Abstract
The puzzle of the absentminded driver combines an unstable decision problem with a version of the Sleeping Beauty problem. Its analysis depends on the choice between “halfing” and “thirding” as well as that between “evidential” and “causal” decision theory. I show that all four combinations lead to interestingly different solutions, and draw some general lessons about the formulation of causal decision theory, the interpretation of mixed strategies and the connection between rational credence and objective chance.
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Notes
For more precise statements, see e.g. Lewis (1981) and Joyce (1999, Chap. 5). While it is not essential for my discussion, I assume that states, acts and outcomes are all entities of the same kind (propositions), and that outcomes can be identified with conjunctions of states and acts; see Joyce (1999, Chap. 2) for discussion. See also Skyrms (1984, Chap. 4) and Joyce (2002) on the causal interpretation of Savage (1954).
There are other ways to understand desire on which desirability isn’t represented by conditional expected utility. Roughly speaking, conditional expected utilities measure the extent to which the agent would be pleased to learn that the relevant proposition is true. Something else is in play when one expresses a positive attitude towards counterfactual scenarios in which Kennedy was not killed by saying, “I wish Oswald hadn’t killed Kennedy”. The causal expected utility \({\textit{EU}}(A)\) of a proposition \(A\) arguably captures this kind of subjunctive desirability of \(A\). See e.g. Etlin (2008, Chap. 2) for discussion.
Proof: I first show that \(P({\textit{Continue}}_1) = P({\textit{Continue}}_2) = c\). Observe that \({{\textit{Second}}}\) entails \({\textit{Continue}}_1\), wherefore \({{\textit{Continue}}}\) entails \({\textit{Continue}}_1\). More specifically, \({\textit{Continue}}_1\) is the disjunction of \({{\textit{Continue}}}\) and \( {{\textit{Second}}} \; \& \; {{\textit{Leave}}}_2\). By uniformity, the latter has probability 0, as does \( {\textit{Continue}}_1 \; \& \; {{\textit{Leave}}}_2\). Hence \( c = P({{\textit{Continue}}}) = P({\textit{Continue}}_1) = P({\textit{Continue}}_1 \; \& \; {\textit{Continue}}_2)\). Since \( {\textit{Continue}}_2 \; \& \; {{\textit{Leave}}}_1\) also has probability 0, it follows that \( P({\textit{Continue}}_2) = P({\textit{Continue}}_2 \; \& \; {\textit{Continue}}_1) = c\).
Now \({\textit{Continue}}_2\) divides into \( {\textit{First}} \; \& \; {\textit{Continue}}_2\) and \( {{\textit{Second}}} \; \& \; {\textit{Continue}}_2\), so \( c = P({\textit{First}} \, \& \, {\textit{Continue}}_2) + P({{\textit{Second}}} \, \& \, {\textit{Continue}}_2)\). By uniformity, it follows that \( c = P({\textit{First}} \, \& \, {\textit{Continue}}_2) + P({{\textit{Second}}} \, \& \, {\textit{Continue}}_1)\). Moreover, by symmetry, \( P({\textit{First}} \, \& \, {\textit{Continue}}_1) = P({{\textit{Second}}} \, \& \, {\textit{Continue}}_1)\), and so \( P({\textit{First}} \, \& \, {\textit{Continue}}_2) = P({{\textit{Second}}} \, \& \, {\textit{Continue}}_1)\) by uniformity. Hence \( P({\textit{First}} \, \& \, {\textit{Continue}}_2) = P({{\textit{Second}}} \, \& \, {\textit{Continue}}_1) = c/2\). \( {{\textit{Second}}} \, \& \, {\textit{Continue}}_1\) is equivalent to \({{\textit{Second}}}\). The remaining possibility \( {\textit{First}} \, \& \, {{\textit{Leave}}}_2\) must then have probability \(1-c\).
Pace Lewis (1981, p. 29f.).
There are also equilibrium states of indecision. In general, any state in which the expected bias is 2/3 is a stable solution.
Neither Aumann et al. nor Piccione and Rubinstein actually mention the connection to evidential and causal decision theory. Piccione and Rubinstein implicitly assume evidential decision theory by setting \(c=b\). In addition, they overlook the fact that the probability of being at the first exit (evidentially) depends on the chosen bias \(b\). Instead they assume that \(P({\textit{First}})\) has a fixed value of 3/5, based on the driver’s prior decision to use bias \(c=2/3\) and the fact that \(P({\textit{First}}) = 1/(c+1)\). Setting \(c=b\) then yields the payoff function \((6b -3b^2 +8)/5\), which has its maximum at \(b = 1/3\). The corrected evidential formula \((8b - 6b^2)/(b+1)\) appears in the appendix of Rabinowicz (2003).
Aumann et al. (1997) object that by setting \(c=b\), Piccione and Rubinstein erroneously represent the driver’s present choice as controlling her choice at the other exit. They suggest that if the driver is certain that she chooses bias \(c\), then the expected utility of Bias \( = b\) is \((4b+4c-6bc)/(c+1)\), which agrees with our formula (21). According to Aumann et al., the optimal bias is then the value of \(b\) that maximises that function when plugged in for \(c\): 2/3.
One disanalogy between the two cases is that the driver’s coin is tossed after she arrives at the first exit, while Beauty’s coin is usually assumed to be tossed before she wakes up on Monday. However, that timing is plausibly inessential to the Sleeping Beauty problem. Since the outcome of the toss has no effect until Tuesday morning, the coin might as well be tossed on Monday instead of Sunday.
This line of thought occurs in Elga (2000) and Lewis (2001), and is further developed e.g. in Bradley (2011), Schwarz (2012) and Schwarz (2013). One complication here is that Beauty may be forced to violate doxastic conservatism if the coin lands tails: for example, her credence in the proposition that the sky is overcast on either Monday or Tuesday will be high on Monday evening, but low on Tuesday morning, although she doesn’t acquire any relevant information. (At least this should be so if we assume, as we did for the driver, that her belief state on Monday morning is identical to her belief state on Tuesday morning.) One might think that this threat of diachronic irrationality somehow undermines the force of doxastic conservatism at the earlier transition from Sunday to Monday. In Schwarz (2013) I argue that it doesn’t.
For another motivation of the thirder assignment, observe that if the experiment were repeated indefinitely, the ratio of occasions where the driver finds herself at an exit and the (first) coin lands tails to such occasions where the coin lands heads would converge to \(2c : (1-c)\).
In the case of Sleeping Beauty, the symmetry assumption is sometimes motivated by a stipulation that Beauty’s two tails awakenings are “subjectively indistinguishable”. Symmetry is then supposed to follow from a general principle of self-locating indifference. Halfers sometimes object to symmetry, suggesting that Beauty ought to be certain on Monday that it is Monday and on Tuesday that it is Tuesday; see e.g. Hawley (2013). This response gets less attractive if it is stipulated, as I have, that the driver has the same beliefs at every exit. (Whether the two situations are otherwise indistinguishable is to my mind irrelevant, but feel free to suppose so if you think it matters).
The driver does in fact not violate the Principle formulated in Lewis (1980), since her credences aren’t “initial”. One might also question whether there even is an objective chance for the centred proposition Tails, as opposed to the uncentred \({\textit{Tails}}_1\) and \({\textit{Tails}}_2\).
For the second exit, we assumed that \( V({{\textit{Second}}} \; \& \; {{\textit{Bias}}}=b) = (1-b)4 + b1 = 4-3b\). This is correct on both halfing and thirding, since on either account the probability of Tails, given \( {{\textit{Second}}} \; \& \; \textit{Bias}=b\), is \(b\).
One noteworthy aspect of the puzzle that I have not discussed concerns the form of the deliberation dynamics. In all versions of the puzzle, the dynamics can’t go by the simple models described in Skyrms (1990). For example, these models would not preserve the symmetry assumption; they also don’t take into account how the driver’s choices at the two exits can become decorrelated through the process of deliberation.
Skyrms’s own formulation is superficially different: he defines \({\textit{EU}}(A)\) as \(\sum _K P(K) \sum _C {\textit{Ch}}_{K}(C/A) V(C)\), where \(K\) ranges over chance hypotheses, \(C\) over “consequences” and \({\textit{Ch}}_K(C/A)\) is the conditional chance of \(C\) given \(A\) according to \(K\).
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Acknowledgments
Ancestors of this paper were presented at the ANU in 2007 and at the Formal Epistemology Workshop in Munich in 2012. I thank the audiences on these occasions, as well as Alma Barner, Rachael Briggs, Kenny Easwaran, Alan Hájek, Daniel Nolan, Michael Titelbaum, David Wiens, and three anonymous referees for comments and discussion.
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Schwarz, W. Lost memories and useless coins: revisiting the absentminded driver. Synthese 192, 3011–3036 (2015). https://doi.org/10.1007/s11229-015-0699-z
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DOI: https://doi.org/10.1007/s11229-015-0699-z