Bradley, Darren

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266 darren bradleySleeping Beauty: a note on Dorr’s argument for 1/3Darren BradleyBeauty is about to be drugged, rendering her unconscious for a long time.During that time she will be awakened briefly, either once (on Monday) ortwice (on Monday and Tuesday). The number of awakenings depends onthe toss of a fair coin: if the result is Tails, she is awakened twice: if Heads,once. The nature of the drug is that she will not remember being awake. Inparticular, when she is awakened, she will not know whether it is Mondayor Tuesday. Upon awakening on Monday, what should her degree of beliefbe that the coin landed Heads?The paradox is that there are compelling arguments for the answers ofboth 1/2 and 1/3. Cian Dorr gives an argument by analogy that the correctAnalysis 63.3, July 2003, pp. 266–68. © Darren Bradley at University of California, Berkeley on August 11, 2010 http://analysis.oxfordjournals.orgDownloaded from sleeping beauty: a note on dorr’s argument for 1/3 267answer is 1/3. I will argue that his analogy is dissimilar in a crucial respectand as such is unhelpful to resolving the problem.Dorr presents the following variation on the story(2002: 293–94):Beauty is about to go to sleep for a long time. During that time she will be awakened briefly, twice, on Monday and Tuesday. A fair coin will be tossed, and what Beauty can remember will depend on theresult. If the coin lands Tails, she will be given the same amnesia-inducing drug as was used in the original experiment. She will wakeup twice, unable to tell if it is the first or second awakening. But if thecoin lands Heads, she will be given a weaker amnesia-inducing drug,which merely delays the onset of memories from the previous day,rather than destroying them completely. If Beauty receives this weakerdrug, the first minute of her awakening on Tuesday will be just as itwould have been if she had received the stronger drug, but after thatthe memories of Monday’s awakening will come flooding back. Shewill realise that it is Tuesday and that the outcome of the toss musthave been Heads.Let P_ be Beauty’s credence function immediately after being woken onMonday in the variant case. We must assign credence for the following 4hypotheses:H1 The coin lands Heads and it’s MondayH2 The coin lands Heads and it’s TuesdayT1 The coin lands Tails and it’s MondayT2 The coin lands Tails and it’s TuesdayHow should this credence be distributed? Dorr claims, and I agree, that theonly plausible answer is: P_(H1) = P_(T1) = P_(H2) = P_(T2) = 1/4.Let P be Beauty’s credence function after a minute has passed onMonday. Assume she has not had the flooding back of memories she wouldhave experienced had H2 been true, so P(H2) = 0. Nothing in her experi-ence during the first minute does anything to discriminate between theother three hypotheses. So the ratio of her credence in H1, T1 and T2 willremain unchanged. Hence P(H1) = P(T1) = P(T2) = 1/3; so P(Heads) =P(H1) = 1/3. I agree with this reasoning.Dorr then argues that there are no relevant differences between hisvariant case and the original case with respect to P(H1). I disagree. Thereis a crucial difference between Beauty after a minute in the variant case andBeauty after a minute in the original case. In the variant case, a certain pos-sibility has been eliminated. It could have turned out that it was Heads andTuesday. Not only could it have been Heads and Tuesday, Beauty couldhave known it was Heads and Tuesday. She would have known this forcertain if she had suddenly got a flood of memories from the previous day. at University of California, Berkeley on August 11, 2010 http://analysis.oxfordjournals.orgDownloaded from 268 darren bradleyThis flood of memories would have made Heads more likely. The absenceof this flood makes Heads less likely. That is, she can conditionalize on theabsence of a flood of memories and conclude that Heads is less likely.While she was waiting for a possible flood of memories in that firstminute, she had no new evidence about how the coin landed. If she got aflood of memories, she would increase the probability that the coin landedHeads (to 1). So in the absence of a flood of memories, she would have todecrease the probability that the coin fell Heads.In the original case, there is no such possibility. There is no possibilitythat Beauty will find out for certain that it fell Heads. So she cannot updateon the lack of such information. Unlike in the variant case, Beauty has nonon-indexical information to conditionalize on.Dorr asks why this delay of a minute, the only difference between theoriginal Sleeping Beauty and the variant case should make any differenceto subjective probabilities. I answer that in the variant case a hypothesisthat could have been eliminated has failed to be (H2). There is no equiva-lent of this is in the original case. The variant case is uninformative.1Stanford UniversityStanford, CA 94305, USAdbradley@stanford.eduReferencesDorr, C. 2002. Sleeping Beauty: in Defence of Elga. Analysis 62: 292–96.1 I would like to thank Branden Fitelson for helpful discussion. at University of California, Berkeley on August 11, 2010 http://analysis.oxfordjournals.orgDownloaded from 

Sleeping beauty: A note on Dorr's argument for 1/3

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Sleeping beauty: A note on Dorr's argument for 1/3

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