I will address examples of the sort that come up in philosophical discussion, including the recent example due to Paolo Santorio.
This post involves the notions introduced in ‘Probabilities of Conditionals: (1)’, and continues the discussion of Set-Up 1 and Set-Up 2.
PROPOSITIONS TRUE IN 0, 1, OR 2 WORLDS
In the previous post I showed that in Set-Up 1 we can construct s(p,.) so that P(p →r) = P(r | p) for any proposition p that has three members. As I also pointed out this cannot be done for propositions with four or five members, since their probability is not a multiple of 1/6. Let us look at what remains.
The empty set Λ is a proposition. We do not really need to define a selection function for it, but if we do, we can set s(Λ, x) = Λ for all worlds x. The result is the same, (Λ → u) = {x: s(Λ, x) is part of Λ} = S, the tautology.
Some propositions are true in one and only one world. Consider now u = {(1)}. It is clear that P(r|u) equals 1 iff u is part of r, and equals 0 otherwise. So we set s(u, x) = {(1)} for all x. Thus (u →{(1)}) = S and (u → r) = Λ if (1) is not in r.
Finally, we still need to consider propositions true in precisely two worlds. Let t = {(1), (2)}. If x is in t we want P({(x)} | t) to equal 1/2. So we need there to be two worlds in ~t whose nearest world in t is (1), and two worlds in ~t whose nearest world in t is (2). Since there are precisely four worlds in ~t, that is easily done.
But note well that here, in Set-Up 1, we cannot have conditionals with either ~u or ~t as antecedents, for those have respectively five and four members.
Thus we have here canvassed all the conditionals that can appear as antecedents in examples of conditionals represented in Set-Up 1. These are the propositions true in just no, one, two, or three worlds. But even here we can find some interesting, even somewhat surprising, consequences.
SOME INTERESTING CONSEQUENCES
As in the preceding post,
p = {(1), (3), (5)}, “the outcome is odd”
q = {(1), (2), (3)}, “the outcome is low”
Previous results: (p→q) = {(1), (3), (2), (4)} and (~p → q) = {(2), (1)
[I] Some nesting of arrows already appear in Set-Up 1
Since (~p → q) = {(2), (1)}, the second-degree proposition (p → (~p → q)) is the proposition (p → {(2), (1)}). This has (1) in it since that is in the intersection of antecedent and consequent. And it has (2) in it, since the nearest p-world of (2) is (1), which also belongs to {(2), (1)}. And that is all, so [p → (~p → q)] is just (~p → q) itself.
This already shows that Import-Export fails, because (p & ~p) → q is the whole of S, since s(p & ~p, x) is necessarily empty for any world x.
[II] An assumption in some ‘triviality’ discussions
Another example that appears in ‘triviality’ discussions is a proposition of the form of [~q → (p → q)], and the assertion that P(p →q| ~q) = 0.
(See Khoo and Mandelkern, page 506, for its role in connection with Lewis’ triviality proof).
In the previous post we saw that (p → q) = {(1), (3), (2), (4)}. We can at once see the conditional probability P(p → q | ~q): it is the proportion of {(1), (3), (2), (4)} in {(4), (5), (6)}, and that equals 1/3, not zero!
Intuitively: since p →q and ~q have world (4) in common, that world (4) must be in [~q →(p → q)]. Therefore the probability of [~q →(p → q)] must be at least as high as the probability of {(4)}, so it cannot be zero.
So it is not the case that in general P(p →q| ~q) = 0.
[III] Deceptive intuitions about conjunction
Another observation concerns conjunctions. Intuitively we would read “The match will light if it is struck, and the match will light if it is not struck” as meaning simply that the match will light. (We might add, “regardless of whether it is struck”).
That intuition may have some value, or it may not, but it does not generalize. In our example here we see that
[(p –> q) ∩ (~p –> q)] = {(1), (2)}, which has probability 1/3
That is not the probability of q, which is 1/2.
Nor is q the same proposition as [(p → q) & (~p → q)].
[IV] The example due to Paolo Santorio
Finally, we can actually treat the probabilities in the example due to Paolo Santorio, which I heard from Branden Fitelson.
A die is tossed and to check Import-Export we investigate whether the following two propositions have the same probability:
If the outcome is either odd or six then (if the outcome is even then it is six)
If the outcome is odd or six, and it is even, then it is six
The second proposition is necessarily true (given our assumptions about the situation) so it has probability 1. But the first proposition has a nested “if .. then” and so is, philosophically, up for grabs. Must it have probability 1? As we will see the answer in CE is NO.
We can handle this example to some extent in Set-Up 1, although it involves a proposition true in four worlds.
a = {(1), (3), (5), (6)} “the outcome is either odd or six”
b = {(2), (4), (6)} “the outcome is even”
c = {(6)} “the outcome is six”
Since the meet of a and b is c, it is clear at once that [(a ∩ b) → c] = S
Here in Set-Up 1 we can already deal with (b → c). In fact b = ~p in our first example, so we can verify, with a quick look at the diagram, that (b → c) = {(5), (6)}.
Thus P(b →c | a) = 1/2.
This will still be correct when we look at it in Set-Up 2: there (b → c) has twenty members, all of which belong to a, which has forty members.
So if [a → (b → c)] has a probability at all it must be, according to the Equation, 1/2 and not 1.
[V] What about Santorio’s conditional [a → (b → c)]?
In Set-Up 2, (b → c) is the set of worlds which is the union of [(5)] and [(6)], and similarly, proposition a is the union of [(1)], [(3)], [(5)], and [(6)].
We need to construct s(a, .) so that the conditional probabilities come out right, as before. So P(a → [(1)] = 1/4, etcetera. This we do by dividing the union of [(2)] and [(4)] evenly:
s(a, <2,x>) = <1,x> for x = 1, …, 5
s(a, <2,x>) = <3,x> for x = 6, …, 10
s(a, <4,x>) = <5,x> for x = 1, …, 5
s(a, <4,x>) = <6,x> for x = 6, …, 10
So [a → (b → c)] is true first of all in all of [(5)] and [(6)], and secondly in all of { <4, x>: x = 1, …, 10}, a total of 30 worlds. The probability of [a →(b →c)] is accordingly 30/60, that is, 1/2 as it should be.
This also shows the failure of Import-Export, now we have represented the example completely in Set-Up 2. For the ‘other’ conditional, [(a ∩ b) → c] = (c → c) is not the same proposition at all, it is a tautology and has probability 1.
REFERENCES
Justin Khoo and Matthew Mandelkern, “Triviality Results and the Relationship between Logical and Natural Languages”. Mind 128 (2019): 485-526.
Bas van Fraassen's philosophy blog 