what "subjective probability" could possibly be is also kinda what the whole puzzle is about. I just thought we could pause and consider the foundations. — Srap Tasmaner

I roll two six sided dice. I tell you the resulting sum is an odd number. What is your "credence"="subjective probability" that the dice landed on a 3 and a 4? Is it 2/36, or 2/18 (there are two ways they can land on 3 and 4)?

This is pertinent to your question if one thinks the answer to the SB problem is 1/2, solely because a fair coin toss has that probability. In that interpretation, the answer here is 2/36.

The puzzle is "about" whether SB receives "new information," similar to "the sum is odd," when she is awake. Halfers say she doesn't, and Thirders say she does. What you choose to call the probability has no relevance. In my opinion, arguments based on what you call it are used because they can't defend their usage of "new information or not," and they know that their counterpart cannot defend such a definition.

She receives new information. The reason some think otherwise is because they think "you sleep through the outcome" means "the outcome doesn't occur." There are four equally-likely experiment states that can arise that are distinct from each other. One of them means the subject is asleep (or not asked for credence, same thing), but it is a state nonetheless. If she is awake, she knows that she is in one of three.

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 happens if the coin lands heads and D happens if the coin lands tails, and the prior probability that a coin will land heads is 1/2. — Michael

1. 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. Even if the coin isn't flipped yet, this is a part of your experiment that you insist others must recognize. So please don't be disingenuous and try to deny it again.
2. (A or B) is indeed certain to happen in the experiment. But since it is possible that SB is in the part labeled D, and part D is not in (A or B), the probability that she is in (A or B) has to be less than one. Do I need to provide a lecture on what probability means? Or are you ready to at least pretend to have an open mind?

But this can all be clarified, by considering my questions. I have little "credence" that you will, because you can't accept those answers.

I’m not asking about your shopping example. — Michael

But I am.

I’m asking about this example:

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

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. And that what does or does not happen in C cannot change SB's credence in A, B, or D. All that matters is that when she knows she is in A, B, or D, she can "rule out" C. You are treating C as if it it isn't a point in the experiment, and it is.

But you would have to address my questions to discuss this, and any truthful answer to them would discredit all you have argued. So you continue to ignore them, and reply only with this same non sequitur.

Which is what? — Michael

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

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. — Michael

Tell me if you remember reading this before: In any experiment, measures of probability define a solution, not the experiment itself. The more you repeat this non sequitur (that your preferred solution can't be applied to my version of the experiment), the more obvious it becomes that you recognize that my experiment is correct.

BUT, Q4 thru Q6 show that there is a "prior P(X) = 1/4 that becomes P(X) = 0 when she’s asked her credence" in your experiment. There is a possibility that she can rule out, and it DOES NOT MATTER whether she would be awake, or asked anything, in that possibility. All that matters is that she knows it is not the current case.

Now, we could discuss all this if you would reply with anything except this same, tired non sequitur. Q1 thru Q3 are an easy start. But they don't involve mentioning prior probabilities.

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. — Michael

She cannot use ""First Time" and "Second Time" in her solution at all, except in the context of the Law of Total Probability. Because she cannot have the slightest clue which "time" (and you don't define if "second time" means the second possible waking, or the second actual waking) it is. So she can "rule out" this case as part of this Law's application.

The entire point of my suggested implementation is that "First Time" and "Second Time" look exactly the same. So the "prior" means the circumstances that govern "This Time," not "How this time relates to the other possible time."

But you still ignore that the details used anybody's solution do not establish whether the implementation of the problem is correct. They can only establish whether a solution is. Your argument here is a non sequitur. The experiment does not require the subject to distinguish between which "time" it is; and in fact, it prevents isolating such a difference.

Let me repeat "the rules of the experiment." From Elga's 2000 paper titled "Self-locating belief and the Sleeping Beauty problem" :

Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

Q1: Do you agree, or disagree, that the procedure I have outlined (with two coins, turning coin C2 over, but asking only for credence in coin C1) correctly implements this?

Q2: Do you agree, or disagree, that the subject can be wakened, and asked for this credence, during the "second time" the waking+interviewing process could occur?

Q3: Do you agree, or disagree, that the subject's answer (when asked) can be based solely on the state of the coins between time A when were examined to see if she should be wakened, and time B when/if she is subsequently wakened and interviewed?

And you also ignored this:

1. Shopping 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 given amnesia and taken shopping.
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. — JeffJo

Q4: Do you agree, or disagree, that she can have a credence in Heads while shopping in step 3 of this variation of what you presented?

Q5: And that she even can be asked for it without affecting her credences at other times?

Q6: And that those credences can't be affected by whether they lied to her about step 3?

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

Yes, it is. The Law of Total Probability says that

Pr(Heads) = Pr(Heads&First Time) + Pr(Heads&Second Time)
But "not ruing out" has not significance." Do this instead:

1. Shopping 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 given amnesia and taken shopping.
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.

If Shopping Beauty is taken shopping, she knows that Pr(Heads)=0, even tho she isn't asked.If she is asked for a credence, she know that one of four possibilities is ruled out.

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

Okay, I missed that.

No prior probability is ruled out when asked. — Michael

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

That is what is wrong with your solution. You disagree with this criticism, but that does not invalidate these steps. In fact, nothing about anybody's solution invalidates anybody's set of steps.

But you have not addressed my steps, or my solution. You have only repeated your solution. A solution I claim is invalid.

OK, I understand your argument now, — Michael

No, you seem to understand the process finally, but your counterargument completely misses the point of the argument.

Specifically, you are saying that by not applying probability theory as you do, in the same way you do, that I am not implementing the problem correctly. I am saying - and nothing you have said challenges this - that:

1. I have implemented the original problem (not the variation of it that Elga solved) exactly.
   
   • I may have added a detail (coin C2), but that is how the specifics of the original problem (one or two wakings) is managed.
   • You (well, Elga) are adding a similar detail with Monday/Tuesday schedule. It also adds a detail, and is correct. But it obfuscates the solution instead of clarifying it.
2. Mine has a trivial solution that establishes what the answer to any correct implementation of the problem is.
3. So any solution that does not get that answer, to a correct implementation, is wrong.

in your experiment the prior probability P(HH) = 1/4 is ruled out when woken — Michael

Which is the entire point. You refuse to recognize that something must be "ruled out" due to the fact that your SB cannot be awake, after Heads, on Tuesday. The reason you do this is because the probability space that governs the situation on your Monday is different than that which governs Tuesday, and you disagree with how a joint probability space is calculated by Thirders.

My entire point is to create a single probability space that applies, unambiguously, to any waking.

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 — Michael

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

From Elga:

The Sleeping Beauty problem:
Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is
Heads?

So if the coin combination is HH then the participant will be asked their credence during the second pass? If so then you are wrong when you said "she also knows that the fact that she is awake eliminates (H,H) as a possibility." — Michael

No. Please pay attention to the highlighted texts.

My implementation of the SB problem, the one I have been describing, is:

But the difference Elga introduced was unnecessary. So don't do it; do this instead:

Tell SB all the details listed here.
Put SB to sleep.
Flip two coins. Call them C1 and C2.
Procedure start:
If both coins are showing Heads, skip to Procedure End.
Wake SB.
Ask SB "to what degree do you believe that coin C1 is currently showing Heads?"
After she answers, put her back to sleep with amnesia.
Procedure End.
Turn coin C2 over, to show its opposite side.
Repeat the procedure.
Wake SB to end the experiment. — JeffJo

What matters is the probability that you will be asked for your credence at least once during the experiment. — Michael

0
What matters is not the question, but that you are awake for an interview. And in my implementation, you are. After careful double-checking I found that you not only linked to the wrong posting, you ignored the part that said:

Now, let the "further details" be that, if this is the first pass thru experiment, the exact same procedure will be repeated. Otherwise, the experiment is ended. Whether or not you were asked the question once before is irrelevant, since you have no memory of it. The arrangement of the two coins can be correlated to the arrangement in the first pass, or not, for the same reason. — JeffJo

I did forget to say that coin C2 is turned over, but that was said before. What I outlined in 3 posts above is identical to the SB problem. What I said in that post you dissected applies to one pass only, and the intent was to have two passes where, if there was no question in the first, there would be in the second.

The lab assistant only asks your credence if the coin combination isn't HH. — Michael

And in the original, on Tuesday after Heads, you are also not asked for a credence. The only difference is that in this subset of my implementation, you started out awake but forget the occurrence. Oh, yeah, it had only one possible waking, so it was not the full SB problem.

The point of extracting this from the full procedure was (A) the probability can be calculated based on the state of the current pass, and (B) only asking a question is important, not waking or sleeping.

My implementation of the SB problem, the one I have been describing, is:

But the difference Elga introduced was unnecessary. So don't do it; do this instead:

Tell SB all the details listed here.
Put SB to sleep.
Flip two coins. Call them C1 and C2.
Procedure start:
If both coins are showing Heads, skip to Procedure End.
Wake SB.
Ask SB "to what degree do you believe that coin C1 is currently showing Heads?"
After she answers, put her back to sleep with amnesia.
Procedure End.
Turn coin C2 over, to show its opposite side.
Repeat the procedure.
Wake SB to end the experiment.


When SB is awake, she knows that she is in the middle of the procedure listed in steps 4 thru 9. Regardless of which pass thru these steps it is, she knows that in step 5 of this pass, there were four equally-likely combinations for what (C1,C2) were showing: {(H,H),(H,T),(T,H),(T,T)}. This is the "prior" sample space. — JeffJo

What you were looking at was one pass thru step 4 to 9.

Probability=1 means "is certain." You said the probability of being asked was 1. That means certain. The subject in my implementation is always asked.

"Prior probability" means "prior to information being given," so "prior probability...if Heads" is vacuous.

I have no idea what you mean by "my example." My implementation is an exact version of your problem #1. I notice that you don't claim otherwise, you just try to say that #2 is different and I implemented #2. #3 is closer, and it is equivalent to #1.

My explanation of how your additional constraints turned #2 into #3 is trivially true. As is how #3 is equivalent to #1.

What credence is asked for does not affect, in any way, how the events themselves might occur. You are arguing with non sequiturs.

All you have identified is how my solution differs from yours, not how my implementation of the problem differs from the actual problem. This is because it is trivially obvious that it does not. But you can't attack my solution, which is also trivially correct, so you attack the implementation.

What matters is that in the Sleeping Beauty problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1 — Michael

What has no significance to the SB problem, is what might be different if event X, or event Y, happens WHEN BOTH ARE CERTAIN TO HAPPEN.

• X = The prior probability of being asked one's credence at least once, equal to 1.
• Y = The prior probability of being asked one's credence at least once if heads, also equal to 1.

Now, you are either a troll, very confused, or expressing yourself poorly. But if you won't consider that I might be right (as I keep assuming about you, until I actually find something wrong) there is no way to discuss anything with you.

Neither participant knows if they are A or B. — Michael

I'm going to ignore the fact that neither A nor B is woken twice, so this isn't the SB problem. What you seem to mean is that the subject is woken once as A if Heads, and once each as A and as B if tails.

Then you have created a third problem that is the equivalent of #1. Literally, it just adds a certain detail about names that is significant in #2 when they apply to different people, but makes it identical to #1 when it is the same person that can have either name. And an event it uses to choose the name - Heads or Tails - is already mentioned in the problems:

1. A is woken once if Heads, twice if Tails.
2. A is woken once if Heads, A and B once each if Tails.
3. SB if woken once in room 100A if Heads, or once each in of the rooms 100A and 100B if tails.

My version is indeed equivalent to both #1 and #3. It is not equivalent to #2 if A and B are different people, which was not given in its definition. In the special case where that is specified, all three are the same problem.

These are two different problems:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails — Michael

Yes they are.

#1 has two subjects, and #2 has two.

In #1, the only subject knows she will be wakened, just not how many times. Her credence in Heads is what we are asked about. Since both Heads and Tails are possible when she is awake, 0<Pr(Heads|Awake)<1.

In #2, A knows she will be wakened and that the coin is irrelevant. So Pr(Heads|Awake)=Pr(Heads)=1/2. B knows that she will only be wakened if Heads. So Pr(Heads/Awake)=1.

In my version, there is one subject who knows she will be wakened, just not how many times. Please, without referring to how you you would solve the problem once it is established, how is this different than #1? And how does the presence of B in #3 make it like mine?

In the original problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1, which is why the answer is 1/2 and why your example isn't comparable. — Michael

So now it isn't that I never asked about two coins ("You toss two coins and don’t ask them their credence if both land heads. That’s what makes your experiment equivalent to my second example where B isn’t asked if heads.")?

Do I need to explain "my version" to you again? You now say it is different because:

1. In the original problem the prior probability of being asked one's credence at least once is 1 ...
   
   • In mine, the prior probability of being asked ones credence (that coin C1 is showing Heads) at least once is also 1.
2. ... and the prior probability of being asked one's credence at least once if heads is 1 ...
   
   • In mine, the prior (?) probability of being asked one's credence at least once if Heads is also 1
   • But this doesn't appear to be a prior probability. That seems to be what "if Heads" means.

So you are still describing how it is the same problem, while claiming that it is different.

... which is why the answer is 1/2 and why your example isn't comparable.

Non sequitur.

Please, show me how

• The existence of two credence=1 events...
• ... which, while the facts that they are certain are valid, do exist as an issue in the problem,

... produce the conclusion "the answer is 1/2."

Do you want to try again? But do try to realize that probabilities used within a solution cannot affect whether the problems are the same. They can only affect the solutions, and credence=1 events do not do that.

I do not ask anybody (for) their credence if both coins landed on Heads. — JeffJo

Exactly. It is precisely because the prior probability of being asked at least once is 3/4 that the probability that the first coin landed heads is 1/3. — Michael

The original problem is about one coin, not two. Asking about two would make it a different problem. Asking about one is what makes it the same problem.

But yes, it is indeed true that the prior probability of 3/4 is what makes the answer 1/3. You identified the wrong event for that prior probability (she is always asked), but it is the fact that this same prior probability applies to any waking, and not different prior probabilities depending on whether the subject is wakened on Monday or Tuesday, that makes it usable in a valid solution.

Thank you for stating, in your own words, why this is so.

You toss two coins and don’t ask them their credence if both land heads. That’s what makes your experiment equivalent to my second example where B isn’t asked if heads. — Michael

I do not ask anybody (for) their credence if both coins landed on Heads. I don't ask anybody about coin C2 at all, although it has to be taken into account.

I ask for a credence in whether coin C1 is currently showing Heads. And the way the problem is set up, that is always the same event where coin C1 landed on Heads. Maybe you do not realize that coin C1 is the coin in the problem? And coin C2 is not?

C2 is just the coin that controls ordering. Since amnesia makes ordering irrelevant to the single subject, what coin C2 is showing cannot affect the answer.

That’s what makes your experiment equivalent to my second example where B isn’t asked if heads. — Michael

There is no B anywhere, as far as I can tell. You don't seem to want to explain the important details, like whether B is a person, a person in a different situation, or (as it seems here) if B is an event that is not a part of the experiment.

AGAIN: I use a single subject. That subject is always wakened, but may be wakened twice. That subject is always asked "if heads" about the only coin that the original problem is concerned with, (It's even possible we could pose the question while she sleeps, but we shouldn't expect an answer.) When she can answer, she knows that there are three equally likely combinations for what the two coins are showing, and in only one is coin C1 showing Heads.

This is not difficult. But you do need to stop trying to contrive a situation where it is wrong. So far, you have not even described a situation that applies.

We have two different experiments:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails

Your version of the experiment is comparable to the second experiment, not the first. — Michael

Um, no.

In "my experiment" I will literally and explicitly wake the single subject once if coin C1 lands on Heads, and twice if it lands on Tails. And there literally and explicitly is no second subject. So it is an exact implementation of 1, not 2.

What you are doing here, is confusing the change in the details that determine how this single subject will be awakened, with her being a different subject. You name the different subjects "subject A" and "subject B." Well, the same thing can be done with Elga's ("the most frequent") implementation:

3. Wake subject C once on Monday if Heads, or wake subjects C and D once each (on Monday and then Tuesday, respectively) if tails.

And the reason that neither 2, nor 3, fits the intended problem is that the subject has to know she will be wakened, but not remember whether it happens one or two times. So that she is the same person, but does not know if she is in "world" (I don't like this word here, but it is how philosophers approach the problem) A, B, C or D.

AGAIN:

The controversy created by Elga's ("the most frequent") implementation, of the same problem that I implement, is this:

• The incarnation of the single subject does not know if she is in world C or D.
• This prevents her from defining a sample space that applies to just her world.
• So she has to combine them into a single world
• Halfers do it by saying they are the same world as the world that contains both, so the single probability space that applies to the combination applies to each individually.
• Thirders do it by saying they are separate worlds, requiring a joint probability space and an assessment of which world she might be in.

And the way my implementation solves this is by creating a simple probability space that applies to just world A, or to just world B, so it no longer matters which world the subject is in.

So we have two different versions of the experiment: — Michael

I'm not quite sure why you quoted me in this, as the two version you described do not relate to anything I've said.

To reiterate what I have said, let me start with a different experiment:

You volunteer for an experiment. It starts with you seated at a table in a closed room, where these details are explained to you:

1. Two coins will be arranged randomly out of your sight. By this I mean that the faces showing on (C1,C2) are equally likely to be any of these four combinations: HH, HT, TH, and TT.
2. Once the combination is set, A light will be turned on.
3. At the same time, a computer will examine the coins to determine if both are showing Heads. If so, it releases a sleep gas into the room that will render you unconscious within 10 seconds, wiping your memory of the past hour. Your sleeping body will be moved to a recovery room where you will be wakened and given further details as explained below.
4. But if either coin is showing tails, a lab assistant will come into the room and ask you a probability question. After answering it, the same gas will be released, your sleeping body will be moved the same way, and you will be given the same "further details."

So you sit in the room for a minute, the light comes on, and you wait ten seconds. A lab assistant comes in (so you weren't gassed, yet) and asks you "What is the probability that coin C1 is showing Heads?

The answer to this question is unambiguously 1/3. Even tho you never saw the coins, you have unambiguous knowledge of the possibilities for what the combinations could be.

Now, let the "further details" be that, if this is the first pass thru experiment, the exact same procedure will be repeated. Otherwise, the experiment is ended. Whether or not you were asked the question once before is irrelevant, since you have no memory of it. The arrangement of the two coins can be correlated to the arrangement in the first pass, or not, for the same reason.

The point of my "alternate version" that I presented above is that it creates what is, to your knowledge, the same probability space on each pass. Just like this one.It exactly implements what Elga described as the SB problem. He changed it, by creating a difference between the first and second passes. The first pass ignores the coin, so only the second depends on it.

What you describe, where the (single) coin isn't established until the second pass, is manipulating Elga's change to emphasize that the coin is ignored in the first pass. It has nothing to do with what I've tried to convey. The only question it raises, is if Elga's version correctly implements his problem.

What if the experiment ends after the Monday interview if heads, with the lab shut down and Sleeping Beauty sent home? Heads and Tuesday is as irrelevant as Heads and Friday. — Michael

Then, in the "scenario most frequently discussed," SB is misinformed about the details of the experiment. In mine, the answer is 1/3.

Your proposed scenario certainly provides an interesting variation, but it doesn't quite correspond to the structure of the situation typically discussed in literature, the one that seems to give rise to a paradox. — Pierre-Normand

You seem to be confused about chickens and eggs, so let me summarize the history:

• The original problem was formulated by Arnold Zuboff, and was shared amongst a small group. Its scenario lasted for a huge number of days (I've seen a thousand and a trillion). Based on the same coin flip we know of today, the subject would be wakened either every day, or once on a randomly selected day in this period.
• The second version of the problem came out when Adam Elga put it in the public domain in 2000. In his statement of the problem (seen above), he reduced Zuboff's number of days to two. But he did not specify the day for the "Heads" waking. So it was implied that the order was still random.
• But he did not address that problem directly. He changed it into a third version, where he did fix the day for the "Heads" waking on Monday.
• He apparently did this because he could isolate the commonality between {Mon,Tails} and {Mon,Heads} by telling SB that it was Monday. And then the commonality between {Mon,Tails} and {Tue,Tails} by telling her that the coin landed on Tails.That is how he got his solution, by working backwards from these two special cases to his third version of the problem.
• You could say that Elga created three "levels" of the probability space. The "laboratory" space, where the coin flip is clearly a 1/2:1/2 chance, the "awakened" space where we seek an answer, and the "informed" spaces where SB knows that {Mon,Tails}, and whichever other is still possible, are equally likely.
• The "situation typically discussed in literature" is about how the informed spaces should relate to the awakened space. Not the problem itself.

All the "situation typically discussed in literature" accomplishes is how well, or poorly, Elga was able to relate these "informed" spaces to the "awakened" space in the third version of the problem. And the controversy has more to do with that, than the actual answer to the second version.

All I did, was create an alternative third version, one that correctly implements the second version. If you want to debate anything here, the issue is more whether Elga's third version correctly implements his second. If it does, then a correct answer (that is, the probability, not the way to arrive at the number) to mine is the correct answer to Elga's. It can tell you who is right about the solution to Elga's third version

That answer is 1/3. And Elga's thrid verion is a correct version of his second, since he could have fixed the Heads waking on Tuesday and arrived at the same answer.

In your scenario, there are four potential outcomes from the experiment, each of which is equally probable. — Pierre-Normand

And in the "scenario most frequently discussed," there is a fourth potential outcome that halfers want to say is not a potential outcome. SB can be left asleep on Tuesday. This is an outcome in the "laboratory" space whether or not SB can observe it. It needs to be accounted for in the probability calculations, but in the "frequent discussions" in "typical literature," the halfers remove it entirely. Rather than assign it the probability it deserves and treating the knowledge that it isn't happening as "new information."

And the reason they prefer to remove it, rather than deal with it, is that there is no orthodox way to deal with it. That's why Elga created the two "informed" spaces; they are "awakened" sub-spaces that do not need to consider it.

That's all I did as well, but without making two cases out of it. I placed that missing outcome, the fourth one you described, in a place where you can deal with it but can't ignore it.

And there are other ways to do this. In the "scenario most frequently discussed" just change {Tue,Heads} to a waking day. But instead of interviewing her, take her on a shopping trip at Saks. Now she does have "new information" when she is interviewed, and her answer is 1/3. My question to you is, why should it matter what happens on {Tue, Heads} as long as it is something other than an interview?

Or, use four volunteers. Three will be wakened each day. Each is assigned a different combination of {DAY, COIN} for when she would be left asleep. The one assigned {Tue, Heads} is the SB in the "scenario most frequently discussed."

One each day, bring the three together and ask each what the probability is, that this is her only waking. None can logically give a different answer than the others, and only one of them fits that description. The answer is 1/3.

And my point is that, while these "frequent discussions" might be interesting to some, there is a way to say which gets the right answer. Rather than arguing about why one should be called the tight solution. The answer is 1/3.

I'm not going to wade through 14 pages. The answer is 1/3, and it is easy to prove. What is hard, is getting those who don't want that to be the answer to accept it.

First, what most think is the problem statement, was the method proposed by Adam Elga (in his 2000 paper "Self-locating belief and Sleeping Beauty problem") to implement the experiment. The correct problem statement was:

Some researchers are going to put you to sleep. During the [experiment], they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you [are awake], to what degree ought you believe that the outcome of the coin toss is
Heads?

The two changes I made do not affect anything in the problem, but they do show that Elga was already thinking of how he would implement his thirder solution. The first change is where he brought up "two days," and confused the continuation of the experiment with continuation of sleep. The second suggested that your (or SB's) information might change while you are (she is) awake, which is how Elga solved the problem piecewise.

But what Elga added made the circumstances of the two (if there are to be two) wakings different. And it is this difference that is the root of the controversy that has occurred ever since.

Patient: Doctor, Doctor, it hurts if I do this.
Doctor: Then don't do that.

But the difference Elga introduced was unnecessary. So don't do it; do this instead:

1. Tell SB all the details listed here.
2. Put SB to sleep.
3. Flip two coins. Call them C1 and C2.
4. Procedure start:
5. If both coins are showing Heads, skip to Procedure End.
6. Wake SB.
7. Ask SB "to what degree do you believe that coin C1 is currently showing Heads?"
8. After she answers, put her back to sleep with amnesia.
9. Procedure End.
10. Turn coin C2 over, to show its opposite side.
11. Repeat the procedure.
12. Wake SB to end the experiment.

When SB is awake, she knows that she is in the middle of the procedure listed in steps 4 thru 9. Regardless of which pass thru these steps it is, she knows that in step 5 of this pass, there were four equally-likely combinations for what (C1,C2) were showing: {(H,H),(H,T),(T,H),(T,T)}. This is the "prior" sample space.

She also knows that the fact that she is awake eliminates (H,H) as a possibility. This is a classic example of "new information" that allows her to update the probabilities. With three (still equally likely) possibilities left, each has a posterior probability of 1/3. Since in only one is coin C1 currently showing Heads, the answer is 1/3.

The reason for the controversy, is that the difference Elga introduced between the first and (potential) second wakings obfuscates the prior sample space. This implementation has no such problem.

But I'm positive that halfers will try to invent one. I've seen it happen too many times to think otherwise.

I'm not sure what your point was, but you can't make a valid one by leaving out half of the quote.

when the researchers looked at the coins, there are four possible arrangements with probability 1/4 each: {HH, HT, TH, TT}. — JeffJo

The state of the system at that time consisted of four equally-likely possibilities. SB knows, from the experiment set up, about these four equally-likely possibilities. But she also knows that the current state of the system does not include HH. It matters not what the state would be (asleep, gone shopping, whatever) if it did include HH, because she knows that it does not.

Except for the unusual setup, this is a classic example of an introductory-level conditional probability problem. Within her knowledge, the conditional probability of HT is now 1/3.

The fallacy in the halfer argument is that they confuse the changing state of the system, with what SB was aware could be possible states when the experiment started on Sunday. They do this because they can't describe the current state in terms of SB's knowledge when she is awake.

The point of my variation is to make the current state of the system functionally the same anytime she is asked the question.

This is why I refuse to use betting arguments.

Suppose that SB gets paid $1 if the coin lands tails, otherwise she must pay $1. Furthermore, suppose that before the experiment begins she is given the choice as to whether or not she will have amnesia during the course of the experiment. According to thirder reasoning, she should choose to have amnesia in order to raise the probability of tails to 2/3 — sime

Is/does she paid/pay this $1 on both days, or on Wednesday after the experiment is over? In the latter case, can she choose not to have amnesia, and then choose "Heads" if she recalls no other waling but change that to Tails if she does?

This is why I tried to use a different setup, where this illogical juxtaposition of two different states is not an issue.

+++++

Probability is not a property of the system, it is a property of what is known about the system. Say I draw a card. After I look at it, I tell Andy that it is a black card, Betty that it is a spade, Cindy that its value is less than 10, and David that it is a seven (all separately). I ask each what they think the probability is that it is the Seven of Spades. Andy says 1/26, Betty says 1/13, Cindy says 1/32, and David says 1/4. All are right, but that does not affect my draw. I had a 1/52 chance to draw it.

The thirder argument is that SB knows of only three possible states. To her knowledge, there is no functional difference between them, so each has the same probability. Since they must sum to 1, each is 1/3. The fault they find with the halfer argument, is that it implies that to her, Monday&Tails and Tuesday&Tails represent the same state of possible knowledge. And they don't. They represent the same future, as determined on Sunday, but not the same state on a waking day.

Given these four possible experimental runs following the four possible initial coin flip results, we find that when Sleeping Beauty awakens, she can certainly rule out HH as the current state of the two coins during that specific awakening episode. However, this does not eliminate the possibility of being in either of the last two experimental runs (in addition to, of course, either of the first two). — Pierre-Normand

The points of the variation are:

• There is an "internal" probability experiment that begins when the researchers look at the two coins, and ends when either they see HH or they put SB back to sleep after seeing anything else.
• The effect of the amnesia drug not just that no information is transmitted between "this" SB and a possible version of herself in another internal experiment. It is that there are two such internal experiments in the full experiment, and each is a well-defined, self-contained experiment unto itself.
• The question in the internal probability experiment is about the state of the coins during it, not whether it is the first or second possible internal experiment. So the issue of "indexical 'today'" is completely irrelevant.

But as you point out, halfers are very good at making up reasons for why the answer they intuitively believe must be correct, could be. The "indexical" argument is one. Your example, where they might try to argue that SB must consider the version of herself in the other internal experiment is another - I've seen it used in the classical version.

So, if you need, make one slight change to mine. Always wake SB. If the coins are showing HH, take her on a shopping spree at Saks Fifth Avenue. If not, ask the question. This gets around your suggestion, since there are always two wakings. It is being in a "question" waking that allows an update: Pr(HH|Q)=0 and Pr(HT|Q)=Pr(TH|Q)=Pr(TT|Q)=1/3.

Now ask yourself whether it matters, to the reasons why these updates are possible, if the researchers had simply said "something other than a waking question will happen." The update is valid because SB has observed that the consequence of HH did not happen, regardless of what that consequence could be.

+++++

When I tell you that the card I drew is a Spade, 39 of 52 outcomes are "eliminated" as possibilities. So you can update Pr(Ace of Spades)=1/52 to Pr(Ace of Spades|Spade)=1/13. But "Diamonds" are still a part of the sample space, and their (prior) probabilities are still used for that calculation.

The difficulty with the classic version of the SB Problem arises when we try to physically remove one possibility from the sample space, not just eliminate it from the current instance of the process. That can't be done when the two "internal" parts of the whole problem are different, but SB has to view them as the same. That is the motivation for my variation, to make the internal parts identical in the knowledge basis used o answer the question.

Being left asleep does not remove Tuesday+Heads from whatever sample space you think is appropriate. It is still a possibility, and being awake constitutes an observation that it did not happen.

No, the thirder answer is not based on "who am I?" That appears to be just an excuse to reject the logic. In fact, in that post you linked, that question is neither asked nor answered. But there are equivalent ways to get the right answer, with only one SB.

Before I describe one, I need to point out that what most people think is the Sleeping Beauty Problem is actually a more contrived modification of it, invented by Adam Elga to enable his thirder solution. And the controversy is entirely about the parts he added, not the original.

This is the problem as it first ever appeared publicly, in the paper "Self-locating belief and the Sleeping Beauty problem". The two modifications alter words that clearly indicate Elga was thinking about his solution, since they have no impact on this text:

Some researchers are going to put you to sleep. During the [time] that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. [While you are awake], to what degree ought you believe that the outcome of the coin toss is Heads? — Adam Elga

Elga made the two wakings occur on Monday and Tuesday, and made them different by having Monday be a mandatory waking, and Tuesday's be optional. The first issue you must face, is to decide whether you think Elga's changes alter the correct response.

But there is another way to implement the original problem, as worded, where there is an easy answer. After putting SB to sleep, flip two coins. Call them C1 and C2. If ether is showing Tails, wake SB and ask her "to what degree do you believe that coin C1 is showing Heads?"

After she answers, put her back to sleep with amnesia and turn coin C2 over to show its other side. Then repeat the steps in the previous paragraph, starting with "If either...".

The issue with the SB problem, is whether to consider the two potential wakings as the same experiment, or different ones. This version resolves that. SB knows that when the researchers looked at the coins, there are four possible arrangements with probability 1/4 each: {HH, HT, TH, TT}. She also knows that, since she is awake, HH is eliminated. She can update her beliefs in the other three to 1/3 each.

Look, we've got your point, Jeffjo. — ssu

No, I really don't think you do. Or at least, you have shown no evidence of it.

The point of the evolution you misinterpret is to determine if the set T is internally consistent. — JeffJo

No. It's the inconsistency between two or more axioms in the axiomatic system, which make the system inconsistent.

And how is that not what I said?

But my point, that you have not shown you understand, is that finding such an inconsistency does not mean any of these two-or-more axioms is false.

There is a thing called the philosophy of mathematics and there are various schools of thought in philosophy of math, you know.

And it is that there are no pre-determined truths, only truths that follow from one's axioms which are assumed to be true without proof.

And in just what category would you put your idea presented here btw

It is a statement about philosophy, not a statement in math. "True" statments in math are either axioms, or theorems that follow from axioms. Unlike what you want here:

Axiom of Infinity is anything but established and self-evidently true. The discussion here ought to show it. — ssu

Even if math follows it's own logic (no pun intended), it's still something that people do and it does evolve. — ssu

Nobody has said otherwise. (Well, other than "what people do" is completely ambiguous.)

What was said, is that Math accepts no absolute truths.The entire point is that there should be no need to discuss what is, or is not, self-evidently true.

Math says "If the statements in the set of axioms A are accepted as true, then the statements in the set of theorems T follow logically from them. The point of the evolution you misinterpret is to determine if the set T is internally consistent. If it is not, then the set A is invalid as a set of Axioms. But no one axiom in A is invalid, and any one of them can be in another set A' that is consistent.

And notice the word "could". Could doesn't have the same meaning as is. I've only said it could be a possibility that in the future it is shown to be inconsistent. — ssu

Did you notice the word "could" also? Anything "could" happen.The sun could explode tomorrow, ending life as we know it. Do you discuss that possibility? I personally am not even considering emptying my bank account, to use it before I lose it. mathematics is not about what "could" be true, it is about what is shown to be true within a given framework of axioms.

And did you understand that finding that a ****SET**** of axioms is inconsistent does not mean that any one axiom in the set is "false"? Or that claiming that an axiom could be false is an oxymoron?

I don't blame the axiom, in my view Infinity (and hence an axiom for it) is an integral part of mathematics. All I've said that we haven't understood infinity well.

And all I've said is that this is a nonsensical statement. You, on the other hand, said:

Axiom of Infinity is anything but established and self-evidently true. The discussion here ought to show it. — ssu

That would be a pointless discussion. An axiom is not, and cannot be, inherently "self-evidently true." We cannot "prove" it, and no amount of discussion will shed any light on it. It is because it cannot be shown to be self-evidently true, or false, that we assume it is self-evidently true. So we can lay the groundwork for a specific field of mathematics.

we have gotten new insights on mathematics in history and our understanding of math has greatly changed from what it was during Ancient times and what it is now.

The insights you refer to all apply within a particular set of axiom, The only insights on mathematics in general that we have gained, are how to demonstrate that a field is internally consistent, and that there is no such thing as truth *in* a field from supposed truths *outside of* that field.

No, the set of axioms are inconsistent when they aren't consistent with each other. You don't compare two different axiomatic systems to each other. — ssu

Which is what I have been saying. When the set of axioms lead to an inconsistency, it is the set is that is inconsistent. No one axiom is inconsistent, or false. Nor is any one axiom inconsistent with another. The set itself is inconsistent.

And I never compared two systems to each other. I listed three different, consistent sets that include three contradictory versions of the parallel postulate. The point wasn't to compare the sets, it was to show that none of these parallel postulates could be called "true" or "false." But I'm beginning to suspect you know this, and are deliberately arguing around the point.

And you still have not demonstrated an inconsistency with Zermelo–Fraenkel set theory, You have supposed it could be inconsistent, and blamed it on the Axiom of Infinity possibly being false. Which is preposterous.

At least I'm trying to understand your point. — ssu

There is no evidence of it.

What I think, is that you only try to see what contradicts your predetermined idea of universal truth. And then you say the first thing that comes to your mind, that seems to imply what I said as wrong. Example:

That geometry is different in two dimensions and more dimensions is evident yes. Yet we do speak of Geometry, even when there is Euclidean and non-Euclidean geometry.

This is a classic example of a strawman argument.

I described three different examples of two-dimensional geometry. They can be expanded into more dimensions. Yet I mentioned no such higher dimensions, and their existence is completely irrelevant. There are ***THREE*** ***DIFFERENT***, mutually contradictory versions of ***PLANE*** geometry. Each is a consistent field of mathematics. And the point is that there is no such thing as this universal, capital-G Geometry as you imply. Just consistent fields of geometry based on different sets of axioms. One axiom in particular is different in each field, making the other two false in that field. Yet none of them is true, or false, outside of a set of axioms that describes it.

If you truly are trying to understand my point, understand that one. And then try to see that your claim is preposterous, because it says the opposite. The claim that the Axiom of Infinity can be considered to be true, or false, outside of a field of mathematics that has either accepts it as true, or as false, without justification or proof.

So how much do you do with "inconsistent" axiomatic systems, or as you wrote, "a *****SET***** of Axioms" that is inconsistent?

?????
I don't "do" any quantity of whatever it is you are implying with them. I also don't suppose that they could be inconsistent because they contradict a "universal truth" that I want to others to accept as blindly as you do, and dismiss an individual Axiom in the set solely on the basis of that unsupported supposition. Which is exactly what you are doing.

So changing the axioms isn't changing the way think about math? — ssu

That's the first thing you've gotten right. And the fact that you will disagree is why you won't ever understand what I am saying.

Axioms don't define "the way we think about math." They define areas ("fields") within the framework of "how we think about math." There are three different fields that use contradictory Parallel Axioms (hyperbolic, elliptic, and Euclidean geometry), yet the way we think about them in math is the same. Each field is valid, and each Parallel Axiom is true within its field, and false in the others.. Each becomes a false statement (not an Axiom) in the others.

So are against something the idea that if something is inconsistent (in math/logic),it is false,...

I can repeat this as often as you ignore it, but I'm running out of ways to make it sound different from what you've ignored before. You keep using the indexical word "something" without indicating what it refers to, the statement or the set. Some part of what you say is clearly wrong each time you use the word, but how it is wrong depends on what you mean. And if you understand the difference.

Even if you prove that a ****SET**** of Axioms is inconsistent, that does not mean that any one Axiom in it is "false." And a ****SET**** can't be false, just inconsistent.

Qualities a *****SET***** of Axioms can have include "consistent" and "inconsistent," but not "true" or "false."

Qualities a ****STATEMENT**** can have include "true in the context of a ****SET**** of Axioms," and "false in the context of a ****SET**** of Axioms," but not "true outside the context of a ****SET**** of Axioms," of "false outside the context of a ****SET**** of Axioms."

An Axiom is a special kind of statement that assumed to be true to define the ****SET***. The only quality an Axiom can have is ""true because we assume it in this ****SET**** of Axioms." Outside the context of any ****SET**** - that is, in the absolute or universal sense you deny you are using - it is neither true nor false. It is merely a set of words.

Quite circular reasoning you have there, Jeffjo. — ssu

Do think you understand the point of Axioms? Maybe you need to explain what you think it is. Because it is your arguments that are circular.

The axiom of infinity could be wrong in the way that it is inconsistent with the other axioms of ZF, for example.

And Santa Claus could visit my house tomorrow night. But I don't draw conclusions from suppositions like that.

It is you that is making the case of some eternal truth ...

You are the one suggesting that statements could be called true, or false, outside of an axiomatic system. All I'm saying th that the AoI can be part of a consistent system, and you can't conclude anything about "Infinity" outside of one.

I'm really not making the case for some universal truth here either.

Yes, you are.

My point is that from the historical perspective we have thought about math one way and because of new theorems or observations we have changed our way of thinking about math.

No, we have not. We may have changed the Axioms.

All I understand is that if something is inconsistent, we can say it's false.

Define what "something" represents here. Because an Axiom, by itself, cannot be this "something" here yet youy keep treating it as though it can.

A ****SET***** of axioms can be inconsistent, which only means that at least one of them disagrees with one or more of the others. Not that any of them is "false." And claiming otherwise is claiming that a universal truth exists.

This is a straw-man argument. — ssu

It is how Mathematics works. Anything that "exists" has to be based on Axioms.

we don't understand Infinity yet clearly.

Now that's a strawman argument. You need the AoI before you can even try to understand this thing you want to call "infinity."

No, the axioms are inconsistent to each other in the defined axiomatic system.

And you have proven this? Or are you just supposing it could be so?

As I said: "I'm not looking for some ultimate truth.

Yes, you did say that. You have also said that the AoI could be "wrong" and that we need to discuss whether it is.These statements contradict each other. This makes your axiomatic system inconsistent, and "false" by your definition.

The question is if a set of axioms, an axiomatic system, is simply consistent. I just happen to be such a logicist that I think that something that is inconsistent in math is in other words false.

Not ultimately false, or absolutely false, but some other kind of "false"? What kind?

Perhaps you didn't understand my point. — ssu

It's clear you don't understand mine. Nor have you tried.

The question is if a set of axioms, an axiomatic system, is simply consistent.

Yes, it is. That is exactly what you have not addressed.

If they aren't consistent, I would in my mind declare then an axiom or axioms to be false

And you would be wrong to do so. All it shows is that the set is inconsistent. Any of the axioms could individually be part of a different, consistent, set. Yet you are calling an axiom, or axioms, "false" in a sense that can only be called "ultimate" or "absolute."

Get this point straight: The Axiom of Infinity cannot be proven to be true, or false, outside of some set of Axioms. I believe your words were that that his discussion should establish whether the AoI is self-evidently true. Nothing is further from the point if this discussion.

Given a consistent set theory that accepts the existence of an infinite set, we require different sizes of such sets. And not believing in infinity cannot change that.

Besides, one shouldn't assume that one school of Mathematical philosophy is correct and another is not. — ssu

And you still haven't grasped the very simple fact that no field of mathematics claims to be "correct", or that another is not. Only that no statement is can be shown to be true without first assuming a set of unsupported Axiom, and proving theorems within that framework.

And it is quite clear that you have no interest in any formalism but your own.

An axiom is a statement - statements are true or false. End of story. — Devans99

Not the end of the story. Definitions are not commutative. An axiom is indeed a "proposition regarded as self-evidently true without proof." That does not mean that every "proposition regarded as self-evidently true without proof" is an axiom.

Looking further at Wolfram, an[ i]Axiomatic System[/i] is a "logical system which possesses an explicitly stated set of axioms from which theorems can be derived." So once again, the statement you claimed was an axiom was never stated as part of such a set, from which theorems could be derived. It was not an axiom, it was a near-religious belief.

Yet what you are stating is a philosophical view of mathematics.

No. What I am saying is that theorems in a field of mathematics need to be based on some set of accepted truths that are called the axioms of that field. Such a set can be demonstrated to be invalid as a set by deriving a contradiction from then, but not by comparing them to other so-called "truths" that you choose to call "self-evident."

What I am explicitly saying is that any arguments not based in the axiomatic system of a field of mathematics cannot not say anything about that field.

I already gave an example of what was thought to be an axiom that wasn't. — ssu

You gave an example of a near-religious belief. It was never an axiom in a consistent mathematics.This is similar to the belief that we can't treat aleph0 as a valid mathematical concept.

And I gave you an example where three different, consistent mathematics have been created based on three different axioms that contradict each other. The point is that neither mathematics, nor the axioms any specific field of mathematics is based on, are intended to represent absolute truth. You cannot win this debate with a nonsense claim like "the axiom is wrong."