2.  Sleeping Beauty goes to sleep on Sunday night. Right before this happens, a fair coin is flipped in her presence. She does not see the outcome of the flip. If heads, she will awaken in a blue room Monday morning and be asked the probability that the flip was heads. If tails, she will awaken in a blue room Monday morning and be asked the probability that the flip was heads, then after she responds, be given a drug that erases her memory of waking up and puts her out for another 24 hours, whereupon she will wake up again in the blue room on Tuesday and again be asked the probability that the flip was heads. We can assume she will not otherwise wake up in a blue room or be asked for such a probability, and she knows all of the above. The question is, what should she answer? Halfers say one half. It is obviously one half before she goes to sleep Sunday night and she hasn’t learned anything. She knew she was going to wake up in a blue room and she knew she would be asked the question. Thirders say one third. Indeed, any answer other than one third would leave her vulnerable to a Dutch book betting scheme…that is, one entailling a sure loss. (The halfer for example would regard as fair a bet of 3 on tails made Sunday night and bets of 2 on heads upon each blue room wakening, a betting scheme that would always result in a loss of 1.) Or, if one imagines the experiment being run ad infinitum, two thirds of the awakenings would be tails awakenings, which supports the thirder. One can even remove the science fiction element from the ad infinitum scenerio: in the next room someone is flipping a coin over and over. If a given flip is heads, he waits 1 minute before flipping again. If tails, he waits 2 minutes. At some arbitrary instant Beauty is asked for the probability that the previous flip was heads. Obviously she answers one third. On the other hand if she is asked for the probability that the 17th flip was heads, she will say one half. And these are true even if the previous flip was the 17th flip. This is no more mysterious than saying that with probability nine tenths, the best basketball player in the school is over six feet tall, while with probability two fifths, the best chess player in the school (who happens to be the same person) is under six feet tall. This points to the thirder being correct under de dicto interpretation and to the halfer being correct under de re interpretation. Which is more natural? Arguably the memory erasure drug (or threat of same) precludes de re knowledge of the coin flip (although there are no other coin flips to confuse it with, Beauty cannot even say whether this flip took place 12 or 36 hours ago). If that’s right it supports (albeit not uncharitably) the thirder.

3. How do you grade a weatherman? The weatherman says it will rain with probability .7 and it rains. Another say probability .6 so the first weatherman wins. The next day they say .64 and .73 respectively and it rains again. The second weatherman wins. Who is now winning the competition? Plainly the weatherman who is less surprised by the outcome is the better weatherman. How to quantify this? The weatherman’s degree of surprise can be measured by the number of bits of information he gains upon learning the result (rain or no rain). This is minus log base 2 of p, where p is the probability he assigned to the actual result. So the original weatherman gains  .5146 bits of information and that is his score for that day. (Lower scores are better.) The next day he scores .6439. The second weatherman scores .7370 and .4540 on the two days. The first weatherman is winning this competition. More than just telling us how to grade rational agents, this tells us how rational agents should assign probabilities to events. Assign them in such a way as to minimize minus log base 2 p, where p is the probability assigned to what actually happens (either the event or its complement in the case where you are dealing with assigning probabilities to two cell partitions though larger partitions are possible as well if you are dealing with, say, a golf tournament rather than a basketball game).

4. What does this scheme tell us about Beauty? Sparing the computations, it tells us the thirder is correct, assuming that Beauty is graded once per blue room awakening. On the other hand, what if Beauty isn’t betting for herself, but sending text messages to her partner Phillip, who isn’t part of the experiment but is allowed to place side bets? Maybe Phillip didn’t see the original coin. He thought it might have been a biased coin, maybe a fair coin, he wasn’t sure. Beauty finds out Sunday night the coin is fair, but is only allowed to text message Phillip during blue room awakenings. What does she send? One third or one half? Plainly one half.

5. What if we take the money out of it, though? And the grading? Remember Joe, who lost two bets but was (arguably) wrong only once. If we take seriously the dictum that Beauty qua rational agent should choose probabilities so as to minimize her expected information gain,  then we ought to make a determination as to whether she really gains information on Monday morning under the tails scenerio. She’s told the result of the flip, presumably, but then her memory is erased. Can that count as information gain? Does it even matter that she is told the result of the flip? Maybe she could be told in all blue room situations except Monday in the tails scenario. It is difficult to imagine that she would care, as her memory is about to be erased anyway. All this suggests there is no actual information gain on Monday tails blue room encounters, which supports the halfer. Perhaps rational agents aren’t out to minimize information gain, though, but out to minimize unfavorable consequences. What if, at every blue room encounter, Beauty has a choice between receiving an unpleasant shock if and only if the coin landed heads, or  receiving an unpleasant shock with probability two fifths? Clearly she’ll go with the coin. Information may go away, but the shock is real, whether or not she remembers it. All this supports the thirder again.

6. Jacob Ross thinks he has an argument against the thirder. What if the coin is flipped until there is a head, and N is the number of flips it took. Beauty will have now 2^N blue room awakenings. According to Ross, the thirder is committed to assigning equal probabilities to “N=1″ and “N=2″. After all, N=1 with probability one half (as measured on Sunday) and N=2 with probability one fourth. Beauty will awaken twice if N=1 and four times if N=2. The thirder, says Ross, assigns subjective probabilities to outcomes in Sleeping Beauty scenarios in such a way as to be proportional to the product of the objective probability and the number of awakenings associated with the outcome. If that is so, then, “N=k” gets assigned the same probability as “N=1″ not just for k=2 but for all natural numbers k. That violates countable additivity (the sum of these events must be 1, hence they cannot all have the same probability) and that is an argument, according to Ross, against the thirder. Is that right?

7. Unfortunately not. The thirder is not committed to Ross’s generalized thirder principle. Why not? It has to do with the way in which one replaces time randomness with space randomness. Say I have a process that spits out an infinite string of letters on a finite alphabet A. I want to jump into the output at a random time and start generating output according to the laws that dictate the sequence. How do I do this? One way would be to simply pick a random time. That isn’t a very good plan, because (again by countable additivity) there isn’t any way to do it uniformly over the available times, and for some other reasons as well. Here’s how you can do it in cases where the asymptotic density of each finite subword exists: just start choosing symbols according to these densities. (The idea here is that since you can’t choose a time uniformly at random from a countably infinite set, you choose a time uniformly at random from an initial set of times {1,2,3,…,T} and let T go to infinity.) For example, if your alphabet is {c,a,t} and those letters occur with densities one third, one sixth, one half respectively, choose your first letter of output with those probabilities. Say you choose “c”. Now you choose your second letter according to the probabilities of two letter subwords beginning with “c”. That is, you condition on the “c” you just chose. And so on. Now if your alphabet is infinite this might work, but having the densities exist isn’t enough now. For a fixed subword length, the sum of the densities of the words of that length must be 1.  And that is just what does not happen in the Ross example. The density of  blue room awakenings for which N=k is zero, for every k. This doesn’t of course contradict countable additivity as densities are not probabilities. And you can’t make the transition to probabilities either, for reasons previously outlined. (The densities don’t sum to 1.) This is all standard from symbolic dynamics…Ross fails to realize that the thirder proportionality principle requires that the densities sum to 1, and so his argument against the thirder is not convincing.

8. Which leaves us where? Well, not surprisingly, it depends on what quantity Beauty is trying to optimize, and if this quantity is marginal information, it depends on whether you count information that is erased after it is given. In other words…it seems that the solution to the Sleeping Beauty problem depends on the context. Which probably won’t surprise anyone.

This argument begins as an argument against the state-description explanation of analyticity, which Quine attributes to Carnap. A state-description is “any exhaustive assignment of truth values to the atomic, or noncompound, statements of the language.” All other statements are taken to be truth-functions of the atomic statements; analytic statements are those that come out true under all state-descriptions. This is all very familiar from truth-tables.

Now, Quine makes the following strange remark. “…this version of analyticity serves its purpose only if the atomic statements of the language are, unlike `John is a bachelor’ and `John is married’, mutually independent. Otherwise there would be a state-description which assigned truth to `John is a bachelor’ and to `John is married’ and consequently `No bachelors are married’ would turn out synthetic….” I confess to not understanding the remark at all.

First an observation: when you have any kind of system with R states, you can exhaust these states by truth value assignments to an independent set of “atomic statements” if and only if R is a power of 2, and even then the set of atomic statements is not unique. For example, if your states are {1,2,3,4} then A={1,2}, B={1,3} does the job. On the other hand, A’={1,4}, B’={3,4} does, too. It may be that early empiricism was infected with a “power of 2” bias (cf. the Tractatus, e.g. 4.27 and 4.42; I’m less familiar with Carnap); this discussion won’t be so infected.

Here’s a simple system. There are a woman, Marge, and two men, John and Adam. The system has three states. (1) Marge is married to John. (2) Marge is married to Adam. (3) Marge is unmarried. No other marriages, and that’s it. Just three states. The intension of “bachelor” is as follows. In (1), {Adam}. In (2), {John}. In (3), {Adam, John}. The intension of “man” is constant, namely {Adam, John}. The intension of “married”: In (1), {John, Marge}. In (2), {Adam, Marge}. In (3), emptyset. It’s now easy to see that “all bachelors are unmarried” comes out true in all three states. In other words, it is analytic.

 Okay, back to Quine’s passage from the second paragraph above. First, what are the “atomic statements”? Any statement can be made non-compound, simply by giving it a label. Even granting that “the” atomic statements exist, and are mutually dependent, how exactly would this guarantee a state description in which “John is a bachelor” and “John is married” come out true? Let M = “Marge is married”, J = “John is married”. Then (1) may be identified with J AND M. (2) may be identified with (NOT J) and M. (3) may be identified with (NOT M) AND (NOT J). J and M are mutually dependent in the sense that there is no state where J and (NOT M). Reading Quine literally, there should therefore be a state-description which assigns truth to `John is a bachelor’ and to `John is married’ and consequently `No bachelors are married’ should turn out synthetic. That’s not right.

No surprise here, as there’s no apparent problem with analyticity in this sort of setting. This is all rather trite, in fact, but it may be that intensional semantics wasn’t always so transparently modelled.

In Language, Truth and Logic, A.J. Ayer attempted to formulate a verifiability condition, according to which there are two types of verifiable statements.

1. Directly verifiable statements.

a. Observation statements: “the coin landed heads”.

b. Statements that, in conjunction with one or more observation-statements, entail observation-statements not deducible from these other premises alone: “if that coin is placed in the fire, it will burn.”

2. Indirectly verifiable statements: statements that, in conjunction with premises known independently to be verifiable, entail directly verifiable statements not deducible from these other premises alone: “if that coin is flammable, and it is placed in water, then it will float”. (Here “that coin is flammable” is shorthand for “if that coin is placed in the fire, it will burn.” Note however that you run into problems trying to derive “it will float” when the observation statement “place in the fire” is used as a premise unmediatedly.)

A. Church killed this formulation, not quite as follows though the idea is more or less the same. Let S=”Either the nickel landed heads, or the quarter landed heads and it is not that case that the absolute is perfect.” This is directly verifiable, since it entails, together with “the nickel landed tails”, that “the quarter landed heads”. But now “the absolute is perfect” together with the directly verifiable premise S entails “the nickel landed heads.” Therefore “the absolute is perfect” is indirectly verifiable (which you don’t want).

It’s a bit easier to see what is going on if you adopt the viewpoint that propositions are sets of possible worlds. So, let P denote the set of possible worlds in which “the absolute is perfect” holds. It seems reasonable to say that “the absolute is perfect” is publically meaningful precisely insofar as P is publically determinable. Let N and Q denote the set of worlds where the nickel and quarter landed heads, respectively, and let S be the set of worlds where either the nickel landed heads, or the quarter landed heads and it is not the case that the absolute is perfect.

Even if we know nothing whatsoever about P, observe that we do know some things about S=N OR (Q AND P’). Namely, that S is an extension of N by a subset of Q.  That is all we know about S, in fact. But, we know nothing about P…so how do we know that P intersect S is contained in N (as we surely do)? Somehow, the addition of the meaningless premise P added information.  

The answer, of course, is that in asserting that S=N OR (Q AND P’) obtains, we do a bit more than simply assert that  the actual world lies in an extension of N by a subset of Q. The more we do is an act of naming. Specifically, we stipulate that if the actual world is not an N-world then it gets the name P’. Now the additional premise P, which says that the actual world does not have the name P’, does contain information–you can immediately infer from it that the actual world is an N-world.

This is one of the most important points of Wittgenstein’s philosophy: acts of naming impart no information on their own. Say there is a certain metal bar in Paris. I say “the Paris bar is one glurg in length”. You have no idea what a glurg is, so my statement imparts no information. Now there is another bar in London. I say “the London bar is one glurg in length”. Equally meaningless on its own, of course, but together, the two statements do impart (relational)  information; namely the information that the Paris bar and the London bar are of the same length.