I agree that we need to remove any dependence on the indexical “today,” but what you propose doesn’t do that. Determining whether “today counts” still depends on it. But there is a way to unequivocally remove this dependence. Use four volunteers, and wake each either once or twice as in the original Sleeping Beauty Problem (OSBP). But change the day and/or coin result that determines the circumstances where each is left asleep.
So one volunteer (call her SB1) will be left asleep on Tuesday, if Heads is flipped, as in the OSBP. Another (SB2) will be left asleep on Monday, also if Heads is flipped. This is essentially the same as the OSBP, since in a non-indexical version the day can’t matter. The other two (SB3 and SB4) will be left asleep if Tails is flipped, one on Monday and one on Tuesday. And if we ask them about their confidence in Tails instead of Heads, the correct answer should be the same as the OSBP, whatever that turns out to be.
The “de-indexicalization” is accomplished by changing the question to an equivalent one. For SB1 and SB2, the truth value of the statement “I am awake now, and it is my only awakening” is the always the same as “Heads was flipped.” For SB3 and SB4, it is the same as “Tails was flipped.”
Note that it can’t matter if you know which of these volunteers you are, or if you are allowed to discuss the question “Is this my only awakening?” as long as you can’t reveal which one you are to the others. One each day of the experiment, exactly three will be awake. For exactly one of those three, it will be her only awakening.
“Determining whether “today counts” still depends on it”—No, you just ask about the (second) coin which determines what day counts and whether it shows heads or tails (for consistency assume that we flip a heads-only coin if the first coin comes up heads). So questions becomes, “What is the chance of the (first) coin being heads given Sleeping Beauty’s non-indexical state of knowledge on Monday if the second coin is heads or Sleeping Beauty’s non-indexical state of knowledge on Tuesday if the second coin is tails?”
If an interview on one day “counts,” while an interview on another day doesn’t, you are using an indexical to discriminate those days. Adding another coin to help pick which day does not count is just obfuscating how you indexed it.
This is why betting (or frequency) arguments will never work. Essentially, the number of bets (or the number of trials in the frequency experiment) is dependent on the answer, so the argument is circular. If you decide ahead of time that you want to get 1⁄3, you will use three bets (or “trials”) that each have a 1⁄2 *prior* probability of happening to Beauty on a single *indexed* day in the experiment. If you want to get 1⁄2, you use two. So, that you get 1⁄2 by your method is not surprising in the slightest. It was pre-ordained.
You need to find a way to justify one or the other that is not a non sequitur, and that isn’t possible. You can’t justify why “ensuring that only one interview ever ‘counts’” solves an issue in the debate. You never tried, you just asserted that it was the thing to do.
Wait for my next post on this topic. Unfortunately, I chose a narrow scope for this post (only explaining the halfer response to the specific Beauty and the Prince objection, not justifying this approach in general) and everyone is posting objections that would require a whole post to answer. But basically, I will argue that there are valid reasons for adopting this formalism that aren’t merely trivial.
And in my reply I will show how you are addressing the conclusion you want to reach, and not the problem itself. No matter how you convolute choosing the sample point you ignore, you will still be ignoring one. All you will be doing, is creating a complicated algorithm for picking a day that “doesn’t count,” and it will be probabilisticly equivalent to saying “Tuesday doesn’t count” (since you already ignore Tue-H). That isn’t the Sleeping Beauty Problem.
But you haven’t responded to my proof, which actually does eliminate the indexing issue. Its answer is unequivocally 1⁄3. I think there is an interesting lesson to be learned from the problem, but it can’t be approached until people stop trying to make the lesson fit the answer they want.
+++++
The cogent difference between halfers and thirders, is between looking at the experiment from the outside, or the inside.
From the outside, most halfers consider Beauty’s awakenings on Mon-T and Tue-T to be the same outcome. They cannot be separated from each other. The justification for this outlook is that, over the course of the experiment, one necessitates the other. The answer from this viewpoint is clearly 1⁄2.
But it has an obvious flaw. If the plan is to tell Beauty what day it is after she answers, that can’t affect her answer but it clearly invalidates the viewpoint. The sample space that considers Mon-T and Tue-T to be the same outcome is inadequate to describe Beauty’s situation after she is told that it is Monday, so it can’t be adequate before. You want to get around this by saying that one interview “doesn’t count.” In my four-volunteer proof, this is equivalent to saying that one of the three awake volunteers “doesn’t count.” Try to convince her of that. Or, ironically, ask her for her confidence that her confidence “doesn’t count.”
But a sample space that includes Tue-T must also include Tue-H. The fact that Beauty sleeps though it does not make it “unhappen,” which is what halfers (and even some thirders) seem to think.
To illustrate, let me propose a slight change to the drugs we assume are being used. Drug A is the “go to sleep” drug, but it lasts only about 12 hours and the subject wakes up groggy. So each morning, Beauty must be administered either drug B that wakes her up and overrides the grogginess, or another dose of drug A. The only point of this change, since it cannot affect Beauty’s thought processes, is to make Tue-H a more concrete outcome.
What Beauty sees, from the inside, is a one-day experiment. Not a two-day one. At the start of this one-day experiment, there was a 3⁄4 chance that drug B was chosen, and a 1⁄4 chance that it was drug A. Beauty’s evidence is that it was drug B, and there is a 1⁄3 chance of Heads, given drug B.
I agree that we need to remove any dependence on the indexical “today,” but what you propose doesn’t do that. Determining whether “today counts” still depends on it. But there is a way to unequivocally remove this dependence. Use four volunteers, and wake each either once or twice as in the original Sleeping Beauty Problem (OSBP). But change the day and/or coin result that determines the circumstances where each is left asleep.
So one volunteer (call her SB1) will be left asleep on Tuesday, if Heads is flipped, as in the OSBP. Another (SB2) will be left asleep on Monday, also if Heads is flipped. This is essentially the same as the OSBP, since in a non-indexical version the day can’t matter. The other two (SB3 and SB4) will be left asleep if Tails is flipped, one on Monday and one on Tuesday. And if we ask them about their confidence in Tails instead of Heads, the correct answer should be the same as the OSBP, whatever that turns out to be.
The “de-indexicalization” is accomplished by changing the question to an equivalent one. For SB1 and SB2, the truth value of the statement “I am awake now, and it is my only awakening” is the always the same as “Heads was flipped.” For SB3 and SB4, it is the same as “Tails was flipped.”
Note that it can’t matter if you know which of these volunteers you are, or if you are allowed to discuss the question “Is this my only awakening?” as long as you can’t reveal which one you are to the others. One each day of the experiment, exactly three will be awake. For exactly one of those three, it will be her only awakening.
The non-indexical answer is 1⁄3.
“Determining whether “today counts” still depends on it”—No, you just ask about the (second) coin which determines what day counts and whether it shows heads or tails (for consistency assume that we flip a heads-only coin if the first coin comes up heads). So questions becomes, “What is the chance of the (first) coin being heads given Sleeping Beauty’s non-indexical state of knowledge on Monday if the second coin is heads or Sleeping Beauty’s non-indexical state of knowledge on Tuesday if the second coin is tails?”
If an interview on one day “counts,” while an interview on another day doesn’t, you are using an indexical to discriminate those days. Adding another coin to help pick which day does not count is just obfuscating how you indexed it.
This is why betting (or frequency) arguments will never work. Essentially, the number of bets (or the number of trials in the frequency experiment) is dependent on the answer, so the argument is circular. If you decide ahead of time that you want to get 1⁄3, you will use three bets (or “trials”) that each have a 1⁄2 *prior* probability of happening to Beauty on a single *indexed* day in the experiment. If you want to get 1⁄2, you use two. So, that you get 1⁄2 by your method is not surprising in the slightest. It was pre-ordained.
You need to find a way to justify one or the other that is not a non sequitur, and that isn’t possible. You can’t justify why “ensuring that only one interview ever ‘counts’” solves an issue in the debate. You never tried, you just asserted that it was the thing to do.
Wait for my next post on this topic. Unfortunately, I chose a narrow scope for this post (only explaining the halfer response to the specific Beauty and the Prince objection, not justifying this approach in general) and everyone is posting objections that would require a whole post to answer. But basically, I will argue that there are valid reasons for adopting this formalism that aren’t merely trivial.
And in my reply I will show how you are addressing the conclusion you want to reach, and not the problem itself. No matter how you convolute choosing the sample point you ignore, you will still be ignoring one. All you will be doing, is creating a complicated algorithm for picking a day that “doesn’t count,” and it will be probabilisticly equivalent to saying “Tuesday doesn’t count” (since you already ignore Tue-H). That isn’t the Sleeping Beauty Problem.
But you haven’t responded to my proof, which actually does eliminate the indexing issue. Its answer is unequivocally 1⁄3. I think there is an interesting lesson to be learned from the problem, but it can’t be approached until people stop trying to make the lesson fit the answer they want.
+++++
The cogent difference between halfers and thirders, is between looking at the experiment from the outside, or the inside.
From the outside, most halfers consider Beauty’s awakenings on Mon-T and Tue-T to be the same outcome. They cannot be separated from each other. The justification for this outlook is that, over the course of the experiment, one necessitates the other. The answer from this viewpoint is clearly 1⁄2.
But it has an obvious flaw. If the plan is to tell Beauty what day it is after she answers, that can’t affect her answer but it clearly invalidates the viewpoint. The sample space that considers Mon-T and Tue-T to be the same outcome is inadequate to describe Beauty’s situation after she is told that it is Monday, so it can’t be adequate before. You want to get around this by saying that one interview “doesn’t count.” In my four-volunteer proof, this is equivalent to saying that one of the three awake volunteers “doesn’t count.” Try to convince her of that. Or, ironically, ask her for her confidence that her confidence “doesn’t count.”
But a sample space that includes Tue-T must also include Tue-H. The fact that Beauty sleeps though it does not make it “unhappen,” which is what halfers (and even some thirders) seem to think.
To illustrate, let me propose a slight change to the drugs we assume are being used. Drug A is the “go to sleep” drug, but it lasts only about 12 hours and the subject wakes up groggy. So each morning, Beauty must be administered either drug B that wakes her up and overrides the grogginess, or another dose of drug A. The only point of this change, since it cannot affect Beauty’s thought processes, is to make Tue-H a more concrete outcome.
What Beauty sees, from the inside, is a one-day experiment. Not a two-day one. At the start of this one-day experiment, there was a 3⁄4 chance that drug B was chosen, and a 1⁄4 chance that it was drug A. Beauty’s evidence is that it was drug B, and there is a 1⁄3 chance of Heads, given drug B.