The subject in my implementation is always asked. — JeffJo

No they're not. I'll quote 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."

The lab assistant only asks for my credence if the coin combination isn't HH.

If you take away this condition and so I am always asked my credence then the answer is 1/2.

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.

And in the original, on Tuesday after Heads, you are also not asked for a credence. — JeffJo

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

In the normal problem you are certain to be asked at least once during the experiment.

In your problem you are not certain to be asked at least once during the experiment.

Your problem isn’t equivalent and so the answer to your problem is irrelevant.

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.

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. — JeffJo

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."

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

perhaps because Tunisians walk in hidden pairs. When you meet a member of a Tunisian pair for the first time, their sibling ensures they are the next one you meet. — Pierre-Normand

I'm gonna come back to this, but I just want to point out that you're now describing a pickpocketing team, a stick and a cannon.

I foresee fun new variations of Sleeping Beauty.

Turn coin C2 over, to show its opposite side — JeffJo

OK, I understand your argument now, and it's what I said to you before: in your experiment the prior probability P(HH) = 1/4 is ruled out when woken; in the normal experiment where I am woken once if heads and twice if tails there is no prior probability P(X) = 1/4 that is ruled out when woken.

To make this fact clearer the experiment will be conducted in the simplest possible form:

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

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".

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


They shouldn’t. 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 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. — Michael

You know 50% is a ratio, right?

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).

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?

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.

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.

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.

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.

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.

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. — Michael

I know what you think you're saying.

But you want to express your degree of belief as a percentage. Fine. But percentages are just a funny way of writing fractions. So what do the numerator and the denominator represent?

If you want to say that you're just describing your confidence, somewhere between 0% representing absolute certainty of falsehood, with respect to some P, and 100% representing absolute certainty of the truth of P, you ought to be able to explain where these numbers come from. So where do they come from?

What exactly does it mean to say you have, say, 75% confidence? That you'll turn out to have been right 3 times out of 4? Out of 4 whats? Universes? --- That's ratio talk, because percentages are ratios. If you want to use percentages, you have to have some such account. So what's yours?

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). — Michael

Actually, I suggested that P(X) could be understood as referring to the ratio of |{X}| to (|{X}| + |{not-X}|) in epistemically identical situations with respect to X. There is some flexibility in defining what the relevant situations are.

In the case where Sleeping Beauty can say "I am experiencing an H-awakening iff I am experiencing an H-run", and there is a one-to-one mapping between H-awakenings and H-runs, we still can't logically infer that P(H-awakening) = P(H-run). This is because one can define P(H-awakening) as |{H-awakening}|/|{awakenings}| and similarly define P(H-run) as |{H-run}|/|{run}| where {awakenings} and {run} are representative sets (and |x| denotes cardinality). For the inference to hold, you would also need a one-to-one mapping between the sets of T-runs and T-awakenings.

So, the grounds for a Thirder's credence P(H-awakening) being 1/3 (where it is defined as |{H-awakening}|/|{awakenings}|) simply comes from the propensity of the experimental setup to generate twice as many T-awakenings as H-awakenings.

You argue that the number of T-awakenings is irrelevant to the determination of Sleeping Beauty's credence P(H-awakening) because her having multiple opportunities to guess the coin toss result when it lands tails doesn't impact the proportion of tails outcomes. However, while it doesn't impact |{H-run}|/|{run}|, it does impact |{H-awakening}|/|{awakenings}|, which is why I argue that Halfers and Thirders are talking past each other.

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). This setup would, in the long run, reverse the ratio of H-awakenings to T-awakenings compared to the original setup. In that case, when Sleeping Beauty awakens, would you still argue that her credence P(H-awakening) remains 1/2? A Thirder would argue that her credence should now increase to 2/3, based on the same frequency-ratio reasoning.

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.

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.

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?

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.

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.

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?

Which is what? — Michael

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

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

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.

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.

What it means for one's credence to be 1/2 rather than 1/3 is a secondary matter. — Michael

For my post too which expressed curiosity about what "1/2" means in this context.

I understand this is a sideline to the Sleeping Beauty discussion, but 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.