I skipped answering the initial question because I’ve always been a thirder. I’m just trying to comment on the reasons people have given. Mostly how many will try to use fuzzy logic—like “isn’t the question just asking about the coin flip?” in order to make the answer that they find intuitive sound more reasonable. I find that people will tend to either not change their answer because they don’t want to back down from their intuition, or oscillate back and forth, without recalling why they picked an answer a few weeks later. Many of those will end up with “it depends on what you think the question is.”
JeffJo
Karma: 4
I try to avoid any discussion of repeated betting, because of the issues you raise. Doing so addresses the unorthodox part of an unorthodox problem, and so can be used to get either solution you prefer.
But that unorthodox part is unnecessary. In my comment to pathos_bot, I pointed out that there are significant differences between the problem as Elga posed it, and the problem as it is used in the controversy. It the posed problem, the probability question is asked before you are put to sleep, and there is no Monday/Tuesday schedule. In his solution, Elga never asked the question upon waking, and he used the Monday/Tuesday schedule to implement the problem but inadvertently created the unorthodox part.
There is a better implementation, that avoids the unorthodox part.
Before being put to sleep, you are told that two coins will be flipped after you are put to sleep, C1 and C2. And that, at any moment during the experiment, we want to know the degree to which you believe that coin C1 came up Heads. Then, if either coin is showing Tails (but not if both are showing Heads):
You will be wakened.
Remember what we wanted to know? Tell us your degree of belief.
You will be put back to sleep with amnesia.
Once this is either skipped or completed, coin C2 is turned over to show its other side. And the process is repeated.
This implements Elga’s problem exactly, and adds less to it than he did. But now you can consider just what has happened between looking at the coins to see if either is showing Tails, and now. When examined, there were four equiprobable combinations of the two coins: HH, HT, TH, and TT. Since you are awake, HH is eliminated. Of the three combinations that remain, C1 landed on Heads in only one.
The same problem statement does not mention Monday, Tuesday, or describe any timing difference between a “mandatory” waking and an “optional” one. (There is another element that is missing, that I will defer talking about until I finish this thought.) It just says you will be wakened once or twice. Elga added these elements as part of his solution. They are not part of the problem he asked us to solve.
But that solution added more than just the schedule of wakings. After you are “first awakened,” what would change if you are told that the day is Monday? Or that the coin landed on Tails (and you consider what day it is)? This is how Elga avoided any consideration, given his other additions, of what significance to attach to Tuesday, after Heads. That was never used in his solution, yet could be the crux of the controversy.
I have no definitive proof, but I suspect that Elga was already thinking of his solution. He included two hints to the solution: One was “two days,” although days were never mentioned again, and that “when first awakened.” Both apply to the solution, not the problem as posed. I think “first awakened” simply meant before you could learn information.
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You point out that, as you are trying to interpret it, SB cannot make the determination whether this is a “first awakening.” But the last element that is usually included in the problem, but was not in what Elga actually asked, is that the question is posed to you before you are first put to sleep. So the issue you raise—essentially, whether the question is asked on Tuesday, after Heads—is moot. The question already exists, as you wake up. It applies to that moment, regardless of how many times you are wakened.
Say I ask you to draw a card and then, without looking at it, show it to me. I tell you that it is an Ace, and ask you for the probability that you drew the Ace of Spades. Is the answer 1⁄52, 1⁄4, or (as you claim about the SB problem) ambiguous?
I think it is clear that I wanted the conditional probability, given the information you have received. Otherwise, what was the point of asking after giving the information?
The “true” halfer position is not that ambiguity; it is that the information SB has received is null, so the conditional probability is the same as the prior probability. The thirder position is that there are four possible observation opportunities of the coin, all equally likely, and one has been eliminated. To see this better, always wake SB on Tuesday, but instead of asking about the coin, take her on a shopping spree if the coin landed on Heads. If she is asked the question, she knows that one observation opportunity is eliminated, and the answer is clearly 1⁄3.
The difference between the halfer and thirder, is that the halfer thinks that sleeping thru Tuesday removes Tuesday from the sample space of the full experiment, while the thirder sees it as something that contradicts observation.
The need for distinguishing between SIA and SSA is not needed in the Sleeping Beauty Problem. It was inserted into Adam Elga’s problem when he changed it from the one he posed, to the one he solved. I agree that they should have the same answer, which may help in choosing SIA or SSA, but it is not needed. This is what he posed:
“Some researchers are going to put you to sleep. During the two days that your sleep will last, 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 first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?”
The need was created when Elga created a schedule for waking, that treated two days differently. So that “existence” became an alleged issue. There is a better way. First, consider this “little experiment”:
Two coins, C1 and C2, will be randomly arranged so that the four possible combinations, HH, HT, TH, and TT, are equally likely.
If either coin is showing Tails, you will be asked for what you believe to be the probability that coin C1 is showing Heads.
If both are showing Heads, you will not be asked a question.
The answer should be obvious: From the fact that you are in step 2, as established by being asked a question, the combination HH is eliminated. What happens in step 3 is irrelevant, since the question is not asked there. In only one of the three remaining combinations is coin C1 showing Heads, so there is a 1⁄3 chance that coin C1 is showing heads.
To implement the experiment Elga proposed—not the one he solved—put SB to sleep on Sunday night, and flip the two coins. On Monday, perform the “little experiment” using the result of the flips. You will need to wake SB if step 2 is executed. What you do in step 3 is still irrelevant, but can include leaving her asleep. Afterwards, if she is awake, put her back to sleep with amnesia. AND THEN TURN COIN C2 OVER. On Tuesday, perform the “little experiment” again, using the modified result of the flips.
SB does not need to consider any other observers than herself to answer the question, because she knows every detail of the “little experiment.” If she is awake, and asked a question, the coins were arranged as described in step 1 and she is in step 2. The answer is 1⁄3.
This paper starts out with a misrepresentation. “As a reminder, this is the Sleeping Beauty problem:”… and then it proceeds to describe the problem as Adam Elga modified it to enable his thirder solution. The actual problem that Elga presented was:
Some researchers are going to put you to sleep. During the two days[1] that your sleep will last, 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.2 When you are first awakened[2], to what degree ought you believe that the outcome of the coin toss is Heads?
There are two hints of the details Elga will add, but these hints do not impact the problem as stated. At [1], Elga suggests that the two potential wakings occur on different days; all that is really important is that they happen at different times. At [2], the ambiguous “first awakened” clause is added. It could mean that SB is only asked the first time she is awakened; but that renders the controversy moot. With Elga’s modifications, only asking on the first awakening is telling SB that it is Monday. He appears to mean “before we reveal some information,” which is how Elga eliminates one of the three possible events he uses.
Elga’s implementation of this problem was to always wake SB on Monday, and only wake her on Tuesday if the coin result was Tails. After she answers the question, Elga then reveals either that it is Monday, or that the coin landed on Tails. Elga also included DAY=Monday or DAY=Tuesday as a random variable, which creates the underlying controversy. If that is proper, the answer is 1⁄3. If, as Neal argues, it is indexical information, it cannot be used this way and the answer is 1⁄2.
So the controversy was created by Elga’s implementation. And it was unnecessary. There is another implementation of the same problem that does not rely on indexicals.
Once SB is told the details of the experiment and put to sleep, we flip two coins: call then C1 and C2. Then we perform this procedure:
If both coins are showing Heads, we end the procedure now with SB still asleep.
Otherwise, we wake SB and ask for her degree of belief that coin C1 landed on Heads.
After she gives an answer, we put her back to sleep with amnesia.
After these steps are concluded, whether it happened in step 1 or step 3, we turn coin C2 over to show the opposite side. And then repeat the same procedure.
SB will thus be wakened once if coin C1 landed on Heads, and twice if Tails. Either way, she will not recall another waking. But that does not matter. She knows all of the details that apply to the current waking. Going into step 1, there were four possible, equally-likely combinations of (C1,C2); specifically, (H,H), (H,T), (T,H), and (T,T). But since she is awake, she knows that (H,H) was eliminated in step 1. In only one of the remaining, still equally-likely combinations, did coin C1 land on Heads.
The answer is 1⁄3. No indexical information was used to determine this. No reference the other potential waking, whether it occurs before or after this one, is needed. This implements Elga’s question exactly; this only possible issue that remains is if Elga’s implementation does.
The “need to throw the second coin” is to make the circumstances underlying any awakening the same. Using a random method is absolutely necessary, although it doesn’t have to be flipped. The director could say that she is choosing her favorite coin face. As long as SB has no reason to think that is more likely to be one result than the other, it works. The reason Elga’s version is debated, is because it essentially flips Tails first for the second coin.
What the coins are showing at the moment are elementary outcomes of the experiment-within-the-experiment. “Causal connection,” whatever you think that means, has nothing to do with it since we are talking about a fixed state of the coins from the examination to the question.
Which statement here to you disagree with?
Just before she was awakened, the possible states of the coins were {HH, HT, TH, TT} with known probabilities {1/4, 1⁄4, 1⁄4, 1⁄4}.
SB knows this with certainty. That is, she did not have to be awake at that moment to know that this was true at that time.
With no change to the coins, they were examined and the next part of the experiment depends on the actual state.
SB knows this with certainty.
Since she is awake, she knows that the possible change is that the state HH is eliminated.
SB knows this with certainty.
The probabilities in step 1 constitute prior probabilities, and her credence in the state is the same as the posterior probabilities
Pr(HT|Awake) = Pr(HT)/[Pr(HT)+Pr(TH)+Pr(TT)] = 1⁄3.
But there are other ways, that have whatever “causal connection” you think is important. That is, they match the way Elga modified the experiment. That’s another point you seem to ignore, that the two-day version differs from the actual question by more than mine does.
Try using four volunteers instead of one, but one coin for all. Each will be wakened on both Monday and Tuesday, except:
SB1 will be left asleep on Tuesday, if the coin landed on Tails. This is Elga’s SB, with the same “causal connection.”
SB2 will be left asleep on Tuesday, if the coin landed on Heads.
SB3 will be left asleep on Monday, if the coin landed on Tails.
SB4 will be left asleep on Monday, if the coin landed on Heads.
This way, on each of Monday and Tuesday, exactly three volunteers will be wakened. And unless you think the specific days or coin faces have differing qualities, the same “causal connections” exist for all.
Each volunteer will be asked for her credence that the coin landed on the result where she would be wakened only once. With the knowledge of this procedure, an awake volunteer knows that she is one of exactly three, and that their credence/probability cannot be different due to symmetry. Since the issue “the coin landed on the result where you would be wakened only once” applies to only one of these three, this credence is 1⁄3.
AINC: “What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday?”
If she has no reason to think this is not one of her awakenings on Monday/Tuesday, then her credence is the same as it would have been then: 1⁄2 if she is a halfer, and 1⁄3 if she is a thirder.
AINC: “If it’s 1⁄2 what is the reason for the change from 1/3?”
The only way it could change from 1⁄3 to 1⁄2, is if she is a thirder and you tell her that it is Wednesday. And the reason it changes is that you changes the state of her information, not because anything about the coin itself has changed.
But if you think her credence should be based on the actual day, even when you didn’t tell her that is was Wednesday, then you have told an implicit lie. You are asking her to formulate a credence based on Monday/Tuesday, but expecting her answer to be consistent with Wednesday.
AINC: The point of the Wednesday question is to highlight, that, what you mean by “credence”, isn’t actually a probability estimate that the coin is Heads.
And the point of my answer, is that it is actually a conditional probability based on an unusual state of information.
My point is that SB must have reason to think that she exists in the “Monday or Tuesday” waking schedule, for her to assign a credence to Heads based on that schedule. If she is awake, but has any reason to think she is not in that situation, her credence must take that into account.
You told her that she would be asked for her credence on Monday and maybe on Tuesday. “What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday?” is irrelevant because you are allowing for the case where that is not true yet you want her to believe it is. That is lying to her, in the context of the information you want her to use.
But she can from an opinion. “My credence should be halfer/thirder answer if Wednesday has not yet dawned, or 1⁄2 if it has. Since the cue I was told would happen—being asked for my credence—has not yet occurred, I am uncertain which and so can’t give a more definitive answer.” And if you give that cue on Wednesday, you are lying even if you promised you wouldn’t.
And yes, my mathematical model corresponds to SB’s reality when she is asked for her credence. That is the entire point. If you think otherwise, I’d love to hear an explanation instead of a dissertation that does not apply.
Whether your knowledge correctly represents the state of the system, or not, is irrelevant. Your credence is based on your knowledge. If they lie to SB and wake her twice after Heads, and three times after Tails? But still tell here it is once or twice? A thirder’s credence should still be 1⁄3.
And those who guess would not be considered the rational probability agent that some versions insist SB must be.
The correct answer is 1⁄3. See my answer to this question.
Ape in the coat: “What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday?”
The snarky answer is “irrelevant.” The assessment of her credence, that she gives on Monday or Tuesday, is based on the information that it is either Monday or Tuesday within the waking schedule described in the experiment. She is not responsible for incorporating false information into her credence.
One thing that seems to get lost in many threads about this problem, is that probability and/or credence is a function of your knowledge about the state of the system. It is not a property of the system itself. You are trying to make it a property of the system.
Halfers are most guilty of not understanding this, since their arguments are based on there being no change to the system despite an obvious change to the knowledge (I’m ignoring how to treat that knowledge). Thirders recognize the change, but keep trying to cast the difference in terms of episiotomy when it really is much simpler.
“However she finds a piece of paper hidden in her pajamas from the past version of herself, experiencing an awakening. This past version managed to cheat and send a message to the future where she claims that her current credence for Heads is 1⁄3. Should she expect to successfully guess Tails with 2⁄3 probability?”
No, since her knowledge is different now than it was then.
“Or suppose that SB is given memory loss drug only before she is awaken on Tuesday so that on Wednesday she always remembers her last awakenings though she doesn’t know whether it was on Monday or Tuesday. Is her credence for Heads still 1/3?”
If I understand this correctly, she knows that it is Wednesday. Her credence is 1⁄2. She can, at the same time on Sunday or Wednesday, make an assessment of the state of her knowledge on Monday or Tuesday, and recognize how it is different. In other words, “my credence now is 1⁄2, but my credence then should be 1⁄3.” The paradox you seem to be fishing for does not exist.
No, much of the debate gets obscured by trying to ignore how SB’s information, while it includes all of the information that will be used it separate the experiment into two observations, also is limited to the “inside information” and she is in just one of those parts. That’s the purpose of the shopping spree I suggested. Correcting what you wrote, if the memory wipe is complete and SB has literally no way of knowin what information might apply to what is clearly a distinct part of the experiment, but she knows there is a different part than the current one, how can it not be “considered (by her) to be a different experiences.”
For the purposes of sampling, there are two parts to the experiment. What you call them is not relevant, only that SB knows that she is in one, and only one, of two parts. In the classic version, as Elga modifided it from the actual problem he proposed, one part must have a waking and one part has either a sleeping or a waking. Are you really claiming that the part where she is left asleep is not a part of the experiment? One that she knows is a possibility that is ruled out by her current state? And so fits the classic definition of “new evidence” that you deny? But if you really believe that, what about my version where she is taken shopping? It is even better as a classic example of evidence.
Or you could look at my answer. There, I explain how the problem you are solving actually is an “alternate formulation.” And, while there are still two parts, they are equivalent so we do not need to treat them separately. The answer is unequivocally 1⁄3.
Or try this: Instead of a coin, roll two six-sided dice. On Monday, ask her for the probability that the resulting sum is 7. On Tuesday, if the sum is odd, ask her the same question. But if it is even, ask her for the probability that the sum is 8.
If the room has a calendar, Beauty should certainly say Pr(S=7)=1/6 on Sunday and Monday. But on Tuesday, in answer to “what is the probability of 7,” she should say Pr(S=7)=1/3. But without a calendar, Pr(S=7)=1/6 can’t be right. Because it might be Tuesday, and Pr(S=7)=1/3 can’t be right. And it might be Monday, when Pr(S=7)=1/6 can’t be right. It has to be something in between, and that something is (2/3)*(1/6)+(1/3)*(1/3)=4/18=2/9. Yes, even though this is never the answer if she knows the day. That is how conditional probability works.
The “valid answers for different questions” position is a red herring, and I’ll show why. But first, here is an unorthodox solution. I don’t want to get into defending whether its logic is valid; I’m just laying some groundwork.
On Sunday (in Elga’s re-framing of the problem he posed—see my answer), the sample space for the experiment seems to be {T,H} with equal probability for each outcome. Both Monday and Tuesday will happen, and so cannot be used to specify outcomes. This is one of your “different questions.”
But when SB is awake, only one of those days is “current.” (Yes, I know some debate whether days can be used this way—it is a self-fulfilling argument. If it is assumed to be true, you can show that it is true.) Elga was trying to use the day as a valid descriptor of disjoint outcomes, so that in SB’s world the outcomes T&Mon and T&Tue become distinct. The prior probabilities of the three disjoint outcomes {T&Mon, T&Tue, H} are each 1⁄2, and the probability for H can be updated by the conventional definition:
Pr(H|T&Mon or T&Tue or H) = Pr(H)/[Pr(T&Mon)+Pr(T&Tue)+Pr(H)]
Pr(H|T&Mon or T&Tue or H) = (1/2)/[(1/2)+(1/2)+(1/2)] = 1⁄3
If you think about it, this is exactly what Elga argued, except that he didn’t take the extreme step of creating a probability that was greater than 1 in the denominator. He made two problems, that each eliminated an additional outcome. I suggest, but don’t want to get into defending, that this is valid since SB’s situation divides the prior outcome T into two that are disjoint. Each still has the prior probability of the parent outcome.
But this is predicated on that split. Which is what I want to show is valid. What if, instead of leaving SB asleep on Tuesday after Heads, wake her but don’t interview her? Instead, we do something quite different, like take her on a $5,000 shopping spree at Saks Fifth Avenue? (She does deserve the chance of compensation, after all). This gives her a clear way to distinguish H&Mon&Interview from H&Tue&Saks as disjoint outcomes. The use of the day as a descriptor is valid, since this version of the experiment allows for it to be observed.
The actual sample space, on Sunday, for what can happen in an “awake” world for SB is {T&Mon, H&Mon, T&Tue, H&Tue&Saks}. Each has a prior probability of 1⁄4, for being what will apply to SB on a waking day. When she is actually interviewed, she can eliminate one.
And now I suggest that it does not matter, in an interview, what happens differently on H&Tue. SB is interviewed in the context of an interview day, so the question is placed in that context. Claiming it has the context of Sunday Night ignores how the Saks option can only be addressed in a single-day context, and can affect her answer.
I can’t tell you whether it is considered to be sound. In my opinion, it is. But I do know that the issues causing the controversy to continue for 23 years are not a part of the actual problem. And so are unnecessary.
If anyone thinks that sounds odd, they should go back and read the question that Elga posed. It does not mention Monday, Tuesday, or that a Tails-only waking follows a mandatory waking. Those were elements he introduced into the problem for his thirder solution.
In the problem as posed, the subject (I’ll call her SB, even though in the posed problem it is “you”) is woken once, or twice, based on the outcome of a fair coin flip (Heads=once, Tails=twice). Elga enacted that description with a mandatory waking on Monday, and an optional one on Tuesday. This way, could create two partial solutions by revealing two different bits of information that removed one of the three possibilities. The unfortunate side effect of this was that the conditions surrounding Monday and Tuesday are different; thirders require Tuesday to be part of a different outcome, and halfers insist it is the same outcome as Monday+Tails.
And that is what is unnecessary. Instead of that Monday/Tuesday schedule, simply flip two coins (call them C1 and C2) after SB is first put to sleep. Then:
If both coins are showing Heads, skip to step 6.
Wake SB.
Ask SB for her credence that coin C1 is showing Heads.
Wait for her answer.
Put her back to sleep with amnesia.
End this stage of the experiment.
Now turn coin C2 over to show its other face, and repeat these steps.
When SB is woken, she knows that: (A) She is in step 2 of a pass thru these six steps. (B) In step 1, there were four equally-likely states for the two coins: HH, HT, TH, and TT. (C) Since she is awake, the state HH is eliminated but the other three remain equally likely. (D) In only one of those states is C1 showing Heads. So she can confidently state that her credence is 1⁄3.
The difference between this, and how Elga enacted the problem he posed, is that here there is no ambiguity about how she arrived at her current, awake situation.
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:
SB is put to sleep.
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.
Observe what the coins are showing:
If either coin is showing Tails:
Wake SB.
Ask her for her degree of belief that coin C1 is showing Heads.
Put SB back to sleep with amnesia.
If both coins are showing heads:
Do something else that is obviously different than option 3a.
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.
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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 the context of the question includes a reward structure, then the correct solution has to be evaluated within that structure. This one does not. Artificially inserting one does not make it correct for a problem that does not include one.
The actual problem places the probability within a specific context. The competing solutions claim to evaluate that context, not a reward structure. One does so incorrectly. There are simple ways to show this.