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

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

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

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

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

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

Note: the conference rooms will be indistinguishable to SB.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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