Sleeping Beauty is testing a hypothesis

Let's look at the third type of case in which credences can come apart from known chances. Consider the following variation of the Sleeping Beauty problem (a.k.a. "The Absentminded Driver"):

Before Sleeping Beauty awakens on Monday, a coin is tossed. If the coin lands tails, Beauty's memories of Monday will be erased the following night, and the coin will be tossed again on Tuesday. If the Monday toss lands heads, no memory erasure or further tosses take place. Beauty is aware of all these facts.

When Beauty awakens on Monday morning and learns that today's toss has landed tails (alternatively: that the Monday toss has landed tails), how should that affect her credence in the hypothesis that the coin is fair?

Let Tails be the hypothesis Beauty would express by "today's toss landed tails"; let Tails-M be "the Monday toss landed tails". Given Beauty's background knowledge, Tails entails Tails-M: if today's toss landed tails then the Monday toss must also have landed tails, for either today's toss is the Monday toss, or today's toss is the Tuesday toss in which case the Monday toss must have landed tails.

Moreover, let's assume that Beauty gives positive credence to the hypothesis that it is Tuesday, and to the hypothesis that the Tuesday toss lands heads (Heads-T). So Cr(Tails-M & Tuesday & Heads-T) > 0, which implies that Cr(Tails-M & ~Tails) > 0. Since Tails entails Tails-M, it follows that Cr(Tails) > Cr(Tails-M).

Hence we get what looks like a counterexample to the Principal Principle. Even if Beauty knows that the coin is fair, either her credence in tails on today's toss or in tails on the Monday toss will not equal what she knows to be the objective chance.

Which of the two will deviate from the known chance depends on whether Beauty obeys halfing or thirding. Let H be the hypothesis that the objective probability ('chance') of tails on any toss is 1/2. If Beauty is a halfer, then Cr(Tails-M / H) = 1/2 but Cr(Tails / H) = 3/8. If Beauty is a thirder, then Cr(Tails / H) = 1/2, but Cr(Tails-M / H) = 2/3. See section 5 of my paper on the absentminded driver for the calculations.

Now assume Beauty doesn't know the bias of her coin, but gives equal credence to a variety of hypotheses about the coin's bias. How should the information that (say) today's toss landed tails affect her credence in the hypothesis that the coin is fair?

If Beauty obeys halfing, then on the supposition that the coin is fair she ought to regard Heads as more likely than Tails (5/8 vs. 3/8), and so she ought to regard Tails as counting more strongly against the assumption that the coin is fair than Heads.

At first sight, that seems strange, but I'm inclined to say that it is correct. After all, the information that today's toss landed tails is relevant to the outcome of two coin tosses: first, it entails that the Monday toss landed tails, and second, it raises the probability that the Tuesday toss landed tails. Imagine someone told you about a series of coin tosses: "the first toss definitely landed tails, and the second probably did so too". Surely that lends some support to the hypothesis that the coin is biased towards tails. By contrast, the information that today's toss landed heads would entail that either there is only one toss overall (on Monday) which landed heads, or there are two tosses the first of which landed tails and the second heads. So the information is "the series started wither with heads, or with tails followed by heads". That also lends some support to the hypothesis that the coin is biased (towards heads), but less so than in the previous case.

Thirders should object that we have implicitly assumed that if the coin is fair, then there's an equal probability of getting heads and tails on the first toss. That's true according to halfers, but not according to thirders: for thirders, Cr(Tails-M / H) = 2/3. The information that today's toss landed tails still contains information about two tosses: it entails that the Monday toss landed tails and it raises the probability that the Tuesday toss landed tails. However, on the hypothesis that the coin is fair, the first piece of information is not all that surprising, since the prior probability of Tails-M was already 2/3. As it turns out, this precisely cancels the fact that Tails contains information about two tosses, so that on the thirder account, H makes Heads and Tails equally probable.

What if Beauty learns not that today's toss landed tails, but that the Monday toss landed tails? For halfers, the subjective likelihood of this outcome given H is 1/2. Not so for thirders, who claim that Cr(Tails-M / H) = 2/3. That also seems strange. How could the information that a certain coin landed tails support the hypothesis that the coin is fair more strongly than the information that the coin landed heads? Thirders will answer that this is because there are more opportunities to learn that the Monday coin landed tails (if it did) than to learn that it landed heads (if it did).

In any case, the upshot is that both halfers and thirders should agree that in cases like Sleeping Beauty, subjective likelihoods Cr(E/H) don't always equal the objective probability assigned to E by H. Subjective Bayesian Confirmation Theorists should say that what matters to confirmation then are -- of course -- the subjective likelihoods, not the objective probabilities.

What if we aren't Subjective Bayesian Confirmation Theorists? Could Objective Bayesians or likelihoodists maintain that the likelihoods that matter are the probabilities assigned to the evidence by the hypotheses? Arguably not. As we've seen, provided that Cr(Tuesday & Tails) > 0, it is impossible for Cr(Tails / H) to equal Cr(Tails-M / H), even though presumably H assigns a probability of 1/2 to both "today's toss landed tails" (Tails) and "the Monday toss landed tails" (Tails-M). That is still true if we read Cr not as degree of belief but as objective degree of evidential support. (I think the best response here is to deny that physical theories about the probability of outcomes in coin tosses directly assign probabilities to propositions such as Tails and Tails-M.)

One more thing. From the above discussion, one might infer that halfers and thirders disagree on questions about evidential support: halfers believe that Heads more strongly confirms that the coin is fair than Tails; thirders disagree. Thirders believe that Tails-M more strongly confirms that the coin is fair than Heads-M; halfers disagree. There is obviously such a disagreement if we assume Subjective Bayesian Confirmation Theory: if degree of support is cashed out in terms of (rational) degree of belief, then disagreement about the latter obviously gives rise to disagreement about the former. But I have some sympathies for Objective BCT, and I believe that thirding even presupposes a notion of evidential support that is distinct from subjective degree of belief. So let's assume some such notion. Should halfers and thirders disagree about it? Should they disagree on whether, say, Tails objectively confirms H more strongly than Heads? I think they shouldn't. I'm a halfer about degrees of belief, but a thirder about objective support.

Comments

# on 30 December 2015, 17:15

There is an easy way to answer the SB problem, without having to answer the more philosophical questions above. Use four volunteers, one coin, the same drugs, and the same two days.

SB1 is wakened both days, unless it is Monday after Heads was flipped. SB2 is wakened both days, unless it is Monday after Tails was flipped. SB3 is wakened both days, unless it is Tuesday after Heads was flipped. SB2 is wakened both days, unless it is Tuesday after Tails was flipped. They are kept in separate rooms, and each is asked for her assessment of the probability that she will be wakened exactly once during the experiment.

A simple examination shows that SB3's situation is the same as our original volunteer's. Her question looks different, but evaluates to the same thing. The other three are in functionally identical situations, with a functionally equivalent question.

Whenever any of these women is awake, she knows that (A) She is one of three women who are awake, and (B) Exactly one of this set three will be wakened exactly once during the experiment. The chances that it is herself can only be 1 in 3.

# on 30 December 2015, 21:32

@JeffJo: I'm generally skeptical about such appeals to different but allegedly analogous scenarios. One really needs to show, rather than just claim, that the differences make no difference. In any case, you also presuppose that the probability Beauty should assign to heads is determined by her evidence upon awakening (by what she knows). As I've assumed in this blog post and argued e.g. in the paper linked to above, I believe this assumption really does support thirding but I also believe that the assumption is problematic.

# on 01 April 2016, 20:35

@wo:

Original SB problem: She is wakened on Monday, and again on Tuesday if the coin flips Tails. She is asked for her credence in the event "Heads."

SB3 in this version: She is wakened on Monday, and again on Tuesday if the coin flips Tails. She is asked for her credence in an event which happens if, and only if, the event "Heads" happens. That is, it represents a different description of the same event.

In what way is this not a proof that they represent the same scenario?

Comparing SB1 thru SB4: Each will be wakened on a day we can call her "required day," and also on a "optional day" if the event "flipped coin matches assigned coin" happens. Since none of them know what the day is, or what the flip is, the problem is identical for each of them.

In what way is this not a proof that they are all solving the same problem, with the same evidence?

Solving SB1 thru SB4: Each awake volunteer has the evidence "I am one of three who are awake." All three know that only one of them corresponds to {required day, mismatch}. Since each has the same evidence that she is the one, how can the credence be other than 1/3?

And yes, I do "presuppose" that if SB has evidence that pertains to the question, she should provide a conditional probability based on that evidence. I don't understand how you could suggest anything else.

Halfers and Thirders both do this - but halfers claim the evidence is null. The whole point of the analogy was to change the experiment from one where the difference between actual evidence and null evidence seems ambiguous, to one where it clearly isn't.

# on 01 April 2016, 21:34

@JeffJo: In your example, SB has the evidence "I am one of three who are awake", which seems to be important for your argument. In the original scenario, SB doesn't have that evidence. So it's not entirely trivial that whatever holds of your scenario carries over to the original.

But as I said, I'm happy to grant the analogy and your argument: I agree that Beauty's evidence supports "Heads" to degree 1/3. I still think her degree of belief should be 1/2 -- essentially by the standard halfer argument that her degree of belief should have been 1/2 on Sunday and that she gained no surprising new evidence on Monday.

# on 02 April 2016, 14:44

You are one of six volunteers who are dealt the numbers "1" thru "6." Whether you are told your numbers is unimportant - just that you know there are six in your group, and each member has a different number.

A die is rolled, but not shown to you. Each of you is asked for your credence that your number matches the die roll. What is the important information here - that you are one of six in this group, or that your assignment is one of six possibilities?

That is, would there be anything different TO YOU about this experiment, if you were the only volunteer and were assigned a random number? Does it matter if you know that number?

What you identified is not "evidence," it is an exposition of a sample space for the experiment using different people to represent each element of that space. Literally, the three volunteers who are awake represent the three elements of the sample space that could agree with the evidence that you are awake.

The same conclusions apply to my SB variation, that applied to the dice experiment. One volunteer is assigned a random day, and a random coin face. Whether she knows these assignments, or doesn't, or is only told after her initial answer, is irrelevant. Her current situation corresponds to one of four that initially were equally likely, but one is eliminated. She is asked for her credence that is is one of the three that remain possible.

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