You hit one point on the head: “it’s really just a math problem.” But what that means is that it is not a philosophy problem. It was originally posed as a problem in philosophy, but the only thing such considerations do is obfuscate the math problem. Without justification. Just the need to put it in the realm of philosophy to match the original intent.
Consider a variation of the problem, that isn’t a variation at all, but I’m sure will get a response as to why it doesn’t fit in with a desired philosophical solution. I know this, because it has happened before. Use the same details with four volunteers, but one coin, and different schedules. One volunteer will be left asleep on Tuesday, after Heads, as in the popular[1] version. Each of the other three will be left asleep on a different one of the other three possible combinations of the day (Monday or Tuesday) and the coin result (Heads or Tails).
So on each day in the experiment, three of the volunteers will awakened. Each will be asked, individually, for the probability[2] that this is her only waking[3]. Then, the three will be brought together and asked to discuss the same question. The only taboo topic is the combinations where each would be left asleep.
In the individual interview, each is solving an problem that is identical to the popular version. In the combined one, each knows that she has a 1⁄3 chance to have the same assigned coin result as the missing volunteer. The only possible issue, is why should the answer change between these interviews.
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[1] It isn’t the original, or even the version posed in the paper that created the popular version.
[2] “Odds” means a ratio, not a probability. Odds of X:Y mean a probability of X/(X+Y). Halfers say the odds are 1:1, or a probability of 1⁄2. Thirders say the odds are 1:2, or a probability or 1⁄3.
[3] This is the question as it was in the original version. It is equivalent to “What is the probability that the coin result means you will be wakened only once?”
Here is a similar problem based on the same Mathematics. I call it Camp Sleeping Beauty. The same sleep and Amnesia details will be used, but at a week-long camp. On the arrival day, Sunday, you will be told all these details.
There are six “camp” activities, and you will partake of a random one on each day: Archery, Boating, Canoeing, Dodge Ball, Exercising, and Free Time.
The random part is that, after you are put to sleep on Sunday, a six-sided die will be rolled. But before you are put to sleep, you will help to randomly populate a six-by-six Camp Calendar with these activities. The board has one column for each of the days, Monday through Saturday, and one row for each die roll.
If you think it necessary, assume the non-random assignment of Monday&1=A, Tuesday&2=B, etc., so that every activity is guaranteed to happen.
When you awake each day, you will not be told the day, but you will be taken to that day’s assigned activity using Sunday’s die roll.
At lunch, you will be interviewed and asked to assign probabilities to each of the six die rolls, based on the knowledge you gained from the activity. You will have access to the calendar you helped to make.
Like I said, this is a Math problem. The probability for each die roll is the number of times today’s activity appears in that row, divided by the number of times it appears on the board. That is, if the activity appears only once on the board—or if every time it appears it is in the same row—there is a 100% chance that that was the die roll. If it appears twice on different rows, there is a 1⁄2 chance for each of those two rolls. Etc.
The important part here is that the arrangement of the other activities, or the actual activities, is completely irrelevant to this answer. “Today” is not an enigmatic value that you need a “universe model” to access, it is a random sampling of this six-by-six array.
Now, change the “Exercise” activity to “Extra sleep,” where you get to sleep all day. That is, no interview. Nothing changes on the days where you do wake up. The answers do not change to the “Halfer” solution of 1⁄6 for each just because you know you would not wake up those days. The information you have about them is exactly the same—they are inconsistent with what happened today.
The popular Sleeping Beauty Problem is just a two-by-two version of this, with three of the day’s activities being “wake up” and one “sleep.” The answer, when you do wake up, is the number of times “wake” appears in the Heads row (one) divided by the number of times it appears on the board (three).
You hit one point on the head: “it’s really just a math problem.” But what that means is that it is not a philosophy problem. It was originally posed as a problem in philosophy, but the only thing such considerations do is obfuscate the math problem. Without justification. Just the need to put it in the realm of philosophy to match the original intent.
Consider a variation of the problem, that isn’t a variation at all, but I’m sure will get a response as to why it doesn’t fit in with a desired philosophical solution. I know this, because it has happened before. Use the same details with four volunteers, but one coin, and different schedules. One volunteer will be left asleep on Tuesday, after Heads, as in the popular[1] version. Each of the other three will be left asleep on a different one of the other three possible combinations of the day (Monday or Tuesday) and the coin result (Heads or Tails).
So on each day in the experiment, three of the volunteers will awakened. Each will be asked, individually, for the probability[2] that this is her only waking[3]. Then, the three will be brought together and asked to discuss the same question. The only taboo topic is the combinations where each would be left asleep.
In the individual interview, each is solving an problem that is identical to the popular version. In the combined one, each knows that she has a 1⁄3 chance to have the same assigned coin result as the missing volunteer. The only possible issue, is why should the answer change between these interviews.
--
[1] It isn’t the original, or even the version posed in the paper that created the popular version.
[2] “Odds” means a ratio, not a probability. Odds of X:Y mean a probability of X/(X+Y). Halfers say the odds are 1:1, or a probability of 1⁄2. Thirders say the odds are 1:2, or a probability or 1⁄3.
[3] This is the question as it was in the original version. It is equivalent to “What is the probability that the coin result means you will be wakened only once?”
Here is a similar problem based on the same Mathematics. I call it Camp Sleeping Beauty. The same sleep and Amnesia details will be used, but at a week-long camp. On the arrival day, Sunday, you will be told all these details.
There are six “camp” activities, and you will partake of a random one on each day: Archery, Boating, Canoeing, Dodge Ball, Exercising, and Free Time.
The random part is that, after you are put to sleep on Sunday, a six-sided die will be rolled. But before you are put to sleep, you will help to randomly populate a six-by-six Camp Calendar with these activities. The board has one column for each of the days, Monday through Saturday, and one row for each die roll.
If you think it necessary, assume the non-random assignment of Monday&1=A, Tuesday&2=B, etc., so that every activity is guaranteed to happen.
When you awake each day, you will not be told the day, but you will be taken to that day’s assigned activity using Sunday’s die roll.
At lunch, you will be interviewed and asked to assign probabilities to each of the six die rolls, based on the knowledge you gained from the activity. You will have access to the calendar you helped to make.
Like I said, this is a Math problem. The probability for each die roll is the number of times today’s activity appears in that row, divided by the number of times it appears on the board. That is, if the activity appears only once on the board—or if every time it appears it is in the same row—there is a 100% chance that that was the die roll. If it appears twice on different rows, there is a 1⁄2 chance for each of those two rolls. Etc.
The important part here is that the arrangement of the other activities, or the actual activities, is completely irrelevant to this answer. “Today” is not an enigmatic value that you need a “universe model” to access, it is a random sampling of this six-by-six array.
Now, change the “Exercise” activity to “Extra sleep,” where you get to sleep all day. That is, no interview. Nothing changes on the days where you do wake up. The answers do not change to the “Halfer” solution of 1⁄6 for each just because you know you would not wake up those days. The information you have about them is exactly the same—they are inconsistent with what happened today.
The popular Sleeping Beauty Problem is just a two-by-two version of this, with three of the day’s activities being “wake up” and one “sleep.” The answer, when you do wake up, is the number of times “wake” appears in the Heads row (one) divided by the number of times it appears on the board (three).