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I've encountered a strange argument in an article, and I don't quite understand what the author means.

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  1. “If A, then if B, then C” and “If A and B, then C” are logically equivalent, and logical equivalence is probability-preserving. And so we have that these probabilities are equal: Pr (If A, then if B, then C ) = Pr(If A and B, then C)
  2. This is an instance of 1), so we have Pr (If ¬B, then if A, then B ) = Pr(If A and ¬B, then B)
  3. SH is defined as: (SH) Pr(If A, B) = Pr(B | A), for all probability functions Pr such that Pr(A) > 0. That is to say, if conditionals express propositions, the probability of the proposition expressed by a given conditional must equal the conditional probability of the conditional’s consequent given its antecedent. Applying SH to 2) gives us Pr(If A, then B | ¬B) = Pr(B | A and ¬B).
  4. Pr(B | A and ¬B) = 0. This is because given A & ¬B, the probability of B is 0.
  5. The probability of Pr(B | A and ¬B) is equal to Pr(If A, then B | ¬B), so it is also 0.
  6. And Pr(If A, then B | ¬B) = 0 iff Pr(¬B |If A, then B) =0 iff Pr(B |If A, then B) = 1 # I have no clue what happened here.

Could someone please help me understand how the author is justified to be making these conclusions?

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  • Are you just asking about that last step 6? Ignoring everything prior, it's not hard to show those three statements are equivalent. In general, P(x|y) = 0 always implies P(y|x) = 0 and vice versa, since if P(y|x) wasn't 0 there would be some trials where x occurred and y occurred as well, which would violate P(x|y) = 0. And in general P(~x|y) = 0 always implies P(x|y) = 1, since on every trial where y occurred, it must be the case either that x occurred or that it didn't, and if there were 0 trials where y occurred but x didn't, then that means x always occurred when y did.
    – Hypnosifl
    Jun 28 at 22:13
  • @Hypnosifl Thank you. Could you also explain how this implies that one cannot be certain of a conditional without being certain of the conditional's consequent?
    – Nick Doe
    Jun 28 at 22:31
  • Think of the example he gives on that page (p. 11) with the statement "if John comes to the party, then Mary will come too"--let B be "Mary comes to the party" and A be "John comes to the party". So he's arguing it's a problem that you get P(B|if A, then B) = 1, since it might be the case that Mary doesn't come to the party in our world (so P(B) can't be 1), yet the modal statement "if A, then B" is true (Mary is guaranteed to come in any nearby possible world where John does).
    – Hypnosifl
    Jun 28 at 22:48
  • My feeling is that this shows a problem with the original assumption labeled (IE) on p. 5, that it doesn't adequately distinguish between modal statements about statistical dependences between events in nearby possible worlds vs. statements about what actually occurs in our world. If you restrict A, B, and C in the original (IE) to all be statements about what's true in our world I don't think you'd get these kinds of problems. But I haven't read Lewis' argument so I don't know if he says something similar.
    – Hypnosifl
    Jun 28 at 22:51
  • Consider for example the statement "if Mary will come to the party if John comes to the party, then Amanda won't come to the party". (IE) would say this is equivalent in meaning to "if Mary comes to the party and John comes to the party, then Amanda won't come". But it could mean that Amanda doesn't like Mary basing her decisions on John this way, so Amanda won't go if she learns that Mary will automatically go if John does; but on the other hand if Mary isn't basing her decision on John, then Amanda might go even in the case where both Mary and John go.
    – Hypnosifl
    Jun 28 at 23:25
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It might help to provide some background here. This discussion is part of an attempt to give an account of the meaning of conditional expressions, particularly those where the antecedent might or might not be true. Stalnaker proposed back in the 1960s that uncertain conditionals should be understood as statements of conditional probability. I.e., the appropriate degree of credence that attaches to "if A then B" is given by P(B|A). This works fine in a lot of cases. It agrees with our simple intuitions about uncertain conditionals, and it is consistent with work done in the logic of uncertainty and in decision theory.

But it has an important consequence. The conditional probability P(B|A) is not the probability of a proposition. There is no categorical proposition X such that P(B|A) = P(X) over a range of distributions of values for P. If you think in terms of the probabilities of events, this can be understood as saying that P(B|A) is not the probability of an event. It is the probability of B within the range of possibilities constrained by A. In more general terms, where we understand P() as a degree of credence rather than a frequency, it means that while the categorical P(X) is the probability of the proposition X being true, P(B|A) is not the probability of any proposition being true.

This appears to have been first proved by David Lewis back in the early 1970s. Some theorists have taken it to punch a hole in the Stalnaker account, since if P(B|A) is meant to represent the probability of "if A then B" and P(B|A) is not the probability of any proposition, then it follows that conditionals are not propositions. Others have embraced this consequence and accept that conditionals are indeed not propositions. Notable examples among these are Ernest Adams, Dorothy Edgington and Richard Bradley.

It is strange at first to think of conditionals as things that are not propositions, and hence don't have truth values. When studying logic 101, the first conditional you are introduced to is material implication, and this does have a truth value and is in fact a simple truth function. But material implication is a narrow and specialised kind of conditional and does not accurately reflect the meaning of most ordinary language conditionals. The position defended by Adams, et al, is sometimes referred to as the suppositional account of conditionals, and it can be thought of as holding that it is not the conditional "if A then B" that has a truth value, but rather that B has a truth value within the suppositional context A. When we say of a conditional that it is true, this is a slightly elliptical way of saying that B is true within the context A.

The whole issue of the logic of conditionals and the pros and cons of different accounts of conditionals has developed into a huge literature in the last 50 or 60 years. Stalnaker himself went on to propose an account of conditionals as statements about possible worlds. There are still many rival theories of conditionals. The SEP articles on conditionals provide an introduction, here, here and here.

Now back to the passage you quote. Douven presents a simplified alternative to Lewis' proof, making use of the import-export principle. This is the principle that the right-nested conditional "if A, then if B then C" is always equivalent to "if A and B, then C". Despite what Douven says, this is not endorsed by virtually all who have thought about it. It is actually a controversial principle to which there are arguably exceptions, especially when we think about conditionals in possible world terms. Nevertheless, if we allow it, then, like the Lewis proof, it shows that conditional probabilities with nested conditionals collapse into triviality.

Your step 6 goes from Pr(If A, then B | ¬B) = 0 to Pr(¬B |If A, then B) = 0, which is an instance of P(X|Y) = 0 entails P(Y|X) = 0. Then it goes to Pr(B |If A, then B) = 1, which is an instance of P(¬X|Y) = 0 entails P(X|Y) = 1. In both cases, assuming X and Y are well-defined and the probability of the antecedent is non-zero.

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  • The section of the first SEP article on conditional probability indicates that one way of interpreting probabilities involves an ensemble of equally-probable possible worlds: "Imagine a partition as carved into a large finite number of equally-probable chunks, such that the propositions with which we are concerned are true in an exact number of them. The probability of any proposition is the proportion of chunks in which it is true."
    – Hypnosifl
    Jun 29 at 17:18
  • (cont.) So in terms of this picture, why can't P(B|A) be interpreted as the probability of a proposition involving such an ensemble like "if you look only at the subset of chunks where A occurred and pick one at random, you will find B occurred in the chunk you chose"?
    – Hypnosifl
    Jun 29 at 17:20
  • Indeed, on some accounts a proposition just is that set of possible worlds in which it is true. But a conditional probability is not a set of possibilities. Its probability is a ratio of the quantitity of possibilities in which A∧B holds, to the quantity of possibilities in which A holds. This cannot be identical to any set of possibilities. To put it another way, for any A, B, we can point to the set of PWs in which A∧B is true, or A∨B, or even A⊃B, but there is no set of PWs in which B|A is true.
    – Bumble
    Jun 29 at 21:04
  • I suppose my notion of "picking one at random" from a set of possible worlds PW={PW1, PW2, PW3, ..., PWN} would require some notion of "second-order" possible worlds based on the original "first-order" set, i.e. "in one second-order possible world I picked randomly from that set of first-order possible worlds and got PW1, in another second-order possible world I picked randomly from that set of first-order possible worlds and got PW2, etc." Presumably this is just not allowed in the particular scheme for translating probability statements into possible world statements that Lewis considers?
    – Hypnosifl
    Jun 29 at 21:16
  • If your second-order worlds are sets of first-order worlds, then that would not work. If they are not, I'm not sure what they are. Lewis follows Frank Jackson in holding that material implication gives the truth conditions of a conditional, but there are also assertability conditions, which are given by the conditional probability. The idea is that we start with P(A⊃B) then subtract off P(A⊃B|¬A) because ¬A does not contribute to the assertability of "if A then B", and we are left with P(B|A). For myself, I find it highly dubious that we should distinguish truth from assertability like this.
    – Bumble
    Jun 29 at 22:04

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