Let us now apply this analysis to a more straightforwardly repeated version of the original Sleeping Beauty problem: The N-fold Sleeping Beauty Problem This is like the original Sleeping Beauty problem repeated N times on consecutive weeks. Beauty knows that the experiment is repeated N times, but she is unable to determine which run of the experiment she is currently in. For N = 1, this reduces to the original Sleeping Beauty problem, and Beauty’s credence in HEADS should be 1/2, both before and after learning that it is Monday. If N is some large number, such as N = 3,000, then the case approximates Three Thousand Weeks, and Beauty’s credence in HEADS should be approximately 1/3 before learning that it is Monday, and 1/2 after being told that it is Monday. (“HEADS” here stands for “My current awakening is in one of the trials where the coin fell heads”.) The larger N is, the more exact will the approximation be. The credences in the 3,000-fold Sleeping Beauty Problem are not exactly equal to those in Three-Thousand Weeks because the total number of awakenings is not strictly fixed. There is, however, a very high chance that there will be roughly 3,000 tails-awakenings and 1,500 heads-awakenings in the 3,000-fold Sleeping Beauty Problem, so it closely approximates Three-Thousand Weeks. Illustration: the hybrid model to the N = 2 case It may be instructive to calculate the exact credences for the N = 2 case. There are four possible outcomes of the coin tosses: heads-heads, heads-tails, tails-heads, and tails-tails. We can represent these four possibilities along with the possible agent-parts they would realize as follows: Week 1 | Week 2 w1: h1 | h2 w2: h3 | t1 t2 w3: t3 t4 | h4 w4: t5 t6 | t7 t8 Each of these four possibilities has an equal chance of occurring (p = 1/4). Since each of these agent-parts are in the same evidential situation, Beauty’s conditional credence, given one of the four possibilities, is divided equally between the agent-parts that that possibility would realize. Hence, her unconditional credence in being any particular possible agent-part is obtained by multiplying this conditional credence with her prior credence in the possibility in question (i.e. 1/4). Thus we get the following assignment of credence to the centered propositions that she is currently a particular agent-part: Week 1 | Week 2 w1: 1/8 | 1/8 w2: 1/12 | 1/12 1/12 proposition when given an agent-part as an argument. In the text, the context should make it clear what is intended in each case. 15 w3: 1/12 1/12 | 1/12 w4: 1/16 1/16 | 1/16 1/16 We obtain P(HEADS) by summing the credences of the centered propositions that imply HEADS (indicated with boldface): P(HEADS) = 1/8 + 1/8 + 1/12 + 1/12 = 5/12 Since P(HEADS | MONDAY ) = P(HEADS & MONDAY) / P(MONDAY), we likewise get P(HEADS | MONDAY) = (5/12) / (17/24) = 10/17 This, however, is not the credence that Beauty should assign to HEADS if she were told that it is Monday. For the same reasons as noted above in the discussion of the original (1-fold) Sleeping Beauty problem, the relevant quantity is instead P+(HEADS | MONDAY). To determine this quantity, we again represent four possibilities, but these now include agent-parts that know that it is Monday (these are the agent-parts in the middle columns, whose names end with the letter ‘m’): Week 1 | Week 2 w1: h1 h1m | h2 h2m w2: h3 h3m | t1 t1m t2 w3: t3 t2m t4 | h4 h4m w4: t5 t3m t6 | t7 t4m t8 Since the number of agent-moments that know that it is Monday is the same in all four possibilities (i.e., two in each case), each of these agent-parts (who are in the same evidential situation) should assign the same credence to being a particular one of these agent-parts, namely (1/4)(1/2) = 1/8, and they should assign zero credence to being some other agent-part. Thus: Week 1 | Week 2 w1: 0 1/8 | 0 1/8 w2: 0 1/8 | 0 1/8 0 w3: 0 1/8 0 | 0 1/8 w4: 0 1/8 0 | 0 1/8 0 To obtain P+(HEADS | MONDAY), we sum the credences of the centered propositions that imply both HEADS and MONDAY (indicated in boldface), and divide this by the sum of the credences that imply MONDAY: P+(HEADS | MONDAY) = (1/8 +1/8 +1/8 +1/8) / 1 = 1/2 16 The hybrid model thus implies that when Beauty learns that it is Monday, she should have credence 1/2 in HEADS. This is so both in the original one-shot version of the Sleeping Beauty problem and in the repeated (“N-fold”) versions where N ≥ 1. Discussion We have argued that the standard arguments for the standard positions on the Sleeping Beauty problem, the 1/2 view and the 1/3 view, are, if not directly question-begging then at least inconclusive in that they rely on eminently deniable premises. To evaluate the standard positions, therefore, we need to seek for further constraints. We presented two such constraints in the form of two thought experiments. The Presumptuous Philosopher thought experiment, in a version adapted for application to the Sleeping Beauty case, strongly suggests that the 1/3-view is wrong. The Beauty the High Roller thought experiment strongly suggests that the 1/2-view is wrong. On these grounds, we concluded that both the standard models for reasoning about self-location are unacceptable. In the second, constructive part of the paper we proposed a new model. This model seeks to combine the most attractive features of the 1/3- and the 1/2-view, so we termed it the hybrid model. It implies that Beauty should not take the fact that she is currently awake as evidence that there are large numbers of awakenings. But it also implies that when Beauty discovers that it is currently Monday, she should not take this as evidence against the hypothesis that there will be many more awakenings in the future. If the hybrid model is correct, it might explain the fact that both the 1/3- and the 1/2-views have some intuitive appeal. According to the hybrid model, both these views get something right. The 1/3-view is right that Beauty’s posterior credence in HEADS after being informed that it is Monday should be one-half. The 1/2-view is right that Beauty’s prior credence in HEADS, after awakening but before learning that it is Monday, should be one-half. The 1/3 view is also right that in the version of the Sleeping Beauty where the experiment is repeated a large number of times, Beauty should (in the infinite limit), upon awakening, assign a prior credence of 1/3 to the centered proposition that the coin fell heads in that particular trial. The hybrid view distinguishes between actual and merely possible agent-parts. In the N-fold Sleeping Beauty problem, for N >> 1, it is (almost certainly) the case that approximately one-third of all actual agent-parts of Beauty are in trials in which the coin fell heads, and the total number of awakenings is (with high probability) approximately determined in advance. By contrast, in the 1-fold version, it is not the case that one-third of all actual agent-parts of Beauty are in a heads-trial. There, either all are, or none. Moreover, in the 1-fold version, the total number of awakenings is strongly correlated with which hypothesis, HEADS or TAILS, is true. The hybrid model corrects for the bias in favor of many awakenings that is inherent in the 1/3 view. (In cases where N is small but larger than 1, the hybrid model gives a prior credence that is intermediate between the that of the 1/3 view and the 1/2 view, thus avoiding any sharp discontinuity. In general, for N ≥ 1, we have 1/3 ≤ P(HEADS) ≤ 1/2.) The main concern about the hybrid model is that it appears to violate Bayesian conditionalization. I argued, however, that this violation is merely apparent. If we pay close attention to the changing indexical information available to different agentsegments, we find that the model does not violate Bayesian conditionalization. A lesson here is that while indexical evidence is irrelevant and can be ignored in most ordinary 17 cases of Bayesian updating, there are special cases – Sleeping Beauty included – where such evidence is relevant. In these special cases, certain implicit assumptions in the common way of applying Bayesian conditionalization are false. In closing, I will address one challenge that could be directed at the hybrid model.19 If Beauty follows this model and agrees to betting odds matching her credence function, she can be Dutch-booked. The Beauty and the Bookie This is like the original one-shot version but with an added bookie, who is put to sleep at the same time as Beauty and given the same amnesia drug. (We put the bookie through this procedure to make sure that he does not have any relevant information that Beauty lacks.) Upon awakening, on both Monday and Tuesday, before either knows what day it is, the bookie offers Beauty the following bet: Beauty gets $10 if HEADS and MONDAY. Beauty pays $20 if TAILS and MONDAY. (If TUESDAY, then no money changes hands.) On Monday, after both the bookie and Beauty have been informed that it is Monday, the bookie offers Beauty a further bet: Beauty gets $15 if TAILS. Beauty pays $15 if HEADS. If Beauty accepts these bets, she will emerge $5 poorer. Since Beauty is able to anticipate the result of accepting all the bets, it is clear that she should not do so. Following the hybrid model, Beauty should have no objection to accepting the second Monday bet. The hybrid model implies that P+(HEADS | MONDAY) = P+(TAILS | MONDAY) = 1/2. Being offered a single straightforward bet on HEADS at even odds, knowing that it is Monday, she has no reason to refuse it. It is the other set of bets that she should reject. The hybrid model implies that Beauty, before learning that it is Monday, assigns P(HEADS | MONDAY) = 2/3. This appears to justify her accepting the bookie’s first offer. But here the situation is more complicated. Since neither party knows whether it is Monday, the Bookie cannot offer this bet only on Monday. He must offer it on both awakenings. This means that the total number of bets will vary depending on how the coin falls: if heads, the first type of bet is offered only once; but if tails, it is offered twice. Moreover, we may assume that Beauty will either accept it on both occasions or reject it on both occasions, as she has no effective way of telling which occasion she is currently encountering.20 So Beauty knows that she would be accepting two bets if TAILS and one bet if HEADS. 19 I’m grateful here to one anonymous referee. A similar Dutch-book argument has recently been advanced in (Hitchcock 2004). 20 If Beauty could opt for a mixed strategy, she could decide to accept the bet at a given occasion with a certain probability. This would complicate the argument but would not affect the conclusion. 18 Now, we already know from other examples that when the number of bets depends on whether the proposition betted on is true, then the fair betting odds can diverge from the correct credence assignment. For instance, suppose you assign credence 9/10 to the proposition that the trillionth digit in the decimal expansion of π is some number other than 7. A man from the city wants to bet against you: he says he has a gut feeling that the digit is number 7, and he offers you even odds – a dollar for a dollar. Seems fine, but there is a catch: if the digit is number 7, then you will have to repeat exactly the same bet with him one hundred times; otherwise there will just be one bet. If this proviso is specified in the contract, the real bet that is being offered you is one where you get $1 if the digit is not 7 and you lose $100 if it is 7. That you should reject this bet is quite unproblematic and does not in any way undermine your original assessment that the probability of the trillionth digit being 7 is 1/10. A similar situation can arise in a more subtle way. We can construct a scenario where, even though no “catch” is explicitly part of the contract, you nevertheless know that you will be put in a position where you will end up betting a hundred times if you are wrong but only one time if you are right. This could happen e.g. if there is a machine that will determine the correct answer and then, on the basis of what this answer is, will decide whether to repeatedly administer an amnesia drug to you that makes you forget whether you have already betted. The machine could do this in such a way that you end up making a larger number of bets if you are wrong. If you believe that you are facing a situation of this kind, you should take corrective action to limit the distortive effects of the memory erasure on your decision-making. In particular, you may decide to reject bets that seem fair to you and that may have been perfectly acceptable in the absence of the forced irrationality constraint. Let us return to the case of Beauty and the Bookie. Beauty knows that she faces the risk of having her memory erased and thus of becoming irrational. (Memory erasure entails a form of irrationality.) For reasons such as those described above, Beauty may therefore reject the bookie’s first set of bets as a form of damage control to minimize the impact of the failures of rationality from which she knows she is at risk. If the deviation of her optimal betting odds from her credence assignment can be justified on these grounds, then she can use the hybrid model and still avoid being Dutch booked. It is interesting that in Beauty and the Bookie, Beauty’s betting odds should deviate from her credence assignment even though the bet that might be placed on Tuesday would not result in any money switching hands. In a sense, the bet that Beauty and the bookie would agree to on Tuesday is void. Nevertheless, it is essential that this bet is included in the example. The bookie is unable to pursue the policy of only offering bets on Monday since he does not know which day it is when he wakes up. If we changed the example so that the bookie knew that is was Monday immediately upon awakening, then Beauty and the bookie would no longer have the same relevant information, and the Dutch book argument would fail. If instead we changed the example so that Beauty as well as the bookie knew that it was Monday immediately upon awakening, then Beauty’s credence in HEADS & MONDAY would be 1/2 throughout Monday, so again she would avoid a Dutch book.21 21 If Beauty would know on Monday that it is Monday, then she would also be able to infer on Tuesday – from the fact that she does not know then that it is Monday – that it is Tuesday. So she would always know what day it is. (We assume that Beauty always know the general setup of the experiment she is in.) 19 In conclusion, the hybrid model combines the comely aspects of the 1/2 view and the 1/3 view while avoiding their faults. The main concern with the hybrid model is that it may appear to violate Bayesian conditionalization. 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Analysis 64(1): 8-10. 22 For comments and discussions, I am grateful to Adam Elga, Bradley Monton, Brian Kierland, Simon Saunders, and anonymous referees.