The Wednesday Sleeping Beauty Problem
In 2009, at the ANU, Mike Titelbaum organized a small workshop on the Sleeping Beauty problem. I gave a talk in which I argued that the answer to the problem depends on whether we accept genuinely diachronic norms on rational belief: if yes, halfing is the most plausible answer; if no, we get thirding. A successor of this talk is now forthcoming in Noûs. Here's a PDF. In this post, I want to discuss a surprisingly hard question Kenny Easwaran raised in the Q&A after my talk:
How confident should Beauty be on Wednesday that the coin has landed heads?
The intuitive answer, I think, is that her credence in Heads should be 1/2. If you're a thirder, this means you have to explain why Beauty's credence should change from 1/2 on Sunday to 1/3 on Monday (and possibly Tuesday) and back to 1/2 on Wednesday.
I'm a halfer, and it's in the context of my account that Kenny asked the question. So let me briefly explain my case for halfing.
We want to know how Beauty should update her beliefs from Sunday to Monday. The answer is tricky because Beauty's belief state on Tuesday morning will be reset to equal her belief state on Monday morning if the coin lands tails.
Begin with a simpler case in which Beauty knows that her Tuesday morning beliefs will be reset to her Monday morning beliefs. (There's no coin toss.) Any update protocol or update disposition she adopts on Sunday night therefore determines both her Monday morning beliefs and her Tuesday morning beliefs. In light of this, an optimal protocol arguably shouldn't let her wake up confident that it is Monday: this would lead to a highly accurate belief state on Monday but to a highly inaccurate belief state on Tuesday. She ought to adopt a protocol that makes her indifferent between Monday and Tuesday.
The memory reset in this version of the story ensures that Beauty's Sunday state has two "doxastic successors", in the sense of Meacham (2010). From an epistemic perspective, the case resembles a case of personal fission, where a single person wakes up as two persons.
The original Sleeping Beauty problem resembles a case of merely possible fission: Beauty's Sunday state has two successors iff the coin lands tails. It's analogous to the case of Fissioning Fred that I discuss in Schwarz (2015):
Fissioning Fred. Fred's home planet, Sunday, is surrounded by two moons, Monday and Tuesday. Tonight, while Fred is asleep, he will be teleported to Monday: his body will be scanned and destroyed and recreated from local matter on Monday. Depending on the outcome of a fair coin flip, he will also be teleported to Tuesday – tails he will, heads he won't.
When Fred wakes up, he should give credence 3/4 to being on Monday, and 1/2 to Heads. The same is true for Sleeping Beauty, who finds herself in a structurally analogous predicament. See my paper for the subtleties.
Now, Kenny observed that this account seems to have a strange consequence for Beauty's Wednesday beliefs: it seems to imply that her Wednesday credence in Heads should be 2/3.
Here's why. According to the fission model, Beauty turns into two epistemic subjects if the coin lands tails. On Tuesday evening, the Tuesday subject goes to sleep and wakes up on Wednesday. What happens to the Monday subject when she goes to sleep on Monday evening? She seems to disappear: she has no doxastic successors.
Let's add analogous assumptions to the story of Fred.
Fission Fred, continued. The following night, if the coin has landed heads, Fred will be teleported again from Monday to another moon, Wednesday. If the coin has landed tails, his Tuesday successor will be teleported to Wednesday and his Monday successor will be destroyed.
Suppose after the first awakening, Fred gives credence 1/2 to Heads ∧ Monday, and 1/4 each to Tails ∧ Monday and Tails ∧ Tuesday, as I think he should. He knows that he will not survive the following night in case of Tails ∧ Monday. When he does wake up again, he should therefore redistribute his beliefs about his previous location to the remaining possibilities, Heads ∧ Monday and Tails ∧ Tuesday: his credence in Heads should increase to 2/3.
I'm inclined to agree that this is how Fred should update his beliefs. What about Sleeping Beauty?
I think there's a disanalogy. When Fred wakes up on Wednesday, he might reasonably say: "Ah, I'm still alive. This was sure to happen if the coin landed heads, but unsure if it landed tails." But when Beauty wakes up on Wednesday, I don't think she could reasonably say this.
We shouldn't get too carried away with the fission analogy. Let's focus on the real question: what update protocol (or disposition) should Beauty adopt before she goes to sleep the second time?
If it is Monday and the coin has landed tails, it doesn't matter what update protocol she adopts, as it won't be implemented anyway. Her post-awakening state is already fixed to equal her Monday morning state. Ideally, her post-awakening state would be sensitive to her pre-awakening state, but it isn't. That's why the Sleeping Beauty case, unlike the case of Fred, is a case of enforced non-ideality (as Arntzenius (2002) pointed out). We could imagine milder versions of the belief reset, where only some of Beauty's beliefs are reset on Monday night. In that case, Beauty should consider the possible effects of the chosen protocol on her Tuesday morning beliefs. In the original case, there are no such effects, so there is nothing to consider. But this doesn't mean that the Tuesday morning state should be treated as belonging to a different person.
This undercuts the argument from the fission analogy. But it doesn't tell us what Beauty's Wednesday credence is Heads should be.
There's another reason to think it should be 2/3. Consider a practical analogy.
Betting I. Your credence is divided between three possibilities, H, T1, and T2. You give credence 1/2 to H and 1/4 each to T1 and T2. I offer you a deal that pays $1 in case of H and $0 otherwise. In case of T2, you simply lose however much you paid for the deal. In case of T1, the deal is annulled and your money refunded. How much would you pay?
If you pay $x for the deal, the expected payoff is 1/2 $(1-x) + 1/4 $-x + 1/4 $0. This is positive as long as x < 2/3. The expected loss in case of ¬H is effectively cut in half.
One might argue that this setup resembles that of Sleeping Beauty, where the choice of update protocol is ignored in one of the two Tails possibilities.
But, again, the analogy isn't perfect. An important difference is that Beauty can be confident that if the coin has landed tails then she adopts the same protocol on Monday and on Tuesday. And the protocol chosen on Tuesday does make a difference.
The answer to the Wednesday Sleeping Beauty problem might turn on whether the (hypothetical) choice of update protocol should be assessed in terms of Evidential or Causal Decision Theory.
Before we look at the two approaches, note that we can simplify the decision problem. An update protocol (or disposition) is a contingency plan specifying a new credence depending on the information Beauty will receive. But Beauty knows that she won't receive any relevant, unforeseen information when she wakes up. Her choice of an update protocol reduces to a choice of what credence she should plan to have when she wakes up.
Using EDT, Beauty could reason as follows.
"Whatever credence I now plan to have is sure to be my credence on Wednesday morning, and at no other time. So the question is, what credence would I like to have on Wednesday morning? I'm 50% confident that the chosen credence will be instantiated in a Heads world, and 50% confident that it will be instantiated in a Tails world. To maximize expected accuracy, the optimal credence is indifferent between Heads and Tails."
With CDT, things are less straightforward. Let's first look at a practical analogy.
Betting II. Your credence is divided between three possibilities, H, T1, and T2. You give credence 1/2 to H and 1/4 each to T1 and T2. I offer you a deal that pays $1 in case of H and $0 otherwise. In case of T2, you simply lose however much you paid for the deal. In case of T1, you lose however much your perfect twin paid for the deal in an identical choice situation. How much would you pay?
Let $y be the (unknown) amount your twin has paid. The expected payoff of paying $x is 1/2 $(1-x) + 1/4 $-x + 1/4 $-y = $(1/2 - 3/4 x - 1/4 y).
This is an unstable decision problem. If you're sure your twin paid $1, you should pay up to $1/3. If you're sure your twin paid $0, you should pay up to $2/3. There's a unique equilibrium solution at x = y = 1/2. So, plausibly, you should pay up to $1/2.
The analogy suggests that the optimal update protocol by CDT lights also yields a Wednesday credence in Heads of 1/2. But I'm not sure how to show this directly.