I don't even have to be put to sleep and woken up to do this. I can just say before the experiment starts that I choose to place 6 bets that the die will land on a 6 instead of 1 bet that it won't. — Michael

I wonder why you are so insistent on this arbitrary payout structure. Why not make an even-money payout on each occasion where she is being awakened and offered the opportunity to bet on the coin toss outcome as it is already determined right now? Would not her expected value exactly mirror—and be governed only by—her credence regarding the hidden die having landed six right now? A six is the most likely outcome, so I'm betting on it. No word games. Immediately maximized expected profit (and guaranteed long term profit as well).

It's not an arbitrary payout structure.

A £100 reward is paid out at 6:00pm for any correct bet on the outcome of a die roll. Sleeping Beauty and Prince Charming each bet that their die will not land on a 6. They are both free to change their bet at any time before 6:00pm, e.g. if something happens to affect their credence in the outcome, and can do so as many times as they like. Neither of them has a watch.

Sleeping Beauty is told that if her die landed on a 6 then she will be put to sleep and woken up at six arbitrary points before 6:00pm, otherwise she will be put to sleep and woken up at one arbitrary point before 6:00pm.

Prince Charming is told before 6:00pm that his die is loaded and that the probability that it landed on a 6 is $6\over11$.

It doesn’t make any sense to argue that Sleeping Beauty (after being put to sleep and woken up) and Prince Charming (after being told that his die is loaded) come to share the same credence in the outcome of their die roll but that only he changes his bet. If she truly shares his credence then she would also change her bet.

A six is the most likely outcome, so I'm betting on it. — Pierre-Normand

A six is the least likely outcome, but has the highest expected return, and so she bets on it. Her reasoning both before being put to sleep and after being woken up is:

P1. If I always bet that the die didn't land on a 6 and it didn't then I will win £100 at the end of the experiment (1 × £100 bet)
P2. If I always bet that the die did land on a 6 and it did then I will win £600 at the end of the experiment (6 × £100 bets)
P3. The probability that the die did land on a 6 is $1\over6$
C1. Therefore, the expected return if I always bet that the die didn't land on a 6 is £83.33
C2. Therefore, the expected return if I always bet that the die did land on a 6 is £100
C3. Therefore, the expected return if I always bet that the die did land on a 6 is $6\over11$
C4. Therefore, I will always bet that the die did land on a 6

Her credence remains committed to P3, else she’d calculate very different expected returns after being put to sleep and woken up.

Her credence remains committed to P3, else she’d calculate very different expected returns. — Michael

P3—"The probability that the die did land on a 6 is 1/6"—is an ambiguous statement since, although it makes reference to the die, it fails to sufficiently specify SB's epistemic situation in relation to the die, which is a consideration that seldom arises explicitly outside of the peculiar context of the the Sleeping Beauty problem.

When asked about her credence, SB could reason: "I am currently in a situation (awakening episode) such that 6 times out of 11, when I find myself in such a situation, the die landed on a 6. If I could place an even money bet now, and get fully paid on that bet, it would therefore be rational for me to bet that the die landed on a 6, in accordance with my higher credence in this specific outcome."

She could equally validly reason: "I am currently in a situation (experimental run) such that 1 time out of 6, when I find myself in such situations, the die has landed on a 6. If I could place an even money bet now and not change my bet in subsequent awakening episodes, and get paid at the end of the current experimental run, it would therefore be rational for me to bet that the die didn't land on a 6, in accordance with my higher credence in this specific outcome (i.e. not-six).

Those two reasonings concern the same dice but two different statements of credence in two different kinds of events/outcomes. How SB chooses which one of those two different sorts of credence (and the duration of the "event" she is now involved in) as an apt explicitation of the ambiguous phrase "The probability that the die did land on a 6" can be guided by pragmatic considerations. In this case, the relevant consideration is the specific payout structure and what kinds of events/outcomes this payout structure was designed to track. In a pair of examples I had designed early in this discussion, the relevant pragmatic considerations were either the need for SB to set up an appointment with her aunt (to get a lift at the end of the experimental run), or choose a tool (plank or torch) for escaping the experimental facility during the current awakening episode.

As stated in the original ambiguous statement of the SB problem, the forced choice between the Halfer or Thirder interpretations of SB's credence is a false dichotomy. Your stance leads you to propound Halfer interpretations/elaborations of the problem, which are valid, and to dismiss Thirder interpretations as misconstruals of your Halfer stance. But they're not misconstruals. They're alternative and equally valid interpretations. Thirders often make the same mistake, believing that their interpretation gets at the fundamental truth regarding SB's credence in the (ill specified) "outcome" or "current state of the die".

Those two reasonings concern the same dice but two different statements of credence in two different kinds of events/outcomes. — Pierre-Normand

This makes no sense. There is only one kind of event; being woken up after a die roll. Her credence in the outcome of that die roll cannot be and is not determined by any betting rules. Maybe she's not allowed to place a bet at all

After waking up, either she continues to believe that the probability that the die landed on a 6 is 1/6, as Halfers say, or she now believes that it is 6/11, as Thirders say.

Only then, if allowed, can she use her credence to calculate the expected returns of placing or changing a bet, accounting for the particular betting rules. And as I believe I showed above, only a credence of 1/6 provides a consistent and sensible approach to both betting scenarios.

This makes no sense. There is only one kind of event; being woken up after a die roll. Her credence in the outcome of that die roll cannot be and is not determined by any betting rules. Maybe she's not allowed to place a bet at all. — Michael

I agree that her credence in the outcome (however this outcome is characterized) isn't determined by the betting rules. The betting rules, though, can make one rather than another characterization of the outcome more natural. It's not true that there is only one kind of event. The relevant event is protracted. Sleeping Beauty could focus on her current awakening as the event where she either is facing a die that landed on six or didn't (and this event is over when she is put back to sleep, while her next awakening, if there is any, will be a separate event). Or she could focus on the current experimental run as the protracted event that her present awakening is, in some cases, only a part of. Nothing in the Bible, in the fundamental laws of nature, or in the mathematical theory of probability, determines what specific event (awakening or experimental run) should be the proper focus of attention. This choice of focus yields different analyses and different credences since those credences target differently individuated events. However, once one analysis has been settled on, and one payout structure has been determined, Halfers and Thirders (almost) always agree on the expected value of a given betting strategy.

After waking up, either she continues to believe that the probability that the die landed on a 6 is 1/6, as Halfers say, or she now believes that it is 6/11, as Thirders say.

Indeed, and, as previously explained, that because Halfers and Thirders are typically talking past each other. They're not talking about the same events.

Only then, if allowed, can she use her credence to calculate the expected returns of placing or changing a bet, accounting for the particular betting rules. And as I believe I showed above, only a credence of 1/6 provides a consistent and sensible approach to both betting scenarios.

I don't think you've shown the Thirder analysis to be inconsistent. You just don't like it. There are scenarios where the Thirder analysis is more more natural. Remember the flip-coin scenario where the singular H-awakenings take place in the West-Wing of the Sleeping Beauty Experimental Facility and the dual T-awakenings are taking place in the East-Wing. The West-Wing is surrounded by a moat with crocodiles and the East-Wing is surrounded by a jungle with lions. On the occasion of her awakening Sleeping Beauty (we may call her Melania) finds a rare opportunity to escape and can either choose to bring a torch (that she can use to scare off lions) or a wooden plank (that she can use to safely cross the moat). A Thirder analysis of the situation is natural in that case since it tracks singular escape opportunities. Her credence that she will encounter lions is 2/3 (as is her credence that the coin landed Tails). Taking the torch is the safest bet and, indeed, two thirds of Sleeping Beauties who make this bet on the rare occasions where this opportunity presents itself to them survive.

On edit: For this analysis to be sound, we must assume that the rare escape opportunities don't convey any significant amount of information that SB didn't already have when she awoke, and hence present themselves with the same (very low) frequency on each awakening occasion.

those credences target differently individuated events — Pierre-Normand

This is where I believe the mistake is made. The question she is asked after being woken up is the same question she is asked before being put to sleep. There is no ambiguity in that first question, and so there is no ambiguity in any subsequent question. There is a single event that is the target of the question before being put to sleep and we are asking if being put to sleep and woken up gives Sleeping Beauty reason to re-consider her credence in that event, much like Prince Charming re-considers his credence in that event after being told that his coin is loaded. Neither Sleeping Beauty nor Prince Charming is being asked to consider their credence in one of two different events of their own choosing.

Indeed, and, as previously explained, that because Halfers and Thirders are typically talking past each other. They're not talking about the same events. — Pierre-Normand

Which is why I think that Thirders have fabricated a problem that doesn't exist and Halfers are right. The problem only arises because it is suggested that Sleeping Beauty's credence in Event A changes after being put to sleep and woken up, despite no new information. All I can gather from your responses is that Thirders say that Sleeping Beauty's credence in Event B is $2\over3$. But we're not interested in Sleeping Beauty's credence in Event B; we're only interested in Sleeping Beauty's continued credence in Event A.

Remember the flip-coin scenario where the singular H-awakenings take place in the West-Wing of the Sleeping Beauty Experimental Facility and the dual T-awakenings are taking place in the East-Wing. The West-Wing is surrounded by a moat with crocodiles and the East-Wing is surrounded by a jungle with lions. On the occasion of her awakening Sleeping Beauty finds a rare opportunity to escape and can either choose to bring a torch (that she can use to scare off lions) or a wooden plank (that she can use to safely cross the moat). A Thirder analysis of the situation is natural in that case since it tracks singular escape opportunities. Her credence that she will encounter crocodiles is 2/3 (as is her credence that the coin landed Tails). Taking the plank is the safest bet and, indeed, two thirds of Sleeping Beauties who make this bet on the rare occasions where this opportunity presents itself to them survive. — Pierre-Normand

That you're more likely to escape if you assume that the coin landed tails isn't that the coin most likely landed tails. You just get two opportunities to escape if the coin landed tails. It's exactly the same as being able to place either two bets on outcome A or one bet on outcome B, and where P(A) <= P(B) but P(B) < 2P(A). It is more profitable to bet twice on the least probable outcome than once on the most probable outcome. You don't need to force yourself to believe that outcome A is more probable to justify placing those bets. The expected return already does that for you.

That you're more likely to escape if you assume that the coin landed tails isn't that the coin most likely landed tails. You just get two opportunities to escape if the coin landed tails. — Michael

She gets two opportunities to escape if the coin landed tails (or rather she is twice as likely to have an opportunity to escape when the coin landed tails) precisely because she twice as often finds herself being awakened when the coin landed tails. This is the reason why, whenever she is awakened, her epistemic relationship to the coin that has been tossed changes. There is a causal relationship between the coin toss result and the number of awakenings (and escape opportunities) she thereby experiences (encounters). It's her knowledge of this causal relationship that she can harness to update her credence in the new epistemic situation she finds herself in when she awakens.

Notice that, in this example, the success of her escape strategy isn't predicated on there being more opportunities when the coin landed tails. The choice being offered to her isn't between escaping or staying put. It's a choice between carrying a plank or a torch. Taking the torch will enable her to survive if and only if she's being housed in the East-Wing. Else, she's going to be eaten by crocs. The success rate of betting on lions (and, correlatively, on the dice having landed tails) is twice as high as the success rate of betting on crocs (and on the dice having landed heads). The success rate of her betting decisions directly track her credence on the specific outcome she is betting on on those occasions.

If a Halfer claims that, when she awakens, SB's credence on the coin having landed tails remains 1/2, and hence likewise for her credence that she is surrounded by lions, there would be no reasons for her when she attempts to escape on this occasion to bring a torch rather than a plank. She could pick either the torch or the plank at random. Half of such Halfer Beauties who make an escape attempt would survive. Two thirds of Thirder Beauties would survive. The Halfers weren't wrong in their credence assessment. But they picked the wrong credence (targeting expected frequencies of runs rather than frequencies of awakenings) for the task at hand.

SB has no unusual "epistemic relationship to the coin," which is what the point of my new construction was trying to point out. That fallacy is based on the misconception that Tuesday somehow ceases to exist, in her world, if the coin lands on Heads. It still exists, and she knows it exists when she addresses the question.

But she also knows that it is not the current situation when she is asked the question. In this new construction, there are N random values (N=2 for the original, N=6 at camp) that determine which row of N days in the schedule is used. So there are N^2 equally likely entries in the schedule, with N possible ways an entry could be observed.

The probability for each random value is the number of times the observed activity appears in the row for that random value, divided the number of times it appears in the schedule. The probability for each day is the number of times the observed activity appears in the column for that day, divided the number of times it appears in the schedule. What the other observations are - even "I wouldn't be able to observe" - are irrelevant since SB knows what was observed, and that the opportunity for the other observations exist regardless of whether the activity is observable.

There is no need to debate a "payout schedule" since the probabilities apply to what SB knows at the time, which depend only which schedule entries are consistent and which are not. To argue otherwise, you have to defend why E="Egg Hunt" gives different answers than E="Extended Sleep," and what that change would be.

You seem to continue to conflate an outcome's expected return with its probability and assert that one's behaviour is only governed by one's credence in the outcome. Neither of these things is true. I've shown several times that the least likely outcome can have the greater expected return and so that this assessment alone is sufficient to guide one's decisions. No number of analogies is going to make either "she wins two thirds of the time if she acts as if A happened, therefore she believes (or ought to believe) that A most likely happened" or "she believes that A most likely happened, therefore she acts (or ought to act) as if A happened" valid inferences.

But the most important part of my previous comment were the first two paragraphs, especially when considering the standard problem.

SB has no unusual "epistemic relationship to the coin," which is what the point of my new construction was trying to point out. That fallacy is based on the misconception that Tuesday somehow ceases to exist, in her world, if the coin lands on Heads. It still exists, and she knows it exists when she addresses the question. — JeffJo

According to a standard Thirder analysis, prior to being put to sleep, SB deems the two possible coin toss outcomes to be equally likely. When she awakens, she could be in either one of three equiprobable situations: Monday&Tails, Monday&Heads and Tuesday&Tails (according to Elga's sensible argument). SB's credence in the truth of the statement "Today is Tuesday" is 1/3. That possibility doesn't cease to exist. Her epistemic relationship to the already flipped coin changes since she is now able to refer to it with the self-locating indexical proposition: "The coin-toss result on the occasion of this awakening episode", which she wasn't able to before.

Before the experiment began, SB could (correctly) reason that is was equally likely that she would be awakened once when the coin toss result is Heads and twice when the coin toss result is Tails. When she is awakened, on any occasion, her epistemic relationship to the coin changes since it's only in the case where the result is Tails that she experiences an awakening twice. In general, events that make it more likely for you to encounter them result in your being warranted to update your credence in them when you do encounter them. This stems from the core rationale of Bayesian updating.

You seem to continue to conflate an outcome's expected return with its probability and assert that one's behaviour is only governed by one's credence in the outcome. — Michael

I've acknowledged this distinction. It's not the credence alone that governs the rational betting behavior. It's the (well defined) credence in combination with the payoff structure that jointly govern the rational betting behavior.

Neither of these things is true. I've shown several times that the least likely outcome can have the greater expected return and so that this assessment alone is sufficient to guide one's decisions.

I've also myself repeatedly made the point that when the payout structure rewards a consistent betting policy (or the last bet being made after being given to opportunity to change it on each awakening occasion) with an even-money bet only once at the end of the experimental run, then, in that case, it's rational to bet on the least likely outcome (namely, a non-six result, which occurs only 5/11th of the times) since this is the betting behavior that maximizes the expected return. In fact, it could be argued that this arbitrary payoff structure is misleading in the present context since it is being designed precisely to incentivise the bettor to bet on the least likely outcome according to their own credence. It's quite fallacious to then charge the Thirder with inconsistency on the ground that they are betting on an outcome that they have the least credence on. When doing so, you are committing the very conflation that you are charging me of doing.

No number of analogies is going to make either "she wins two thirds of the time if she acts as if A happened, therefore she believes (or ought to believe) that A most likely happened" or "she believes that A most likely happened, therefore she acts (or ought to act) as if A happened" valid inferences.

The analogies are being offered for the sake of illustration. They don't aim at proving the validity of Thirder stance, but rather its pragmatic point. By the same token, your own analogies don't prove the validity of the Halfer stance. Remember that I am not a Halfer or a Thirder. My main goal rather was to show how different situations make salient one rather than another interpretation of SB's "credence" as being pragmatically relevant to specific opportunities: highlighting specific kinds of events one gets involved in and that one wishes to track the long term frequency of as a guide to rational behavior.

But the most important part of my previous comment were the first two paragraphs, especially when considering the standard problem.

So, I'll address this separately.

I'm coming back to one of the two paragraphs you had flagged as the most important part of your comment.

This is where I believe the mistake is made. The question she is asked after being woken up is the same question she is asked before being put to sleep. There is no ambiguity in that first question, and so there is no ambiguity in any subsequent question. There is a single event that is the target of the question before being put to sleep and we are asking if being put to sleep and woken up gives Sleeping Beauty reason to re-consider her credence in that event, much like Prince Charming re-considers his credence in that event after being told that his coin is loaded. Neither Sleeping Beauty nor Prince Charming is being asked to consider their credence in one of two different events of their own choosing. — Michael

I assume that the singular event that is the target of the question is, according to you, the coin toss event. And the question is: what is SB's credence in the outcome of this coin toss? Of course, the question is indeed about this unique event, and remains so after she awakens. However, when asked about her credence regarding this specific outcome, SB has to consider some determinate range of possible outcomes, and what makes it more likely in her current epistemic situation that one of those possible outcomes is actual. Any piece of information SB acquires upon awakening that is conditionally dependent on the target outcome provides her with the means to update her credence (using Bayes' theorem). It's also often alleged (e.g. by David Lewis) that no such new information becomes available to her when she awakens, which is true albeit misleading since it neglect a more subtle change in her epistemic situation.

One particular way in which one can acquire information about a specific outcome T occurs when the occurrence of T biases the probability of one encountering this outcome. For instance, if a bunch of fair pennies fall on the ground but, due to reflectivity and lighting conditions, pennies that landed Tails are more noticeable from a distance, then, on the occasion where I notice a penny shining in the distance, my credence that this penny landed tails is increased. (How silly and point missing would a "Halfer" objection be: "It was not more likely to land Tails, you were just more likely to notice it when it did land Tails!")

The SB setup is a very close analogy to this. Coins landing Tails play a similar causal role. Just replace "increased proclivity to being noticed by a passerby" with "increased proclivity to awaken a random test subject in the Sleeping Beauty Experimental Facility".

Of course, one salient disanalogy between this penny drop analogy and the SB problem is that, in the standard SB problem, each coin is being tracked separately and noticed at least once, on Monday. But I don't think this disanalogy undermines the main point. It's because tail-outcomes causally increase the proportion of awakening episodes at which SB would encounter them that, on each occasion where she encounters them, SB can update her credence that the coin landed Tails. That this rational ground for Bayesian updating remains valid even in cases of singular experimental runs with amnesia (as in the original SB problem) is something that I had illustrated by means of a Christmas gift analogy (see the second half of the post).

According to a standard Thirder analysis, prior to being put to sleep, SB deems the two possible coin toss outcomes to be equally likely.

According to the often-misrepresented, original Thirder analysis by Adam Elga, there are two independent random elements: the coin toss, and the day. They combine in four (not three) ways. But I suppose Elga suspected how obtuse halfers would be about them, so he only considered the two overlapping pairs of two that you think constitute the entire sample space.

When she awakens, she could be in either one of three equiprobable situations: Monday&Tails, Monday&Heads and Tuesday&Tails (according to Elga's sensible argument)

That's not the reasoning.

• If (upon awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2. Since being in T1 is subjectively just like being in T2, and since exactly the same propositions are true whether you are in T1 or T2, even a highly restricted principle of indifference yields that you ought then to have equal credence in each. But your credence that you are in T1, after learning that the toss outcome is Tails, ought to be the same as the conditional credence P(T1|T1 or T2), and likewise for T2. So P(T1|T1 or T2) = P(T2|T1 or T2), and hence P(T1) = P(T2).
• If (upon awakening) you were to learn that it is Monday, that would amount to your learning that you are in either H1 or T1. Your credence that you are in H1 would then be your credence that a fair coin, soon to be tossed, will land Heads. It is irrelevant that you will be awakened on the following day if and only if the coin lands Tails — in this circumstance, your credence that the coin will land Heads ought to be 1/2. But your credence that the coin will land Heads (after learning that it is Monday) ought to be the
  same as the conditional credence P(H1|H1 or T1). So P(H1|H1 or T1) = 1/2, and hence P(H1) = P(T1).
• Combining results, we have that P(H1) = P(T1) = P(T2). Since these credences sum to 1, P(H1) = 1/3.

SB's credence in the truth of the statement "Today is Tuesday" is 1/3.

.
Wrong. Elga's credence in the truth of the statement "Today is Tuesday and the coin landed Tails" is 1/3. That's what Elga's T2 means.

You are doing exactly what I said halfers do - denying the existence of what you would call Tuesday&Heads, and that Elga would call H2. The prior pronbability, of a day that is "H2 or T2", is 1/2. Because H2 and T2 are independent results, and the prior pronbabilities are P(H2)=P(T2)=1/4.

But because she is awake, SB receives the "new information" [see note] that H2 is eliminated, leading to Elga's result.

Before the experiment began, SB could (correctly) reason that is was equally likely that she would be awakened once when the coin toss result is Heads and twice when the coin toss result is Tails.

As far as I know, SB can tell that this day is just one of those three possibilities. Please explain if you think otherwise. And before the experiment begins (this is called the "prior" to those who understand probability), she also knows that the experiment exists on Tuesday&Heads, even though she will not be awake to observe it. So when she is awake, she eliminates that possibility.

Note: The way Baysean Updating works is that you define a sample space (called the prior) comprising all possible outcomes of a procedure. This procedure is doing something with SB on a single day based on a coin flip and the index of the day - the 1 or 2 in Elga's notation. So the red herring about indexicals does not apply. The prior probabilities of these outcomes should sum to 1.

Once you have that, you make an observation about an outcome - WHICH IS WHAT HAPPENS ON A SINGLE DAY. This is called "new information" because some of those outcomes in the prior sample space can be eliminated. Not because what you know about what did happen was a surprise. This is another red herring halfers use. Once you "eliminate" any outcomes in the prior that are inconsistent with the observation, you update the prior probabilities so those that are consistent sum to 1.

In the SB experiment, there are four things that can happen ON A SINGLE DAY. They are H1, T1, H2, and T2. Prior probabilities are 1/4 each. When SB is awake, she knows that she is in A SINGLE DAY but it must be H1, T1, or T2. Not H2. So she updates these probabilities to 1/3 each.

This is a trivial conditional probability problem. The reason I posed the "Camp Sleeping Beauty" version, is that it exposes the red herrings. And I assume that is the reason you ignore it, and how the red herrings are exposed.

This is a trivial conditional probability problem. The reason I posed the "Camp Sleeping Beauty" version, is that it exposes the red herrings. And I assume that is the reason you ignore it, and how the red herrings are exposed. — JeffJo

I didn't ignore your post. I read it and referred to it in a reply to Michael as a more aposite (than his) elucidation of the Thirder position. It's true that I now depart somewhat from the sorts of analyses of the problem that were favored by Elga and Lewis since I think the problem can be demystified somewhat by focusing not on the updating of priors regarding predefined situations SB can potentially find herself in at a future time but rather on the shift in her epistemic situation in relation to the coin-toss outcome on any occasion when she awakens. Also, I no longer see Thirder and Halfer interpretations of Sleeping Beauty's epistemic condition to be mutually exclusive responses to a well defined problem but rather each being motivated by complementary interpretations of the sort of event her "credence" in the coin-toss outcome is supposed to be about. If you can't see what a sensible rationale for a Halfer interpretation might be, you can refer to my Aunt Betsy variation laid out here (and following post).

it's rational to bet on the least likely outcome (namely, a non-six result, which occurs only 5/11th of the times) since this is the betting behavior that maximizes the expected return. In fact, it could be argued that this arbitrary payoff structure is misleading in the present context since it is being designed precisely to incentivise the bettor to bet on the least likely outcome according to their own credence. — Pierre-Normand

The multiple bets structure is the misleading structure, and where one is betting on the least likely outcome. If you are offered the opportunity to place six bets that the die landed on a 6 or one bet that it didn’t, what do you do? You place six bets that the die landed on a 6 even though your credence that it did is $1\over6$. Nothing changes after being made to forget before any bet and so you remain committed to what you knew before being put to sleep.

The single bet structure (why do you call it “arbitrary”?) is the appropriate structure to properly assess the problem: does being put to sleep and woken up change her credence in the die roll, like Prince Charming being told that his die is loaded? If it did then she would follow his lead and change her bet, and we would have a genuine paradox (although she'd lose money). If she doesn’t then her credence hasn’t changed and the problem is resolved in the Halfer’s favour (more on this below).

It's the (well defined) credence in combination with the payoff structure that jointly govern the rational betting behavior. — Pierre-Normand

Yes, so consider the previous argument:

P1. If I keep my bet and the die didn't land on a 6 then I will win £100 at the end of the experiment
P2. If I change my bet and the die did land on a 6 then I will win £100 at the end of the experiment
P3. My credence that the die landed on a 6 is $6\over11$
C1. Therefore, the expected return at the end of the experiment if I keep my bet is £$n$
C1. Therefore, the expected return at the end of the experiment if I change my bet is £$m$

What values does she calculate for $n$ and $m$?

She multiplies her credence in the event by the reward. Her calculation is:

C1. Therefore, the expected return at the end of the experiment if I keep my bet is £45.45
C2. Therefore, the expected return at the end of the experiment if I change my bet is £54.55

This is exactly what Prince Charming does given his genuine commitment to P3 and is why he changes his bet.

So why doesn’t she change her bet? Your position requires her to calculate that $n \gt m$ but that’s impossible given P1, P2, and P3. She can only calculate that $n \gt m$ if she rejects P3 in favour of “my credence that the die landed on a 6 is $1\over6$”.

I’ll respond to the other comment this evening after work.

I didn't ignore your post. — Pierre-Normand

You didn't respond to a single point in it. You only acknowledged its existence, while you continued your invalid analysis about changing bets and expected runs. None of which can answer the questions I raised. Using this table

. M T W H F S
1 A D C F E B
2 F B B C C F
3 A B C D F D
4 F E B D B C
5 C D F C E E
6 E E F C C F

• On each single day, after activity X, what is the probability/SB's confidence that the die roll was d, for d=1 to 6?
• On each single day, after activity X, what is the probability/SB's confidence that the day is Y, for Y in {M,T,W,H,F,S}?
• Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads.

I use "single day" because each day is an independent outcome to SB.

We do not need payout schemes to determine this. We do not need to know what SB expects over the row. We do not need to address "indexicals" The correct answers are:

• COUNT(CELL=X in row d)/COUNT(CELL=X in table)
• COUNT(CELL=X in column Y)/COUNT(CELL=X in table)
• No.

The SB setup is a very close analogy to this. Coins landing Tails play a similar causal role. Just replace "increased proclivity to being noticed by a passerby" with "increased proclivity to awaken a random test subject in the Sleeping Beauty Experimental Facility". — Pierre-Normand

This, I think, shows the fallacy. You're equivocating, or at least begging the question. It's not that there is an increased proclivity to awaken in this scenario but that waking up in this scenario is more frequent.

In any normal situation an increased frequency is often explained by an increased proclivity, but it does not then follow that they are the same or that the latter always explains the former – and this is no normal situation; it is explicitly set up in such a way that the frequency of us waking up Sleeping Beauty does not mirror the probability of the coin toss (or die roll).

If you are allowed to place 6 bets if the die lands on a 6 but only 1 if it doesn't then it is both the case that winning bets are more frequently bets that the die landed on a 6 and the case that the die is most likely to not land on a 6.

You didn't respond to a single point in it. You only acknowledged its existence, while you continued your invalid analysis about changing bets and expected runs. — JeffJo

I didn't provide a detailed response to your post because you didn't address it to me or mention me. I read it and didn't find anything objectionable in it. If you think my own analyses are invalid, then quote me or make reference to them and state your specific objections. I'll respond.

Apologies for doing this as a second post. I did mean to include this earlier but miss-clicked.

Of course, one salient disanalogy between this penny drop analogy and the SB problem is that, in the standard SB problem, each coin is being tracked separately and noticed at least once, on Monday. But I don't think this disanalogy undermines the main point. It's because tail-outcomes causally increase the proportion of awakening episodes at which SB would encounter them that, on each occasion where she encounters them, SB can update her credence that the coin landed Tails. That this rational ground for Bayesian updating remains valid even in cases of singular experimental runs with amnesia (as in the original SB problem) is something that I had illustrated by means of a Christmas gift analogy (see the second half of the post). — Pierre-Normand

I think your comment sidestepped the issue I was raising (or at least misunderstood it, unless I'm misunderstanding you), but this reference to Bayesian probability will make it clearer.

Everyone agrees that $P(H) = {1\over2}$.

Halfers claim that $P(H|Awake) = {1\over2}$ and Thirders claim that $P(H|Awake) = {2\over3}$.

You claim that both Halfers and Thirders are right because they are referring to different events, which I understand to mean that the $H$ in $P(H|Awake) = {1\over2}$ and the $H$ in $P(H|Awake) = {2\over3}$ do not designate the same event, which means that one or both do not designate the same event as the $H$ in $P(H) = {1\over2}$.

The problem is only a problem (or paradox) if the $H$ in $P(H|Awake) = {2\over3}$ designates the same event as the $H$ in $P(H) = {1\over2}$, and so it cannot be that both Halfers and Thirders are right. One may be "right" in isolation, but if used in the context of this paradox they are equivocating, and so are wrong in the context of this paradox.

Yes, so consider the previous argument:

P1. If I keep my bet and the die didn't land on a 6 then I will win £100 at the end of the experiment
P2. If I change my bet and the die did land on a 6 then I will win £100 at the end of the experiment
P3. My credence that the die landed on a 6 is 6/11
C1. Therefore, the expected return at the end of the experiment if I keep my bet is £
C1(sic). Therefore, the expected return at the end of the experiment if I change my bet is £

What values does she calculate for and ?

She multiplies her credence in the event by the reward. Her calculation is:

C1. Therefore, the expected return at the end of the experiment if I keep my bet is £45.45
C2. Therefore, the expected return at the end of the experiment if I change my bet is £54.55

This is exactly what Prince Charming does given his genuine commitment to P3 and is why he changes his bet.

So why doesn’t she change her bet? Your position requires her to calculate that > but that’s impossible given P1, P2, and P3. She can only calculate that > if she rejects P3 in favour of “my credence that the die landed on a 6 is 1/6”. — Michael

While Thirders and Halfers disagree on the interpretation of SB's credence expressed as "the likelihood that the die didn't land on a six", once this interpretations is settled, and the payout structure also is settled, they then actually agree on the correct betting strategy, which is a function of both.

The Thirder, however, provides a different explanation for the success of this unique (agreed upon) betting strategy. The reason why SB's expected return—from a Thirder stance—is higher when she systematically bets on the least likely coin toss result (i.e. 'non-six' which end up being actual only five times on average in eleven awakenings) than when she systematically bets on the most likely one (i.e. 'six' which ends up being the actual result six times on average in eleven awakenings) is precisely because the betting structure is such that in the long run she only is being rewarded once with £100 after betting eleven times on the most likely result ('six') but is rewarded five times with £100 after betting eleven times on the least likely result ('non-six'). On that interpretation, when SB systematically bets on the least likely outcome, she ends up being rewarded more because instances of betting on this outcome are being rewarded individually (and cumulatively) whereas instances of betting on the more likely events are rewarded in bulk (only once for six successful bets placed.) This is the reason why SB, as a Thirder, remains incentivized to bet on the least likely outcome.

Your calculation of her expected return spelled out above was incorrect. It's not simply the result of multiplying her credence in an outcome with the potential reward for this outcome. It's rather the result of multiplying her credence in an outcome with the average reward for this outcome. Since she is only being rewarded with £100 for each sequence of six successful bets on the outcome 'six', her expected value when she (systematically) changes her original bet is:

C2: credence('six') * 'average reward when bet successful' = (6/11) * (£100/6) = £9.091

And her expected value when she doesn't change her bet is

C1: credence('non-six') * 'average reward when bet successful' = (5/11) * £100 = £45,45

She thereby is incentivized to systematically bet on 'non-six', just like a Halfer is.

Notice also that, at the end of an average experimental run, where the number of betting opportunities (i.e. awakening episodes) is 11/6 on average, her calculated expected return is (11/6) * £45,45 = £83.3, which matches the expecting return of a Halfer (who is winning £100 five times out of six runs) as expected.

I think your comment sidestepped the issue I was raising (or at least misunderstood it, unless I'm misunderstanding you), but this reference to Bayesian probability will make it clearer.

[...]

it cannot be that both Halfers and Thirders are right. One may be "right" in isolation, but if used in the context of this paradox they are equivocating, and so are wrong in the context of this paradox. — Michael

I agree with your Bayesian formulation, except that we're more used to follow with Elga's convention, and predicate two awakenings on Tails, such that it's P(T|Awake) that is 2/3 on the Thirder interpretation of this credence.

To be clear about the events being talked about, there is indeed a unique event that is the same topic for discussion for both Halfers and Thirders: namely, the coin toss. However, even after the definition of this unique event has been agreed upon, there remains an ambiguity in the definition of the credence that SB expresses with the phrase "the probability that the coin landed Tails." That's because her credence C is conceptually tied with her expectation that this event will be repeated with frequency C, in the long run, upon repeatedly being placed in the exact same epistemic situation. Thirders assert the the relevant epistemic situation consist in experiencing a singular awakening episode (which is either a T-awakening or a H-awakening) and Halfers assert that the relevant epistemic situation consist in experiencing a singular experimental run (which comprises two awakenings when it is a T-run). So, there are three "events" at issue: the coin toss, that occurs before the experiment, the awakenings, and the runs.

Since it's one's subjective assessment of the probability of the unique event (either H or T) being realized that is at issue when establishing one's credence, one must consider the range of epistemic situations that are, in the relevant respect, indistinguishable from the present one but that one can reasonably expect to find oneself into in order to establish this credence. The Thirders insist that the relevant situations are the indistinguishable awakening episodes (being generated in unequal amounts as a result of the coin toss) while the Halfers insist that they are the experimental run (being generated in equal amounts as a result of this toss). I've argued that both stances yield sensible expressions of SB's credence, having different meanings, and that the choice of either may be guided by pragmatic considerations regarding the usefulness of either tracking relative frequencies of awakenings types or of experimental run types for various purposes.

This, I think, shows the fallacy. You're equivocating, or at least begging the question. It's not that there is an increased proclivity to awaken in this scenario but that waking up in this scenario is more frequent.

In any normal situation an increased frequency is often explained by an increased proclivity, but it does not then follow that they are the same or that the latter always explains the former – and this is no normal situation; it is explicitly set up in such a way that the frequency of us waking up Sleeping Beauty does not mirror the probability of the coin toss (or die roll). — Michael

I’m with you on the distinction. "Proclivity" and "frequency" aren’t the same thing. The only point I’m making is simple: in my shiny-penny story, a causal rule makes certain observations show up more often, and Bayes lets us use that fact.

In the shiny-penny case, fair pennies have a 1/2 chance to land Tails, but Tails pennies are twice as likely to be noticed. So among the pennies I actually notice, about 2/3 will be Tails. When I notice this penny, updating to (2/3) for Tails isn’t smuggling in a mysterious propensity; it’s just combining:

1) the base chance of Tails (1/2), and
2) the noticing rates (Tails noticed twice as often as Heads).

Those two ingredients, or proclivities, generate the observed 2:1 mix in the pool of "noticed" cases, and that’s exactly what the posterior tracks. No amnesia needed; if you were really in that situation, saying "My credence is 2/3 on Tails for the penny I’m looking at" would feel perfectly natural.

If you are allowed to place 6 bets if the die lands on a 6 but only 1 if it doesn't then it is both the case that winning bets are more frequently bets that the die landed on a 6 and the case that the die is most likely to not land on a 6.

Right, and that’s the clean way to separate the two perspectives:

1) Per run: most runs are 'non-six', so the per-run credence is P(6)=1/6 (the Halfer number).
2) Per awakening/observation: a 'six-run' spawns six observation-cases, a 'non-six' run spawns one. So among the observation-cases, 'six' shows up in a 6/5 ratio, giving P('six'|Awake)=6/11 (the Thirder number).

Once you say which thing you’re scoring, runs or awakenings, both beliefs lead to the same betting strategy and the same expected value under any given payout scheme. Different grains of analysis, same rational behavior.

I didn't provide a detailed response to your post because you didn't address it to me or mention me. — Pierre-Normand

It's "addressed" to what I thought was a discussion forum. You know, to discuss this problem and the approach to its solution. And more specifically, to the unnamed in the discussion who try to obfuscate what is a simple conditional probability problem.

Using this table, in the Camp Sleeping Beauty setup:

. M T W H F S
1 A D C F E B
2 F B B C C F
3 A B C D F D
4 F E B D B C
5 C D F C E E
6 E E F C C F

• On each single day, after activity X, what is the probability/SB's credence that the die roll was d, for d=1 to 6?
• On each single day, after activity X, what is the probability/SB's credence that the day is Y, for Y in {M,T,W,H,F,S}?
• Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads. in the popular version?

I use "single day" because each day is an independent outcome to SB. SB does not know anything about any other days, only what is happening in this day. Even if it is possible that the same thing could happen on other days. She can place this day within the context of the set of all days, and the subset where this day's outcomes can happen. BUT SHE ONLY KNOWS ABOUT ONE DAY.

We do not need payout schemes to determine this. We do not need to know what SB expects over the row. We do not need to address "indexicals" The correct answers are:

• COUNT(CELL=X in row d)/COUNT(CELL=X in table).
• COUNT(CELL=X in column Y)/COUNT(CELL=X in table).
• No.

I use "single day" because each day is an independent outcome to SB. — JeffJo

I had misunderstood your original post, having read it obliquely. I had thought you meant for the participants to experience, over the duration of one single day, all six activities in the table row selected by a die throw, and be put to sleep (with amnesia) after each activity. In that case, their credence (on the occasion of any particular awakening/activity) in any given die throw result would be updated using the non-uniform representation of each activity in the different rows. This would have been analogous to the reasoning Thirders make in the original Sleeping Beauty problem. But the variation that you actually propose, when only one activity is being experienced on any given day, yields a very straightforward Bayesian updating procedure that both Halfers and Thirders will agree on. I don't see how it bears on the original problem where the new evidence being appealed to for purposes of Bayesian updating isn't straightforwardly given—where, that is, all the potential awakening episodes are subjectively indistinguishable from Sleeping Beauty's peculiar epistemic perspective.

the variation that you actually propose, when only one activity is being experienced on any given day, yields a very straightforward Bayesian updating procedure that both Halfers and Thirders will agree on. — Pierre-Normand

Thank you for that. But you ignored the third question:

• Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads. in the popular version?

I don't see how it bears on the original problem where the new evidence being appealed to for purposes of Bayesian updating isn't straightforwardly given — Pierre-Normand

Then you don't want to see it as straightforward. Tuesday still exists if the coin lands Heads. It is still a single day, with a distinct activity, in the experiment. Just like the others in what you just called straightforward.

And what makes it "a very straightforward Bayesian updating procedure" is observing that it does not match what is happens on the day SB is experiencing. That is the straightforward Bayesian methodology.

I don't see how it bears on the original problem — Pierre-Normand

Then try this schedule:
. M T W H F S
1 A E E E E E
2 A A E E E E
3 A A A E E E
4 A A A A E E
5 A A A A A E
6 A A A A A A

Here, A is "awake and interview."

If E is "Extended Sleep," the Halfer logic says Pr(d|A)=1/6 for every possible roll, but I'm not sure what Pr(Y|A) is. Halfers aren't very clear on that.

But if E is anything where SB is awoken but not interviewed, then the straightforward Bayesian updating procedure you agreed to says Pr(d|A)=d/21, and if Y is an index for the day, Pr(Y|A)=Y/21.

My issue is that, if A is what SB sees, these two cannot be different.

Thank you for that. But you ignored the third question:

Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads. in the popular version?

"I don't see how it bears on the original problem where the new evidence being appealed to for purposes of Bayesian updating isn't straightforwardly given"
— Pierre-Normand

Then you don't want to see it as straightforward. Tuesday still exists if the coin lands Heads. It is still a single day, with a distinct activity, in the experiment. Just like the others in what you just called straightforward. — JeffJo

Oh yes, good point. I had overlooked this question. Indeed, in that case your variation bears more directly on the original SB thought experiment. One issue, though, is that if is E is just another activity like the other ones, then SB should not know upon awakening on that day that her scheduled activity is E, just like the original problem, when SB wakes up on Tuesday, she isn't informed that she is experiencing a Tuesday-awakening. So, you haven't quite addressed the issue of the indistinguishability of her awakening episodes.

Then try this schedule:
. M T W H F S
1 A E E E E E
2 A A E E E E
3 A A A E E E
4 A A A A E E
5 A A A A A E
6 A A A A A A

Here, A is "awake and interview."

If E is "Extended Sleep," the Halfer logic says Pr(d|A)=1/6 for every possible roll, but I'm not sure what Pr(Y|A) is. Halfers aren't very clear on that. — JeffJo

Halfers don't condition on the propostion "I am experiencing an awakening". They contend that for SB to be awakened several times, rather than once, in the same experimental run (after one single coin toss or die throw) has no incidence on her rational credence regarding the result of this toss/throw.

But if E is anything where SB is awoken but not interviewed, then the straightforward Bayesian updating procedure you agreed to says Pr(d|A)=d/21, and if Y is an index for the day, Pr(Y|A)=Y/21.

My issue is that, if A is what SB sees, these two cannot be different.

Yes, I agree with the cogency of this Thirder analysis. Halfers, however, interpret SB's credence, as expressed by the phrase "the probability that the coin landed Tails" to be the expression of her expectation that the current experimental run, in which she is now awakened, (and may have been, or will be, awakened another time,) is half as likely to be a T-run or a H-run, which also makes sense if she doesn't care how many times she may be awakened and/or interviewed in each individual run. Her credence tracks frequencies of runs rather than (in Thirder interpretations of the problem) awakening episodes.

Halfers don't condition on the propostion "I am experiencing an awakening". — Pierre-Normand

Right. And this is they get the wrong answer, and have to come up with contradictory explanations for the probabilities of the days. See "double halfers."

They contend that for SB to be awakened several times, rather than once, in the same experimental run

Right. Which, after "Tails," requires SB's observation to be happening on both days, simultaneously. See "on a single day."

Halfers, however, interpret SB's credence, as expressed by the phrase "the probability that the coin landed Tails" to be the expression of her expectation that the current experimental run,

Why? How does something that is not happening, on not doing so on a different day, change her state of credence now? How does non-sleeping activity not happening, and not doing so on a different day, change her experience on this single day, from an observation of this single day, to an "experimental run?"

You are giving indefensible excuses to re-interpret the experiment in the only way it produces the answer you want.

How about this schedule:

. M T W H F S
1 A E E E E E
2 A B E E E E
3 A B A E E E
4 A B A B E E
5 A B A B A E
6 A B A B A B

When A happens, if E is treated as only an unseen portion of a different "experimental run," should not B be also considered that? What happened to "straightforward Bayesian updating procedure"?

OR, what if E is "If a coin flipped in Sunday landed Heads, leave SB asleep. But if it landed Tails, wake SB and take her to Disneyworld without an interview." Now you have to use a two models - not different probabilities, two different probability models, for each E day.

How about this schedule:

. M T W H F S
1 A E E E E E
2 A B E E E E
3 A B A E E E
4 A B A B E E
5 A B A B A E
6 A B A B A B

When A happens, if E is treated as only an unseen portion of a different "experimental run," should not B be also considered that? What happened to "straightforward Bayesian updating procedure"?

OR, what if E is "If a coin flipped in Sunday landed Heads, leave SB asleep. But if it landed Tails, wake SB and take her to Disneyworld without an interview." Now you have to use a two models - not different probabilities, two different probability models, for each E day.