Since SB doesn't remember Monday, she cannot feel the difference but the structure of the experiment KNOWS the difference.So if she is asked twice, Monday and Tuesday, that only happens with tails outcome. Even without memory, her credence may shift, but because the setup itself is informative. — Kizzy

It's also part of the protocol that although SB knows that she is being awakened a second time on Tuesday if and only if the coin landed tails, on each occasion where she is being awakened, she isn't informed of the day of the week. As specified in the OP (and in the Wikipedia article), she doesn't know if her awakening is happening on Monday or Tuesday (though she does know, or rather can infer, that it's twice as likely to be occurring on Monday). Hence, the relevant information available to her for establishing her credence is the same on each occasion of awakening.

I think I understand and see my mistake, I was confused from the start as I misunderstood the problem because of the way its laid out and my brain,[thing to note208am] and since I was and am assuming SB is using prior knowledge plus new information Tuesday, without amnesia Sunday to Monday. I just find it to be seemingly difficult to not use some sort of self locating knowledge to assign degree of belief in something, that something being a coin heads up with or without amnesia Sunday and the chance for belief to update come Tuesday.

Since I didn't account for the amnesia from Sunday to Monday, only Monday to Tuesday. My pov, though in error for mistaking PROTO, understood the OG problem (see wiki for Sleeping Beauty Paradox*) to mean after she goes to sleep Sunday, the experiment does not start until Monday, where she goes into the experiment asleep and awakened [inevitable], and asked her credence, put to sleep and either asked again Tuesday with amnesia or sleeps through it. awakening Wednesday and experiment is officially over. .

I do not put it passed me, that even though I just very well may have completely misinterpreted the problem before my own eyes, to be standing corrected, as I have to in order to carry on...

However, now knowing Sunday to Monday she is given amnesia, something about knowing the experiment duration and rules tells me self locating knowledge must come from somewhere, from something in time and our relation to it is relevant to the possible knowing, yes being awake is evidence of some thing but what else gives?

Thanks for your contributions to this thread. And also to your AI exploration experience and work that you've shared here on the forum as well. Of interest...

BUT even though, I was and still am off, in this way:

What else besides knowing what she signed up for, prior know ledge and new intel [ i am awake now] does not alter her credence Monday, only verifies indirect info about SB maybe? If she was given amnesia Sunday and put to bed, awakened Monday then what awakening is she forgetting? *see quote from wiki below, note bold words that are throwing me.....

Anyways, either way -

OH never mind, OF course if she knew it was Monday she wouldn't say 1/3, but what if she was off...and Tuesday comes around and it changes to 0? the chance to change or update belief still exists if tails and asked twice. On Monday she does not know for certain if heads or tails only gives her degree of belief in heads, knowing nothing Wednesday when experiment ends, tomorrow she will be awakened or sleep through the day, she can still guess reasonably participating, I think? I don't know, perhaps I am in over my head here...again!

* per Wikipedia, "On Sunday she will be put to sleep. Once or twice, during the experiment, Sleeping Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening." https://en.wikipedia.org/wiki/Sleeping_Beauty_problem

OH never mind, OF course if she knew it was Monday she wouldn't say 1/3, but what if she was off...and Tuesday comes around and it changes to 0? the chance to change or update belief still exists if tails and asked twice. On Monday she does not know for certain if heads or tails only gives her degree of belief in heads, knowing nothing Wednesday when experiment ends, tomorrow she will be awakened or sleep through the day, she can still guess reasonably participating, I think? I don't know, perhaps I am in over my head here...again! — Kizzy

You'll more easily wrap your head around the problem if you don't overcomplicate things (even though it will remain a tough problem). The purpose of the drug merely is to make it impossible for Sleeping Beauty on any occasion of awakening to know if this occasion was a first or a second one in the experiment (which she could otherwise deduce if she had a memory of the previous one or the lack thereof). This makes all three possibilities—Monday&Heads, Monday&Tails and Tuesday&Tails—indistinguishable from her subjective perspective although she knows at all times that over the course of the experiment all three of those situations could be experienced by her (without knowing which one it is whenever she's experiencing one of them). You can now place yourself in her shoes and start pondering what the chances are that the coin landed tails.

(I'm glad you're enjoying my AI experiment reports!)

SB knows that Monday waking is guaranteed, no matter what the outcome of the coin toss, if so how can she eliminate the sleeping day and update the probabilities or her credence to 1/3 — Kizzy

SB does not know if a waking day is a Monday. Only that it is a waking day. She can eliminate the sleeping day because she knows this is a waking day.

Compare two versions:

• Three days where she is wakened and interviewed, and a fourth where she sleeps.
• Three days where she is wakened and interviewed, and a fourth where she is wakened and taken to DisneyWorld.

On a waking day in the second version, she clearly can eliminate the DisneyWorld day and the probability of Heads. Why is that different? Whether on not she would be awake on that "fourth day" is irrelevant. The important fact is not being able to observe it when it happens, it is being able to observe that it is not happening when it does not.

disagreements arise regarding the meaning of Sleeping Beauty's "credence" about the coin toss result when she awakens, and also about the nature of the information she gains (if any) when she is awakened and interviewed. — Pierre-Normand

Well, isn't this exactly that I tried to say about this being about information?

Should Sleeping Beauty express a 1/2 credence, when she is being awakened, that the coin landed heads? Should it be 1/3, or something else? — Pierre-Normand

Isn't the only the she can say simply that she's participating in the experiment... and she cannot know if its monday or tuesday. Information has an effect on the probability (as in the Monty Hall). Without the information, the probability cannot be accurately defined by her when waking up.

Since SB doesn't remember Monday, she cannot feel the difference but the structure of the experiment KNOWS the difference.So if she is asked twice, Monday and Tuesday, that only happens with tails outcome. Even without memory, her credence may shift, but because the setup itself is informative. — Kizzy

You are one of four volunteers gathered on Sunday Night. You see the combinations "Monday and Heads," "Monday and Tails," "Tuesday and Heads," and "Tuesday and Tails" written on four different note cards. They are turned over, shuffled, and distributed between you, but you can't look. You are told that after you go to sleep, a single fair coin will be flipped. Then, on Monday and again on Tuesday, three of you will be wakened asked some questions. The one who is left out will be the one whose card says the actual coin flip result, and the current day. Afterwards, you will be put back to sleep with amnesia.

Some time later, you find yourself awake and sitting in a room where you can see two of the other three volunteers on TV monitors (you are instructed to not try to communicate through them). One is labeled "Monitor A," and the other "Monitor B." You, and these other two, have their card face-down card on the table in front of them.

• Not knowing what your card says, you are asked for your credence that the coin result written on you card is the actual coin result. AND, the same question about your credence for it matching A's card, and B's card.
• Once you all have provided an answer (unseen by the others, of course), you are told to look at your card, without revealing it, and answer the same questions.

I say the answers in #1 cannot be anything but 1/3. You have the exact same information about each, and they have to add up to 1. If you disagree, please explain how it is possible. Note that the "structure of the experiment KNOWS" that there is a day, an a coin face, that apply. The importangt part is that yolu don't know these.

I say the answers in #2 can't change. Knowing the specific names applied to you "sleep day" does not change their existence what the "structure of the experiment KNOWS," in any way. You seem to think it can; that what the "structure of the experiment KNOWS" changes for you.

But the same applies to A and B. If it changes the same way, your answer for them changes the same way and eveybody's is 1/2. This is a paradox.

And if it changes in a different way for A and B, allowing you to say 1/4 for them, how did it change differently?

The Halfer's run-centered measure just is a way to measure the space of probabilities by partitioning the events that Sleeping Beauty's credence — Pierre-Normand

<Sigh.> I can repeat this as often as you ignore it.

The experiment, when viewed from the outside, consists of two possible runs. The experiment that SB sees is one day, from one run, and to her that one day is independent of whichever run she is in. Since she cannot know which run she is in, that is not information that is useful to her. Inside the experiment, an outcome consists of one "day" only. The only point that is significant to SB is that she can tell that an interview day is not a sleeping day. This constitutes "new information" in probability.

In fact, "new information" is not defined in probability. The information that allows for probability updates is whatever eliminates outcomes that exist in the prior sample space, but are inconsistent with that information. Yes, this usually means a positive fact about the outcome, that does not apply to all, hence some call it "new information." But being "new" isn't what is important, it is the elimination. H&Tue is a member of the prior sample space. It is eliminated when she is awoken and interviewed.

And you can check this is several ways, all of which you ignore. One, you can change the sleeping day to one where she is awakened, but not interviewed. I'll stick with the example that you take her to DisneyWorld. Now one "day" in what you call the "Heads run" is eliminated when she is interviewed. Since the "Heads run" has a 50% probability, and she is can't be in all of the probability-weight of the "Heads Run,", her credence in Heads must be less than 50%/

But it cannot matter what happens on H&Tue. What affects SB's credence is that she knows that the current "day" is not H&Tue. Which she knows whatever happens on H&Tue.

Or you could address the Camp Sleeping Beauty version with more than just "it illustrates the thirder view." You could try to apply the "six day run" theory to Camp Sleeping Beauty. And you will not be able to do so consistently.

• Is SB's credence in each die roll the number of times today's activity appears in that row, divided by the number of times it appears in the 6x6 calendar. If you disagree, please say what it is based on the "six day run" theory.
• If one of the activities is replaced with "sleep through this day," does that change her credence in any way? HOW, and TO WHAT?

SB knows that Monday waking is guaranteed, no matter what the outcome of the coin toss, if so how can she eliminate the sleeping day and update the probabilities or her credence to 1/3 — Kizzy

Because her current knowledge and existence in the experiment is fully limited to one day. Knowing that she will always be awakened on Monday does not change that. See above.

I do think this related to the Monty Hall problem where information affects probabilities. Information does affect probabilities, you know. — ssu

This is called conditional probability.

It's easier indeed to understand the Monty Hall when there's a lot more doors

What that does is make it more intuitive. Since there is a 99.9999% chance Monty Hall picked that one door for the specific reason that it has the car, and a 0.0001% chance that he picked a goat door randomly, it makes sense top go with the 99.9999%. This is harder to wee then the numbers are 66.7% and 33.3%.

So yes, there is similarity in that the information that allows conditional probability to be used is hard to see. But the reasons are quite different. In Sleeping Beauty, it is because philosophers want to propose inconsistent ways to view information.

<Sigh.> I can repeat this as often as you ignore it.

The experiment, when viewed from the outside, consists of two possible runs. The experiment that SB sees is one day, from one run, and to her that one day is independent of whichever run she is in. — JeffJo

SB doesn't have the magical power to make the other awakenings, or their mutual causal relationships, drop out of existence on the occasion where she awakens. She still knows that the two potential T-awakenings live on the same timeline (and hence that when she's experiencing one of them, she will also go on to experience, or will have experienced, the other one in the same run).

Since she cannot know which run she is in, that is not information that is useful to her. Inside the experiment, an outcome consists of one "day" only. The only point that is significant to SB is that she can tell that an interview day is not a sleeping day. This constitutes "new information" in probability.

The fact the the information isn't new to her doesn't make the possibility of there being other potential awakenings in the same run irrelevant. She already has information about those possibilities (and long run frequencies) since she was told about them before the experience began. The Halfer stance, just like the Thirder (equally valid) stance, does not depend on her learning anything new when she awakens since it merely depends on her knowledge of the relative frequencies of H-runs to T-runs.

You're saying that when she awakens, she learns that an interview day is not a sleeping day. But she already knew that interview days never are sleeping days. She can't be asleep and awake at the same time. She knew before the experiments began that the awakenings she would potentially experience in the future would equally as often turn out (merely unbeknownst to her at the time) to have been T-Mon, T-Tue and H-Mon and hence that, when she experiences any of them, those three possibilities would be equally likely from her epistemic standpoint. The Halfer-credence isn't either based on anything new that she learns upon awakening but it is about a differently partitioned relative frequency of events.

To recap what I had said earlier: When SB, as a Thirder, says that the odds that the coin landed tails are 2/3, what she means is that that her current awakening episode just is one from a set of indistinguishable awakening epodes that, in the long run, will turn out to have been T-awakenings 2/3 of the time. When SB, as a Halfer, says that the odds that the coin landed tails are 1/2, what she means is that her current awakening episode is part of a set of indistinguishable runs that, in the long run, will turn out to have been T-runs one half of the time.

Just as you view it as irrelevant to your Thirder claim that T-mon and T-tue belong to the same run, which it indeed is, a Halfer views it as irrelevant to their claim that T-runs spawn more than one awakening episode, which it indeed is. The Halfer and yourself simply are talking about different things.

SB doesn't have the magical power to make the other awakenings, or their mutual causal relationships, drop out of existence on the occasion where she awakens. — Pierre-Normand

Exactly. That is the opposite side of the ability you claim she could have, to make one "other awakening" selectively pop into significance based on know;edge she does not possess. That is, to treat the "other Tails awakening" when the coin landed Tails differently than the "Heads awakening."

Again: the prior sample space comprises FOUR combinations of Coin+Day. In the prior, each is equally likely to apply at the moment the lab techs decide whether or not to awaken her. If they do, to entirety of her information about it is that it is one of the THREE combinations that correspond to an awakening. To her, there is no more, or less, of a connection to the "other" day in this two-day run that indicates, to her, whether it is Monday or Tuesday, if the coin landed Heads or Tails, or which "run" she is in. If you think otherwise, I'd be glad to hear why. An explicit reason why, on T&Mon, she could be more or less likely to think it is H&Mon. This requires knowledge of whether she is in a Heads or Tails run, not the knowledge that such runs are possibilities.

When SB, as a Halfer, says that the odds that the coin landed tails are 1/2, what she means is that her current awakening episode is part of a set of indistinguishable runs that, in the long run, will turn out to have been T-runs one half of the time. — Pierre-Normand

Indistinguishable? You contradict yourself here, because in the long run you do distinguish them.

But you use this argument to once again evade answering the direct questions I have asked several times. One of them is "If the sleeping day is changed to a non-interview waking day, what should her answer be on an interview waking day?" It can't be 1/2, because that would not allow here to have 100% credence in Heads in the non-interview waking day. So she must answer 1/3. But if she answers 1/3, what is different in her knowledge on an interview waking day in the original version?

And if you try to hand-wave a difference, how does in work in the Camp Sleeping Beauty version when each run can contain a different number of waking days?

But I've given up the silly notion that you will address these questions. Which probably means you can't.

Indistinguishable? You contradict yourself here, because in the long run you do distinguish them. — JeffJo

No. I just mean that when she awakens she isn't able to tell if she's in a T-run anymore than she can tell if she's in a T-Monday-awakening or any other possible awakening. That's why the best she can express is a rational credence. She distinguishes runs, and awakenings, and coin toss results, as distinct possibilities that are realized with frequencies determined by the experiment's protocol. If those possibilities were irrelevant, then her knowledge of the protocol that sets their long run frequencies would also be irrelevant. But it's clearly relevant to both Halfers and Thirders.

No. I just mean that when she awakens she isn't able to tell if she's in a T-run anymore than she can tell if she's in a T-Monday-awakening or any other possible awakening. — Pierre-Normand

And I'm saying that this is the exact reason why she cannot base credence on what may, or may not, be the other part(s) of the "run" she is in. I'm saying that all she can base credence on is the one day she can see. And this is trivial to confirm, by addressing the questions you refuse to acknowledge.

And I'm saying that this is the exact reason why she cannot base credence on what may, or may not, be the other part(s) of the "run" she is in. I'm saying that all she can base credence on is the one day she can see. And this is trivial to confirm, by addressing the questions you refuse to acknowledge. — JeffJo

I've made quite a few points that you've never acknowledged, some of them in responses to questions of yours that I responded to more than once. But some of the objections you raise are so vague and bear so little relationship to what I've said that the best I can do in response to them is to try to reiterate my own view more clearly. You repeatedly claimed that I'm disallowed to make reference to any awakening opportunity Sleeping Beauty isn't currently experiencing. But how do you yourself arrive at a credence of 2/3 without making reference to the fact that there are three possible awakening opportunities in total and not just the single one that she is experiencing?

What the SB problem amounts to is a Reductio ad absurdum against the principle of indifference being epistemically normative, a principle that in any case is epistemically inadmissible, psychologically implausible, and technically unnecessary when applying probability theory; a rational person refrains from assigning probabilities when ignorant about frequency information; accepting equal odds is not a representation of ignorance (e.g Bertrand's Paradox).

- It is commonly falsely argued by thirders, that halvers are suspect to a Dutch-book argument, by virtue of losing twice as much money if the coin lands tails, than they gain if the coin lands heads (since the dutch-book is defined as an awoken SB placing and losing two bets, each costing her $1 in the case of tails, one on monday and one on tuesday, versus her placing and winning only one bet rewarding her with $1 on Monday if the coin lands heads). But this dutch book argument is invalidated by the fact that it it equivalent to SB beingapriori willing to win $1 in the case of heads and losing $2 in the case of tails, i..e. SB knowingly accepting a Dutch Book with an expected loss of 0.5x1 - 0.5x2 = -$0.5 before the experiment begins, given her prior knowledge that P(H) = 0.5. So the Dutchbook argument is invalid and is actually an argument against the thirder position.

The (frankly unnecessary) lesson of SB is that meaningful probabilities express causal assumptions, and not feelings of indifference about outcomes.

You repeatedly claimed that I'm disallowed to make reference to any awakening opportunity Sleeping Beauty isn't currently experiencing. — Pierre-Normand

You can refer to any part of the experiment you want. Sleeping Beauty knows all of the parts (*), but has no means to relate her current awake period to any others. You are saying halfers base their answer on doing that. They can't.

But how do you yourself arrive at a credence of 2/3 without making reference to the fact that there are three possible awakening opportunities in total and not just the single one that she is experiencing?

Are you really that obtuse? As I indicated with the (*), she knows all of the parts. That's what establishes the prior sample space. All four possibilities, with equal probabilities. Since she is awake, she eliminates the one she sleeps through.

And as I have said, betting arguments don't work because you have to agree on how many bets are placed. But there is no logical fallacy in a direct probability analysis, as I have done.

There is nothing vague about my questions, unless you refuse to understand it.

• Compare two versions of the popular problem; one where she stays asleep on H+Tue, and one where she is awakened but taken to Disney World instead of being interviewed. In the halfer, two-runs model, does her credence in Heads change between these two versions? What is her credence in Heads when she goes to DisneyWorld?
• In my Camp Sleeping Beauty version, is her credence in die roll D (# times today's activity appears in row D)/(# times today's activity appears in table), as thirders would claim, or is it 1/6 as halfers would claim? How does the halfer's answer change if today's activity does not appear in all rows?

What the SB problem amounts to is a Reductio ad absurdum against the principle of indifference being epistemically normative, a principle that in any case is epistemically inadmissible, psychologically implausible, and technically unnecessary when applying probability theory; a rational person refrains from assigning probabilities when ignorant about frequency information; accepting equal odds is not a representation of ignorance (e.g Bertrand's Paradox). — sime

I don't see any questionable appeal to the principle of indifference being made in the standard Thirder arguments (though JeffJo may be making a redundant appeal to it, which isn't needed for his argument to go through, in my view.) Sleeping Beauty isn't ignorant about frequency information since the relevant information can be straightforwardly deduced from the experiment's protocol. SB doesn't infer that her current awakening state is a T-awakening with probability 1/3 because she doesn't know which one of three indistinguishable states it is that she currently is experiencing (two of which are T-awakenings). That would indeed be invalid. She rather infers it because she knows the relative long run frequency of such awakenings to the 2/3 by design.

The (frankly unnecessary) lesson of SB is that meaningful probabilities express causal assumptions, and not feelings of indifference about outcomes.

I don't think that is the salient lesson from the thought experiment but I agree with your claim.

Regarding the Dutch-book arguments, they represent specific payout structures. They highlight why it's rational for Halfers to be indifferent between betting on H or T when only one betting opportunity and payout is afforded to them in one run of the experiment. They also highlight why Thirders are not likewise indifferent between betting on H or T when one betting opportunity and payout is afforded to them on any awakening occasion.

I don't see any questionable appeal to the principle of indifference being made in the standard Thirder arguments (though JeffJo may be making a redundant appeal to it, which isn't needed for his argument to go through, in my view.) Sleeping Beauty isn't ignorant about frequency information since the relevant information can be straightforwardly deduced from the experiment's protocol. SB doesn't infer that her current awakening state is a T-awakening with probability 1/3 because she doesn't know which one of three indistinguishable states it is that she currently is experiencing (two of which are T-awakenings). That would indeed be invalid. She rather infers it because she knows the relative long run frequency of such awakenings to the 2/3 by design. — Pierre-Normand

But the SB experiment is only assumed to be performed once; SB isn't assumed to have undergone repeated trials of the sleeping beauty experiment, let alone have memories of the previous trials, but only to have been awoken once or twice in a single experiment, for which no frequency information is available, except for common knowledge of coin flips. So the SB is in fact appealing to a principle of indifference as per the standard explanation of the thirders position, e.g. wikipedia.

In any case, a frequentist interpretation of P(Coin is Tails) = 0.5 isn't compatible with a frequentist interpretation of P(awoken on Tuesday) = 1/3.

For sake of argument, suppose P(Coin is Tails) = 0.5 and that this is a frequential probability, and that inductive reasoning based on this is valid.

Now if P(awoken on Tuesday) = 1/3, then it must also be the case that

P(awoken on Tuesday | Coin is Tails) x P(Coin is tails) = 1/3, as typically assumed by thirders at the outset. But this in turn implies that

P(awoken on Tuesday | Coin is tails) = (1/3)/0.5 = 2/3.

Certainly this isn't a frequential probability unless SB having undergone repeated trials notices that she is in fact woken more times on a tuesday than a monday in cases of Tails, in contradiction to the declared experimental protocol. Furthermore, this value doesn't even look reasonable as a credence , because merely knowing apriori that the outcome of the coin is tails shouldn't imply a higher credence of being awoken on tuesday rather than Monday.

Credences are a means of expressing the possession of knowledge without expressing what that knowledge is. To assign consistent credences requires testing every implied credence for possible inconsistencies. Thirders fail this test. Furthemore, credences should not be assigned on the basis of ignorance; a rational SB would not believe that every possible (day, coin-outcome) pair has equal prior probability, rather she would only assume was is logically necessary - namely that one of the pairs will obtain with either unknown or undefined probability.

The SB problem is a classic illustration of confusing what probability is about. It is not a property of the system (the coin in the SB problem), it is a property of what is known about the system. That is, your credence in an outcome is not identically the prior probability that it will occur. Example:

• I have a coin that I have determined, through extensive experimentation, is biased 60%:40% toward one result. But I am not going to tell you what result is favored.
• I just flipped this coin. What is your credence that the result was Heads?

Even though you know that the probability-of-occurrence is either 60% or 40%, your credence in Heads should be 50%. You have no justification to say that Heads is the favored result, or that Tails is. So your credence is 50%. To justify, say, Tails being more likely than Heads, you would need to justify Tails being more likely to be the favored result. And you can't.

And the reason I have not responded to many of Pierre's comments, is that they try to justify answers that directly contradict the answers to the questions I have asked, but have gone unanswered. Because he is trying to convince me with unsupported logic that would be dismissed if he answers mine. Since there is no end in sight to the carousel of unanswered questions, I am going to assert the answers to mine.

I'm going to describe several alternate scenarios that encompass my point. All include amnesia at the end of each day. What I would like to see, is either agreement with these assertions; or disagreement, with reasons. And if reasons are given, I will respond to them.

• Heads & Monday: Wake SB and interview her in conference room A.
• Tails & Monday: Wake SB and interview her in conference room B.
• Heads & Tuesday: Wake SB and take her to Disney World.
• Tails & Tuesday: Wake SB and interview her in conference room D.
• When SB is interviewed, she is asked fro her credence in each line item.

Note: the conference rooms will be indistinguishable to SB.

Assertion #1: These are different outcomes, regardless of SB's ability to distinguish them.

Assertion #2: As she wakes up, SB's credence in today being each line item should be 25%.

Assertion #3: SB cannot make use of runs. That is, if she is being interviewed in conference room A there will be a trip to Disney World. If she is being interviewed in D, she can't make use of the fact that she had been interviewed in C. Such knowledge is of no use to her.

Assertion #4: Since her credence in each outcome is 25%, and she cannot utilize "runs," when she is interviewed her credence in each of the three "interview" outcomes updates to 33%.

• Heads & Monday: Wake SB and interview her in conference room A.
• Tails & Monday: Wake SB and interview her in conference room A.
• Heads & Tuesday: Wake SB and take her to Disney World.
• Tails & Tuesday: Wake SB and interview her in conference room A.

Assertion #5: There is no difference, that can affect SB's credence, in this scenario. Whatever "identifies" an interview has nothing to do with the room where it occurs, it is the circumstances under which it occurs. But not "runs."

• Heads & Monday: Wake SB and interview her.
• Tails & Monday: Wake SB and interview her.
• Heads & Tuesday: Leave SB asleep
• Tails & Tuesday: Wake SB and interview her.

Assertion #6: There is no difference, that can affect SB's credence in an interview, in this scenario. Not being able to observe H&Tue does not remove it from the set of outcomes she knows can happen. Those are determined by the plan described on Sunday, not SB being able to observe it.

Assertion #7: Not being able to observe H&Tue does not make allow SB to utilize the difference between a "Heads run" and a "Tails run."

• The Camp Sleeping Beauty setup, with six distinguishable activities named A, B, C, D, E, and F. Each day in the six-day-by-six-die-rolls camp calendar is randomly assigned one.
• After participating in each day's activity, SB is asked for her credence about the possible die rolls.

Assertion #8 (Thirder version): Her credence in die roll D should be the number of days that today's activity occurs in row D, divided by the number of times it appears on the calendar.

Assertion #8A (Impossible Halfer Version): Her credence in each die roll should be 1/6, even for die rolls where today's activity does not appear. (It is impossible since SB knows the die roll can't be for a row where today's activity does not appear.)

Assertion #8B (Inconsistent Halfer Version): Her credence in each die roll where today's activity does not appear has to be 0. Fore those where it does, it should be 1/N, where N is the number of rolls where it appears. (It is inconsistent since it contradicts the halfer concept that her credence can't be updated.)

Assertion #9: It does not matter if one of the activities is "sleep all day and skip the question."

Assertion #10: The halfer logic is inconsistent. The correct answer is the thirder's.

The SB problem is a classic illustration of confusing what probability is about. It is not a property of the system (the coin in the SB problem), it is a property of what is known about the system. — JeffJo

Then you are referring to subjective probability which is controversial, for reasons illustrated by the SB problem. Aleatory probability by contrast is physical probability and directly or indirectly refers to frequencies of occurrence.

That is, your credence in an outcome is not identically the prior probability that it will occur. Example:

I have a coin that I have determined, through extensive experimentation, is biased 60%:40% toward one result. But I am not going to tell you what result is favored.
I just flipped this coin. What is your credence that the result was Heads?
— JeffJo

It is correct to point out that credence does not traditionally refer to physical probability but to subjective probability. It is my strong opinion however, that credence ought to refer to physical probability. For example, my answer to your question is say that my credence is exactly what you've just told me and nothing more, that is my credience is 60/40 in favour of heads or 60/40 in favour of tails.

Even though you know that the probability-of-occurrence is either 60% or 40%, your credence in Heads should be 50%. You have no justification to say that Heads is the favored result, or that Tails is. So your credence is 50%. To justify, say, Tails being more likely than Heads, you would need to justify Tails being more likely to be the favored result. And you can't. — JeffJo

I definitely would not say that my credence is 50/50, because any statistic computed with that credence would not be reflective of the physical information that you have provided.

I definitely would not say that my credence is 50/50, because any statistic computed with that credence would not be reflective of the physical information that you have provided. — sime

Then what would you say it is? If you say Q, then your credence in Tails must be 1-Q, and you have a paradox.

Then what would you say it is? If you say Q, then your credence in Tails must be 1-Q, and you have a paradox. — JeffJo

If you insist that credence must be expressed as a number Q, then in general I would refuse to assign a credence for that reason - cases like SB in which credences are artificially constrained to be single probability values, doesn't merely result in harmless paradoxes but in logical contradictions (dutch books) with respect to causal premises. Likewise, I am generally more likely to bet on a binary outcome when I know for sure that the aleatoric probability is 50/50, compared to a binary outcome for which I don't know the aleatoric probability.

In order to avoid unintented inferences, the purpose for assigning credences needs to be known. For example, decisions are often made by taking posterior probability ratios of the form P(Hypothesis A | Observation O )/ P(Hypothesis B | Observation O). For this purpose, assigning the prior probability credence P(Hypothesis A = 0.5) is actually a way of saying that credences don't matter for the purpose of decision making using the ratio, since in that case the credences cancel out in the posterior probability ratio to produce the likelihood ratio P(observation O | Hypothesis A)/P(Observation O | Hypothesis B) that only appeals to causal (frequential) information. This is also the position of Likelihoodism; a view aligned with classical frequential statistics, that prior probabilities shouldn't play a part in decision making unless they are statistically derived from earlier experiments.

An acceptable alternative to assigning likelihoods, which often cannot be estimated as in single experiment situations, is to simply to list the possible outcomes without quantifying. Sometimes there is enough causal information to at least order possibilities in terms of their relative likelihood, even if quanitification of their likelihoods isn't possible or meaningful.

Thanks for the tempered reply, @JeffJo. I appreciate the reset. Let me try to restate where I think we converge, then where we part…

In my view, when considering how Sleeping Beauty's credence is to be interpreted, Halfers and Thirders are talking past one another. When considering their specific justifications for their views, though, I think they hold complementary pieces of the puzzle. It's important to put those pieces together.

@sime (Halfer) and @JeffJo (Thirder) appear to me to each own a crucial half of the puzzle.

Sime's piece of the puzzle: The grounding of SB's credence is aleatoric. The fair coin doesn't just draw the map, it drives the frequencies the map will realize across many runs (or, justify expectations over one single run). If we look at the physical world's output, it churns out Heads-runs and Tails-runs in a 1:1 pattern and, because the protocol ties one interview to Heads-runs and two to Tails-runs, it thereby churns out interviews in a 1:2 ratio. Those are physical frequencies, not an appeal to a principle of indifference.

JeffJo's piece of the puzzle: SB's credence is about where she should place herself given what she knows. That's an epistemic task. Where I part ways is only here: we don't need fresh "evidence" to update. We only need to be clear which frequency ensemble her question ranges over.

Put together, the protocol + fair coin fixes the layout and its long-run frequencies (or, equivalently, the single-run expectation). The credence question (that is, its interpretation) fixes which of those frequencies constitutes SB's "credence." Once we say what we're counting, the ratio is forced. No indifference needed.

Here is a variation structurally identical to the original problem: Sleeping Beauty's Garden, where the protocol fixes the topology and the question determines how Sleeping Beauty is conceived to pace through it (the topometry).

Before Beauty sleeps, the attendants lay out a garden she knows everything about:

At the gate there's a fair fork: Heads-path or Tails-path (that's the aleatoric 1/2 vs 1/2).

Along the paths they place the stopping spots (interviews): one lit spot on Heads, two lit spots on Tails.

Amnesia just means each lit spot looks the same when she stands under it.

That's the topology plus its built-in frequencies. What remains open, and defines SB's "credence," is how she conceives of herself pacing through the garden given those frequencies. There are two perfectly sensible ways:

Gate-pacing (the Halfer question).

"One step per run. Which coin toss result did the run that this very awakening belongs to produce?"

SB's answer: "Across repeats, the fair coin makes Heads-runs and Tails-runs equally common. And the bare fact that I am awake now doesn't tilt those run chances: each kind of run guarantees at least one awakening. So, speaking about this awakening: the run that it belongs to will turn out to have been a T-run about half the time. Hence my credence that the current coin toss result is Tails is 1/2."

Lamp-pacing (the Thirder question).

"One step per interview. Which lamp is this?"

SB's answer: "Because the protocol ties one lamp to Heads-runs and two lamps to Tails-runs, among the awakenings that actually occur across repeats, the lamp I'm under now will have turned out to be a T-lamp about two times out of three. So my credence that the current coin toss result is Tails is 2/3." (A biased coin would change these proportions; no indifference is assumed.)

The coin's fairness fixes the branches and the long-run frequencies they generate. The protocol fixes how many stopping points each branch carries. Beauty's "what are the odds?" becomes precise only when she specifies what it is that she is counting.

Note on indifference: The Thirder isn't cutting the pie into thirds because the three interview situations feel the same. It's the other way around: SB is indifferent because she already knows their long-run frequencies are equal. The protocol plus the fair coin guarantee that, among the awakenings that actually occur, the two T-awakenings together occur twice as often as the single H-awakening, and within each coin outcome the Monday vs Tuesday T-awakenings occur equally often. So her equal treatment of the three interview cases is licensed by known frequencies, not assumed by a principle of indifference. Change the coin bias or the schedule and her "indifference" (and her credence) would change accordingly.

Note on one run versus many runs: The fair coin doesn’t just draw the map, it fixes the frequencies the map tends to realize. We can read those either as long-run proportions or as the single-run expectation for Beauty’s present case. In both readings, the protocol—not indifference—does the work.

SB's answer: "Because the protocol ties one lamp to Heads-runs and two lamps to Tails-runs, among the awakenings that actually occur across repeats, the lamp I'm under now will have turned out to be a T-lamp about two times out of three. So my credence that the current coin toss result is Tails is 2/3." (A biased coin would change these proportions; no indifference is assumed.)

The coin's fairness fixes the branches and the long-run frequencies they generate. The protocol fixes how many stopping points each branch carries. Beauty's "what are the odds?" becomes precise only when she specifies what it is that she is counting.

Note on indifference: The Thirder isn't cutting the pie into thirds because the three interview situations feel the same. It's the other way around: SB is indifferent because she already knows their long-run frequencies are equal. The protocol plus the fair coin guarantee that, among the awakenings that actually occur, the two T-awakenings together occur twice as often as the single H-awakening, and within each coin outcome the Monday vs Tuesday T-awakenings occur equally often. So her equal treatment of the three interview cases is licensed by known frequencies, not assumed by a principle of indifference. Change the coin bias or the schedule and her "indifference" (and her credence) would change accordingly. — Pierre-Normand

Thirders who argue their position on the basis of frequential probabilities are conflating the subject waking up twice in a single trial (in the case of Tails) for two independent and identically distributed repeated trials, but the subject waking up twice in a single trial constitutes a single outcome, not two outcomes. Frequentist Thirders are therefore overcounting.

There is only one alleatorically acceptable probability for P(Head | Monday OR Tuesday) (which is the question of the SB problem) :

P(Head | Monday OR Tuesday) =
P(Monday OR Tuesday | Head) x P(Head) / P(Monday Or Tuesday)

where

P(Head) = 0.5 by assumption.
P(Monday Or Tuesday) = 1 by assumption.

P(Monday OR Tuesday | Head) = P(Monday | Head) + P(Tuesday | Head) = 1 + 0 = 1.

P(Head | Monday OR Tuesday) = 1 x 0.5 / 1 = 0.5.

Sime's piece of the puzzle: The grounding of SB's credence is aleatoric. The fair coin doesn't just draw the map, it drives the frequencies the map will realize across many runs (or, justify expectations over one single run) — Pierre-Normand

When SB N is awake, while she is aware of the map, she has no information that she can use to place herself in that map. IT IS IRRELEVANT.

In any way that SB can assess her credence, that does not reference her position in the map, the answer is 1/3.

• Using four volunteers, where each sleeps though a different combination in {H&Mon, T&Mon, H&Tue, T&Tue}? On any day, the credence assigned to each of the three awake volunteers cannot be different. and they must add up to 1. The credence is 1/3.
• Use the original "awake all N days, or awake on on one random day in the set of N" problem? N+1 are waking combinations, only one corresponds to "Heads." The credence is 1/(N+1).
• Change the "sleep" day to a non-interview day? It is trivial that the answer is 1/3.

I'm sure there are others. The point is that the "halfer run-based" argument cannot provide a consistent result. It only works if you somehow pretend SB can utilize information, about which "run" she is in, that she does not and cannot posses.

But to consider a halfer's objections, I have to get that halfer to address the direct questions I have asked about how a halfer argument applies in other scenarios that must use similar arguments. So far, I can't get that to happen.

I'm sure there are others. The point is that the "halfer run-based" argument cannot provide a consistent result. It only works if you somehow pretend SB can utilize information, about which "run" she is in, that she does not and cannot posses. — JeffJo

The only information she makes use of about the run she's in is that the fair coin decided it. It's the exact same information that she's also making use of in your own "four-cards" variation since it's on the basis of this information that she knows that getting a T-Monday, a T-Tuesday or a H-Monday card are equally likely outcomes (1/4 each before updating, on your account). If she knew the coin to be biased towards Tails, say, she would need to adjust her credence accordingly, right?

To clarify the nature of the information that, according to you, she only gets upon awakening, let me make my SBG (Sleeping Beauty's Garden) analogy closer to yours. Rather than there being 2 lit spots (or lampposts) on the T-path and 1 lamppost on the H-path, let us put two lampposts on each path, but only the first one of those two is lit on the H-path. This is equivalent to sleeping on Tuesday. So, on your view, when she happens upon a lit lamppost, she can exclude H-second-lamppost (equivalent to excluding H-Tuesday upon awakening). This, according to you, enables SB, when she happens upon a lit lamppost, to update her priors from [1/4 T-first; 1/4 T-second; 1/4 H-first; 1/4 H-second] to [1/3 T-first; 1/3 T-second; 1/3 H-first; 0 H-second]

Would it matter at all if the original experiment were modified in such a way that when the coin lands Tails, Sleeping Beauty is awakened at two separate random times on Monday, or else is only awakened at one single random time still on Monday? This would be equivalent to my original SBG experiment where there are two lamppost on the T-path and only one on the H-path, and they all are lit (and occasions for an interview) and we don't care at all about the exact location of the lampposts along the paths. There is therefore no information to be had about sleeping days or unlit lampposts. How do you account for her gaining information upon awakening (or when happening upon a lamppost) in those scenarios?

In any way that SB can assess her credence, that does not reference her position in the map, the answer is 1/3.
Using four volunteers, where each sleeps though a different combination in {H&Mon, T&Mon, H&Tue, T&Tue}? On any day, the credence assigned to each of the three awake volunteers cannot be different. and they must add up to 1. The credence is 1/3.
Use the original "awake all N days, or awake on on one random day in the set of N" problem? N+1 are waking combinations, only one corresponds to "Heads." The credence is 1/(N+1).
Change the "sleep" day to a non-interview day? It is trivial that the answer is 1/3.

I'm sure there are others. The point is that the "halfer run-based" argument cannot provide a consistent result. It only works if you somehow pretend SB can utilize information, about which "run" she is in, that she does not and cannot posses. — JeffJo

No, the Halfer position doesn't consider SB to have any information that she could utilize when awakened, due to the fact that SB's knowledge that it is either Monday or Tuesday doesn't contribute new information about the coin, which she only observes after the experiment has concluded.

Also, your reasoning demonstrates why we shouldn't conflate indifference with equal credence probability values. Yes, an awakened SB doesn't know which of the possible worlds she inhabits and is indifferent with regards to which world she is in and rightly so. No, this doesn't imply that she should assign equal probabilty values for each possible world: For example, we have already shown that if an awakened SB assigns equal prior probabilities to every possible world that she might inhabit, then she must assign unequal credences for it being monday versus tuesday when conditioning on a tails outcome.

To recap, if P(Monday) = 2/3 (as assumed by thirders on the basis of indifference with respect to the three possible awakenings), and if P(Tails | Monday) = 1/2 = P(Tails) by either indiffererence or aleatoric probability, then

P(Monday | Tails) = P(Tails |Monday) x P(Monday) / P(Tails) = (1/2 x 2/3) / (1/2) = 2/3.

So let's assume that SB is awakened on monday or tuesday and is told, and only told, that the result was Tails. According to the last result, if SB initially assigns P(Monday) = 2/3 on the basis of the principle of indifference as per thirders, then she must infer having learned of the tails result that monday is twice as likely as tuesday, in spite of mondays and tuesdays equally occcuring on a tails result.

As this demonstrates, uniform distributions have biased implications. So if SB insists on expressing her state of indifference over possible worlds in the language of probabilty, she should only say that any probability distribution over {(Monday,Heads),(Monday,Tails) ,(Tuesday,Tails)} is compatible with her state of indifference, subject to the constraint that the unconditioned aleatoric probability of the coin is fair.

However, if she really must insist on choosing a particular probability distribution to represent her state of indifference, then she can still be a halfer by using the prinicple of indifference to assert P(Monday | Tails) = P(Tuesday | Tails), and then inferring the unconditioned credence that it is monday to be P(Monday = 1/2), which coheres with the halver position.