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A Dutch book for CDT thirders

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Abstract

I give a Dutch book argument for CDT thirders in the context of a generalized Sleeping Beauty scenario. In combination with the Briggs (2010) Dutch book for EDT thirders, this amounts to a purely decision-theoretic argument for halfing in Sleeping Beauty. In combination with the Hitchcock (2004) Dutch book for CDT halfers, this amounts to a Dutch book argument against CDT. The combined Dutch book against CDT invites a plausible diagnosis of the reasons for CDT’s failure: CDT is incompatible with the Dutch book requirement because it allows the values of epistemically impossible outcomes influence one’s decision.

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Notes

  1. I would like to thank the two anonymous Synthese reviewers for their very substantial comments. Thanks also to Wolfgang Schwarz and Jonathan Peeters for helpful discussion of early drafts.

  2. And therefore that Briggs’ claim (c) above is mistaken. It will not be rewarding to discuss Briggs’ argument for (c) here, given that what I will argue for is the stronger claim that there is a Dutch book for \(\hbox {CDT}^{1/3}\).

  3. Assume that Sleeping Beauty is risk-neutral, and her utilities are linear in money.

  4. Here, it is crucial that we offer Sleeping Beauty a choice to press one of the buttons, rather than a choice to accept or reject a bet. See Sect. 2.1 for an explanation. The same key idea is used by Conitzer (2015a).

  5. For reasons given in Sect. 2.1, it is best to prevent Sleeping Beauty from using a randomizing device in case of ties. We can do this by simply stipulating that she always chooses Accept in such cases.

  6. See Piccione and Rubinstein (1997) for the original discussion. The tree shows the coin toss taking place after Sleeping Beauty’s Monday decision.

  7. The Briggs book consists of a Sunday bet of \(15+\epsilon \) if Heads, \(-15+\epsilon \) if Tails, followed by a Monday/Tuesday bet of \(-20+\epsilon \) if Heads, \(5+\epsilon \) if Tails. The \(EDT^{1/3}\)-following Sleeping Beauty then calculates as follows (here, following Briggs 2010, A is the centered proposition that Sleeping Beauty accepts the presently offered bet, and D is the centered proposition that Sleeping Beauty accepts the bet on a different day):

    figure b

    and

    figure c

    And so the \(EDT^{1/3}\)-following Sleeping Beauty accepts the Monday/Tuesday bets, but, in combination with the Sunday bet, is guaranteed to lose \(\$5-2\epsilon \) if Heads and \(\$5-3\epsilon \) if Tails.

    The Hitchcock book consists of a Sunday bet of \(-15+\epsilon \) if Heads, \(15+\epsilon \) if Tails, followed by a Monday/Tuesday bet of \(10+\epsilon \) if Heads, \(-10+\epsilon \) if Tails. So the \(CDT^{1/2}\)-following Sleeping Beauty calculates as follows (the partition is H, T&D, T& \(\lnot \)D):

    figure d

    and

    figure e

    and so

    $$\begin{aligned} U_{CDT^{1/2}}(A)-U_{CDT^{1/2}}(\lnot A)=\epsilon \end{aligned}$$

    And so the \(CDT^{1/2}\)-following Sleeping Beauty accepts the Monday/Tuesday bets, but, in combination with the Sunday bet, is guaranteed to lose \(\$5-2\epsilon \) if Heads and \(\$5-3\epsilon \) if Tails.

  8. For an overview of Dutch book arguments, see Vineberg (2011), Hájek (2008) and Pettigrew (2019). For possible worries about the bookie using information unavailable to Sleeping Beauty, see Briggs (2010).

  9. This condition corresponds to betting at 1:2 odds or better if Sleeping Beauty is offered explicit bets. Cf. Arntzenius (2002).

  10. For more on deliberational dynamics, see Skyrms (1990), Joyce (2012), Joyce (2018).

  11. Hare and Hedden (2016) call these cases of ‘self-reinforcing decision-dependence’.

  12. \( p(T \& D)=p(D)=p(A \& T)=p(T|A)p(A)\), so \( p(A)=p(T \& D)\times 3/2\).

  13. I thank the anonymous referee for bringing this complication to my attention. In fact, allowing a third equilibrium at \(p(A)=3/4\) would not affect the final result, since Sleeping Beauty knows that unless she starts out with \(p(A)=3/4\), she will never reach that credence. But the case is cleaner if we avoid this complication.

  14. There have been a number of interesting arguments against CDT in recent years. This is not the place for a serious survey and comparison of the results. But see Ahmed (2014) and Joyce (2016) on Newcomb Insurance, and Hare and Hedden (2016) and Bales (2018) on Three Crates. But one point is worth stressing: it is one thing to come up with a scenario in which CDT makes intuitively the wrong call (e.g. Egan cases from Egan 2007), but it is another thing to show that CDT fails to satisfy the Dutch book criterion.

  15. It should be noted that Conitzer’s WBG book, unlike the GSB, Briggs, and Hitchcock books, makes a substantive assumption about belief revision: the argument only goes through on the assumption that \(p(H)=1/3\) in the WBG scenario. But this seems a rather safe assumption: even compartmentalized conditionalizers would agree (see, e.g. Meacham 2008).

  16. This search is well under way (see, e.g. Stern 2017 and Yudkowsky and Soares 2017), but outside the scope of this paper. Of course whether this search will be successful is an open question. For example, a recent attempt by Gallow (2020), meant specifically to deal with decision instability problems, suggests that we should maximize improvement news, as follows:

    figure g

    where \( U_{CDT}(B|A)=\sum _i p(K_i|A)V(B \& K_i)\), \(\{K_i\}\) a Lewis partition, and X ranges over all the options. In the GSB case, we will have:

    figure h

    —the two stable decisions recommend themselves. And so on Gallow’s proposal accepting and rejecting tie, and so Gallow’s Beauty may accept, and so has no immunity from the GSB book.

  17. In the EDT case, there is of course only one stable option to accept.

  18. ‘All but’ because we could, for example, have a weighting function f that assigns zero weight to positive outcomes and non-zero weight to negative outcomes. But if f is independent of the value of outcomes, then \(f(A, w_1)=f(A, w_3)\) is the only option.

  19. It should be noted that the result Conitzer proves does not seem to be fully general: it is not clear whether it follows that \(\hbox {CDT}^{1/3}\) is immune in all additive scenarios. For example, it is an assumption of his argument that the number of awakenings is independent of Sleeping Beauty’s actions, and this leaves out cases like the absent-minded driver. But Conitzer’s result does cover GSB cases, which is what matters here.

  20. Conitzer suggests that the intuitive difference between additive and non-additive scenarios is that in the latter Sleeping Beauty has to coordinate with her future and past selves, since the contribution of each individual decision to the total payout depends on what other decisions Sleeping Beauty makes. This is a helpful idea, but it does not help explain why CDT should fail to give good advice in cases where such coordination is not possible, as in the GSB book.

  21. Of course a decision theory may have no need of deliberational dynamics. For example one can have CDT without dynamics (that’s the version Egan 2007 investigated).

  22. Thanks to an anonymous reviewer for suggesting this point.

  23. This is of course compatible with the idea that some plans are not feasible, and so not options. The proposal strictly speaking covers only the Sleeping Beauty scenario, and is compatible with the thought that in other scenarios, to the contrary, only individual actions may be options.

  24. Note that this is so only if Sleeping Beauty converges on \(p(A)=1\); if Sleeping Beauty converges on \(p(A)=0\), then she prefers to reject if she considers individual actions, but still prefers to accept if she considers plans (since the valuation of plans remains the same).

  25. For a clear development of this thought see, e.g., Joyce (2018).

  26. Of course one of the two Dutch books (GSB or Hitchcock’s) is undermined by the true theory of belief updates, but the point I want to make is independent of which Dutch book really does the work against CDT.

  27. The GSB book added the further stipulation that \(u_2=u_4=0\). I am now considering the slightly more general case.

  28. And recall the tree:

    figure i
  29. It may at first seem puzzling that a non-zero value for \(u_2/u_4\) makes \(\hbox {CDT}^{1/3}\) immune from Dutch books in the original Sleeping Beauty scenario. The intuition would be that if the problem with CDT is that it fails to ignore the value of \(u_2/u_4\), then the one case in which it should be immune is precisely the case where \(u_2/u_4=0\). But there is no mystery here: CDT fails to ignore the value of \(u_2/u_4\) by allocating a positive probabilistic weight to these outcomes. What happens in the original scenario is that the ‘mis-allocated’ probability alloted to \(u_2/u_4\) is exactly compensated by the non-zero values assigned to \(u_2/u_4\). (Thanks to the anonymous reviewer for bringing up this concern.)

  30. It is perhaps worth pointing out that Lewis’ formulation of CDT, \( U_{Lewis}(A)=\sum _i p(K_i)V(A \& K_i)\), does not allow for this phenomenon, since \( V(A \& K_i)\) is a sum over outcomes in epistemically possible worlds. And so Lewis’ CDT is unusable in the Sleeping Beauty scenario (and so this and other discussions of CDT in the Sleeping Beauty context de facto use some counterfactual version of CDT, such as the Joyce formulation above).

  31. One may worry, at this point, that if the lesson is that one must disregard epistemically dead possibilities, then one is forced into the hands of EDT, which, precisely, forces us to ignore the value of \(u_4\), and so respects the epistemic constraint imposed by the Dutch book criterion. But this is not so. There are plenty of possible views that respect the lesson I am recommending without collapsing into EDT. Here is a quick example.

    Let’s start with the selection function formulation above (Eq. 25). With appropriate constraints on similarity that make sure that f captures the counterfactual similarity relation, this gives us a formulation of CDT. And now let us exchange f for its epistemic cousin, \(f^e\), while keeping the underlying closeness measure: let \(f^e_A(v, w)=1\) iff w is the closest epistemically possible A-world to v. In the GSB-scenario, we can make do with four epistemically possible worlds: \(w_1\), \(w_3\), \(w_5\), \(w_6\) (each \(w_i\) witnessing the history that results in the payoff \(u_i\)). The decision theory \(\hbox {CDT}^e\) that uses \(f^e\) instead of f will give us \(f^e_A(w_1, w_1)=1\), \(f^e_A(w_3, w_3)=1\), \(f^e_A(w_5, w_1)=1\), and \(f^e_A(w_6, w_3)=1\). And so we have:

    figure j

    to be contrasted with:

    figure k

    So \(\hbox {CDT}^e\) gives the same value to A as EDT, and ignores the values of \(w_4\) and \(w_2\). But \(\hbox {CDT}^e\) obviously diverges from EDT in the standard Newcomb scenario (by recommending two-boxing).

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Korzukhin, T. A Dutch book for CDT thirders. Synthese 198, 11925–11941 (2021). https://doi.org/10.1007/s11229-020-02841-7

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