As always, comments are much appreciated.

Let P an agent's credence function at time t1, P' the credence function at t2, and E the evidence received at t2. Since E is a centered proposition, it can be true at multiple points within a world. Suppose, however, that the agent assigns probability 0 to worlds at which E is true more than once. Then to compute P', first conditionalise P on the uncentered fragment of E -- i.e. the strongest uncentered proposition entailed by E. This rules out all worlds at which E is true nowhere. Second, move the center of each remaining world to the (unique) point at which E is true.

Call this rule "IC", for "Indirect Conditionalisation". With minor variations, this is the rule defended in Piccione and Rubinstein (1997), Halpern (2004), Elga (2007), Meacham (2008), Titelbaum (2008), Kim (2009) and Briggs (2010), and probably other places as well. Apart from details of the presentation, the variations mostly concern what to do if the new evidence is not certain to be true at most once per world.

In my paper on updating self-locating beliefs, I present a Dutch Book argument against IC. Rachael Briggs, in her 2010 article, gives a proof that IC is immune to Dutch Books. At least one of us must be wrong.

Here is a simple version of my Dutch Book against IC. Assume at t1, an agent's credence is divided between three uncentered worlds: w1, w2 and w3. For each of these, there is exactly one place where she might be, according to her beliefs. At t2, the agent will receive either evidence E or E', and she knows that this is the case. E is true at a unique point in w1 and w2, but nowhere in w3. E' is true at a unique point in w1 and w3, but nowhere in w2. (The point where E is true in w1 is different from the point where E' is true.) Initially, the agent gives credence 1/2 to w1 and credence 1/4 to each of w2 and w3. She therefore regards as fair a bet that pays $-3 in case of { w1 } and $3 otherwise. Now suppose she learns E and updates by IC. The uncentered fragment of E rules out w3, but not w1 and w2. The new probabilities are therefore 2/3 for w1 and 1/3 for w2. Similarly, if she learns E', the new probabilities are 2/3 for w1 and 1/3 for w3. Either way, she will regard as fair a bet that pays $2 in case of { w1 } and $-4 otherwise. Hence, whether { w1 } is true or not, and whatever evidence she receives, the agent is certain to make a loss.

Now for Rachael's argument why there cannot be any such Dutch Book. One thing worth pointing out is that Rachael only allows bets on uncentered propositions. In my paper, I give a more general Dutch Book argument in support of a different update rule. This argument makes use of centered bets. However, the Dutch Book just presented only involves the uncentered proposition { w1 } and its negation, so this can't be the relevant point of disagreement.

Rachael's argument relies on a proof in Teller (1973) showing that an agent with uncentered beliefs who updates by conditionalising on uncentered evidence is immune to diachronic Dutch Books. Rachael's argument then is that if there were a Dutch Book B against an agent following IC, we could convert it into a Dutch Book B' for an imaginary agent with uncentered beliefs who updates by conditionalisation. By Teller's result, this is impossible.

For my Dutch Book in place of B, Rachael's argument is actually quite simple. The variant B' is B itself. So imagine an agent with only uncentered beliefs, who gives credence 1/2 to w1 and 1/4 to w2 and w3. Like our original agent, this agent initially regards as fair the bet that pays $-3 in case of { w1 } and $3 otherwise. Moreover, just as our original agent updates by conditionalising on either E or E', the imaginary agent updates by conditionalising on either u(E) or u(E'), where u(X) is the strongest uncentered proposition entailed by X. Since u(E) is true at w1 and w2, and u(E') is true at w1 and w3, either way the imaginary agent ends up with credence 2/3 in w1. So now they regard as fair the bet that pays $2 in case of { w1 } and $-4 otherwise. We have converted the Dutch Book for the original agent into a Dutch Book for the imaginary agent. Yet by Teller's theorem, the imaginary agent can't be Dutch Booked!

In fact, the converted Dutch Book is not a genuine Dutch Book. The reason is that u(E) and u(E') are not mutually exclusive: they both contain w1. And then we can't assume that the (imaginary) agent is certain at t1 that their total evidence is going to be either u(E) or u(E'). By contrast, E and E' are mutually exclusive. The converted Dutch Book does satisfy conditions (a) and (b) on p.19 of Rachael's paper. But it does not follow, as Rachael assumes, that it constitutes a Dutch Book against the imaginary agent.

This, I think, is the mistake in Rachael's argument: uncentering a partition of centered evidence does not always result in a partition of uncentered evidence.

Call the information in the previous paragraph E, and suppose it's all you know about the situation. How confident are you that Alice has the disorder?

Letting our subjective probabilities be guided by the stated frequencies, we can use Bayes' Theorem to figure out that P(disorder | positive) = P(positive | disorder) * P(disorder) / (P(positive | disorder) * P(disorder) + P(positive | ~disorder) * P(~disorder)) = 0.99 * 0.0001 / (0.99 * 0.0001 + 0.01 * 0.9999) = 0.0098. Assume then that your degree of belief is about 0.01.

Now suppose you learn that my degree of belief in the hypothesis that Alice has the disorder, based on the same evidence E, is 0.9. How should that affect your own confidence?

Note first that you now have a new piece of evidence about the situation that you didn't have before: the information that my degree of belief in Alice having the disorder, based on evidence E, is 0.9. Call this information E2. So we may ask to what extend the conjunction of E and E2 supports the hypothesis that Alice has the disorder. The answer, it seems to me, is clear: a rational prior credence function conditionalised on E & E2 would still assign probability ~0.01 to Alice having the disorder. The fact that I take E to strongly support the hypothesis merely shows that I have incorrectly assessed the evidence. It lends no support at all to the hypothesis.

On the other hand, you may not be certain that you yourself have assessed the evidence correctly. It seems to you that this is a case where Bayes' Theorem applies, and that it yields something around 0.0098, but you know that humans get easily confused in such cases, and it may well be, for all you know, that you even made a calculation mistake.

If this is your situation, we have to distinguish two things: the result of the method by which you assessed the evidence, and your actual credence. Suppose, for example, you are only 90 percent confident that you have assessed the evidence correctly, and the remaining 10 percent of your confidence go to the claim that the evidence makes the hypothesis probable to degree 0.9 (just as I claim). Then surely you would not bet at very high stakes against the hypothesis. Your actual credence will be nowhere near 0.01.

If this isn't obvious, consider a more extreme case where you take the evidence to entail a certain hypothesis, so that the probability of the hypothesis given the evidence is 1. Suppose again that you are not absolutely certain that you have assessed the evidence correctly. That is, you are not certain whether the evidence really does entail the hypothesis. It's a live possibility, you say, that the evidence is true and the hypothesis false. Then clearly your credence in the hypothesis is not 1.

If you're not certain whether your assessment of the evidence is correct, then your credence should be a mixture of the different possible assessments of the evidence, weighted by your confidence that they are the correct assessment. For example, if you are 90 percent confident that the evidence makes the hypothesis probable to degree 0.01, and 10 percent confident that it makes the hypothesis probable to degree 0.9, then your credence should be something like 0.9 * 0.01 + 0.1 * 0.9 = 0.099.

Now return to the original question. What happens when you hear that my degree of confidence in Alice having the disorder is 0.9? If you don't think that you are in general much better at assessing the evidence than I, this should lower your confidence that the correct assessment yields a probability of 0.01. Your credence should therefore rise from 0.099 to something closer to 0.9. How much it should rise depends on how likely it seems to you that I might have been better at assessing the evidence -- i.e. to what extent the distribution of weights over assessments that underlies my credence gives higher weight to more correct assessments than your own distribution.

So we have two answers: the new evidence is evidentially irrelevant to the hypothesis, and yet you ought to change your credence in response to it!

What's going on here is that we have a norm for ideal agents, and a secondary norm for agents who fail that ideal. The norm for ideal agents is to "proportion their belief to the evidence". The norm for non-ideal agents is to be cautious about their own assessments and weigh the outcome of their assessment with their degree of confidence that they didn't make a mistake. This makes the question whether or not they made a mistake relevant to the resulting credence assigned to the hypothesis. Since evidence of disagreement supports the assumption that they made a mistake, such evidence becomes relevant.

Another use of 'P(A/B)', more common in statistics than in philosophy, is where B gives the value of a parameter of the probability distribution P. Call this the parameter usage. For example, the probability of k heads in n tosses of a coin may be stated as

Here x stands for the probability of heads on an individual toss. P(k / x) is not to be identified with the ratio P(k & x)/P(x). In frequentist statistics, both numerator and denominator of this ratio are deemed nonsensical. Likewise for the inverted probability P(x / k). Unlike on the ratio usage, 'P(A/B)' on this usage therefore does not satisfy ordinary rules for conditional probabilities, such as Bayes's Theorem.

A third kind of use is one in which B in 'P(A/B)' denotes the setup, or condition under which A has the relevant probability. Call this the setup usage. Thus we might find 'P(heads/toss)', or 'P(future state is Y / present state is X). On many interpretations of probability, this too must be sharply distinguished from the ratio usage. The problem is best known for the propensity interpretation, where conflating the two readings leads to "Humphreys' Paradox". (But the problem also arises for other accounts such as that of von Mises.)

On the propensity interpretation, 'P(A/B)' measures the degree to which setup B is disposed to causally produce A. Again, this magnitude has little to do with the ratio of P(AB) over P(B), which may both be deemed nonsensical. And again, when P(A/B)=x, which means that there is a certain tendency x of B to produce A, then there is generally no tendency for A to produce B (nor even to have been produced by B); so P(B/A) is undefined, or at least has a value quite unlike what one would expect on the raio usage.

Humphreys, along with Fetzer and others, conclude from the fact that the setup usage does not satisfy the laws governing the ratio usage that propensity theorists should reject and revise the standard rules for conditional probabilities. Others, like Wesley Salmon, take this fact to be an argument against the propensity interpretation.

It would be better, I think, to not use the same notation for so many different things. On the propensity view, probabilities are only defined relative to a given setup. Each setup S brings with it a probability function P_S. These setup-relative probabilities obey the standard probability calculus, and nothing stops us from introducing standard conditional probabilities, governed by the standard ratio principles. Thus P_{toss}(2 / even) would denote the propensity of a die toss to produce the outcome 2 conditional on producing an outcome with an even number. This may be defined as P_{toss}(2 & even)/P_{toss}(even), or it could be taken as primitive to allow for conditional probabilities with zero-probability conditions.

Consider an example of Humphreys's. Our setup S consists of a photon source and a half-silvered mirror placed at some distance such that the chance of the photon hitting the mirror is 0.5. Let I be the event that the photon impinges upon the mirror, and T the event that the photon is transmitted through the mirror. Now the propensity of the photon hitting the mirror is causally independent of whether or not it afterwards gets transmitted: if we change the mirror's opacity, the probability of I remains unchanged. Thus

(1) P(I/TS) = P(I/~TS) = P(I/S).

On the other hand, the photon can only be transmitted if it hits the mirror: T entails I. So

(2) P(I/TS) = 1.

But now it follows that P(I/S) = 1, contradicting the assumption that the chance of I in S is 0.5.

What went wrong? Humphreys accepts (1) and rejects (2). McCurdy accepts (2) and rejects (1). What's really going on, I think, is that two completely different readings of the conditional stroke are being mixed together. (1) might better be written

(1') P_{TS}(I) = P_{~TS}(I) = P_{S}(I).

The claim is that I is just as probable to be produced under conditions S as under conditions ST or S~T. (These are somewhat odd "conditions" because they lie partly before and partly after the time of I. Assuming there is no backwards causation, only the part of the condition before that time is causally relevant to the production of I, which is why (1') holds. The point could also be made by instead reading S~T as a modification of S in which the mirror is opaque, and setting aside P_{TS}(I).) So here we have a straightforward setup usage.

(2) is meant to express a ratio-style conditional probability in setup S:

(2') P_S(I/T) = 1.

The claim is that under conditions S, the probability that the photon hits the mirror given that it later passes through it, is 1. This, in turn, might be understood as the ratio of the propensity of S to produce an outcome in which the photon impinges and transmits over the propensity of S to produce an outcome in which the photon transmits.

The general point is that the ratio notion of conditional probability, as well as perhaps the parameter notion, can be very useful even for propensity theorists. And there is no reason to choose between the three. If we stop equivocating, there is no paradox here. (1') and (2'), for example, do not entail anything that conflicts with the original description of the scenario. And the claim that P_S(I) may be defined while P_I(S) is undefined is as unproblematic as the claim that on time-relative notions of chance, P_t(A) can be defined while P_A(t) is nonsense.

First of all, they are not intersubstitutable in attitude reports and speech reports. I don't think this is very problematic because such reports are partly quotational, and of course expressions with the same meaning aren't always intersubstitutable inside quote marks. But 'S is true' and S are also not intersubstitutable in simple intensional contexts, as witnessed by examples like

(1) If 'pig' meant bird, then it would be the case that 'pigs can fly' is true.

(2) If 'pig' meant bird, then it would be the case that pigs can fly.

On a straightforward reading, (1) is true and (2) false.

Similarly, suppose at some point in history, a certain part of the African mainland was called 'Madagaskar', but some influencial cartographer erroneously put the name 'Madagaskar' on an island off the coast. Since his maps were widely used, the name came to denote the island. Then (3) is true but (4) false, at least on one sensible reading.

(3) The cartographer's error brought it about that 'Madagaskar is an island' is true.

(4) The cartographer's error brought it about that Madagaskar is an island.

So in English, 'S is true' does not have the same meaning as the unembedded sentence S. Setting aside paradoxes, I guess one could have a purely disquotational predicate 'true*' for which 'S is true*' and 'S' are intersubstitutable in every (non-quotational) context. But the English predicate 'true' is not disquotational in this sense.

There is a more general lesson here. In an intensional language, you can't give the meaning of a term by merely stating analytic biconditionals, or by giving introduction and elimination rules. This might fix the diagonal of the Kaplanian character, but it doesn't say how the rest of the table is to be filled in. 'It is true that S', 'now S' and 'actually S' arguably have the very same introduction and elimination rules, but very different compositional meanings.

Things are even worse in languages like English that are capable of triple-indexing. Then even filling in the whole character won't be enough to determine an expression's compositional meaning.