JeffJo

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Betting arguments - including the "expected value of the lottery ticket" I saw when skimming this - are invalid since it is unclear whether there is exactly one collection opportunity, or the possibility of two. You can always get the answer you prefer by rearranging the problem to the one that gets the answer you want.

But the problem is always stated incorrectly. The original problem, as stated by Adam Elga in the 2000 paper "Self-locating belief and the Sleeping Beauty problem," was:

"Some researchers are going to put you to sleep. During [the experiment], they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are [awakened], to what degree ought you believe that the outcome of the coin toss is Heads?"

Elga introduced the two-day schedule, where SB is always wakened on Monday, and optionally wakened on Tuesday, in order to facilitate his thirder solution. You can argue whether his solution is to the same problem or not. But if it is not, it is the variation problem that is wrong. And it is unnecessary.

First, consider this simplified experiment:

  1. SB is put to sleep. 
  2. Two coins, C1 and C2, are arranged randomly so that the probability of each of the four possible combinations, {HH, HT, TH, TT} has a probability of 1/4.
  3. Observe what the coins are showing:
    1. If either coin is showing Tails:
      1. Wake SB.
      2. Ask her for her degree of belief that coin C1 is showing Heads.
      3. Put SB back to sleep with amnesia.
    2. If both coins are showing heads:
      1. Do something else that is obviously different than option 3a.
      2. Make sure SB is asleep, and can't remember option 3b happening.

Note that if SB is asked the question in option 3a, she knows that the observation was that at least one coin is showing Tails. It does not matter what would happen - or if anything would happen - in 3b. Her answer can only be 1/3.

We can implement the original problem by flipping these two coins for one possible awakening in the original problem, and then turning coin C2 over for another.

Why does she get paid only once, at the end? Why not once for each waking?

This is the problem with all betting arguments. They incorporate an answer to the anthropic question by providing one, or #wakings, payoffs. 

There is a simple way to answer the question without resorting anthropic reasoning. Then you can try to make the anthropic reasoning fit the (correct) answer.

Flip two coins on Sunday Night, a Dime and a Quarter. Lock them in a glass box showing the results. On Monday morning, perform this procedure: Look at the two coins. If either is showing Tails, wake SB, interview her, and put her back to sleep with the amnesia drug. If both are showing Heads, leave her asleep.

On Monday night, open the box, turn the Dime over, and re-lock it. On Tuesday morning, repeat the same procedure described for Monday.

In the interview(s), ask SB for her credence that the Quarter is currently showing Heads. Since the Quarter is never changed, it is always showing the same face that was the result of the flip. SB knows that when the locked box was examined, there are four equally probably possibilities for {Dime,Quarter}. They are HH, HT, TH, and TT. Since SB is now awake, she knows that HH is eliminated as a possibility. Her credence for each of TH, HT, and TT are each 1/3.

The only difference between this version, and the original, is that we don't need to say anything about what day it is.

Imagine an only slightly different problem: You volunteer for an experiment where you may be wakened once, or twice, on Monday and/or Tuesday. The administrators of the experiment will flip a coin to decide whether it will be once, or twice; but they do not tell you what coin result determines that the number of wakings. And they do not tell you whether the day you will be left asleep, if only one waking is to occur, will be Monday or Tuesday. Oh, and you will be given the amnesia drug on Monday, if you are wakened on that day.

Whenever you are awake, you will be asked to assess, from your perspective based on these procedures, the probability that you will be wakened only once during the experiment. Surely (call this assertion A) the answer to this question must be the same as in the original Sleeping Beauty Problem.

But when you are awakened, you are told that you are one of four volunteers undergoing the exact same procedures based on the same coin flip; with one exception. The choices for the coin results, and the days, are different for each of the four volunteers. (Since there are four possible combinations, each is used.) You are not told which choices were made for you, or introduced to the others. Surely (Assertion B) this gives you no information about your own situation, so it cannot affect your answer.

But it does allow you to evaluate your perspective with a bit more clarity. Of the four volunteers, you know that one must still be asleep, and it isn't you. That volunteer will be wakened only once. Of the other three, including you, two will be wakened on both days, and one will be wakened only once. Surely (Assertion C) this means that, from your perspective, the probability that you will be wakened once is 1/3.

Unless you can find a flaw with one of my assertions, this means that the answer to the original problem is also 1/3. There are valid, if unorthodox, mathematical proofs of this. But those who trust their intuition over mathematics want the answer to be 1/2. The so-called "double halfer" approach is just a rationalization for ignoring valid mathematics.

The rationalization: an indexical is an identifier for a value relative to some index. If no way to associate the indexical with actual values is provided, as in "the probability it will rain tomorrow," then the indexical cannot be used to assess your perspective. But if such a means is provided, EVEN AS A PROBABILITY DISTRIBUTION, then it can be assessed. For example, say a different volunteer is told she will be wakened once or twice based on the roll of two standard dice; 1=Monday, 2=Tuesday, etc. If both dice results the same, she will be wakened only once. Upon being awake, she can determine that the probability that it is Tuesday is 1/6. "Tuesday" may ne an indexical, but context is provided to index it.

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.

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.

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.

The problem with the Sleeping Beauty Problem, is that probability can be thought of as a rate: #successes per #trials. But this problem makes #trials a function of #successes, introducing what could be called a feedback loop into this rate calculation, and fracturing our concepts of what the terms mean. All of the analyses I've seen struggle to put these fractured meanings back together, without fully acknowledging that they are broken. MrMind comes closer to acknowledging it than most, when he says "'A fair coin will be tossed,' in this context, will mean different things for different people."

But this fractured terminology can be overcome quite simply. Instead of one volunteer, use four.

Each will go through a similar experience where they will be woken at least once and maybe twice, on Monday and/or Tuesday, depending on the result of the same fair coin flip.

All four will be wakened both days with the following exceptions: SB1 will be left asleep on Monday if Heads is flipped. SB2 will be left asleep on Monday if Tails is flipped. SB3 will be left asleep on Tuesday if Heads is flipped. And SB4 will be left asleep on Tuesday if Tails is flipped. Note that SB3's schedule corresponds to the original version of the problem.

This way, three of the volunteers will be wakened on Monday. Two of those will be wakened on again Tuesday, while the third will be left asleep and be replaced by the one who slept through Monday. And each has the same chance to be wakened just once.

Put the three in a room together, and allow them to discuss anything EXCEPT the coin result and day that they would sleep through. Ask each for their confidence in the assertion that she will be wakened just once during the experiment.

No matter what day it is, or how the coin landed, the assertion will be true for one of the three awake volunteers, and false for the other two. So their confidences should sum to 1. No matter what combination of day and result each was assigned to sleep through, each has the same information upon which to base her confidence. So their confidences should be the same.

The only possible solution is that the confidences should all be 1/3. If, instead, SB3 is just told about the other three volunteers, but never meets them, she can still reason the same way and get the answer 1/3. And since "I, SB3, will be wakened only once" is equivalent to "the fair coin landed Heads," our original volunteer can give the same answer.

these situations are mutually exclusive, indistinguishable and exhaustive.

No, they aren't. "Indistinguishable" in that definition does not mean "can't tell them apart." It means that the cases arise through equivalent processes. That's why the PoI applies to things like dice, whether or not what is printed on each side is visually distinguishable from other sides.

To make your cases equivalent, so that the PoI applies to them, you need to flip the second coin after the first lands on Heads also. But you wake SB at 8:00 regardless of the second coin's result. You now have have six cases that the PoI applies to, counting the "8:00 Monday" case twice, and each has probability 1/6.

Coscott’s original problem is unsolvable by standard means because the expected number of wakings is infinite, so you can’t determine a frequency. That doesn’t mean it is unanswerable – we just need an isomorphism. After informing SB of the procedure and putting her to sleep the first time:

1) Set M=0. 2) Select a number N (I’ll discuss how later). 3) Flip a coin. a. If this (odd-numbered) flip lands heads, wake SB N times and end the experiment. b. If this flip (odd-numbered) lands tails, continue to step 4. 4) Flip the coin again. a. If this (even-numbered) flip lands heads, wake SB 3*N times and end the experiment; b. If this (even-numbered)flip lands tails, set M=M+1 go to step #2.

In Coscott’s version, we start with N=1 and multiply it by 9 each time we choose a new one; that is, N=9^M. But does the answer depend on N in any way? Halfers don’t think the answer depends on the number of wakings at all, and thirders think it depends only on the ratio of wakings in step 3a to those in step 4a, not the specific values.

So I maintain that my problem is the same as coscott’s, except in scale, no matter how we choose N. We can answer the original question by choosing N=1 every time.

There is a 2/3 chance of ending after an odd number of flips, and a 1/3 chance of ending after an even number. A halfer should claim SB gains no new knowledge by being awake, so P(odd|awake)=2/3 and P(even|awake)=1/3. A thirder should say there are four possible situations that awake SB could be in , and she cannot differentiate between them. Since 3 of them correspond to an even number of flips, P(odd|awake)=1/4 and P(even|awake)=3/4.

But like coscott’s, this variation, by itself, sheds no light on the original problem. We can even change numbers to something else:

1) Flip a coin. a. If this (odd-numbered) flip lands heads, wake SB N times and end the experiment. b. If this flip (odd-numbered) lands tails, continue to step 2. 2) Flip the coin again. a. If this (even-numbered) flip lands heads, wake SB M times and end the experiment; b. If this (even-numbered)flip lands tails, go to step #1.

You can decide for yourself whether you think the answer should depend on M and N, but I suspect most people will decide that based on whether they are halfers (“it can’t depend on M and N!”) or thirders (“it must depend on M and N!”), rather than what makes mathematical sense. (I’m not saying they will ignore mathematical, I’m saying they will define it by getting the answer they prefer.)

But as long as we are accepting the possibility of infinite wakings, what happens if we hold N constant and let M approach infinity? Halfers will still say the answers don’t change, thirders will say P(odd)=N/(M+N) and P(even)=M/(M+N).

But is it, or is it not, the same if we hold M constant, at a very large number, and let N approach 0? Because at N=0, P(odd)=0, which it can’t be if the halfers are right.

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