Here’s what I think is a new argument against the ‘halfer’ solution to the Sleeping Beauty puzzle. (I assume here a lot of familiarity with the puzzle!) I’m particularly interested in responding to this kind of argument for the halfer view, versions of which you can find in papers by David Lewis, Carrie Jenkins and Joe Halpern.
- On Sunday, the rational credence in Heads is ½.
- Nothing surprising happens between Sunday and Monday, so Beauty doesn’t learn anything new.
- If nothing surprising happens between Sunday and Monday, Beauty’s credence in Heads shouldn’t change.
- So on Monday, the rational credence in Heads is ½.
I think premise 2 in this argument fails, for reasons set out by Bob Stalnaker in Our Knowledge of the Internal World. But Stalnaker’s positive theory is controversial; it would be good to have independent arguments against premise 2 here.
Start by modifying the story. Add in a new character, who I’ll call Prisoner X (or X for short). X will wake once, on either Monday or Tuesday. When he wakes, he’ll be kept awake for exactly as long as Beauty is awake. He’ll be told, truthfully, that Beauty is awake, then returned to sleep. There won’t be any messing around with his memories in any way; he’ll wake on Wednesday with memory intact.
Which day X wakes on will be determined by the following algorithm:
- If Heads, then X will wake on Monday.
- If Tails, X will wake on Monday with probability ½, and Tuesday with probability ½.
X knows all these facts before he goes to sleep on Sunday, as does Beauty. Now for the argument against the halfer position.
- When X wakes up, his rational credence in heads is ½.
- When X wakes up, X’s evidence is just Beauty’s evidence plus the fact that X is awake.
- If 1 and 2, then when she wakes up, Beauty’s credence in Heads conditional on X being awake should be ½.
- When Beauty wakes up, her credence in Heads conditional on X being asleep is 0.
- When Beauty wakes up, her credence in X being asleep should be greater than 0.
- So, when Beauty wakes up, her credence in Heads should be less than ½.
I hope the validity of the argument is obvious. In general, if Pr(A | B) = x, and Pr(A | ¬ B) = 0, and Pr(¬ B) > 0, then Pr(A) < x. That’s all we’re appealing to in the argument.
Premise 1 seems intuitively plausible and hard for the halfer to deny. If Beauty isn’t surprised by anything, then neither is Prisoner X. X knows on Sunday that he’ll wake up once, and knows that Beauty will be awake when he does. So on Sunday he can think about the one and only waking he’ll have during the experiment. (That’s in sharp contrast to Beauty, who can’t think that.) And on Sunday he can say that conditional on Beauty being awake when he is, the probability of Heads is ½. So when he conditionalises on what he is told on waking, his credence in Heads is ½, as suggested.
The motivation for premise 2 comes from thinking about what it would take to get Beauty and X into the same evidential situation. It seems that simply telling Beauty that X is awake would be enough. X knows that he and Beauty are both awake. Beauty knows that she is awake. So that X is awake is enough evidence to put Beauty in the same position.
Premise 3 follows from a form of evidentialism. Rational credence supervenes on evidence. X’s evidence is Beauty’s evidence plus the fact that X is awake. So conditional on X being awake, Beauty’s credences should be just like X’s. And the first two premises (which are, note, the antecedents of premise 3) imply that X should have credence ½ in Heads. So, if everything is working so far, should Beauty.
Premise 4 follows immediately from the setup of the problem. The only way Beauty can be awake while X is asleep is if Tails.
Premise 5 is an instance of a regularity principle. Regularity principles say we should always give positive probability to epistemic possibilities. They are in general a little strong. In cases where there are infinitely many epistemic possibilities, we should give probability 0 to many of them. But in this case, where in some good sense there are only 3 possibilities of the same kind as the one being considered (the possibilities being Heads, Tails and X wakes on today, Tails and X wakes on the other day Beauty is awake), having 0 credence in any of them that is a live possibility seems a little close-minded.
So I think this argument is sound. Of course, one could argue that the presence of X changes things. Perhaps Beauty should have credence ½ in Heads in the original example, but not in this one. But I don’t see how that could be motivated. Beauty doesn’t find out anything about what happens to X on Monday or Tuesday. Her experiences are just like they would be if X didn’t exist, and she knows this at the start. So this looks like a good argument against having credence ½ in Heads in the original puzzle.
Posted by Brian Weatherson in Uncategorized