I agree with the author of this article. After having done a lot of research on the Sleeping Beauty Problem as it was the topic of my bachelor's thesis (philosophy), I came to the conclusion that anthropic reasoning is wrong in the Sleeping Beauty Problem. I will explain my argument (shortly) below:
The principle that Elga uses in his first paper to validate his argument for 1/3 is an anthropic principle he calls the Principle of Indifference:
"Equal probabilities should be assigned to any collection of indistinguishable, mutually exclusive and exhaustive events."
The Principle of Indifference is in fact a more restricted version of the Self-Indication Assumption:
"All other things equal, an observer should reason as if they are randomly selected from the set of all possible observers."
Both principles are to be accepted a priori as they can not be attributed to empirical considerations. They are therefore vulnerable to counterarguments...
The counterargument:
Suppose that the original experiment is modified a little:
If the outcome of the coin flip is Heads, they wake Beauty up at exactly 8:00. If the outcome of the first coin flip is Tails, the reasearchers flip another coin. If it lands Heads they wake Beauty at 7:00, if Tails at 9:00. That means that when Beauty wakes up she can be in one of 5 situations:
Heads and Monday 8:00
Tails and Monday 7:00
Tails and Monday 9:00
Tails and Tuesday 7:00
Tails and Tuesday 9:00
Again, these situations are mutually exclusive, indistinguishable and exhaustive. Hence thirders are forced to conclude that P(Heads) = 1/5.
Thirders might object that Beauty's total credence in the Tails-world would still have to equal 2/3, as Beauty is awakened twice as many times in the Tails-world as in the Heads-world. They are then forced to explain why temporal uncertainty regarding an awakening (Monday or Tuesday) is different from temporal uncertainty regarding the time (7:00 or 9:00 o’clock). Both classify as temporal uncertainties within the same possible world, what could possibly set them apart?
An explanation could be that Beauty is only is asked for her credence in Heads during an awakening event, regardless of the time, and that such an event occurs twice in the Tails-world. That is, out of the 4 possible observer-moments in the Tails-world there are only two in which she is interviewed. That means that simply the fact that she is asked the same question twice is reason enough for thirders to distribute their credence, and it is no longer about the number of observer moments. So if she would be asked the same question a million times then her credence in Heads would drop to 1/1000001!
We can magnify the absurdity of this reasoning by imagining a modified version of the Sleeping Beauty Problem in which a coin is tossed that always lands on Tails. Again, she is awakened one million times and given an amnesia-inducing potion after each awakening. Thirder logic would lead to Beauty’s credence in Tails being 1/1000000, as there are one million observer-moments where she is asked for her credence within the only possible world; the Tails-world. To recapitulate: Beauty is certain that she lives in a world where a coin lands Tails, but due to the fact that she knows that she will answer the same question a million times her answer is 1/1000000. This would be tantamount to saying that Mt. Everest is only 1m high when knowing it will be asked 8848 times! It is very hard to see how amnesia could have such an effect on rationality.
Conclusion:
The thirder argument is false. The fact that there are multiple possible observer-moments within a possible world does not justify dividing your credences equally among these observer-moments, as this leads to absurd consequences. The anthropic reasoning exhibited by the Principle of Indifference and the Self-Indication Assumption cannot be applied to the Sleeping Beauty Problem and I seriously doubt if it can be applied to other cases...
When it comes to probability, you should trust probability laws over your intuition. Many people got the Monty Hall problem wrong because their intuition was bad. You can get the solution to that problem using probability laws that you learned in Stats 101 -- it's not a hard problem. Similarly, there has been a lot of debate about the Sleeping Beauty problem. Again, though, that's because people are starting with their intuition instead of letting probability laws lead them to understanding.
The Sleeping Beauty Problem
On Sunday she is given a drug that sends her to sleep. A fair coin is then tossed just once in the course of the experiment to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a second dose of the sleeping drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday, without flipping the coin again. The sleeping drug induces a mild amnesia, so that she cannot remember any previous awakenings during the course of the experiment (if any). During the experiment, she has no access to anything that would give a clue as to the day of the week. However, she knows all the details of the experiment.
Each interview consists of one question, "What is your credence now for the proposition that our coin landed heads?"
Two popular solutions have been proposed: 1/3 and 1/2
The 1/3 solution
From wikipedia:
Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3.
Yes, it's true that only in a third of cases would heads precede her awakening.
Radford Neal (a statistician!) argues that 1/3 is the correct solution.
This [the 1/3] view can be reinforced by supposing that on each awakening Beauty is offered a bet in which she wins 2 dollars if the coin lands Tails and loses 3 dollars if it lands Heads. (We suppose that Beauty knows such a bet will always be offered.) Beauty would not accept this bet if she assigns probability 1/2 to Heads. If she assigns a probability of 1/3 to Heads, however, her expected gain is 2 × (2/3) − 3 × (1/3) = 1/3, so she will accept, and if the experiment is repeated many times, she will come out ahead.
Neal is correct (about the gambling problem).
These two arguments for the 1/3 solution appeal to intuition and make no obvious mathematical errors. So why are they wrong?
Let's first start with probability laws and show why the 1/2 solution is correct. Just like with the Monty Hall problem, once you understand the solution, the wrong answer will no longer appeal to your intuition.
The 1/2 solution
P(Beauty woken up at least once| heads)=P(Beauty woken up at least once | tails)=1. Because of the amnesia, all Beauty knows when she is woken up is that she has woken up at least once. That event had the same probability of occurring under either coin outcome. Thus, P(heads | Beauty woken up at least once)=1/2. You can use Bayes' rule to see this if it's unclear.
Here's another way to look at it:
If it landed heads then Beauty is woken up on Monday with probability 1.
If it landed tails then Beauty is woken up on Monday and Tuesday. From her perspective, these days are indistinguishable. She doesn't know if she was woken up the day before, and she doesn't know if she'll be woken up the next day. Thus, we can view Monday and Tuesday as exchangeable here.
A probability tree can help with the intuition (this is a probability tree corresponding to an arbitrary wake up day):
If Beauty was told the coin came up heads, then she'd know it was Monday. If she was told the coin came up tails, then she'd think there is a 50% chance it's Monday and a 50% chance it's Tuesday. Of course, when Beauty is woken up she is not told the result of the flip, but she can calculate the probability of each.
When she is woken up, she's somewhere on the second set of branches. We have the following joint probabilities: P(heads, Monday)=1/2; P(heads, not Monday)=0; P(tails, Monday)=1/4; P(tails, Tuesday)=1/4; P(tails, not Monday or Tuesday)=0. Thus, P(heads)=1/2.
Where the 1/3 arguments fail
The 1/3 argument says with heads there is 1 interview, with tails there are 2 interviews, and therefore the probability of heads is 1/3. However, the argument would only hold if all 3 interview days were equally likely. That's not the case here. (on a wake up day, heads&Monday is more likely than tails&Monday, for example).
Neal's argument fails because he changed the problem. "on each awakening Beauty is offered a bet in which she wins 2 dollars if the coin lands Tails and loses 3 dollars if it lands Heads." In this scenario, she would make the bet twice if tails came up and once if heads came up. That has nothing to do with probability about the event at a particular awakening. The fact that she should take the bet doesn't imply that heads is less likely. Beauty just knows that she'll win the bet twice if tails landed. We double count for tails.
Imagine I said "if you guess heads and you're wrong nothing will happen, but if you guess tails and you're wrong I'll punch you in the stomach." In that case, you will probably guess heads. That doesn't mean your credence for heads is 1 -- it just means I added a greater penalty to the other option.
Consider changing the problem to something more extreme. Here, we start with heads having probability 0.99 and tails having probability 0.01. If heads comes up we wake Beauty up once. If tails, we wake her up 100 times. Thirder logic would go like this: if we repeated the experiment 1000 times, we'd expect her woken up 990 after heads on Monday, 10 times after tails on Monday (day 1), 10 times after tails on Tues (day 2),...., 10 times after tails on day 100. In other words, ~50% of the cases would heads precede her awakening. So the right answer for her to give is 1/2.
Of course, this would be absurd reasoning. Beauty knows heads has a 99% chance initially. But when she wakes up (which she was guaranteed to do regardless of whether heads or tails came up), she suddenly thinks they're equally likely? What if we made it even more extreme and woke her up even more times on tails?
Implausible consequence of 1/2 solution?
Nick Bostrom presents the Extreme Sleeping Beauty problem:
This is like the original problem, except that here, if the coin falls tails, Beauty will be awakened on a million subsequent days. As before, she will be given an amnesia drug each time she is put to sleep that makes her forget any previous awakenings. When she awakes on Monday, what should be her credence in HEADS?
He argues:
The adherent of the 1/2 view will maintain that Beauty, upon awakening, should retain her credence of 1/2 in HEADS, but also that, upon being informed that it is Monday, she should become extremely confident in HEADS:
P+(HEADS) = 1,000,001/1,000,002
This consequence is itself quite implausible. It is, after all, rather gutsy to have credence 0.999999% in the proposition that an unobserved fair coin will fall heads.
It's correct that, upon awakening on Monday (and not knowing it's Monday), she should retain her credence of 1/2 in heads.
However, if she is informed it's Monday, it's unclear what she conclude. Why was she informed it was Monday? Consider two alternatives.
Disclosure process 1: regardless of the result of the coin toss she will be informed it's Monday on Monday with probability 1
Under disclosure process 1, her credence of heads on Monday is still 1/2.
Disclosure process 2: if heads she'll be woken up and informed that it's Monday. If tails, she'll be woken up on Monday and one million subsequent days, and only be told the specific day on one randomly selected day.
Under disclosure process 2, if she's informed it's Monday, her credence of heads is 1,000,001/1,000,002. However, this is not implausible at all. It's correct. This statement is misleading: "It is, after all, rather gutsy to have credence 0.999999% in the proposition that an unobserved fair coin will fall heads." Beauty isn't predicting what will happen on the flip of a coin, she's predicting what did happen after receiving strong evidence that it's heads.
ETA (5/9/2010 5:38AM)
If we want to replicate the situation 1000 times, we shouldn't end up with 1500 observations. The correct way to replicate the awakening decision is to use the probability tree I included above. You'd end up with expected cell counts of 500, 250, 250, instead of 500, 500, 500.
Suppose at each awakening, we offer Beauty the following wager: she'd lose $1.50 if heads but win $1 if tails. She is asked for a decision on that wager at every awakening, but we only accept her last decision. Thus, if tails we'll accept her Tuesday decision (but won't tell her it's Tuesday). If her credence of heads is 1/3 at each awakening, then she should take the bet. If her credence of heads is 1/2 at each awakening, she shouldn't take the bet. If we repeat the experiment many times, she'd be expected to lose money if she accepts the bet every time.
The problem with the logic that leads to the 1/3 solution is it counts twice under tails, but the question was about her credence at an awakening (interview).
ETA (5/10/2010 10:18PM ET)
Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3.
Another way to look at it: the denominator is not a sum of mutually exclusive events. Typically we use counts to estimate probabilities as follows: the numerator is the number of times the event of interest occurred, and the denominator is the number of times that event could have occurred.
For example, suppose Y can take values 1, 2 or 3 and follows a multinomial distribution with probabilities p1, p2 and p3=1-p1-p2, respectively. If we generate n values of Y, we could estimate p1 by taking the ratio of #{Y=1}/(#{Y=1}+#{Y=2}+#{Y=3}). As n goes to infinity, the ratio will converge to p1. Notice the events in the denominator are mutually exclusive and exhaustive. The denominator is determined by n.
The thirder solution to the Sleeping Beauty problem has as its denominator sums of events that are not mutually exclusive. The denominator is not determined by n. For example, if we repeat it 1000 times, and we get 400 heads, our denominator would be 400+600+600=1600 (even though it was not possible to get 1600 heads!). If we instead got 550 heads, our denominator would be 550+450+450=1450. Our denominator is outcome dependent, where here the outcome is the occurrence of heads. What does this ratio converge to as n goes to infinity? I surely don't know. But I do know it's not the posterior probability of heads.