But the time period after the coin is flipped still exists, and the coin can be Heads during that time. — JeffJo

She is put to sleep on day 0.

If the coin lands heads then a 14-sided dice is rolled. She is woken that many days later and asked her credence.

If the coin lands tails then a 7-sided dice is rolled. She is woken that many days later, asked her credence, and put back to sleep. The dice is rolled again. She is woken that many days later and asked her credence.

Whether heads or tails she can wake on any day between 1 and 14 days after she is first put to sleep.

Nothing is ruled out when she wakes. She doesn't rule out heads and day 1, she doesn't rule out heads and day 2, ... and she doesn't rule out heads and day 14.

The specific days she’s woken or kept asleep are irrelevant. The only things that matter are that she has either one or two interviews – determined by a fair coin toss – and that she doesn’t know if she’s already had one. Everything else is a red herring.

While in each case the biconditionals "I am now in an H-awakening iff I am now (and will be) in an H-run" or (on Wednesday) "I was in an H-awakening iff I am now in an H-run" hold, the probabilities don't necessarily match due to the two-to-one mapping between T-awakenings and T-runs. — Pierre-Normand

I know I referred you to one of my previous posts, but I’ll respond to this directly too.

We’re discussing credence.

If I am certain that A is true if and only if B is true and if I am pretty sure that A is true then ipso facto I am pretty sure that B is true.

Given the biconditional one’s credence in the left hand side must match one’s credence in the right hand side, even if there is this “two-to-one mapping”.

I am certain that my hand is one of two hands I have if and only if I have two hands.

I am pretty sure that my hand is one of two hands I have.

Therefore I am pretty sure that I have two hands.

It doesn’t make sense to say that you’re pretty sure (P = 2/3) that your current interview is one of two but that you’re on the fence (P = 1/2) as to whether or not you have been assigned two interviews.

Either you are pretty sure that you have been assigned two interviews or you are on the fence as to whether or not your current interview is one of two.

The specific days she’s woken or kept asleep are irrelevant. — Michael

I agree, with a caveat. The specific details of whether she is woken at a specific point ("day") in the experiment do matter. You can test this in Elga's solution. If she is told that the current day is Tuesday, then she knows Pr(Heads) must decrease to 0. If she is told that it is not Tuesday, the Law of Total Probability actually requires it to go up. It goes from 1/3 to 1/2 in the Thirder soluton, and from 1/2 to 2/3 in the Halfer solution. This is quite relevant, even if you think it is wrong AND CAN PROVIDE A VALID REASON

But it isn't the day name that matters, it is the details associated with that day name. And since these details are different on different "days," we can track them by naming the "days."

Now, you could argue about why those details might matter; but so far you have refused to. You have just asserted they don't (in spite of evidence like I just presented). But naming them cannot affect the correct answer, no matter how those details affect it. SO THERE IS NO REASON TO NOT NAME THE "DAYS." And even the possibility that they might have an effect makes them relevant.

In other words, you are proffering the red herring here. You are insisting that we must ignore a piece of potential information, because you think it has no affect. If so, there is no harm in including it.

The only things that matter are that she has either one or two interviews – determined by a fair coin toss – and that she doesn’t know if she’s already had one. Everything else is a red herring. — Michael

I agree (well, if you add "or if she will have one later") about "only things." But the "red herring" you claim exists, is trying to utilize information that you say has no affect while refusing to address affects that are clearly present.

But if you truly believe in this, then it should be easy to address this:

1. She is put to sleep. A coin, call it C1, is tossed to determine if "she has either one or two interviews."
2. But then a second coin is tossed, call it C2.
   
   1. Call the state of the coins at this moment S1.
   2. If either coin is showing Tails in S1, wake her with amnesia and interview her.
   3. If both are showing Heads, do SOMETHING ELSE.
   4. If she isn't asleep at this point, put her to sleep again.
3. Turn coin C2 over to show its other side.
   
   1. Call the state of the coins at this moment S2.
   2. If either coin is showing Tails in S2, wake her with amnesia and interview her.
   3. If both are showing Heads, do SOMETHING ELSE.
   4. If she isn't asleep at this point, put her to sleep again.
4. Wake her and send her home.

This procedure creates what, in your words, are "the only things that matters." She has either one, or two, interviews and does not know if another will/did happen. The only thing that is different, is that your possible two interviews occur under different circumstances; one is mandatory, and one is optional. What we disagree about is whether the part that is missing - the non-interview when the option is not taken - matters.

Here they occur under identical circumstances. That is, either steps 2.1 thru 2.4, or steps 3.1 thru 3.4. And those circumstances can be used to answer the question. I did "name" the details by calling them state S1 or S2, but since they are identical to SB she can call them state S.

There are three possible combinations of the two coins in state S, and they are equally likely. Her credence in state S=(H,T) is 1/3.

This has nothing to do with what may or may not get "ruled out" in a solution to your version of the experiment. That difference is the red herring in the "most frequent" presentation of the problem. This is a self-contained experiment with a trivial answer.

But it's an answer you don't like. So you will either ignore it, or repeat the non sequitur that it includes the "ruling out of a 1/4 probability" that we are debating about above, which is circular logic.

So tell me what prior probability is ruled out in my experiment above.

If there isn't one then there isn't one in the ordinary experiment either. You may have other reasons for believing that the answer to the problem is $1\over3$, but those reasons can't be that some prior probability of $1\over4$ is ruled out when woken.

So tell me what prior probability is ruled out in my experiment above. — Michael

This is getting tiresome. The answer follows trivially from what I have said before - have you read it? But I will not continue to dangle on your string. I'll point it out after you tell me:

1. Whether you agree that procedure I just described implements "The only things that matter are that she has either one or two interviews – determined by a fair coin toss – and that she doesn’t know if she’s already had one [or will have another]."
2. If not, what "thing that matters" is not implemented.
3. What her credence in Heads should be.

The answer follows trivially from what I have said before - have you read it? — JeffJo

Nothing in your above post tells me what prior probability is ruled out when she's woken in this experiment.

In your experiment the prior probability HH = $1\over4$ is ruled out when woken.

If you cannot tell me what prior probability is ruled out when woken in my experiment then I have no reason to accept that your experiment is at all equivalent.

I most certainly can. You just won't try to understand it. Or answer any question placed before you, if you don't want to accept the answer. So once again, just answer mine first and I'll answer yours.

I agree, with a caveat. The specific details of whether she is woken at a specific point ("day") in the experiment do matter. You can test this in Elga's solution. If she is told that the current day is Tuesday, then she knows Pr(Heads) must decrease to 0. If she is told that it is not Tuesday, the Law of Total Probability actually requires it to go up. It goes from 1/3 to 1/2 in the Thirder soluton, and from 1/2 to 2/3 in the Halfer solution. This is quite relevant, even if you think it is wrong AND CAN PROVIDE A VALID REASON

But it isn't the day name that matters, it is the details associated with that day name. And since these details are different on different "days," we can track them by naming the "days."

Now, you could argue about why those details might matter; but so far you have refused to. You have just asserted they don't (in spite of evidence like I just presented). But naming them cannot affect the correct answer, no matter how those details affect it. SO THERE IS NO REASON TO NOT NAME THE "DAYS." And even the possibility that they might have an effect makes them relevant.

In other words, you are proffering the red herring here. You are insisting that we must ignore a piece of potential information, because you think it has no affect. If so, there is no harm in including it. — JeffJo

When I said that the only things that matter are:

1. She has either one or two interviews determined by a fair coin toss and
2. She doesn’t know if she’s already had one

I was referring to her just waking up, not being told any further information.

If she's told that the coin landed tails then 1 is no longer relevant and she just considers 2. If she's told that this is now her first interview then 2 is no longer relevant and she just considers 1.

And I don't see why having to consider 2 affects her consideration of 1. Knowing or not knowing that this is now her first interview only affects her credence that this is now her first interview. It doesn't affect her credence that she will have two interviews.

This procedure creates what, in your words, are "the only things that matters." She has either one, or two, interviews and does not know if another will/did happen. The only thing that is different, is that your possible two interviews occur under different circumstances; one is mandatory, and one is optional. What we disagree about is whether the part that is missing - the non-interview when the option is not taken - matters.

Here they occur under identical circumstances. That is, either steps 2.1 thru 2.4, or steps 3.1 thru 3.4. And those circumstances can be used to answer the question. I did "name" the details by calling them state S1 or S2, but since they are identical to SB she can call them state S.

There are three possible combinations of the two coins in state S, and they are equally likely. Her credence in state S=(H,T) is 1/3.

This has nothing to do with what may or may not get "ruled out" in a solution to your version of the experiment. That difference is the red herring in the "most frequent" presentation of the problem. This is a self-contained experiment with a trivial answer.

But it's an answer you don't like. So you will either ignore it, or repeat the non sequitur that it includes the "ruling out of a 1/4 probability" that we are debating about above, which is circular logic. — JeffJo

In this example when woken she rules out the prior probability HH = $1\over4$.

No prior probability is ruled out here when woken so your example isn't equivalent.

I accept 6 and reject 5. My credence that my current interview is a heads interview isn't equal to the fraction of interviews which are heads interviews.

My argument is:

P1. My credence is the degree to which I believe that a proposition is true
P2. My current interview is a heads interview iff I have been assigned one heads interview
C1. Therefore my credence that my current interview is a heads interview is equal to my credence that I have been assigned one heads interview (from P1 and P2)
P3. If I have been assigned at random by a fair coin toss either one heads interview or two tails interviews then the probability that I have been assigned one heads interview is 1/2
P4. I have been assigned at random by a fair coin toss either one heads interview or two tails interviews
C2. Therefore the probability that I have been assigned one heads interview is 1/2
(from P3 and P4)
P5. My credence that I have been assigned one heads interview is equal to the probability that I have been assigned one heads interview
C3. Therefore my credence that I have been assigned one heads interview is 1/2
(from C2 and P5)
C4. Therefore my credence that my current interview is a heads interview is 1/2
(from C1 and C3) — Michael

The issue arises from a conflation of two distinct ways of individuating events and counting probabilities. We can see this more clearly if we distinguish between the 'timeline perspective' and the 'episodic perspective'. Your propositions P1 through P4 and C1 though C4 above frequently shift between those two perspectives, which vitiates the validity of some inferences.

Let's consider the shift from the 'timeline perspective' to the 'episodic perspective'. In the timeline perspective, there are two possibilities ("possible worlds"): an H-timeline and a T-timeline, each with an equal chance of 1/2. The T-timeline, however, comprises two distinct awakening episodes ("centered possible worlds"). This does not create more exclusive events sharing the probability space; rather, it articulates the unique structure of the T-timeline.

Therefore, when shifting to the episodic perspective, it would be a mistake is to divide the probability of the T-timeline (1/2) between the two T-awakenings, suggesting each has a probability of 1/4. This line of thinking presumes these awakenings to be exclusive events within the T-timeline (as they might be if a second coin would be tossed to determine which one of the two would occur). However, that's not the case. In the T-timeline, the two awakenings are guaranteed to occur concurrently if at all; the only unknown is which of them Sleeping Beauty currently finds herself in.

The shift from the timeline perspective to the episodic perspective is not a straightforward Bayesian update on new information. Instead, it's a shift in how we count the alternatives. This shift happens automatically when Sleeping Beauty awakes, because she can't tell apart the two T-episodes and what were concurrent possibilities become exclusive possibilities.

Once we've dealt with the faulty reasoning that made it appear like the T-first-awakening and T-second awakening were lowered from 1/2 to 1/4 when shifting to the episodic perspective, we can now see how the equiprobability between the H-awakening and T-first-awakening must also be retained when we shift perspectives. When Sleeping Beauty wakes up and doesn't know whether it's Monday or Tuesday, this doesn't change the equiprobability of a H-awakening or a T-first-awakening that she would express her credence in were she to know that it's Monday. Instead, her ignorance about the day of the week introduces an additional possibility—that of her being in a T-second-awakening—which in turn increases the total probability of being in a T-awakening.

So, since the shift to the episodic perspective preserves both the equiprobabilities P(T-first-awakening) = P(T-second awakening) and P(T-first-awakening) = P(H-awakening), and all three outcomes are exclusive from this perspective, the probabilities must sum up to 1 and therefore must shift from 1/2, 1/2, and 1/2 to 1/3, 1/3 and 1/3.

Your propositions P1 through P4 and C1 though C4 above frequently shift between those two perspectives, which vitiates the validity of some inferences. — Pierre-Normand

So which premises are false or which conclusions do not follow?

Therefore, when shifting to the episodic perspective, it would be a mistake is to divide the probability of the T-timeline (1/2) between the two T-awakenings, suggesting each has a probability of 1/4. This line of thinking presumes these awakenings to be exclusive events within the T-timeline — Pierre-Normand

My current interview being the first or the second T-awakening are exclusive events.

But I brought up something like this here:

1. Sleeping Beauty is given amnesia
2. She is asked her credence that a coin has been tossed
3. A coin is tossed
4. If the coin lands tails then:
4A. She is given amnesia
4B. She is asked her credence that a coin has been tossed

Thirder reasoning is that because step 2 is twice as likely to occur as step 4B then I am twice as likely to be in step 2 as step 4B.

Halfer reasoning is that because step 2 is twice as likely to occur as step 4B and that because if 4B will occur then I am equally likely to be in step 2 as step 4B then I am three times as likely to be in step 2 as step 4B.

I know I referred you to one of my previous posts, but I’ll respond to this directly too.

We’re discussing credence.

If I am certain that A is true if and only if B is true and if I am pretty sure that A is true then ipso facto I am pretty sure that B is true. — Michael

When I previously addressed this inference of yours, I conceded that it is generally valid, but I also pointed out that it involved a possible conflation of two meanings of the predicate P(). The problem I identified wasn't with the validity of the inference (within the context of probability calculus), but rather with the conflation that could occur when the expression P(A) appears twice in your demonstration.

What makes you "pretty sure" that A is true is the expectation that A is much more likely to occur than not-A. As such, this probabilistic judgment is implicitly comparative. It is therefore dependent on how you individuate and count not only A events but also not-A events. As I've argued elsewhere, a shift in epistemic perspective can alter the way you count not-Heads events (i.e., Tails events), transforming them from non-exclusive to exclusive. For example, when you move from considering possible world timelines to specific awakening episodes, what were concurrent alternatives (not-H events) become exclusive possibilities. This change in perspective modifies the content of your comparative judgment "H is much more likely to occur than not-H," and consequently affects your credence.

When I previously addressed this inference of yours, I conceded that it is generally valid, but I also pointed out that it involved a possible conflation of two meanings of the predicate P(). The problem I identified wasn't with the validity of the inference (within the context of probability calculus), but rather with the conflation that could occur when the expression P(A) appears twice in your demonstration. — Pierre-Normand

There is only one meaning I'm using: "the degree to which I believe that the proposition is true". It's stated as much in P1.

If I am certain that A is true if and only if B is true then the degree to which I believe that A is true is equal to the degree to which I believe that B is true. This is true for all As and Bs.

It isn't rational for me to believe more strongly in one side of a biconditional that I am certain is true.

What makes you "pretty sure" that A is true is the expectation that A is much more likely to occur than not-A. As such, this probabilistic judgment is implicitly comparative. It is therefore dependent on how you individuate and count not only A events but also not-A events. As I've argued elsewhere, a shift in epistemic perspective can alter the way you count not-Heads events (i.e., Tails events), transforming them from non-exclusive to exclusive. For example, when you move from considering possible world timelines to specific awakening episodes, what were concurrent alternatives (not-H events) become exclusive possibilities. This change in perspective modifies the content of your comparative judgment "H is much more likely to occur than not-H," and consequently affects your credence. — Pierre-Normand

Given the above meaning, and as I said before, these cannot all be true:

1. My current interview is a heads interview iff I have been assigned one heads interview
2. The fraction of interviews which are heads interviews is $1\over3$
3. The fraction of experiments which have one heads interview is $1\over2$
4. The degree to which I believe that my current interview is a heads interview is equal to the fraction of interviews which are heads interviews
5. The degree to which I believe that I have been assigned one heads interview is equal to the fraction of experiments which have one heads interview

You seem to assert that 4 and 5 are true by definition, but they're not. Given the truth of 1, 2, and 3, it must be that one or both of 4 and 5 is false.

There is only one meaning I'm using: "the degree to which I believe that the proposition is true".

If I am certain that A is true if and only if B is true then the degree to which I believe that A is true is equal to the degree to which I believe that B is true. This is true for all As and Bs. — Michael

This overlooks the issue that your credence can change over time when your epistemic perspective changes. If your separate uses of the expression P(H) don't take into account the epistemic perspective within which they're intended to be evaluated, you risk equivocation.

Given the above, as I said before, these cannot all be true:

1. My current interview is a heads interview iff I have been assigned one heads interview
2. The fraction of interviews which are heads interviews is 1/3
3. The fraction of experiments which have one heads interview is 1/2
4. My credence that my current interview is a heads interview is equal to the fraction of interviews which are heads interviews
5. My credence that I have been assigned one heads interview is equal to the fraction of experiments which have one heads interview

You seem to assert that 4 and 5 are true by definition, but they're not. Given the definition of the term "credence", and given the truth of 1, 2, and 3, it must be that one or both of 4 and 5 are false. — Michael

4 and 5 aren't true by definition; rather, they are definitions. Definitions specify how terms are to be used in a given context, and it's the subsequent argumentation that can be evaluated as true or false in light of those definitions. In this case, it appears that you intend for "My credence that my current interview is a heads interview" and "My credence that I have been assigned one heads interview" to both represent Sleeping Beauty's episodic perspective, yet the defining clause in 5 would be more appropriate for a timeline perspective. This potentially equivocal definition could lead to confusion. If, however, 5 is meant to convey a timeline perspective, then 4 and 5 are both reasonable, complementary definitions addressing distinct questions.

So simply asserting that "the fraction of interviews which are heads interviews is 1/3, therefore my credence that my current interview is a heads interview" is a non sequitur.

Granted, such an argument would be a bit quick and likely an enthymeme. However, I didn't present it in that way. I provided more explicit steps and premises in my previous post, explaining how attending to the distinction between the two epistemic perspectives (timeline and episodic) allows us to conclude that, in the episodic perspective, P(H)=P(T-first), P(T-first)=P(T-second), and since all three events are mutually exclusive in this perspective, they must each have a probability of 1/3.

This overlooks the issue that your credence can change over time when your epistemic perspective changes. — Pierre-Normand

It doesn’t. If my credence in A changes then my credence in B will change along with it (or my credence in A iff B will change).

4 and 5 aren't true by definition; rather, they are definitions. — Pierre-Normand

I have since changed the wording (although it makes no difference, this wording is just more exact).

4. The degree to which I believe that my current interview is a heads interview is equal to the fraction of interviews which are heads interviews
5. The degree to which I believe that I have been assigned one heads interview is equal to the fraction of experiments which have one heads interview

These are neither definitions nor true by definition.

One or both must be false given 1, 2, or 3.

These mean two different things:

1. My credence favours this being a tails awakening rather than a heads awakening
2. There are more tails awakenings than heads awakenings

You can argue that 1 is true because I believe that 2 is true, but then my argument above shows that this is unreasonable.

The fraction of awakenings which are tails awakenings is the wrong fraction to base one’s credence that this is a tails awakening on.

My current interview being the first or the second T-awakening are exclusive events. — Michael

They indeed are. As you get involved in the experiment and your perspective shifts from the timeline (before the experiment begins) to the episodic one (whenever you are awakened), relative to the current interview, the two possible outcomes T-first and T-second now are exclusive. However, this doesn't mean that their probabilities shift from 1/2 each (as they were in the timeline perspective) to 1/4 each (as they would if a second coin were tossed to chose only one to be actualized between them). Instead, your finding yourself in a T-awakening episode doesn't exclude the other one from the current timeline; it merely shifts it to the other concurrent episode in this timeline.

The difference between the Thirder and Double-halfer reasoning can be illustrated this way:

If a coin lands heads, you are allowed to pick one ball from an unlabeled "H-bag" containing one blue ball. If the coin lands tails, you are allowed to pick two balls, one at a time, from a "T-bag" containing two red balls. Therefore, there are three possible ball picking episodes: B, R1, and R2. We assume, as usual, that when the opportunity arises to pick a ball from a bag, you forget if it's your first or second opportunity.

A Double-halfer would reason that since you were equally likely to have been presented with a T-bag and, in that case, you are equally likely to be experiencing R1 or R2, P(R1)=P(R2)=1/4.

In contrast, a Thirder would point out that picking a red ball doesn't exclude the other one from the current timeline but rather guarantees it. The implication of this is that the additional opportunity to pick a second red ball from the T-bag does not reduce P(R1) relative to P(B) but rather increases P(R) = P(R1 or R2) by providing a second opportunity. This is especially apparent from the timeline perspective, where P(B)=P(R1)=1/2 regardless of how many more red balls there might be in the T-bag for you to all pick consecutively. The equiprobability between P(R1) and P(B) doesn't change when we shift from the timeline perspective to the episodic perspective, because on each picking occasion, although you can't know if it's the first one, you do know that P(R|first) = P(B|first). In other words, if you were to ask the experimenter presenting you the bag if this is your first pick, and receive a truthful positive answer, you would know that P(B)=P(R1)= 1/2. The next step in the argument is the straightforward inference from P(R|first) = P(B|first) to P(R1) = P(B). Given that, on all accounts, P(R1) = P(R2), and that all three outcomes are exclusive from the episodic perspective, it follows that they all have a probability of 1/3

But I brought up something like this here: — Michael

Indeed, I intended to address this separately but haven't yet gotten round to doing it. This will be my next order of business.

These mean two different things:

1. My credence favours tails awakenings
2. There are more tails awakenings than heads awakenings

I don’t think we can move forward if you insist that they mean the same thing. — Michael

It's precisely because they mean different things that I've provided detailed arguments for deducing 1 from 2 (alongside with other premises). However, the truth of 2 certainly is relevant to the deduction of 1. Nobody would be a Thirder in a scenario where coins lading tails would generate as many awakenings as coins landing heads.

It's precisely because they mean different things that I've provided detailed arguments for deducing 1 from 2 (alongside with other premises). However, the truth of 2 certainly is relevant to the deduction of 1. Nobody would be a Thirder in a scenario where coins lading tails would generate as many awakenings as coins landing heads. — Pierre-Normand

If they mean different things then 4 and 5 are neither definitions nor true by definition, and given that 1, 2, and 3 are true (and that "the degree to which I believe that" means the same thing in 4 and 5), one or both of 4 and 5 are false.

The degree to which I believe that my current interview is my only interview is equal to the degree to which I believe that I have been assigned only one interview.

When I said that the only things that matter are:

1. She has either one or two interviews determined by a fair coin toss and
2. She doesn’t know if she’s already had one

I was referring to her just waking up, not being told any further information. — Michael

And the only point of mentioning new information, was to show that the information you ignore has meaning. Not to solve the problem or alter the problem. But you knew that.

No prior probability is ruled out here when woken so your example isn't equivalent. — Michael

And again, you keep using circular logic. You deny that events with non-zero prior probability are "ruled out" in your solution. So you claim that my solution, which does "rule out," must be wrong. This is a fallacy; your presumption that you are right is your only defense. You have never argued for why you think they aren't events.

So, lets try this variation of what you most recently proposed. The only changes are to make the conditions you require, as in that quote above, easier to describe with both words and probabilities:

1. SB is put to sleep on day 0, and a coin is flipped.
2. The counting of days is facilitated by a tear-away calendar that is posted on the door to her room at midnight. It starts at Day 1, and as the top page is torn off at midnight every day thereafter, the day numbers are increased by 1.
3. If the coin landed Heads, then an N-sided die (N>=2) is rolled and placed on a shelf by the door.
4. If the coin landed Tails, then two N-sided dice are rolled and placed on the shelf. (If they both land on the same number, the roll is repeated until they are different.)
5. Over the next N days, if the calendar number matches a die number, she is woken and asked for her credence.

Whether Heads or Tails, she can wake once or twice over the next N days. The prior probability of a waking is the same on every day (your method didn't make this true). And she will never know if it is the first waking, the second waking, or the only waking.

Do you agree that this does the same things your latest experiment does (if N=14) with the unimportant exception that the prior probability distribution over the N days varies in your version?

And again, you keep using circular logic. You deny that events with non-zero prior probability are "ruled out" in your solution. So you claim that my solution, which does "rule out," must be wrong. This is a fallacy; your presumption that you are right is your only defense. You have never argued for why you think they aren't events. — JeffJo

I can't prove a negative. If there is some prior probability that is ruled out when woken then tell me what it is.

If you can’t then I have every reason to accept that there isn’t one.

@Michael

Consider Leonard Shelby's journey through the "Sleeping Beauty Zoo". In this zoo, each fork in the path presents two options - one path (H-path) leads to a hippo enclosure, while the other (T-path) leads to a tiger enclosure followed by a toucan enclosure. Each path ends with a new similar fork until the zoo exit is reached. Due to Leonard's condition of anterograde amnesia, he forgets his previous encounter (with an enclosure or fork) whenever he reaches a new enclosure. Despite his memory loss, Leonard knows that typically, due to the zoo peculiar layout, he encounters each of the three animals with equal frequency, on average, during his visits.

Now, let's look at a particular moment of Leonard's visit. As he walks, before reaching a new enclosure, he might reason this way: "Since each fork in the path gives an equal chance of leading to a T-path or an H-path, there is a 50% chance that the next enclosure I'll see will have a hippo." Thus, when he approaches an enclosure, he might conclude there is a 25% chance of it being a tiger enclosure, and a 25% chance of it being a toucan enclosure.

Is this reasoning accurate? Not quite, because it neglects a crucial shift in perspective from "timeline" to "episodic".

To clarify, let's imagine Leonard uses a series of coin tosses to predetermine his path at each fork and records the sequence in his notebook. On average, his predetermined path will lead him to an equal number of hippo, tiger, and toucan enclosures. So, whenever he approaches an enclosure during his visit (even if he exits the zoo after only one path segment), he can reasonably assume a 1/3 chance that it is a hippo, tiger, or toucan enclosure.

One might ask: Isn't there a contradiction between Leonard's initial assumption of a 50% chance of being on a H-path and his subsequent conclusion of a 1/3 chance of seeing a hippo in the enclosure?

The answer is no. Although Leonard travels an equal number of H-path and T-path segments, he encounters twice as many T-enclosures. Because of the way the T-enclosures map (two to one) to the T-path segments, Leonard encounters hippos and tigers with equal frequency after taking any fork (which makes P(hippo) = P(tiger), and he encounters tigers and toucans with equal frequency while traveling any T-path (which makes P(tiger) = P(toucan). Since all three possibilities are exclusive from the episodic perspective, he expects all three animals to appear with equal probability at any enclosure, even though T-paths exclude H-paths, and tiger-enclosures and toucan-enclosures occur concurrently.

Keeping this in mind, let me now address you "flowchart" argument:

1. Sleeping Beauty is given amnesia
2. She is asked her credence that a coin has been tossed
3. A coin is tossed
4. If the coin lands tails then:
4A. She is given amnesia
4B. She is asked her credence that a coin has been tossed

Thirder reasoning is that because step 2 is twice as likely to occur as step 4B then I am twice as likely to be in step 2 as step 4B.

Halfer reasoning is that because step 2 is twice as likely to occur as step 4B and that because if 4B will occur then I am equally likely to be in step 2 as step 4B then I am three times as likely to be in step 2 as step 4B. — Michael

From the timeline perspective, the following scenarios are equiprobable:

1. SB asked P(second-awakening), coin lands heads, exit
2. SB asked P(second-awakening), coin lands tails, SB asked P(second-awakening), exit

where P(second-awakening) is Sleeping Beauty's credence, on the occasion of an awakening, that it is her second awakening within the current experimental run.

From the episodic perspective, Sleeping Beauty knows that conditionally on her present awakening being the first, it is equally probable that it is a H-awakening (and that the coin will land heads) or that it is a T-first-awakening (and that the coin will land tails). She also knows that in the event the coin will land (or has landed) tails, it is equiprobable that she is experiencing a T-first-awakening or a T-second awakening. Since the three possible outcomes are exclusive from her episodic perspective, their probabilities must sum up to 1 and since P(H-awakening) = P(T-first-awakening) and P(T-first-awakening) = P(T-second awakening), all three possible outcomes must have probability 1/3.

The Halfer error lies in incorrectly treating the T-first-awakening and T-second-awakening as equally probable and mutually exclusive alternatives within the T-timeline. This leads to an improper division of the T-timeline's probability between these two events. In reality, the T-first-awakening and T-second-awakening are part of the same timeline and their probabilities should not be divided, but understood as part of the cumulative likelihood of being on the T-timeline. This is similar to the presence of as many tiger or toucan enclosures along Leonard's overall (or average) path as there are hippo enclosure increasing the overall probability of him encountering a T-enclosure on each occasion he approaches one to 2/3 (i.e. P(T-enclosure) = (P(tiger or toucan)/P(tiger or toucan or hippo) = 2/3).

whenever she awakens, the coin landed (or will land) tails two times out of three — Pierre-Normand

This is not true. There are three possible awakenings, Monday-Heads, Tuesday-Heads, Tuesday-Tails, and SB's job on awakening is determine the probability that she is experiencing each of these. The coin has a 50% chance of landing heads, and if it does, the awakening will be on Monday 100% of the time. Therefore, P(Monday-Heads) = 50%. The coin has a 50% chance of landing tails, and if it does, the awakening will be on Monday 50% of the time, and Tuesday 50% of the time. Therefore, P(Tuesday-Heads) = P(Tuesday-Tails) = 25%. If this is true, and I don't see how it can be reasonably argued against, on each awakening the coin is equally likely to be heads and tails.

This is not true. There are three possible awakenings, Monday-Heads, Tuesday-Heads, Tuesday-Tails, and SB's job on awakening is determine the probability that she is experiencing each of these. The coin has a 50% chance of landing heads, and if it does, the awakening will be on Monday 100% of the time. Therefore, P(Monday-Heads) = 50%. The coin has a 50% chance of landing tails, and if it does, the awakening will be on Monday 50% of the time, and Tuesday 50% of the time. Therefore, P(Tuesday-Heads) = P(Tuesday-Tails) = 25%. If this is true, and I don't see how it can be reasonably argued against, on each awakening the coin is equally likely to be heads and tails. — hypericin

Are you evaluating the probabilities of the three possible outcomes occurring from the point of view of an external observer or from Sleeping Beauty's own epistemic perspective whenever she awakens?

From an external observer perspective, each awakening—H-awakening, T-first-awakening, and T-second-awakening—have an equal chance of 1/2 to occur. Note that the sum of these probabilities exceeds 1 because the two T-awakenings aren't mutually exclusive but rather, they are concurrent within the same timeline.

If you are evaluating the probabilities from Sleeping Beauty's own epistemic perspective when she awakens, then a careful analysis of the situation shows that the probability of each is 1/3.

Consider first the two possible outcomes conditional on today being Monday. Since Sleeping Beauty always is awakened on Monday regardless of the coin toss result, P(Monday-Heads) = P(Monday-Tails). Consider next the two possible outcomes conditional on the coin having landed tails. Since in that case Sleeping Beauty is awakened once on Monday and once on Tuesday, P(Monday-Tails) = P(Tuesday-Tails), which is something that the Thirders, Halfers and Double-halfers all agree on. We therefore have that P(Monday-Heads) = P(Monday-Tails) = P(Tuesday-Tails). Lastly, since Sleeping Beauty isn't inquiring about the probabilities that any of those three outcomes will occur at least once during her current experimental run, but rather about the probability that her current awakening episode is the realization of one of those three outcomes, the three possibilities are exclusive and exhaustive, and their probabilities must therefore sum up to 1. They therefore all three are 1/3, and P(Tails) = P(Monday-Tails) + P(Tuesday-Tails) = 2/3.

This argument is illustrated in a more picturesque way in my variation: Leonard Shelby Visits the Sleeping Beauty Zoo.

Now, let's look at a particular moment of Leonard's visit. As he walks, before reaching a new enclosure, he might reason this way: "Since each fork in the path gives an equal chance of leading to a T-path or an H-path, there is a 50% chance that the next enclosure I'll see will have a hippo." Thus, when he approaches an enclosure, he might conclude there is a 25% chance of it being a tiger enclosure, and a 25% chance of it being a toucan enclosure.

Is this reasoning accurate? — Pierre-Normand

1. The next enclosure is the toucon enclosure iff I first turned right at the fork (P = $1\over2$) and then passed the tiger enclosure.

2. My credence that the next enclosure is the toucon enclosure is equal to the probability that the first event happened multiplied by the probability that the second (dependent) event happened.

Having amnesia is no excuse to reject (2). Even with amnesia I know that (1) is true.

If my credence that the next enclosure is the toucon enclosure is $1\over3$ then the probability that I passed the tiger enclosure, if I turned right at the fork, is $2\over3$, but then my credence that the next enclosure is the tiger enclosure is $1\over6$.

If the probability that I passed the tiger enclosure, if I turned right at the fork, is $1\over2$, then my credence that the next enclosure is the toucon enclosure is $1\over4$ (as is my credence that the next enclosure is the tiger enclosure).

Either way my credence that the next enclosure is the hippo enclosure is $1\over2$.

From the episodic perspective, Sleeping Beauty knows that conditionally on her present awakening being the first, it is equally probable that it is a H-awakening (and that the coin will land heads) or that it is a T-first-awakening (and that the coin will land tails). She also knows that in the event the coin will land (or has landed) tails, it is equiprobable that she is experiencing a T-first-awakening or a T-second awakening. Since the three possible outcomes are exclusive from her episodic perspective, their probabilities must sum up to 1 and since P(H-awakening) = P(T-first-awakening) and P(T-first-awakening) = P(T-second awakening), all three possible outcomes must have probability 1/3. — Pierre-Normand

This appears to be repeating Elga's argument:

P(T₁|T₁ or T₂) = P(T₂|T₁ or T₂)
∴ P(T₁) = P(T₂)

P(H₁|H₁ or T₁) = P(T₁|H₁ or T₁)
∴ P(H₁) = P(T₁)

∴ P(H₁) = P(T₁) = P(T₂)

Those first two inferences need to be justified.

I can't prove a negative. — Michael

Yet that is the basis of your argument. You even reiterate it here. And it is part of your circular argument, which you used this non sequitur to divert attention from:

1. M's solution is right.
2. J's solution gets a different answer, so it must be wrong.
3. J's solution "rules out" an event.
4. M's does not. In fact, it it says that event doesn't exist (the "negative" you claim).
5. That must be the error, since there must be an error (see #2).

If there is some prior probability that is ruled out when woken then tell me what it is.

I have. You ignore it. But this is a fallacious argument. Claiming I did something different does not prove the way you handles the different thing is right and mine was wrong.

If you can’t then I have every reason to accept that there isn’t one.

Quite an ultimatum, from one who never answers questions and ignores answers he can't refute. Since you haven't proven why the event "Heads&Tuesday" doesn't exist - and in fact can't, by your ":can't prove a negative" assertion, I have every reason to accept that it does exist.

+++++

Here is a version of your latest procedure. The only non-cosmetic change is that I made it easier to determine the prior probabilities.The "prior probabilit[ies] that [are] ruled out" are easily identified. I just need you to agree that it is equivalent to your procedure first. IF YOU DON'T, I HAVE EVERY REASON TO ACCEPT THAT IT IS.

1. She is put to sleep on day 0.
2. A coin is flipped.
3. If the coin landed on Heads, then an N-sided die is rolled, where N>=2. She is woken on day D1 - that is, D1 days after day 0 - where D1 is the result of this roll, and asked her credence.Then she is put back to sleep.
4. If the coin landed Tails then two N-sided dice are rolled. If they land on the same number, repeat the roll until they are different.She is woken on day D1 and day D2, and sked her credence. On the first of these days, she is put back to sleep with amnesia.

Notes on the changes; all but one are cosmetic only:

• By numbering day 0, you allow numbering all days.
• You left out the coin flip, but we all know where it has to go.
• The singular of "dice" is "die."
• Your tenses don't mesh with stating when the coin is flipped.
• There is nothing special about 14. Any number works - it doesn't even have to be even - that allows two wakings.
• Rolling two N-sided dice at the same time, rather than two N/2 dice sequentially, makes the distribution of wakings over the N days uniform rather than a complicated function. So it simplifies any calculations based on it, if needed.

Whether the coin landed on Heads or Tails she can wake on any day between Day 1 and day N.

Something is indeed is ruled out when she wakes. You just don't recognize how the sample space you described doesn't recognize it.

I just need you to agree that it is equivalent to your procedure first. — JeffJo

It's not. You say:

"If the coin landed on Heads, then an N-sided die is rolled, where N>=2. She is woken on day D1 - that is, D1 days after day 0 - where D1 is the result of this roll, and asked her credence. Then she is put back to sleep."

In my example she isn't put back to sleep. The experiment just ends. The same with her second tails interview. So we have no idea how many days the experiment will last. It could be anywhere between 1 and N days.

Something is indeed is ruled out when she wakes. — JeffJo

What is ruled out?

The next enclosure is the toucon enclosure iff I first turned right at the fork (P = 1/2) and then passed the tiger enclosure.

2. My credence that the next enclosure is the toucon enclosure is equal to the probability that the first event happened multiplied by the probability that the second (dependent) event happened. — Michael

(I had assumed that the H-path was the left path at half of the forks, but this is inconsequential since Leonard always forgets which path he took.)

Your reasoning is not entirely misguided. Let's consider a typical path Leonard might navigate through the zoo, guided by four pre-tossed coins:

Tiger, Toucan -- Hippo -- Hippo -- Tiger, Toucan -- Exit

On this path, Leonard will encounter six enclosures within four path segments. The frequency of encounters with each type of enclosure are typically equal, regardless of the manner in which they are divided among different path segments.

First, let's revisit the Thirder argument regarding any enclosure Leonard might approach:

1. Conditionally on its being a first encounter on a path segment, P(Tiger) = P(Hippo)
2. Conditionally on Leonard being on a T-path segment, P(Tiger) = P(Toucan)
3. The three possible outcomes are exhaustive and mutually exclusive
4. Therefore, P(Tiger) = P(Hippo) = P(Toucan) = 1/3

This makes sense because, on a typical journey through the zoo, like the one illustrated above, Leonard can expect to encounter each type of enclosure with the same frequency, variance notwithstanding.

The Double-halfer analysis, on the other hand, posits Leonard on either a H-path or a T-path segment with equal probability. The justification for this comes from assuming that Leonard considers his current path segment as equally likely to be a H-path or a T-path from the time he took the previous fork. This reasoning is relevant and accurate until Leonard approaches an enclosure and has the opportunity to infer which path segment it is located on. If Leonard were to ignore the type of enclosure and only guess the path he's on, confirming his guess only when he reaches the next fork (or zoo exit), regardless of the number of times he might make this guess along the path, then he would find that he was on a H-path half of the time, supporting the Halfer thesis. (This, however, wouldn't vindicate the Double-halfer thesis since encountering tigers does not exclude, but rather guarantees, his also encountering toucans on the same path, and vice versa.)

However, if we acknowledge that each encounter with an enclosure is an occasion for Leonard to be located on a H-path, and the layout of the zoo (and path navigation process) ensures that these occasions occur 1/3 of the time, we realize that, from this new episodic perspective, the probability that Leonard is on a H-path when he approaches an enclosure isn't independent of the number of such encounters. This is because the likelihood of the encountered enclosure being a H-enclosure (and thus Leonard being on a H-path) isn't solely determined by the process that placed Leonard on this path (the fork and coin toss) but also by the proportion of occasions Leonard has to be on such a path.

I’m only considering one fork as only that is comparable to the Sleeping Beauty problem. What’s true of multiple forks isn’t true of one fork, as evidenced by (1):

1. The next enclosure is the toucon enclosure iff I first turned right at the fork (P = $1\over2$) and then passed the tiger enclosure.

This is only true if there is one fork.

So what is wrong about my analysis of one fork?

1. Conditionally on its being a first encounter on a path segment, P(Tiger) = P(Hippo)
2. Conditionally on Leonard being on a T-path segment, P(Tiger) = P(Toucan)
3. The three possible outcomes are exhaustive and mutually exclusive
4. Therefore, P(Tiger) = P(Hippo) = P(Toucan) = 1/3 — Pierre-Normand

As it stands your conclusion is a non sequitur. You need to justify this inference:

P(Hippo|Hippo or Tiger) = P(Tiger|Hippo or Tiger)
Therefore P(Hippo) = P(Tiger)

You need to prove this inference:

P(Hippo|Hippo or Tiger) = P(Tiger|Hippo or Tiger)
Therefore P(Hippo) = P(Tiger) — Michael

This inference follows if we consider what is excluded by the condition "Hippo or Tiger". The case where Leonard is seeing a second enclosure on his path (which always contains toucans) is excluded. Since this second encounter is guaranteed whenever Leonard sees tigers, adding this extra encounter doesn't affect the relative probabilities of P(Hippo) and P(Tiger). However, it does reduce the total probability of him facing either a hippo or a tiger enclosure, i.e., P(Hippo or Tiger).

This reasoning becomes more intuitive if we consider Leonard's entire visit to the zoo and adopt a frequency approach. I understand that you dislike this method, but bear with me as it may illuminate a blind spot in your understanding. Leonard can reason that since he is traversing as many H-path segments as T-path segments on average, he is encountering as many tiger enclosures as he is hippo enclosures. This is because each type of path segment contains precisely one enclosure of each kind, respectively. The presence of toucan enclosures on T-path segments doesn't diminish the number of tiger enclosures in proportion to the hippo enclosures he encounters, but it does increase the total number of T-enclosures (or average number, if we consider only one single fork) relative to the number of hippo enclosures he encounters.

Just like in the Sleeping Beauty case, each fork (or coin toss) can be seen as a random generator of T- and H-events, producing twice as many of the former. Creating more than one T-awakening when the coin lands tails (or more than one encounter with an enclosure when a T-path is taken) dilutes the probability of all the individual events (since they are being experienced one at a time) but increases the total probability of the T-events. Lastly, since the additional T-events being generated aren't generated to the exclusion of the first one, but rather in addition to it, they don't alter the relative probabilities of an H-awakening and a T-first-awakening (or of Hippo relative to Tiger).

I’m only considering one fork as only that is comparable to the Sleeping Beauty problem. What’s true of multiple forks isn’t true of one fork, as evidenced by (1);

1. The next enclosure is the toucon [sic] enclosure iff I first turned right at the fork (P = 1/2) and then passed the tiger enclosure.

This isn’t true if there are two forks.

So what is wrong about my analysis of one fork? — Michael

The same reasoning about probabilities and frequencies that applies to multiple forks (or repeated Sleeping Beauty experiments) also holds for a single fork (or a single Sleeping Beauty experiment).

Consider a brief visit to the zoo where Leonard only takes the first fork, with the intention of taking the exit-shortcut at the next fork. In such cases, half of the time, he sees hippos, and the other half, he sees both tigers and toucans. Given this, Leonard can reason that since his brief zoo visits will put him in front of hippo, tiger, or toucan enclosures with equal probabilities (namely 1/2 each), the probability that the current enclosure he is seeing is a toucan enclosure is 1/3.

The Double-halfer reasoning errs by treating the Tiger and Toucan events as mutually exclusive—as if a second coin toss generates a second, probabilistically independent event—when in fact they both occur on the same timeline whenever either of them occurs.

Your claim that "The next enclosure is the toucan enclosure iff I first turned right at the fork (P = 1/2) and then passed the tiger enclosure," is an assumption that can't be put forward without begging the question against the Thirder. You need to substantiate, rather than presuppose, that when you're nearing an enclosure, there's a 1/2 chance the path you're on is a T-path.

We can, however, agree on the following: P(Hippo|H-path) = 1, and P(Toucan|T-path) = P(Tiger|T-path) = 1/2. The Thirder argument, though, posits that whenever Leonard faces an enclosure, P(H-path) = 1/3, and consequently,

P(Hippo) = P(Hippo|H-path)P(H-path)+P(Hippo|T-path)P(T-path) = 1/3.

Likewise,

P(Toucan) = P(Toucan|T-path)P(T-path)+P(Toucan|H-path)P(H-path) = (1/2)(2/3) = 1/3.

The justification for P(H-path) = P(Hippo) = 1/3 was provided above. The creation of additional T-encounters when a T-path is taken dilutes the probability of individual T-encounters, but raises the cumulative probability of T-encounters. This doesn't change the relative probabilities of Hippo and the first Tiger encounter, since the extra encounters are not produced at the exclusion of the first one, but are in addition to it. It's this crucial point that differentiates the Thirder and Double-Halfer perspectives.

Consider first the two possible outcomes conditional on today being Monday. Since Sleeping Beauty always is awakened on Monday regardless of the coin toss result, P(Monday-Heads) = P(Monday-Tails). Consider next the two possible outcomes conditional on the coin having landed tails. Since in that case Sleeping Beauty is awakened once on Monday and once on Tuesday, P(Monday-Tails) = P(Tuesday-Tails), which is something that the Thirders, Halfers and Double-halfers all agree on. We therefore have that P(Monday-Heads) = P(Monday-Tails) = P(Tuesday-Tails). Lastly, since Sleeping Beauty isn't inquiring about the probabilities that any of those three outcomes will occur at least once during her current experimental run, but rather about the probability that her current awakening episode is the realization of one of those three outcomes, the three possibilities are exclusive and exhaustive, and their probabilities must therefore sum up to 1. They therefore all three are 1/3, and P(Tails) = P(Monday-Tails) + P(Tuesday-Tails) = 2/3. — Pierre-Normand

This is a fallacy:

If Monday, P(Monday-Heads) = P(Monday-Tails)
If Tails, P(Monday-Tails) = P(Tuesday-Tails)
Therefore, P(Monday-Heads) = P(Monday-Tails) = P(Tuesday-Tails)

The conclusion doesn't follow, because the first two equalities depend on the conditionals being true.

You can see this by observing that

P(Monday-Heads) = 1/2
P(Monday-Tails) = 1/4
P(Tuesday-Tails) = 1/4

Also satisfies the two conditional statements, without satisfying the conclusion