You mis-characterize what Elga does. He never directly formulates the state M1, where Beauty is awake. Instead, he formulates two states that are derived from information being added to M1. I’ll call them M2A (Beauty learns the outcome is Tails) and M2B (Beauty learns that it is Monday). While he may not do it as formally as you want, he works backwards to show that three of the four components of a proper description of state M1 must have the same probability. What he skips over, is identifying the fourth component (whose probability is now zero).
What it seems Elga was trying to avoid—as everybody does—is that Beauty still “exists” on Tuesday, after Heads. She just can’t observe it. But it is a component you need to consider in your more formal modeling. To illustrate, here’s a simple re-structuring of your steps that changes nothing relevant to the question she is asked:
On Sunday the steps of the experiment are explained to Beauty, and she is put to sleep.
On Monday Beauty is awakened. She has no information that would help her infer the day of the week. Later in the day she is interviewed. Afterwards, she is administered a drug that resets her memory to its state when she was put to sleep on Sunday, and puts her to sleep again.
On Tuesday Beauty is awakened. She has no information that would help her infer the day of the week. The experimenters flip a fair coin. If it lands Tails, Beauty is interviewed again; if it lands Heads, she is not. In either case, she is then administered a drug that resets her memory to its state when she was put to sleep on Sunday, and puts her to sleep again.
On Wednesday Beauty is awakened once more and told that the experiment is over.
In the interview(s), Beauty is asked to give a probability for her belief that the coin in step 3 lands Heads.
I’m sure you can make this more formal, so I’ll be brief: State M, on Sunday, requires only proposition C describing what Beauty thinks the coin result is (*not* for what it actually is, which becomes deterministic at different times in different versions of the problem). There is no information in state M that favors either result, so the Principle of Indifference applies and the probability for each is 1⁄2.
State M1, when Beauty is first awakened, requires another proposition: D, for what day Beauty thinks it is (*not* for what day it actually is, which is deterministic). Due to the memory-reset drug, the same state M1 applies on both Monday and Tuesday. Since there is no information in state M1 that favors either result, the Principle of Indifference applies and the probability for each is 1⁄2. And (what seems to be overlooked by denying the existence of Tuesday when Beauty sleeps through it) D and H are independent. So M1 comprises four possible combinations of D and H that all have a probability of 1⁄4.
State M2 applies when Beauty is interviewed. The information that takes Beauty from M1 to M2 is that one of the four combinations is ruled out. The remaining three now have probability 1⁄3.
State M1 applies to your version of the problem at the point in time just before Beauty could be wakened, in either step 2 or step 3. It applies, and can be determined later when Beauty is awake, whether or not Beauty is awake at that time. Elga’s solution is essentially the same as mine, except he does it in two parts by adding more information to each. It just avoids identifying the component of the state that Beauty sleeps through.
Your whole analysis rests on the idea that “it is Monday” is a legitimate proposition. I’ve responded to this many other places in the comments, so I’ll just say here that a legitimate proposition needs to maintain the same truth value throughout the entire analysis (Sunday, Monday, Tuesday, and Wednesday). Otherwise it’s a predicate. The point of introducing R(y,d) is that it’s as close as we can get to what you want “it is Monday” to mean.
Well, I never checked back to see replies, and just tripped back across this.
The error made by halfers is in thinking “the entire analysis” spans four days. Beauty is asked for her assessment, based on her current state of knowledge, that the coin landed Heads. In this state of knowledge, the truth value of the proposition “it is Monday” does not change.
But there is another easy way to find the answer, that satisfies your criterion. Use four Beauties to create an isomorphic problem. Each will be told all of the details on Sunday; that each will be wakened at least once, and maybe twice, over the next two days based on the same coin flip and the day. But only three will be wakened on each day. Each is assigned a different combination of a coin face, and a day, for the circumstances where she will not be wakened. That is, {H,Mon}, {T,Mon}, {H,Tue}, and {T,Tue}.
On each of the two days during the experiment, each awake Beauty is asked for the probability that she will be wakened only once. Note that the truth value of this proposition is the same throughout the experiment. It is only the information a Beauty has that changes. On Sunday or Wednesday, there is no additional information and the answer is 1⁄2. On Monday or Tuesday, an awake Beauty knows that there are three awake Beauties, that the proposition is true for exactly one of them, and that there is no reason for any individual Beauty to be more, or less, likely than the others to be that one. The answer with this knowledge is 1⁄3.
You mis-characterize what Elga does. He never directly formulates the state M1, where Beauty is awake. Instead, he formulates two states that are derived from information being added to M1. I’ll call them M2A (Beauty learns the outcome is Tails) and M2B (Beauty learns that it is Monday). While he may not do it as formally as you want, he works backwards to show that three of the four components of a proper description of state M1 must have the same probability. What he skips over, is identifying the fourth component (whose probability is now zero).
What it seems Elga was trying to avoid—as everybody does—is that Beauty still “exists” on Tuesday, after Heads. She just can’t observe it. But it is a component you need to consider in your more formal modeling. To illustrate, here’s a simple re-structuring of your steps that changes nothing relevant to the question she is asked:
On Sunday the steps of the experiment are explained to Beauty, and she is put to sleep.
On Monday Beauty is awakened. She has no information that would help her infer the day of the week. Later in the day she is interviewed. Afterwards, she is administered a drug that resets her memory to its state when she was put to sleep on Sunday, and puts her to sleep again.
On Tuesday Beauty is awakened. She has no information that would help her infer the day of the week. The experimenters flip a fair coin. If it lands Tails, Beauty is interviewed again; if it lands Heads, she is not. In either case, she is then administered a drug that resets her memory to its state when she was put to sleep on Sunday, and puts her to sleep again.
On Wednesday Beauty is awakened once more and told that the experiment is over.
In the interview(s), Beauty is asked to give a probability for her belief that the coin in step 3 lands Heads.
I’m sure you can make this more formal, so I’ll be brief: State M, on Sunday, requires only proposition C describing what Beauty thinks the coin result is (*not* for what it actually is, which becomes deterministic at different times in different versions of the problem). There is no information in state M that favors either result, so the Principle of Indifference applies and the probability for each is 1⁄2.
State M1, when Beauty is first awakened, requires another proposition: D, for what day Beauty thinks it is (*not* for what day it actually is, which is deterministic). Due to the memory-reset drug, the same state M1 applies on both Monday and Tuesday. Since there is no information in state M1 that favors either result, the Principle of Indifference applies and the probability for each is 1⁄2. And (what seems to be overlooked by denying the existence of Tuesday when Beauty sleeps through it) D and H are independent. So M1 comprises four possible combinations of D and H that all have a probability of 1⁄4.
State M2 applies when Beauty is interviewed. The information that takes Beauty from M1 to M2 is that one of the four combinations is ruled out. The remaining three now have probability 1⁄3.
State M1 applies to your version of the problem at the point in time just before Beauty could be wakened, in either step 2 or step 3. It applies, and can be determined later when Beauty is awake, whether or not Beauty is awake at that time. Elga’s solution is essentially the same as mine, except he does it in two parts by adding more information to each. It just avoids identifying the component of the state that Beauty sleeps through.
Your whole analysis rests on the idea that “it is Monday” is a legitimate proposition. I’ve responded to this many other places in the comments, so I’ll just say here that a legitimate proposition needs to maintain the same truth value throughout the entire analysis (Sunday, Monday, Tuesday, and Wednesday). Otherwise it’s a predicate. The point of introducing R(y,d) is that it’s as close as we can get to what you want “it is Monday” to mean.
Well, I never checked back to see replies, and just tripped back across this.
The error made by halfers is in thinking “the entire analysis” spans four days. Beauty is asked for her assessment, based on her current state of knowledge, that the coin landed Heads. In this state of knowledge, the truth value of the proposition “it is Monday” does not change.
But there is another easy way to find the answer, that satisfies your criterion. Use four Beauties to create an isomorphic problem. Each will be told all of the details on Sunday; that each will be wakened at least once, and maybe twice, over the next two days based on the same coin flip and the day. But only three will be wakened on each day. Each is assigned a different combination of a coin face, and a day, for the circumstances where she will not be wakened. That is, {H,Mon}, {T,Mon}, {H,Tue}, and {T,Tue}.
On each of the two days during the experiment, each awake Beauty is asked for the probability that she will be wakened only once. Note that the truth value of this proposition is the same throughout the experiment. It is only the information a Beauty has that changes. On Sunday or Wednesday, there is no additional information and the answer is 1⁄2. On Monday or Tuesday, an awake Beauty knows that there are three awake Beauties, that the proposition is true for exactly one of them, and that there is no reason for any individual Beauty to be more, or less, likely than the others to be that one. The answer with this knowledge is 1⁄3.