On each day in the range 1 to N, the prior probability that she will be woken on that day AND the coin landed on Heads is 1/N. — JeffJo

Pr(Heads & Day = D) = 1/2 * 1/N.

That aside, using your above reasoning, in the normal problem the prior probability that she will be woken on Tuesday and the coin landed on Heads is 0, and the prior probability that she will be woken on Monday and the coin landed on Heads is 1/2.

So when she "rules out" Pr(Heads & Day = Tuesday) she's "ruling out" some Pr = 0, not some Pr = 1/4. "Ruling out" some Pr = 0 after waking does nothing because it was already "ruled out".

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. — Pierre-Normand

My wording may have been imprecise.

Both of these are true (note the tense):

1. To reach the toucon enclosure I must first turn right at the fork and then pass the tiger enclosure
2. The probability that I will turn right at the fork is $1\over2$

When I wake and consider my credence that the next enclosure is the toucon enclosure I consider what must have happened (or not happened) for the next enclosure to be the toucon enclosure. I know that I must have first turned right at the fork (A) and then passed the tiger enclosure (B).

P(A, B) = P(A) × P(B|A)

My claim is that the probability of having turned right at the fork is equal to the probability of turning right at the fork, i.e. $1\over2$.

Your claim is that the probability of having turned right at the fork is equal to the fraction of all encountered enclosures which are right-side enclosures, i.e. $2\over3$.

I don't think your claim makes any sense. The probability of the first event having happened isn't determined by what could happen after that first event happens. The probability of the first event having happened is determined only by the probability of that first event happening.¹

Your conclusion only applies if I'm dropped into an enclosure at random, perhaps via parachute. This is your so-called "episodic perspective" (where it’s not the case that I turned right at the fork; it’s only the case that I’m on the right-side path). But given that this isn't what happens to me, I shouldn't reason this way.

¹ _{Where no new relevant information is learned.}

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)

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?

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.

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.

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.

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.

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.

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.

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 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.

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.

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.

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.

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.

Since I now know that I will soon rationally infer that this note was written during an H-awakening with probability 1/3 (on the basis of no new information), I can already infer this right now. — Pierre-Normand

Then before the experiment starts the thirder will say "since I now know that I will soon rationally infer that the coin will have landed heads with probability 1/3 (on the basis of no new information), I can already infer this right now, before the coin is tossed."

But I think this is wrong.

Making n large makes Sleeping Beauty's epistemic situation on Wednesday, when she receives a note, nearly identical to her situation when she wrote the note, since the Bayesian updating she can perform on the basis of the note being unique is negligible. — Pierre-Normand

They're not nearly identical. On Wednesday she knows that she only had the opportunity once. When she wrote the note she didn't know that it was her only opportunity. So contrary to the above, there is new information on Wednesday.

Note that when Sleeping Beauty doesn't receive a note on Wednesday, her credence P(H) = 1/2 doesn't merely differ in value from her credence P(H) = 1/3 during awakenings; the predicates P() also have different meanings. During awakenings, P(H) refers to the odds that her current awakening episode is occurring during a coin toss that landed heads. On Wednesday, P(H) refers to the odds that the experimental run she is exiting from was an H-run. 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 address this here.

the prior probability for any event is based set of all possibilities that could occur — JeffJo

And her being woken a second time if the coin lands heads can't occur, which is why its prior probability is 0, not $1\over4$.

And that is important here, because you are insisting that a two-day collection of events (I'll call your two passes Monday and Tuesday since only the order matters to anything). You are calling Monday+Tails and Tuesday+Tails the same event. But to SB, who can only observe one at a time, they are distinct events that each have half the prior probability that you assign to the combination. — JeffJo

There aren't two days in my example.

When I will read again the note that I am currently writing, on Wednesday, I will be able to rationally infer that it is twice as likely that this note was written by me on the occasion of a T-awakening. — Pierre-Normand

That depends on the probability that you will be given the opportunity to write a note. If that probability is 1/2 then it won't be rational on Wednesday to infer that it is twice as likely that this note was written by me on the occasion of a T-awakening.

That was only in the specific case where n = 2. As n grows larger, P(H) tends towards 1/3. — Pierre-Normand

And as it grows smaller, P(H) tends to 1. I don't understand the relevance of any of these three answers.

Why is the correct answer given by any of these situations, let alone by the situation where n is arbitrarily large?

Yes, I got the maths wrong there (and deleted my post before you replied; apologies).

Though I don't see why I should accept your claim that if "she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note."

As you said yourself, if the probability of her writing a note is 1/2 then if she finds exactly one note on Wednesday then her credence in Heads is 1/2.

As the occasions to write a note become rarer (e.g. 1/n with n >> 1), the frequency of those overlapping notes become negligible (n times as many single notes are received as double notes) and Sleeping Beauty's epistemic state (i.e. the value of her credence) approaches asymptotically her epistemic state as she was writing the note. And, as I had suggested in my previous post, this is because when she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note. — Pierre-Normand

If heads and n = 100 then the probability of writing a note is 1/100

If tails and n = 100 then the probability of writing exactly one note is 1/100.

So if she finds exactly one note on Wednesday then her credence in heads is 1/2.

@Pierre-Normand

These cannot all be true:

1. Credence "is a statistical term that expresses how much a person believes that a proposition is true"
2. My current interview is a heads interview iff I have been assigned one heads interview
3. The fraction of interviews which are heads interviews is $1\over3$
4. The fraction of experiments which have one heads interview is $1\over2$
5. My credence that my current interview is a heads interview is equal to the fraction of interviews which are heads interviews
6. My credence that I have been assigned one heads interview is equal to the fraction of experiments which have one heads interview

Propositions like 5 and 6 might usually be true, but they are not true by definition.

Given 1 and 2, my credence that my current interview is a heads interview is equal to my credence that I have been assigned one heads interview.

Therefore given 1, 2, 3, and 4, one or both of 5 and 6 is false.

So how will your reasoning let you choose between 5 and 6 without begging the question?

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. 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
P2. I am certain that my current interview is my only interview if and only if I have been assigned only one interview
C1. Therefore 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 (from P1 and P2)
P3. I am certain that if I have been assigned at random by a fair coin toss either one or two interviews then the probability that I have been assigned only one interview is $1\over2$
P4. I am certain that I have been assigned at random by a fair coin toss either one or two interviews
C2. Therefore I am certain that the probability that I have been assigned only one interview is $1\over2$ (from P3 and P4)
P5. The degree to which I believe that I have been assigned only one interview is equal to what I am certain is the probability that I have been assigned only one interview
C3. Therefore the degree to which I believe that I have been assigned only one interview is $1\over2$ (from C2 and P5)
C4. Therefore the degree to which I believe that my current interview is my only interview is $1\over2$ (from C1 and C3)
P6. I am certain that my current interview is my only interview if and only if the coin landed heads
C5. Therefore the degree to which I believe that the coin landed heads is $1\over2$ (from P1, C4, and P6)

Suppose we update the protocol so that on rare occasions, which present themselves with equal probability on each awakening episode, Sleeping Beauty is able to write down a note saying "I have now been awakened and interviewed." She can retain this note and read it again on Wednesday. Upon rereading the note on Wednesday, she can reason that it is twice as likely that such a note was produced if the coin landed tails since she would have been twice as likely to write it during such an experimental run. — Pierre-Normand

Not necessarily.

Assume a probability of 1/2 each time. The probability of writing it if the coin landed heads is 1/2. The probability of writing it (at least once) if the coin landed tails is 3/4. It is 3/2 times as likely to have been tails.

Assume a probability of 2/3 each time. The probability of writing it if the coin landed heads is 2/3. The probability of writing it (at least once) if the coin landed tails is 8/9. It is 4/3 times as likely to have been tails.

But notice that as the probability of writing a note each time approaches 1 the "greater likelihood" of it having been tails gets smaller, approaching 1.

Also, apply the principle of this reasoning to this.

A fair coin toss is equally likely to be heads as tails, she is less likely (not guaranteed) to wake if tails, therefore if she does wake then she reasons that it's less likely to be tails. In waking she rules out TTT.

If the answer to this problem is 4/7 then the answer to the normal problem is 1/2. If the answer to the normal problem is 1/3 then the answer to this problem is 1/2.

You are asking if it has occurred when you know it hasn't — JeffJo

Sleeping Beauty doesn't know that it hasn't occurred. She has amnesia.

"Prior" refers to before information revealed — JeffJo

So after waking, and before new information is revealed, the prior probability that the coin landed heads and that she is being woken for a second time is 1/4? Then what new information is revealed after waking that allows her to rule out that prior probability? It can’t be “being awake” because that isn’t new information. And it can’t be “being asked her credence” because if she has just woken then she knows with certainty that she is about to be asked her credence.

The simplest answer is the correct one. The prior probability that the coin will or did land heads and that she is being or will be woken for a second time is and always was 0. That’s just a rule of the experiment.

Nothing is ruled out when woken or asked her credence that wasn’t already ruled out before the experiment started.

Even Elga understood this:

Before being put to sleep, your credence in H was 1/2. I’ve just argued that when you are awakened on Monday, that credence ought to change to 1/3. This belief change is unusual. It is not the result of your receiving new information

The prior probabilities, for an awakened SB, are 1/4 for each. — JeffJo

It makes no sense to say that when she wakes there is then a prior probability that she’s “asleep” of $1\over4$ that is immediately ruled out. If some prior probability is ruled out when she wakes then it must be that the prior probability is established before she wakes. But the prior probability that she will wake a second time if the coin lands heads is 0, not $1\over4$.

If she’s awake then it is just the case that either the coin hasn’t been tossed or it landed tails.

I can set out an even simpler version of the experiment with this in mind:

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

The prior probability that step 2 will happen is 1 and the prior probability that step 4B will happen is $1\over2$.

When Sleeping Beauty is given amnesia she knows that she is in either step 1 or step 4A. When asked her credence she knows that she is in either step 2 or step 4B.

No prior probability is ruled out when she is given amnesia or asked her credence.

You have to say that in step 2 her credence is $1\over3$. I have to say that in step 2 her credence is $1\over4$. Of note is that neither of us can just apply the principle of indifference and say that in step 2 her credence is $1\over2$.

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.

There is no theory of when prior probabilities are established. But if there were, it would be fom the start, not before the start. — JeffJo

So when is this alleged P(X) = 1/4 prior established if not before the experiment starts?

It cannot be when she is asked her credence as you’ve said that in being asked her credence this prior is reduced to 0.

But it also cannot be before because there is no “current interview” before she’s asked her credence.

So this alleged prior just makes no sense.

But this is the entire controversy behind the Sleeping Beauty Problem. One that I have shown can be trivially removed. And that is why you ignore it. — JeffJo

I'm not ignoring it. I'm showing you that your version with the P(HH) = 1/4 prior that is ruled out when asked is not the same as the normal problem because the normal problem doesn't have an equivalent P(X) = 1/4 prior that is ruled out when asked.

And the prior probability that the current waking, is a step-1 waking, is 1/2. — JeffJo

Prior probabilities are established before the experiment starts, so there is no “current waking” prior because there is no “current waking” before the experiment starts.

We’re talking about prior probabilities, i.e the probabilities as established before the experiment starts.

The prior probability that step 1 will happen is 1.
The prior probability that step 2 will happen is 1.
The prior probability that step 3 will happen is 1/2.
The prior probability that step 4 will happen is 1/2.

There is no prior probability equal to 1/4.

When she is asked her credence she cannot rule out any of these prior probabilities, and she is being asked her credence that step 3 occurs.

Your question "what is your credence the coin will/did land on Heads" is asking SB to distinguish between the cases where your coin will/did land on Heads, and will/did land on Tails. So cases A and B, which depend on the same distinction, must be distinct outcomes to SB. — JeffJo

She’s being asked her credence that step 3 happens.

Step 1 just isn’t two events with a prior probability of 1/4 each. It’s one event with a prior probability of 1. Step 2 has a prior probability of 1 and steps 3 and 4 each have a prior probability of 1/2.

And the reason for the shopping example is pointing out that the four parts that I highlighted and labeled A, B, C, and D each have a prior of 1/4. — JeffJo

No they don’t. Your “A or B” isn’t two separate things but one thing with prior probability 1. C and D each have a prior probability of 1/2; C will happen if the coin lands heads and D will happen if the coin lands tails, and the prior probability that a coin will land heads is 1/2.

And why do you count being sent home in step 3 as part of the probability space but not being sent home in step 4 as part of the probability space”? You’re being inconsistent.

That 1/4 chance that she would have been taken shopping. — JeffJo

I’m not asking about your shopping example. I’m asking about this example:

1. Sleeping Beauty is given amnesia and asked her credence that the coin will or did land heads
2. The coin is tossed
3. If the coin lands heads then she is sent home
4. If the coin lands tails then she is given amnesia, asked her credence that the coin will or did land heads, and sent home

there is a "prior P(X) = 1/4 that becomes P(X) = 0 when she’s asked her credence" in your experiment. — JeffJo

Which is what? What prior P(X) = 1/4 becomes P(X) = 0 when she’s asked her credence?

In your experiment the prior probability P(HH) = 1/4 becomes P(HH) = 0 when she’s asked her credence. But there is no prior P(X) = 1/4 that becomes P(X) = 0 when she’s asked her credence in my simplified form of the experiment.

Hence your experiment is not equivalent and your solution doesn’t apply.

My claim is simply regarding what the word means.

Credence "or degree of belief is a statistical term that expresses how much a person believes that a proposition is true."

Therefore, if I believe that A iff B and if my credence in A is 1/2 then my credence in B is 1/2.

What it means for one's credence to be 1/2 rather than 1/3 is a secondary matter. My only point here is that it cannot be that each of these is true:

1. I believe that A iff B
2. My credence in A is 1/2
3. My credence in B is 1/3

Either both my credence in A and my credence in B is 1/2 or both my credence in A and my credence in B is 1/3.

So now we have an issue:

1. I believe that H-run iff H-interview
2. My credence in H-run is equal to my credence in H-interview
3. 1/2 of all runs are H-runs
4. 1/3 of all interviews are H-interviews

Given (2) it cannot be that my credence in H-run is equal to the fraction of all runs that are H-runs and that my credence in H-interview is equal to the fraction of all interviews that are H-interviews.

It may be that one's credence is a reflection of the ratio in some reference class, but given the above, (3) and/or (4) are the wrong reference class to use.

So we go back to my previous argument:

P1. If I have been assigned at random either one heads interview or two tails interviews then the probability that I have been assigned one heads interview is $1\over2$
P2. I have been assigned at random either one heads interview or two tails interviews
C1. Therefore my credence that I have been assigned one heads interview is $1\over2$
P3. My current interview is a heads interview iff I have been assigned one heads interview
C2. Therefore my credence that my current interview is a heads interview is $1\over2$

P1, P2, and P3 are true, C1 follows from P1 and P2, and C2 follows from C1 and P3. Therefore C2 is true.

My credence that my current interview is a heads interview is equal to the fraction of runs assigned one heads interview, not the fraction of interviews which are heads interviews.

Consider an alternative experiment setup where Sleeping Beauty awakens less often rather than more often when the coin lands tails. For instance, we could eliminate Tuesday awakenings altogether and ensure that Sleeping Beauty awakens once on Monday when the coin lands heads, and only half the time when it lands tails (by tossing a second coin, say). — Pierre-Normand

I consider a similar situation here.

Halfer reasoning gives an answer of 4/7 and thirder 1/2. I think 4/7 is more reasonable.

I’m less likely to wake if tails, heads and tails are equally likely, and so if I do wake my credence is that it’s less likely to be tails.

Even though in the long run the number of heads- and tails-awakenings are equal.

Pr(Heads) = Pr(Heads&First Time) + Pr(Heads&Second Time) — JeffJo

The prior P(Heads & Second Time) = 0 as established by the rules of the experiment. She will never be asked a second time if the coin lands heads. So there's nothing for her to later rule out when she's asked her credence.

When she is asked the second time, the "prior probability" of heads is ruled out. — JeffJo

No it's not. She doesn't know that she's being asked a second time. She can't rule out heads.

This does not implement the original problem. She is wakened, and asked, zero tomes or one time. — JeffJo

She’s asked once in step 1 and then, optionally, again in step 4.

No prior probability is ruled out when asked.

You know 50% is a ratio, right? — Srap Tasmaner

Pierre-Normand is saying that P(X) refers to the ratio of Xs to non-Xs in some given reference class.

I'm saying that P(X) refers to the degree to which I believe X to be true.

If P(X) refers to the degree to which I believe X to be true, and if I believe that A iff B, then P(A) = P(B).

The conclusion doesn't follow because, while the biconditional expressed in P3 is true, this biconditional does not guarantee a one-to-one correspondence between the set of T-interviews and the set of T-runs (or "T-interview sets"). Instead, the correspondence is two-to-one, as each T-run includes two T-interviews. — Pierre-Normand

There is a one-to-one correspondence between the set of H-interviews and the set of H-runs.

H-interview iff H-run
P(H-run) = 1/2
Therefore, P(H-interview) = 1/2

Credences … can be thought of as ratios — Pierre-Normand

One’s credence is the degree to which one believes a thing to be true. Often one’s credence is determined by the ratios but this is not necessary, as shown in this case.

If one believes that A iff B and if one is 50% sure that A is true then one is 50% sure that B is true. That just has to follow.

So:

P1. If I have been assigned at random either one heads interview or two tails interviews then the probability that I have been assigned one heads interview is $1\over2$
P2. I have been assigned at random either one heads interview or two tails interviews
C1. Therefore my credence that I have been assigned one heads interview is $1\over2$
P3. My current interview is a heads interview iff I have been assigned one heads interview
C2. Therefore my credence that my current interview is a heads interview is $1\over2$

You would have to argue:

P4. If I have been assigned at random a heads interview, a first tails interview, or a second tails interview then the probability that I have been assigned a heads interview is $1\over3$
P5. I have been assigned at random a heads interview, a first tails interview, or a second tails interview
C3. Therefore my credence that my current interview is a heads interview is $1\over3$

But P5 is false.

I think you have chosen the wrong reference class. Given the manner in which the experiment is conducted, as shown by this Venn diagram, the reference class of all interviews is of no use to Sleeping Beauty. With respect to individual interviews the only relevant reference classes are "heads", "tails", "first", and "second".