Sleeping Beauty and the absentminded driver Jean Baratgin Institut Jean Nicod GREQUAM - PDF document

Sleeping Beauty and the absent-minded driver Jean Baratgin (Institut Jean Nicod & GREQUAM, jean.baratgin@univmed.fr) Bernard Walliser (ENPC-EHESS, walliser@mail.enpc.fr) Abstract: The Sleeping Beauty problem is presented in a formalized framework which justifies the underlying probability structure. The two rival solutions proposed in the literature by Elga (2000) and Lewis (2001) differ by a single parameter and can be easily compared. They rest in two different ways on the ‘no news no change’ principle asserting that subjective belief only changes in case of new information. But they differ by the cognitive attitude attributed to Sleeping Beauty, who is ‘fuzzy-minded’ for Elga and ‘blank-minded’ for Lewis. Moreover, the traditional absent-minded driver problem is reinterpreted in this framework and sustains Elga’s solution. Keywords: absent-mindedness, belief revision, Sleeping Beauty problem 1. The formalized Sleeping Beauty problem Elga (2000) introduced a problem that has triggered a significant controversy about probabilistic belief revision. An experimenter is going to put a Beauty to sleep on Sunday. During the two days that her sleep will last, he will briefly awake her either once on MONDAY or twice on MONDAY and TUESDAY, depending on the toss of a fair coin (once if HEADS; twice if TAILS). After any AWAKENING, he will put Sleeping Beauty back to sleep with a drug that makes her forget that she was awake. Two questions are considered. When she is (first) awakened, to what degree should Sleeping Beauty believe that the outcome of the coin toss is HEADS? When she is moreover told that it is MONDAY, to what degree should Sleeping Beauty believe that the outcome of the coin toss is HEADS? Sleeping Beauty faces a problem which is characterized by two main features. First, she becomes forgetful in the sense that she does not know some elements of her latter situation. Second, she receives no new information from outside about the change of situation. The debate involved is to know if she nevertheless may revise her beliefs. Let P_, P and P [dans le texte latex, si + est mis en exposant, il faut en faire de meme pour -] be respectively Sleeping Beauty’s probability distribution just before being put to sleep, just after being awakened and just after the experimenter told her that it is Monday. Call H, T, M, T*and A the respective events HEADS, TAILS, MONDAY, TUESDAY and AWAKENING. Some assumptions are usually defined, either explicitly or implicitly: A1. Sleeping Beauty knows the procedure. Especially, she knows that the coin is fair and endorses it: P_(H) = P_(T) = 1/2. She also knows when she will be awakened: P_(A/H&M) = P_(A/T&M) = P_(A/ T&T*) = 1 ; P_(A/H&T*) = 0. A2. Sleeping Beauty applies an indifference principle between Monday and Tuesday in case of Tails: P_(T&M/T) = P_(T&T*/T), hence P_(T&M) = P_(T&T*) = P_(T)/2 ; it follows that: P_(M/T) = P_(T&M)/P_(T) = 1/2 A3. Sleeping Beauty applies Bayes’ rule in order to shift from P_ to P: P(H) = P_(H/A) = P_(H)P_(A/H)/[P_(H)P_(A/H)+P_(T)P_(A/T)] = P_(A/H)/[ P_(A/H)+1] (by A1) [supprimer sur le texte la ligne intermédiaire] A4. Sleeping Beauty applies the Bayes’ rule in order to shift from P to P: P(H) = P(H/M) = P(H)P(M/H)/[P(H)P(M/H)+P(T)P(M/T)] = 2P(H)/[P(H)+1] (by A2) Consider now the probability space: {H, T} x {M, T*} x {A, ¬A} [mettre les crochets dans le texte] We use the traditional notation A&B for A B , A for the complement of A and P(A/B) for the probability of A given B [je pense qu’il faut mettre partout les barres / pour les probas conditionnelles, sinon c’est peu lisible]. The assumption A1 is generally stated for the probability distribution P, but holds for P_ according to A3 The assumption A2 is generally stated for the probability distribution P, but holds for P_according to A3. . It is able to represent the problem as long as the two events of tossing the coin and unrolling the days are independent. This is true since the toss of the coin can take place before the awakening of Sleeping Beauty either on Monday or on Tuesday (Sleeping Beauty is awakened in any case on Monday). The third event, the awakening of Sleeping Beauty, is a direct consequence of the two preceding ones. In a typical Bayesian way, the prior probability distribution P_ on elementary events is given by a bimatrix (where, for instance, a = P_(H&M&A)): A ¬A H T M a a’ b b’ T* c c’ d d’ Table 1. The probability space According to A1, the conditions of fair coin imply: a + a’ + c + c’ = b + b’ + d + d’ = 1/2. Likely, the awakening conditions imply: a’ = b’ = c = d’ = 0. According to A2, the indifference between Monday and Tuesday implies: b = d = 1/4. The preceding probability distribution P_ becomes perfectly defined except for a unique parameter: A ¬A H T M a 0 1/4 0 T* 0 1/2- a 1/4 0 Table 2. The P_ probability distribution According to A3, the preceding distribution, conditional to A, leads to the distribution of P: A ¬A H T M 2a/(2a+1) 0 1/2/(2a+1) 0 T* 0 0 1/2/(2a+1) 0 Table 3: The P probability distribution According to A4, the preceding distribution, conditional to M, leads to the distribution of P: A ¬A H T M 4a/(4a+1) 0 1/(4a+1) 0 T* 0 0 0 0 Table 4: The P+ probability distribution The general solution gives: P(H) = 2a/(2a+1) [dans le texte, il faut l’exprimer sous forme de fraction comme le second] and P(H) = 4a/(4a+1) 2. Usual solutions The remaining problem is to fix the parameter a by introducing some further assumptions. The two basic solutions proposed in the literature share the axioms A1 to A4 and correspond precisely to specific values of the parameter a. Elga (2000) justifies his solution by the further assumption: A5: Sleeping Beauty, once she learns that it is Monday (and knows that the coin may be tossed afterwards), considers that she learned nothing with reference to what she already knew on Sunday: (H) = P_(H). It follows that: a = 1/4 , P(H) = 1/3 and P(H) = 1/2 Lewis (2001) justifies his solution by an alternative assumption: A5’: Sleeping Beauty, once she is awakened, considers that she learned nothing with reference to Sunday since she knew for sure that this will happen in the future : P(H) = P_(H) It follows that: a = 1/2, P(H) = 1/2 and P(H) = 2/3 The Sleeping Beauty problem is grounded on the fact that Sleeping Beauty becomes forgetful in the sense that she does not know some elements of her earlier history. The two solutions share a further assumption but applied to different periods. This assumption called the ‘no news, no change’ principle, asserts that Sleeping Beauty should not revise her beliefs as long as she gets no new information. It follows trivially from the principle of ‘strict conditionalization’ (Teller, 1973) or of ‘temporal conditionalization’ (Talbott, 1991) which asserts that a belief is updated in a Bayesian way if and only if a new message is received. It follows likewise from the ‘strong principle of conservation’ (Walliser-Zwirn, 2002) which asserts that, when a message is already known according to the prior probability, the posterior probability is unchanged. One solution is generally selected by just favouring the assumption A5 or A5’, by changing moreover some of the assumptions A1 to A4 and by eventually generalizing the problem (Dorr 2002; Monton 2002; Hitchcock 2004; Horgan 2004; Weintraub 2004). In this note, we stress that the two usual solutions differ in fact by the interpretation given to the effect of the drug. Two alternative interpretations are possible leading to different results. The validity or not of the ‘no news, no change’ assumption as it is respectively applied follows. According to Elga, the drug creates some ‘confusion’ in Sleeping Beauty’s mind. She is no more able to distinguish between Monday and Tuesday when awakened, but she knows on Tuesday that Monday may have happened. We say that Sleeping Beauty is ‘fuzzy-minded’ since she cannot make precise what her history was (and she knows this happens). More precisely, the P_ probability of Monday and Tuesday is the same. Hence, there are four equiprobable events: H&M, H&T*, T&M, T&T*. The event ‘Heads and Tuesday’ is as valuable as the others, even if Sleeping Beauty is not awakened. Moreover, for Sleeping Beauty, learning on Monday that it is Monday really brings her no new information with regard to the initial situation. But being awakened means that her history is actually on the way and that she forgot where she is. According to Lewis, the drug creates some ‘omission’ in Sleeping Beauty’s mind. She forgets on Tuesday that she was already awakened on Monday, hence thinks on Tuesday (as on Monday) that it is still Monday (the actual Monday did not happen). We say that Sleeping Beauty is ‘blank-minded’ since a piece of her history has simply been suppressed (and she knows this may happen). More precisely, the P probability of being awakened on Monday if Heads is twice the probability of being Lewis uses the term ‘memory erasure’ in order to express that a whole time slice has been suppressed. awakened on Monday if Tails. From an other point of view, the P_ probability of the event ‘Heads and Tuesday’ is zero since to be awakened or not involves no change (when awakened, it is always Monday in Sleeping Beauty’s mind). Moreover, for Sleeping Beauty, learning on Monday that she is awakened really brings her no new information. But learning that it is Monday gives her the illusion to get some information since she may theoretically be on Tuesday. The two solutions may receive a frequentist interpretation when the situation is considered as repeated. Elga’s solution is easy to obtain (Delahaye 2003). Sleeping Beauty is awakened two times out of three on Monday and one time out of three on Tuesday. It follows that P(M/A) = 2/3, hence a = 1/4. Lewis’ solution is a bit harder to justify. When considering all awakenings for which Monday is confirmed to her, two of three correspond to Heads and one on three to Tails. It follows that (H/M)=2/3, hence a =1/2. 3. The decision-theoretic interpretation Another puzzle devoted to the absent-minded driver is proposed by Piccione and Rubinstein (1997) and Aumann, Hart and Perry (1997a, 1997b). A drunk driver starts from a bar and takes the highway. At the first intersection in Montown, he can either exit to finish on the beach or continue to the intersection in Tuestown. When he arrives there, he can again either exit to his home or continue to a hotel. Moreover, when at an intersection, he does not know where he is. In a first application, he decides what strategy he should choose according to payoffs associated to each possible end point. In a second application, he chooses at each intersection in accordance with the toss of a (different) coin, Heads for exit and Tails for continue; but he can buy a lottery ticket which provides a payoff according to the end point. The second application is formalized by a game tree where the two nodes corresponding to the intersections are in a same information set. Bar [mettre bar en minuscule] H (1/2) MONTOWN — beach T(1/2) H’ (1/2) TUESTOWN — home T’(1/2)
 \n\r\n\n Hotel [petit h, pas d’accent circonflexe] The first application is similar, but the toss of the coin is replaced by the choice of a mixed strategy. It consists in the probability of choosing ‘exit’ and ‘continue’, which is the same at each intersection. The problem is a decision-theoretic one, but it can be studied in its epistemic aspect. It neutralizes part of the temporal structure of the Sleeping Beauty problem. Monday and Tuesday are no more successive dates, but correspond to geographical places. hey are nevertheless reached in a given order. Moreover, in order to be fully similar to the Sleeping Beauty problem (as concerns the first question), we keep the second interpretation and we assume that ‘continue’ at Tuestown leads to his home too. Hence the second toss can be replaced by a sure move, even if the two intersections are no more alike. bar [bar en minuscule] H (1/2) MONTOWN — beach T(1/2) TUESTOWN Home [h minuscule] Aumann, Hart and Perry (1997 a,b) compute the probability of being at the intersection in Montown when being at an intersection and obtain P(M)= 2/3. They just observe that the probability of being at Montown is twice the probability of being at Tuestown, in the same spirit than with the frequency interpretation. Hence, they compute ‘naturally’ Elga’s solution. In fact, Elga’s solution is justified since a drunk driver does not know where he is when being at an intersection. To be absent-minded means to be fuzzy-minded rather than blank-minded. The notion of information set induces this solution since it is grounded on the undiscernability of the nodes inside. The authors assume that the driver’s choice may change from the bar to the intersection, as a result of a change in his belief. But the authors claim that the driver’s choice is dynamically consistent, in the sense that he decides at the intersection what he intended to do at the intersection when at the bar. This is a consequence of the ‘reflection principle’ (van Fraassen, 1984) which states that the belief made at the bar as concerns the belief made at the intersection coincides with the belief at the intersection. Lewis’s solution with its specific cognitive interpretation may be applied for games too. But, since it departs from imperfect information, it needs another representation than just an information set. To be absent-minded may again be conceived in different ways. Acknowledgements We thank M. Cozic for fruitful remarks References Aumann, R.,Hart, S., Perry, M. (1997a): The absent-minded driver, Games and Economic Behavior, 20: 102-116. Aumann, R., Hart, S., Perry, M. (1997b): The forgetful passenger, Games and Economic Behavior, 20: 117-120. Delahaye, J. P. (2003) : La Belle au bois dormant, la fin du monde et les extraterrestres. Pour la Science, 309, 98-103 Dorr, C. (2002): Sleeping Beauty: In defence of Elga. Analysis 62, 292-296 Elga, A. (2000): Self-locating Belief and the Sleeping Beauty problem. Analysis 60, 143-147. Hitchcock, C. (2004): Beauty and the Bets. Synthese 139, 405-420. Horgan, T. (2004): Sleeping Beauty awakened: new odds at the dawn of the new day. Analysis 64, 10-21 Lewis, D. (2001): Sleeping Beauty: Reply to Elga. Analysis 61, 171-176 Monton, B. (2002): Sleeping Beauty and the forgetful Bayesian. Analysis 62, 47-53 Piccione M., Rubinstein, A. (1997): On the interpretation of decision problems with imperfect recall, Games and Economic Behavior, 20:3-24. Talbott, W.J. (1991): Two principles of Bayesian epistemology, Philosophical studies, 62, 135-150. Teller, P. (1973): Conditionalization and observation, Synthese, 26, 218-258. van Fraassen, B. (1984): Belief and the will, Journal of Philosophy, 81:235-256. Weintraub, R. (2004): Sleeping Beauty: A simple solution. Analysis 64, 8-10 9 9 1010