.
Phil: Could I get you to illustrate how RH (c) will work in some concrete examples? Suppose, for example, I make the prediction “very probably if in Phoenix the overnight min temp > 83 F, the next day’s high > 110 F.” This on the basis of my long-time personal experience with summer weather here, but without consulting any official weather data. And my prediction turns out to be correct except for a very cases, after we check the weather records for the last 50 years. Or suppose I’m doing stylometrics and I argue “ very probably if Plato wrote the VII Epistles, then he also wrote the II Epistle.” I’m curious to see how you’d construct X in these and assign the prob X.
Alan: Those are fairly complicated examples, and I’m not sure I’m up to the task of tackling them just yet. Let me try to work with something simpler. Let’s take the conditional “If I were to roll this die repeatedly, then each number (1-6) would occur with a frequency of approximately 1/6.” Here p=”I roll this die repeatedly” and q=”Each number (1-6) occurs with a frequency of approximately 1/6.” Now we want an X that, together with p, will entail q without packing anything more than necessary into X. Matters are complicated by the vague word “approximately”, but let’s set that aside. It seems to me that X has got to be something like “this is a fair die and will remain so for the duration of the experiment”. In normal cases of dice rolling, that’s a reasonably safe assumption, so I would say that P(X) is fairly high and that, therefore, P(If p then q) is fairly high. Matters are not quite that simple, however, for any statistician will point out that even p+X doesn’t entail q, unless we project the experiment into the infinite long run. But then X becomes implausible – any normal die will break down and start becoming ‘unfair’ way short of infinity. So we’ve got to idealize the situation to a high degree to make the formula work. This may show that (RH) needs to be modified, and perhaps complicated considerably, to handle more realistic cases. It make still be useful as a rough estimation tool, however.