Thesis: P(if A then C) = P(C/A).

On the account I like, if you say 'P(if A then C)' in English, you almost inevitably end up saying something that denotes the conditional probability P(C/A), rather than the unconditional probability of some proposition expressed by 'if A then C'.

So it's interesting that Vann McGee and Stefan Kaufmann have found intuitive counterexamples to Stalnaker's Thesis. One of Kaufmann's examples in "Conditioning against the grain" goes as follows. There are two bags. In bag X, most balls are red, and most of the red balls have black spots. In bag Y, few balls are red, and few of those balls have black spots. You are 75% confident that the bag in front of you is bag Y. Now consider the statement:

(1) If you pick a red ball, it will not have black spots.

Many people apparently intuit (1) to have fairly high probability. I take that to mean that they would assent to

(1') Probably, if you pick a red ball, it will not have black spots.

This contradicts the Thesis, because getting a red ball is evidence that the bag in front of you is bag X, in which case it is rather likely that the ball has black spots.

As Kaufmann observes, if these facts are made salient -- if one points out that picking a red ball is much more likely if it's bag X rather than Y, and that most red balls in bag X have spots -- then people's intuitions switch and they deem (1) to have low probability. So it looks like the Thesis is right about some contexts, but not about others.

Kaufmann's explanation is that there are two ways of evaluating conditional probabilities, one "local" and one "global". Globally, 'P(if A then B)' denotes P(B/A); locally, 'P(if A then B)' denotes the expectation of P(B/A) relative to a certain parition, here the partition of bags { X, Y }:

(L) P(if A then B) = P(B/AX)P(X) + P(B/AY)P(Y).

The idea, which sounds plausible, is that when we judge (1) to be probable, we hold fixed that P(Y)=0.75 and note that P(No Spots / Red & Y) is high, which by (L) means that the probability of (1) is high.

But why would we use (L) to evaluate conditional probabilities? The "global" evaluation that conforms to Stalnaker's Thesis is predicted by the general Kratzer-style semantics of 'if' and 'probability'. Where does the "local" reading come from?

Kaufmann suggests that the two evaluations corresponds to different ways of supposing A, and also that the local evaluation can be understood as giving the expectated conditional chance of B given A, since chance is credence conditionalised on the true member of a relevant partition. Both of these remarks suggest that (L) could give the subjunctive conditional probability of B given A, P(A\B), rather than the indicative conditional probability P(A/B). Indeed, the kind of compartmentalised conditioning that figures in (L) is precisely what Lewis uses in "Causal Decision Theory" to define the imaging function for subjunctive conditional probabilities.

So maybe that's what's going on: when people judge (1) to be probable, they read the conditional as subjunctive. This isn't too implausible, I think, because in English the distinction between the subjunctive and indicative reading is usually only marked in the past tense. Read subjunctively, the intuitive judgement about (1) is correct, as can be seen if one enforces this reading by saying "if you were to pick a red ball, it would not have black spots".

The hypothesis that the subjunctive reading is in play might also be supported by the fact that the intuition about (1) becomes much weaker -- I think -- if the sentence is put into the past. Suppose you've drawn a ball but haven't looked at it yet. Consider:

(1'') If you picked a red ball, then it does not have black spots.

The hypothesis also fits the phenomenon that people's intuitions flip when it is pointed out that picking a red ball makes it more likely that it's bag X than bag Y: this context, where the topic is what is evidence for what, makes the indicative reading salient.

So far, so good. Unfortunately, the present story does not work for McGee's examples. Here is one Kaufmann discusses as well. Initially, you believe that Murdoch died in an accident. Then somebody who you think is probably Sherlock Holmes says that Murdoch was killed, that Brown is probably the murderer, and that in any case

(2) If Brown didn't kill Murdoch, then someone else did.

According to McGee, most people now regard (2) as highly probable. However, if it turns out that Brown didn't kill Murdoch, then you'd lose your confidence that the speaker is Holmes, and thus return to your judgment that Murdoch died in an accident. So the (indicative) conditional probability corresponding for (2) is low.

Kaufmann doesn't find this problematic, since it conforms to his local evaluation rule (L), this time using the partition { he's Holmes, he's not Holmes }. But this application of (L) cannot plausibly be taken to give the subjunctive conditional probability of someone else killing Murdoch given that Brown didn't kill him. The subjunctive probability is surely low. If you think that Brown probably killed Murdoch, you will not judge it very probable that if Brown hadn't killed him then someone else would have. Moreover, it is anyway implausible that people are reading (2) subjunctively, because it is in the past tense.

The reason why Kaufmann's rule (L) here doesn't yield subjunctive conditional probability is that it uses a bad partition { Holmes, not Holmes }. (This also makes it implausible to describe (L) as computing expected conditional chance.) Roughly speaking, the cells of a good partition would say enough about the the world and its causal structure so that, combined with either the assumption that Brown did kill Murdoch or that he didn't, each cell would entail whether someone else killed Murdoch. Applying (L) to such a partition yields a low conditional probability.

(L) is partition-dependent: the "local" probability of a conditional depends on the chosen partition. By choosing a suitable partition, we can let the local probability have almost any value we like. Kaufmann stresses that not all partitions are acceptable for (L), and that the right partitions must somehow encode the "causal structure of the scenario" [p.598]. But it isn't clear why this makes { Holmes, not Holmes } acceptable.

Let's redescribe Kaufmann's first example with a different partition. Again, you get to draw a ball from either bag X or bag Y; X contains mostly red balls with mostly black spots, Y has few red balls, few of which have black spots; based on your evidence, you are 75% certain that the bag in front of you is bag Y. If the contents of the bags are precisely specified (as Kaufmann does), it is possible to calculate your probability for the hypothesis that you draw a red ball from bag Y. Let this hypothesis be called RY. Given your evidence, the probability of RY is quite low, say 0.05. So you're very confident that not-RY is true. Moreover, if not-RY is indeed true and you draw a red ball, then the ball can only come from bag X, in which case it probably has black spots. Now consider

(1) If you pick a red ball, it will not have black spots.

I suspect many would judge (1) to have low probability in this context, lower than P(No Spots/Red) and much lower than the subjunctive P(No Spots\Red). But the scenario is exactly the same as Kaufmann's -- I've just made a different partition salient.

Here is one lesson we might draw. There aren't just two kinds of conditional probabilities, indicative and subjunctive, but infinitely many, one for each choice of a partition. Every partition induces an imaging function and thereby a type of subjunctive supposition. We could then also fold indicative conditional probability into the subjunctive kind, induced by the single-membered partition. Context usually determines which partition is salient for statements about conditional probability (i.e. for statements that look like statements about the probability of a conditional).

Maybe. But if that's true, I'd like it to follow from the general semantics of 'if' and 'probability'. Neither of these, by itself, seems to be sensitive to the contextually salient partition -- at least not to the extent required for the present proposal to work.

I prefer another, perhaps more obvious, explanation: people who intuit that (2) is probable and (1) very improbable (in the revised context) have made a mistake.

Where does the mistake come from? In part, it may come from the fact that the (standard, indicative) conditional probability is a bit hard to determine in these cases, because one has to keep track of two factors that pull in opposite directions. For example, in the case of (2), the hypothesis that Brown didn't kill Murdoch supports that someone else did it within the "Holmes" cell of the partition, but simultaneously lowers the probability of that cell and thereby the probability that Murdoch was killed.

More importantly, I think the mistake comes from the grammatical illusion that a question about the the probability of a conditional is a question about the probability of a certain proposition. If A is a proposition and { X, Y } a partition, then of course

P(A) = P(A/X)P(X) + P(A/Y)P(Y).

So we can always evaluate the probability of a proposition by considering its probability under different hypotheses and then take the weighted average. The result never depends on the chosen partition. When asked about the probability that if A then B, we mistakenly apply the same recipe, not realising that 'the probability that if A then B on the assumption that X' denotes P(B/AX) rather than something of the form P(A->B/X).

Consider another of McGee's examples. Quantum mechanics entails that

(3) If all atoms in this table decay within the next second, then Z amount of energy is released,

for some particular value Z. McGee suggests that if we trust quantum mechanics, then we will assign high probability to (3). However, P(Z released / table decays) is low, since finding the table suddenly decay would dramatically lower our confidence in quantum mechanics.

If the probability of (3) is the probability of a certain proposition that's entailed by quantum mechanics, then it is clear why trusting quantum mechanics requires assigning high probability to (3). But on the Kratzerian account, there is no such proposition, at least not if the conditional is read indicatively. (It could also be read as a nomologically strict conditional, in which case the failure of Stalnaker's Thesis is unproblematic.) On the indicative reading, there is no proposition that is (i) entailed by quantum mechanics and (ii) whose probability is in question when we ask about the probability of (3). Perhaps it is the prima facie plausibility that there is a proposition satisfying (i) and (ii) that explains why we mistakenly think the probability of (3) must be high, even on the indicative reading.

1. maximally specific: T contains either P or ~P, for every proposition P;
2. consistent: T does not contain P and ~P, for any proposition P;
3. closed under conjunction and logical consequence: if T contains both P and Q, then it contains their conjunction P & Q, and if T contains P, and P entails Q, then T contains Q.

It is often useful to go in the other direction and identify propositions with sets of possible worlds. We can then analyse entailment as the subset relation, negation as complement and conjunction as intersection. Of course, we may not want to say that a world is a (non-empty) set of (consistent) propositions and also that a consistent proposition is a non-empty set of worlds, since these sets should eventually bottom out. But that doesn't seem very problematic, and it is easily fixed as long as there is a simple 1-1 correspondence between worlds and logically closed, consistent and maximally specific theories. In particular, one might suspect that on the present definitions, every logically closed, consistent and maximally specific theory uniquely corresponds to a possible world, namely the sole member of the intersection of the theory's members.

But it looks like this is false. Since there are infinitely many worlds, one can show (e.g. in ZFC) that there are sets of sets of worlds that are logically closed, consistent and maximally specific, but do not single out any particular world: the non-principal ultrafilters on the space of worlds. The non-principal ultrafilters contain the negation of { W } for every world W. So these theories are true at no world whatsoever. They are nevertheless consistent, since they don't contain any proposition together with its negation.

This is odd. I would like to say that although I sometimes define theories as sets of propositions and propositions as sets of worlds, one can (if one wants) just as well go in the other direction and define possible worlds as logically closed, consistent and maximally specific theories. But the two definitions don't seem to line up. I somehow need to exclude the non-principal ultrafilters, without talking about their set-theoretic construction (which would presuppose my own order of definition). I suppose this could be done by strengthening the closure condition, e.g. by saying that whenever T contains some propositions, then it also contains the (possibly infinite and uncountable) conjunction of those propositions. Would that work? Is there a better response?

Evidential decision theory says that you should take only the opaque box, while causal decision theory says you should take both. Since you already know whether your ordinary payment is $100 or $1000, your decision between the boxes has neither causal nor evidential impact on today's ordinary payment. It does have an impact on how much you will get tomorrow, but you don't care about tomorrow. What's left is a standard Newcomb problem.

The upshot is that if you follow evidential decision theory, you have $200 to spend most days, and occasionally only $100 (if the prediction went wrong). If you follow causal decision theory, you have $1010 most days, and occasionally $1110.

In this scenario, the two-boxers might start to ridicule the one-boxers. "Look how much money we got again today, and how many beautiful things we could buy. You, on the other hand, have barely enough for food and rent. Wouldn't you rather be one of us?"

The ridicule is, of course, unfair. When the two-boxers take their two boxes, they don't care what effect this has on their payment the following day. They choose two boxes only because this is guaranteed to give them $10 more today than whatever they would get if they chose only one box. It just happens that their choice also makes them get $1000 the following day. The payment setup rewards two-boxers and punishes one-boxers -- but not because two-boxers are making the better choice. The setup could just as well have been the other way round, so that one-boxers would have been much better off than two-boxers.

Nevertheless, the ridicule is quite similar to the ridicule two-boxers often get from one-boxers in the standard Newcomb problem. "If you're so smart", say the one-boxers, "why ain'cha rich? Look how much money we got; wouldn't you rather be one of us?" We two-boxers reply that we're poor not because we made the wrong choice, but because the Newcomb setup rewards one-boxers and punishes two-boxers: one-boxers mostly get to choose between $100 (in the opaque box) and $110 (in both boxes together), while two-boxers are given a choice between $0 (in the opaque box) and $10 (together). That's why one-boxers end up with more money -- not because they made the better choice. (Indeed, if they had taken both boxes, they would be even richer.)

It would be nice if we could add that the setup might just as well have been the other way round, rewarding two-boxers rather than one-boxers. However, it is difficult to come up with such a reversed setup. Lewis (in "Why ain'cha rich?") conjectured that it can't be done. The story above at least seems to come close.

(The story can also be adapted to several agents rather than a single agent at different times. Suppose infinitely many copies of you have been created and assigned to the numbers ...-3,-2,-1,1,2,3,..., with the number 0 reserved for yourself. Each of you first gets either $100 or $1000. Then you face a Newcomb problem as above. If player i takes both boxes, player i+1 gets $1000, otherwise they get $100. All you care about is your own personal profit, and thus all your copies likewise only care about their own personal profit. If you're a one-boxer (or more generally an evidential decision theorist), you end up with either $200 or $100. If you're a two-boxer (a causal decision theorist), you get either $1010 or $1110.)

As always, comments are much appreciated.