Sly Pete and Mr. Stone are playing poker on a Mississippi riverboat. It is now up to Pete to call or fold. My henchman Zack sees Stone's hand, which is quite good, and signals its content to Pete. My henchman Jack sees both hands, and sees that Pete's hand is rather low, so that Stone's is the winning hand. At this point, the room is cleared. A few minutes later, Zack slips me a note which says "If Pete called, he won," and Jack slips me a note which says "If Pete called, he lost." I know that these notes both come from my trusted henchmen, but do not know which of them sent which note. I conclude that Pete folded.

One puzzle raised by this scenario is that it seems perfectly appropriate for Zack and Jack to assert the relevant conditionals, and neither Zack nor Jack has any false information. So it seems that the conditionals should both be true. But then we'd have to deny that 'if p then q' and 'if p then not-q' are contrary.

Frank Jackson (in conversation) pointed out that Gibbard's passage raises another puzzle that is commonly overlooked. That puzzle is about confirmation.

Let C→W be the conditional 'if Pete called, he won'.

Let E1 be Zack's information -- more specifically, the information that Pete knows Mr. Stone's hand.

Let E2 be Jack's information -- specifically, that Mr. Stone has the better hand.

Intuitively,

(1) E1 strongly supports C→W.

(2) E2 strongly supports ~(C→W).

(3) E1 doesn't strongly support ~E2.

(4) E2 doesn't strongly support ~E1.

But if we read "strongly support" as "making highly probable" then these four assumptions are probabilistically inconsistent. (The proof is left as an exercise.)

You might question (3) or (4). Here's a simpler example where (3) and (4) are not in doubt.

We toss two independent, fair coins. There are four possible outcomes: { H1,T1 } x { H2,T2 }.

Let Same be the proposition (H1 & H2) v (T1 & T2).

Let E1 be Same.

Let E2 be T2.

Let H1→Same be the conditional 'if H1 then Same'.

Intuitively,

(1) E1 strongly supports H1→Same: P(H1→Same/E1) > 0.8 (say).

(2) E2 strongly supports ~(H1→Same): P(~(H1→Same)/E1) > 0.8.

But the following is easily provable:

(3) E1 doesn't strongly support ~E2: P(E2/E1) = 1/2.

(4) E2 doesn't strongly support ~E1: P(E1/E2) = 1/2.

(1)-(4) are probabilistically inconsistent. So (1) and (2) can't be true: either E1 doesn't make H1→Same highly probable or E2 doesn't make ~(H1→Same) highly probable (or both).

The lesson is that our intuitions about whether some piece of evidence supports a given conditional cannot be trusted.

The usual contextualist responses to Gibbard's puzzle seem to be of no help here. The only way to block the lesson would be to give up probabilistic measures of evidential support. But even then we retain the lesson that we can't trust intuitions about whether some evidence renders some conditional probable.

The lesson generalizes. If we can't trust these intuitions, then we
also can't trust intuitions about the probability of a conditional in
a given hypothetical scenario -- for that just *is* an intuition
about the extent to which the assumptions of the scenario makes the
conditional probable. And then we plausibly also can't trust outright
intuitions about the probability of a conditional, since that's the
probability of the conditional given our total evidence.

The lesson is more or less the same as the lesson taught by Lewisian triviality results. But the Gibbard-Jackson route is different from Lewis's route. In particular, we have never assumed that the intuitive probability of a conditional is the corresponding conditional probability.

That said, there is also a way of turning the Gibbard-Jackson argument into an argument against "Stalnaker's Thesis", that for any rational credence function P, P(A→B) = P(B/A). Here is how.

Return to the coin toss scenario. It is easy to see that

(5) P(Same/H1) = 1/2,

(6) P(Same/H1 & Same) = 1

(7) P(Same/H1 & T2) = 0

By Stalnaker's Thesis, it follows that

(8) P(H1→Same / Same) = 1 and

(9) P(H1→Same / T2) = 0,

since P(*/Same) and P(*/T2) are rational credence functions.

(8) and (9) are stronger versions of (1) and (2), and we know that these can't be true. So Stalnaker's Thesis is also false.

]]>So the 'ought' of objective consequentialism evaluates acts "causally", rather than "evidentially". This provides some (intuitive) motivation for using a causal evaluation for the decision-theoretic 'ought' as well. Can we strengthen this observation? How bad would it be to combine objective consequentialism with evidential decision theory?

Here's one attempt to bring out a tension. Imagine an agent whose personal utility function orders possible states of the world in just the way some form of objective consequentialism does, giving highest utility to the "best" states and lowest to the "worst" ones. Suppose also the agent has perfect information about which state would result from each of the options presently available to her. Intuitively, what this agent ought to do in light of her beliefs and desires is precisely what she ought to do according to objective consequentialism. That is, the subjective 'ought' of decision theory and the objective 'ought' of objective consequentialism should here coincide.

In fact, however, the two oughts plausibly do coincide even in evidential decision theory. That's because, as Lewis pointed out in "Causal Decision Theory", conditional on any particular dependency hypothesis (about what the available options would bring about), evidential expected utility and causal expected utility are plausibly equivalent.

So we need a different case to bring out the tension. Here's such a case, inspired by "Jack Spencer and Ian Wells.

Consider a Newcomb Problem in which the outcomes are measured not in dollars but in consequentialist utilities. As before, assume the agent facing the problem has subjective utilities that match the consequentialist utilities.

It is clear what the agent ought to do, from the perspective of objective consequentialism: she ought to take both boxes. (Recall that the 'ought' of objective consequentialism evaluates acts causally, by looking at the outcomes the acts would bring about, given all relevant facts about the world -- known and unknown. One relevant fact is the content of the opaque box. If the opaque box is in fact empty, then one-boxing would lead to zero consequentialist utilities and two-boxing to a thousand; if the opaque box is non-empty, then one-boxing would lead to 1 million utilities and two-boxing to 1 million and 1 thousand. Either way, two-boxing would lead to the better state.)

Now here we have an agent with perfectly consequentialist values
who *knows* that she ought to two-box, in the objective
sense. Yet evidential decision theory says it would be irrational for
her to two-box! That's not a logical contradiction. But it surely
sounds unappealing. It would be better to have a decision theory on
which it can't happen that a morally perfect agent is irrational for
choosing an act of which she knows that she morally ought to choose
it.

The argument generalizes. For one thing, it generalizes beyond evidential decision theory to other decision theories that recommend one-boxing, such as ""timeless decision theory", ""disposition-based decision theory", "Spohn's recent spin on causal decision theory, and whatever decision theory Teddy Seidenfeld thinks is right.

The argument also generalizes beyond objective consequentialism, given that almost every (sensible) moral theory can be consequentialised. In general, if you think the notion of an objective moral ought is coherent, you probably shouldn't say that one-boxing is the rational choice in Newcomb's Problem.

]]>I am puzzled about these efforts, for two reasons.

First, as Lewis and Kratzer pointed out in the 1970s and 80s,
if-clauses often (according to Kratzer, always) function as
restrictors of quantificational and modal operators. So when we see an
if-clause in the vicinity of a modal like 'the probability that', the
*first* thing we should consider is whether the if-clause
restricts the modal.

How would an if-clause restrict a probability modal? Well, what is the probability of B restricted by A? An obvious answer is that it's the probability of B given A. So if the if-clause in 'the probability that if A then B' restricts the probability modal, then the expression denotes the conditional probability of B given A. Which is just what we find.

In other words, there is independent evidence about if-clauses suggesting that 'the probability that if A then B' should be analysed as 'P(B/A)' rather than 'P(if A then B)'. If that's correct, then what's expressed by

(*) the probability that if A then B equals the conditional probability of B given A

is the trivial identity 'P(B/A) = P(B/A)'. There's no need to make a big effort trying to make (*) true.

The case of subjunctive conditionals is parallel. We have the intuition that

(**) the probability that if A were the case then B would be the case equals the conditional probability of B on the subjunctive supposition that A.

Again, the first thing we should check is whether the if-clause restricts the modal. And, plausibly, subjunctive if-clauses restrict probability modals by subjunctive supposition (aka imaging). And then (**) expresses the trivial 'P(B//A) = P(B//A)'.

When people try to give a semantics motivated by (*) or (**), they practically never explain what's wrong with the simple and obvious explanation of (*) and (**) that I've just given.

That's one reason why I'm puzzled by these efforts. Here's a second reason.

For concreteness, let's look at subjunctive conditionals, which I'll write 'A > B'. As Lewis shows towards the end of "Probabilities of Conditionals and Conditional Probabilities", if you want to validate 'P(A > B) = P(B//A)', you have to assume a Stalnaker-type semantics for '>' on which, for any world w and any proposition A, there is a unique A-world that is "closest" to w; 'A > B' is true at w iff B is true at the closest A-world.

But if we assume a Stalnaker-type semantics of would counterfactuals 'A > B', then what should we say about might counterfactuals, 'if A were the case then B might be the case' -- for short, 'A *> B'?

Clearly, 'A *> B' can't be the dual of 'A > B', otherwise the two would be equivalent. The only option I can think of is to say, with Stalnaker, that 'A *> B' must be analysed as Might(A > B).

But that's unappealing, especially in the present context.

For one, the idea that 'A *> B' means 'Might(A > B)' is incompatible with a broadly Kratzerian treatment of 'if' and 'might'.

Moreover, syntactically, 'would' and 'might' seem to play similar roles in 'if A then would B' and 'if A then might B'. One would at least like to see some more evidence that 'might' scopes over the conditional and 'would' does not. Relatedly, (as I mentioned in an earlier post), it seems to me that

What if A were the case? It might be that B

is equivalent to 'if A were the case then it might be that B'. But surely 'might' in the second sentence doesn't somehow scope over 'if' in the first.

Moreover, let's look at the probability of might counterfactuals. Assuming that 'Might' in 'Might(A > B)' is epistemic, 'Might(A > B)' is true relative to an information state s iff s is compatible with A > B. What is the probability that s is compatible with A > B, relative to s? Unless the information state is unsure about itself, it will be either 0 or 1. Specifically, we get the prediction that P(A *> B)) = 1 if P(A > B) > 0 and P(A *> B) = 0 if P(A > B) = 0. But intuitively, 'the probability that if A then might B' is not always 1 or 0.

So what you have to say, if you want to analyse 'A *> B' as 'Might(A > B)', is that despite surface appearance, the expression 'the probability that if A then might B' does not denote the probability of the embedded might counterfactual 'if A then might B'. Perhaps the two epistemic modals merge and the expression denotes the probability of 'if A then would B'. Or whatever. But in the present context, it's funny that you have to say such a thing, given that your whole approach is motivated by your commitment to the idea that 'the probability that if A then would B' denotes the probability of the embedded would counterfactual.

]]>But it's hard to spell out how exactly the argument is meant to go. In fact, I'm not aware of any satisfactory statement. Here's my attempt.

For concreteness, I'll focus on the argument for probabilism, but the case of conditionalization is similar.

The argument begins with an uncontroversial mathematical fact, the
*Dutch Book Theorem*:

Let aunit beton a proposition A be a deal that pays $1 if A is true and otherwise $0. Suppose for any proposition A, an agent is prepared to buy a unit bet on A for up to $Cr(A) – the dollar value corresponding to her credence in A – and she is prepared to sell a unit bet on A for $Cr(A) or more. If her credences do not satisfy the axioms of non-negativity, normalization, and finite additivity, she will then be prepared to buy and/or sell unit bets in such a way that if she makes all these transactions she incurs a guaranteed loss.

How do we get from here to an argument that rational credences should conform to the probability axioms? A few problems immediately stand out. (See "Hajek 2008.)

First, the theorem seems to say nothing about people who aren't prepared to trade bets in accordance with their expected monetary payoff. Surely epistemic rationality does not require having a utility function that is linear with respect to monetary payoff. An epistemically rational agent need not care about money at all.

Worse, even if an agent does care only about money, and her utility function is linear with respect to monetary payoff, she ought not to be prepared to buy a unit bet on any proposition A for up to the dollar value corresponding to her credence in A. For example, let A be the proposition that the agent will not buy any bets today. An agent's credence in A may well be high, yet she ought not to pay much for the corresponding bet, since doing so would render A false.

Even if the relevant propositions are unaffected by the considered bets, there can be interference effects between different bets. For example, what if our agent has earlier bought a high-stakes bet on the proposition that she will not buy any more bets today? Then she may not be prepared to buy a unit bet on any proposition whatsoever. Relatedly, the Dutch Book argument for finite additivity involves at least three bets; if a probabilistically incoherent agent cares about the net outcome of all her transactions, rather than myopically about the isolated outcome of whatever transaction she currently considers, it is not clear why she ought to make all three transactions. (This is the "package principle objection". Interestingly, it seems not to arise for the case of conditionalisation, as "Skyrms 1993 shows.)

Stepping back, why is the mere possibility of making a sure loss normatively relevant? After all, as Lewis said, "there aren't so many sneaky Dutchmen around".

Finally, why is the possibility of financial loss a sign of
*epistemic*, rather than *practical* irrationality?

A neat way to get around most of these problems, which I haven't seen in the literature, is to invoke some broadly Humean principles about the independence of belief and desire. In outline, the idea is that for any probabilistically incoherent agent X there is a possible agent Y who (1) has the same credences as X, (2) only cares about the monetary payoff of whatever transaction they presently consider, and (3) is offered the relevant bets that make up a Dutch Book. Y then makes a sure (and avoidable) loss, despite trying to get as much money as possible. Something has gone wrong. But the fault must lie in Y's beliefs, for neither her utilities nor her decision process is faulty. So Y's beliefs are irrational. But Y's beliefs are identical to X's. So X's beliefs are irrational.

That's the outline. Let's fill in the details.

Let X be an arbitrary agent whose credences violate one of the probability axioms. Our aim is to show that X is epistemically irrational.

Let Y be a possible counterpart of X with the same (centred) credences. But Y has strange desires. Whenever Y is offered a monetary gamble, she only myopically cares about the net amount of money she will make through the present transaction. Specifically, if Y has the option to buy a unit bet on some proposition A for some amount $x, then the only thing she cares about is whether she will eventually (i) win $1 after having paid $x, or (ii) not win after having paid $x, or (iii) not win after not having bought the bet; the utility she assigns to these outcomes are, respectively, (i) $1-$x, (ii) -$x, (iii) $0. Similarly, mutatis mutandis, if Y has the opportunity to sell a unit bet.

I'll also stipulate that when faced with a choice, Y always chooses an option with maximal expected utility.

All this still doesn't ensure that Y is prepared to pay up to $Cr(A) for a unit bet on A because her credence in A may be affected by getting an offer to buy the bet or even by the act of buying (as when A is the proposition that she won't buy any bets today). If we want Y to accept a Dutch Book that involves several transactions, we must also ensure that, say, buying the first bet does not affect the expected utility of buying the second.

I'm not sure how best to get around these problems. Here's a brute force response.

Let's say that X's (and Y's) credence function is *stable* with
respect to some propositions A,B,...,N iff Y regards $Cr(A) as the
fair price for a unit bet on A, $Cr(B) as the fair price for a unit
bet on B conditional on having bought/sold a unit bet on A, and so
on. That is, a credence function is stable with respect to a list of
propositions if the credence in each proposition on the list is not
affected by whether a bet on that proposition or a proposition earlier
in the list has been bought or sold.

If we assume that there is a list of propositions for which X's credences are stable and violate the probability axioms, we can stipulate that Y is made the relevant offers and gets caught in a Dutch Book.

So we need to assume that X's probabilistic incoherence isn't restricted to unstable parts of her credence function. To get a general argument for probabilism, we'll need the following premise.

Premise 1. If any restriction of an agent's credence function to stable propositions should satisfy the probability axioms, then so should her entire credence function.

(Here and throughout, a violation of the probability axioms means that either (i) some proposition has negative probability, or (ii) the tautology does not have probability 1, or (iii) there are disjoint propositions whose disjunction has a probability that is not the sum of the probability of the disjuncts. Boolean closure is not treated as an axiom.)

The motivation for Premise 1 is that the probability axioms are supposed to be general consistency requirements on rational belief. They are meant to hold for beliefs or any kind, not just for beliefs with a specific content.

Jeffrey makes a similar move in *Subjective Probability: The Real
Thing* (pp.4f.):

If the truth [of a proposition about distant planets] is not known in my lifetime, I cannot cash the ticket even if it is really a winner. But some probabilities are plausibly represented by prices, e.g., probabilities of the hypotheses about athletic contests and lotteries that people commonly bet on. And it is plausible to think that the general laws of probability ought to be the same for all hypotheses – about planets no less than about ball games. If that is so, we can justify laws of probability if we can prove all betting policies that violate them to be inconsistent.

Jeffrey here assumes that there's a reasonably wide set of propositions for which our credences match our betting prices. I assume something much weaker. But I'd still like to know how to do better.

(One reassuring fact to keep in mind is that probabilistic incoherence is infectious: if, for example, your credence in an exclusive disjunction A v B is not the sum of your credences in A and B, there will be lots of other propositions for which you'll violate additivity. So it requires some fine-tuning to limit incoherence to unstable fragments of a credence function.)

Moving on, recall that X was an arbitrary agent whose credences fail to satisfy the probability axioms. We want to show that X is epistemically irrational. By Premise 1, we can assume without loss of generality that there are some propositions with respect to which X's credence function is stable but fails to satisfy the probability axioms.

The next premise is that all the differences between X and Y are irrelevant to whether the credence function shared by X and Y is epistemically rational. More precisely:

Premise 2. If X is epistemically rational, then so is Y.

The basic idea is that whether someone's beliefs are epistemically rational does not depend on her goals or desires. If we want to know whether it is epistemically rational (as opposed to practically useful) for an agent to have such-and-such beliefs, we don't need to know anything about her goals or desires.

I've also assumed that Y is an expected utility maximizer, which X may not be. But again, arguably the epistemic rationality of someone's beliefs would not be undermined by finding that they are an expected utility maximizer.

Finally, Premise 2 implies that it does not affect the epistemic rationality of an agent's beliefs if they are about to be offered a series of bets. (That's the final difference between X and Y.)

Premise 2 looks fairly good to me.

Now the Dutch Book theorem tells us that there are certain transactions that Y is prepared to make that would amount to a guaranteed loss. Let's stipulate that Y is made the relevant offers and thus really does make a sure loss.

The next premise states that something has then gone wrong.

Premise 3. It is irrational of Y to make choices that together amount to a sure loss (a loss she could have avoided by making different choices).

Here the guiding idea is that it is irrational for an agent whose sole aim is to maximize monetary profit to knowingly and avoidably enter transactions that are logically guaranteed to cost her money.

Premise 3 relies on the Converse Dutch Book Theorem: that probabilistically coherent agents cannot be Dutch Booked.

I'm not entirely happy with Premise 3. The problem is that, by
assumption, Y does not care about her net wealth. When offered a
series of choices, she only cares about the net outcome of the
*present* choice. It would be nicer if we could stipulate that Y
cares about the net payoff of all the choices she's about to
make. This would make Premise 3 quite compelling, I think. But then
we'd need to explain why Y accepts the individual deals that together
constitute a Dutch Book. (Skyrms offers such an explanation in his
1993 paper on conditionalization, but the argument sadly doesn't
generalize to the case of finite additivity.)

The rest of the argument is simple.

Premise 4. If an agent makes irrational choices, then either she is epistemically irrational or her desires are irrational or her acts don't maximize expected utility.

Premise 5. Y's desires are not irrational.

Y's myopic desires are admittedly weird, but on a suitably weak notion of rationality, I think they should pass. Since Y maximizes expected utility, we can conclude that Y's credences are irrational. Intuitively, Y misjudges the profitability of the relevant bets.

So Y is epistemically irrational. By Premise 2, it follows that X is epistemically irrational. QED.

]]>The question I address is simple: how should we model the impact of perceptual experience on rational belief? That is, consider a particular type of experience – individuated either by its phenomenology (what it's like to have the experience) or by its physical features (excitation of receptor cells, or whatever). How should an agent's beliefs change in response to this type of experience?

Why care? A few reasons:

First, the question is closely related to several traditional issues in epistemology. Intuitively, many of our beliefs are justified because they are suitably connected to relevant perceptual experiences. But what is that connection? How does a "nonconceptual" experience support a "conceptual" belief? How do we accommodate the holism of confirmation? How can an experience justify a belief about an external world if one could have the same experience even if there were no external world? A good model of how perceptual experiences should affect belief would help to make progress on these issues.

Second, the question is important for the "interpretationist" (formerly known as functionalist) approach to belief. On this approach, what makes a physical state a belief state with such-and-such content is that it plays a certain causal role – the role characteristic of the relevant beliefs. Part of this role links the beliefs (and desires) to choice behaviour. Another part of the role links them to perceptual experience: it specifies how beliefs tend to change through perceptual experience. In the paper, I'm effectively trying to spell out that part of the role.

Third, the question plays an import role in cognitive science. People in artificial intelligence have models of how incoming perceptual stimuli should affect a belief system. Similar models have proved fruitful in the neuroscience of perception. The model I propose looks a lot like these models from neuroscience and artificial intelligence. But the models have strange features that call for philosophical comment. In particular, they seem to imply a form of sense datum theory: perceptual experiences are supposed to provide infallible information about a special realm of sense data. How should we understand these sense data? Aren't there decisive philosophical arguments against sense datum theories?

Fourth, the question might provide the key to the hard problem of consciousness. In the paper I suggest that the appearance of irreducibly non-physical properties in perceptual experience is a predictable artefact of the way our brain processes sensory information.

Fifth, the question is interesting because it's really hard to answer. A common idea in the philosophy of perception seems to be that (1) perceptual experiences represent the world as being a certain way, and that (2) in the absence of defeaters, having the experience makes it rational to believe that the world is that way. But that's hardly a full answer. For one, how does an experience – individuated, say, by its physical features – come to represent the world as being a certain way? Moreover, how should the rest of an agent's belief system change if there is no defeater? How should the agent's beliefs change if there is a defeater? (Surely it should still change in some way.)

In the paper, I assume a Bayesian framework. So the question becomes: how should a given type of experience affect an agent's subjective probabilities? The classical Bayesian answer assumes that perceptual experiences make the agent certain of a particular proposition, so that her probabilities can be updated by conditionalization. But that doesn't seem right. Richard Jeffrey proposed an alternative which allows experiences to convey less-than-certain information. But the relevant less-than-certain information in Jeffrey's model is not just a function of the experience; it also depends on the agent's prior probabilities. So how do the prior probabilities together with an experience determine the input to a Jeffrey update? No-one knows. In fact, I argue that it is impossible to know, because the effect an experience should have on a belief system is not fixed by the experience and the (prior) belief system at all. It depends on a further aspect of the agent's cognitive state.

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