Molinism and probabilistic explanation
January 15, 2013 — 14:06

Author: Alexander Pruss  Category: Molinism  Tags: , ,   Comments: 5

Suppose Molinism is true. We know the truth values of some Molinist counterfactuals because we know that their antecedent and consequent are true. But we also have reason to believe many other Molinist counterfactuals. Absent further evidence, if P(A|C) is high, and C is an appropriate antecedent for a Molinist counterfactual C→A, that gives me reason to believe C→A. It certainly gives me reason to believe C→A if I know C is actually true; for if I know C is true, then if P(A|C) is high, P(A) will be fairly high as well, and so A is probably true, and hence C→A is probably true. But I also have reason to think C→A is true in cases where C is false. For instance, if Jones is the sort of person likely to accede to my minor requests, then I have reason to believe that were I to make such-and-such a minor request, he’d accede to it, and I have reason to believe the conditional whether or not I make the request (at least assuming Molinism is true so that the conditional has non-trivial truth-value).

This suggests that if the objective probability of A on C is high, then the objective probability of C→A is also high. So the Molinist conditional C→A, assuming it’s true, doesn’t seem to be a mere brute fact. It is a fact subject to meaningful probabilistic assignments. But if it’s not a mere brute fact, it seems reasonable to look for an explanation of it. What is that explanation?

Well, maybe we have a probabilistic explanation. Maybe the fact that C makes A probable explains why C→A. But this is weird. It seems that probabilistic explanation is a species of causal explanation (with probabilistic causation). But there is surely no causal explanation of why C→A, at least in worlds where C is not true. (What would the cause be? The truthmaker of C? But C is not true and has no truthmaker.)

I’ll leave it as a puzzle: How is a Molinist to explain the connection between P(A|C) and the probability of the conditional C→A?

  • The intuition behind this post is closely tied to the Adams Thesis that P(A→B)=P(B|A). That Thesis is largely taken to have been disproved by Lewis. But in the case of counterfactuals, it is a mistake to take it to have been disproved.

    January 17, 2013 — 10:49
  • Mike Almeida

    It does not hold for counterfactuals either, Alex. It is not in general true that if P(A|B) is high, then P(B -> B) is high or that if P(A|B) is low, then P(B -> B) is low, for ‘->’ read as a counterfactual. There are lots of counterexamples. Suppose that Jones is out of shape and would never enter a 10k race unless he were in top shape and likely to win. There is a 10k race today. The probability that Jones *actually* wins the 10k race, given that he in fact enters it is very low. So, P(W|E) is low. But the probability that he would win the 10k race were he to enter it is much higher. P(E -> W) is high. This is because in the closest worlds to ours in which he does enter that race, he is in top shape.
    Consider the first case, that if P(A|B) is high, then P(B -> B) is also high. Same situation, except imagine that Jones is in top shape and would never enter a race unless he were in really bad shape. Jones likes odd challenges. Its probable that if JOnes enters the race, then he wins. But it is improbable that if he were to enter the race, then he would win.

    January 18, 2013 — 10:43
  • Mike Almeida

    replace all ‘P(B -> B)’ with ‘P(B -> A)’

    January 18, 2013 — 10:45
  • Mike:
    Do you mean things the other way around in the Jones case? It seems that P(W|E) is high, since E is true mainly in worlds where he is fit. But P(E→W) is low, which is why he doesn’t actually enter.
    Maybe. One can also reason about “E→W” as follows: probably, were he to enter, this would only be if he is fit, and so he’d win.
    In any case, these counterexamples aren’t going to apply in the Molinist case, since in Molinist conditionals the antecedent includes all the relevant information prior to the free action, and hence will include what kind of shape Jones is in.

    January 18, 2013 — 16:59
  • Mike Almeida

    Do you mean things the other way around in the Jones case? It seems that P(W|E) is high, since E is true mainly in worlds where he is fit
    I’m pretty sure that’s not the way to read that conditional The conditional P(W|E) tells is the probability that HE ACTUALLY WINS given that HE ACTUALLY ENTERS. If he actually enters, he is going to lose, since he is actually unfit. We ask in this case “will he win if he does enter”, and the answer to that is no. The conditional P(E -> W) tells us the probability that HE WOULD WIN given that HE WERE TO ENTER. Here we are wondering whether in the closest (non-actual) worlds in which he does enter, he wins. And the answer in this case is yes. In the first case, we know that he won’t enter the race, since he is unfit, but we ask, what if he decides to enter anyway. If he does enter, he is not suddenly going to become fit. In the second case we ask what would happen, were he to enter. We know that, he would enter only if he were fit. So, two conditional claims are true here.
    1. If he does enter, then he will not be fit
    2. If he were to enter, then he would be fit.
    There are other cases, if you don’t like that one. How about this. There is poison in the glass and you know it. Ask: what will happen if you drink it? Surely the answer is: you will die. P(D | R) is high. But what would happen, were you to drink it? Suppose you would never drink it unless you knew the glass contained un-poisoned water. So, the closest worlds in which you do drink it, you live. P(R -> D) is low.

    January 18, 2013 — 17:45
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