I'm just following the principal principle. If I can figure out what the objective chances are, so can Beauty, and she can set her credences accordingly. — Srap Tasmaner

That is of no use to her. When awakened, she doesn't know whether she is in an awake state that she should assign a probability of 1/2 to or 1/4 to. She can only condition on being awake and thus assign 1/3.

That is of no use to her. When awakened, she doesn't know whether she is in an awake state that she should assign a probability of 1/2 to or 1/4 to. — Andrew M

I see the problem.

Yes, I argued recently for "discounting", basically the model that Lewis presents.

Now I think that's wrong. There is no discounting. None of the 1/2's should be reduced to 1/4's. Monday is not 1/2:1/4 either.

When Beauty is asked, "What is your credence that the coin landed heads?" she knows there's a chance the experiment is using the heads protocol, in which case this is her one and only interview, and a chance that it is using the tails protocol, in which case this may be her first interview, last, or one of many, depending. By stipulation, there is no evidence she can use to distinguish one interview from another; all she has to go on is her knowledge of the experiment's design. So the right answer is that there is a 1/2 chance of heads protocol -- which is the chance of heads -- and a 1/2 chance of tails protocol. It's 1/2 everywhere, all the time. When she is told is Monday, this makes no difference. The number of interviews conducted in the tails protocol also makes no difference. They are all interviews about the same solitary outcome.

And the question of "updating" never arises because she never does.

I'll take one more stab at this. I did it with marbles above, but here's the application.

Which protocol to use is determined by the toss of a fair coin. Each protocol is an interview pool; when they awaken Beauty before Wednesday, they select an interview from the pool. Each pool has a single member.

This is where Lewis goes wrong, and where I went wrong when I first came around to the halfer position. It's natural to imagine the tails interview pool as a collection of 2 interviews or 1000 interviews or whatever, in which case you end up with each interview being "discounted", as I put it. If there are 2 tails interviews, they each have a 1/2 chance of being selected; since the pool as a whole has a 1/2 chance, they're each 1/4.

There are two problems with this view: (a) there are absurd consequences, like the 2/3 heads advantage on Monday, the likelihood of a second interview being 1/4 instead of 1/2, etc; (b) it does not represent how the experiment is conducted. Remember me frustratedly asking, back when I was a thirder, when anyone ever randomly selects between the two tails interviews? No one ever does, not even Beauty.

The key for me was to recognize that there is only a single interview in each pool, but under the heads protocol you select that interview from the pool (100% chance on heads) without replacement, but under the tails protocol you select that interview from the pool (100% chance on tails) with replacement. Thus the chance of that tails interview is 1/2, just as it was for the heads interview, and the next time you do a tails interview, its chance is once again 1/2, and it's always 1/2, as often as you go back to the tails pool and select that interview again.

(As a thirder I argued for conditioning this 50:50:50 to 33:33:33, but that's also wrong. The heads and tails interviews are never part of the same pool; it's one or the other. The wagering payoffs make it clear that this conditioning does not happen: every 1/2 stays a 1/2. The chance of a second interview is clearly 1/2, not 1/3. Now I understand how all this is possible.)

The question Beauty needs to answer is, which protocol is in force? Each has a 1/2 chance. That's it. Either she's being interviewed once about a heads or repeatedly about a tails, but the chances of heads and tails remain 1/2 regardless.

Now I think that's wrong. There is no discounting. None of the 1/2's should be reduced to 1/4's. Monday is not 1/2:1/4 either. — Srap Tasmaner

When Beauty is being interviewed, what probability should she assign to Monday and Tails? If 1/2 then Tuesday and Tails would be 0 which doesn't seem right.

No. I'm not sure how to formalize this (@fdrake help!), but I think if we want to do this as a table, it will be n+1-dimensional, where n is the number of tails interviews. We're going to multiply at each step, but that's just multiplying by 1 since each interview is a certainty. At the front, we're multiplying by 1/2 for the outcome of the toss.

(We usually write about chains of independent events "combinatorially" -- HHHH, HHHT, HHTH, etc. We could do that here: "H" and "T" toss outcomes, "h" and "t" interviews, and then we're choosing between Hh and Ttt. Each has a chance of 1/2. Other permutations are eliminated by the rules: there is no Ht, no Tht, and so on.)

There is no uncertainty in the interview pools themselves. This I have tried to express by having a single interview available to be selected. In the case of tails, that selection is with replacement, so you can repeat that same certain selection indefinitely. The only uncertainty here is in the outcome of the coin toss.

Consider that from the experimenter's point of view, there is never any doubt about which interview comes next. There is uncertainty for Beauty, of course, but again if she can figure out the objective chances, that's what she sets her credences to. Knowing which of, say, 1000 tails interviews she's in would be useless to her -- either she knows it's a tails interview or not, and she'll never care which one it is. Knowing that it's not Monday would be useful, but by stipulation she can't.

I've not been following the thread so I'll be of no help.

Here's a story where Lewis's table seems to make sense:

Beauty wonders to herself whether she's already been interviewed, and whether she'll be interviewed again. She reasons that there's a 1/2 chance of tails, and then a 1/2 chance that this is the first of two interviews, for 1/4; and there's also a 1/2 chance that this is the second of two of interviews, for another 1/4. The chance of either one of those being the case is 1/4 + 1/4 = 1/2, so she concludes that, since the chances this is not her only interview are the chances of tails, then the chances of tails are 1/2. Huzzah!

But this is pretend reasoning. She starts out knowing the chances of tails are 1/2.

No. I'm not sure how to formalize this — Srap Tasmaner

I don't think the halfer view ultimately flies. I think it conflates the question of the nature of fair coins (which we all agree come up heads half the time) with the question of what an agent knows about the outcome of a particular fair coin toss.

For a straightforward example where those two answers differ, suppose that the experimenter will toss two fair coins in sequence. Alice (in Wonderland) will not know the outcome of the coin tosses until after the experiment has completed. However the rule of the experiment, which Alice knows beforehand, is that she will only be interviewed if the outcome is not a double-header. If she is interviewed, she will be asked the probability that the first coin came up heads.

When being interviewed, Alice should condition on being interviewed which results in a sample space of { HT, TH, TT } and an answer of 1/3.

I think the Sleeping Beauty experiment is analogous to this. The theater is just that she is interviewed twice if the first coin toss comes up tails and amnesia is added to make the interviews indistinguishable.

what an agent knows about the outcome of a particular fair coin toss. — Andrew M

Which in Beauty's case is zilch, isn't it?

I agree about your double-header example, but don't see the similarity to SB at all. Interviewing here clearly gives you information. In SB it does not.

Which in Beauty's case is zilch, isn't it? — Srap Tasmaner

Beauty knows beforehand that she will be awakened and interviewed. But P(Heads|Awake) does not become relevant until she is actually awakened in the experiment, in which case P(Heads) consequently changes from 1/2 to 1/3 for her. Before and after the experiment, P(Heads) = 1/2.

I agree about your double-header example, but don't see the similarity to SB at all. Interviewing here clearly gives you information. — Srap Tasmaner

Yes, but what is actually relevant is that the interview condition has obtained. It is on that basis that P(First coin heads|Interviewed) = 1/3 becomes relevant and P(First coin heads) consequently changes from 1/2 to 1/3 for Alice.

Both Beauty and Alice can calculate all the applicable conditionals beforehand. The only difference is that Beauty knows with certainty that the interview condition will obtain in her future whereas Alice does not. But what makes the conditional relevant is whether the condition currently obtains for the agent, not whether it constitutes new or old information.

Here are the rules:

• I will flip a coin.
• On heads I will give you a box with 1 red marble in it.
• On tails I'll give you a box with either 1 blue marble in it or 2 blue marbles in it.
• You cannot tell the difference between a box with 1 marble in it and a box with 2.
• You don't see the outcome of the toss.
• You don't get to look in the box.

I have tossed the coin and given you a box.
What are the chances there's a red marble in the box?

When Beauty is asked, "What is your credence that the coin landed heads?" she knows there's a chance the experiment is using the heads protocol, in which case this is her one and only interview, and a chance that it is using the tails protocol, in which case this may be her first interview, last, or one of many, depending. By stipulation, there is no evidence she can use to distinguish one interview from another; all she has to go on is her knowledge of the experiment's design.

It is indeed true that Beauty has no evidence that she can use to distinguish Monday from Tuesday. This does not mean that such evidence does not exist, only that she does not have it.

Monday is a different day than Tuesday. You are treating them as the same event from Beauty's point of view, and they are not. They are within the overall experiment, but Beauty sees only half - one day of two - of what the overall experiment encompasses.

I provided an un-refuted (and irrefutable) demonstration that the answer must be 1/3 here. But you apparently need another.

So, do everything as in the original experiment, except don't tell Beauty that she might sleep though a day. Tell her that she will be interviewed on Monday, but only on Tuesday if Tails was flipped. If it was Heads, Something Other Than An Interview (SOTAI) will happen (you can even give an example: say that last week's volunteer was taken to DisneyWorld). BUT, the stipulation that she cannot distinguish the interviews still applies; only Interview/SOTAI are distinguishable.

You were right before that some "discounting" needs to be done. You just did it incorrectly. Beauty does not see the entire experiment, she sees just one day of it. Using SOTAI demonstrates how she should do this discounting: To Beauty, there aren't just two protocols, there are four:

Monday+Tails = I am to be Interviewed,
Monday+Heads = I am to be Interviewed,
Tuesday+Tails = I am to be Interviewed, and
Tuesday+Heads = SOTAI.

On Sunday Night, she knows that each of these is equally likely to occur. But from the point of view of the overall experiment, the events that happen on different days are not disjoint; each has a probability of 1/2 that it will occur in the future. There is no inconsistency, since two will happen.

But when she is wakened, she sees only one day and not the entire experiment. She has evidence only of the present, not the future or past. So Monday+Tails and Tuesday+Tails are just as much a different presents, as are Monday+Heads and Tuesday+Heads.

If SOTAI happens, she knows the sub-protocol was based on Tuesday+Heads. So the probability of Heads is 100%. This should be a big clue to Halfers, since an increased probability in some circumstances requires that it decrease in others.

If she is interviewed, she can deduce that one of these equiprobable sub-protocols can't be the protocol responsible for today. The other three remain equiprobable, and only one includes Heads. So the probability of Heads is 1/3.

Finally, note that it does not matter what SOTAI is, just that when she is in an interview she knows that SOTAI is not happening. Her logic is based entirely on the fact that the three possible interviews are indistinguishable from each other, but distinguishable from SOTAI. So, let SOTAI be "don't wake her up."

when she is in an interview she knows that SOTAI is not happening — JeffJo

If I condition on ~(Tuesday & HEADS), I exclude neither the heads protocol nor the tails protocol, as neither included it. This helps me not at all.

I have tossed the coin and given you a box.
What are the chances there's a red marble in the box? — Srap Tasmaner

1/2.

Yeah that one was too easy. If I could have come up with a way for you not to know the difference between getting one box and getting two, I would have done that. Every now and then I think if I can find just the right analogy, I'll convince you!

I'm disappointed that I'm still struggling with this, but it's a chance to learn.

Yeah that one was too easy. — Srap Tasmaner

It happens! Here's how I see it. Lewis' halfer view fails because it gives an absurd result of 2/3 when conditioning on Monday. But the double-halfer view also fails because on informing Beauty that it is Monday the result of 1/2 violates conditionalization.

Whereas accepting conditionalization and reasoning from the Monday result of 1/2 leads naturally to the thirder view. So that's one argument in favor of it.

Note also that all the real action takes place after the Monday interview since the coin toss can occur that evening. So suppose the Monday interview were removed from the experiment entirely. If Beauty were interviewed at all, then she would know that the outcome is tails (i.e., she conditions on being interviewed on Tuesday).

If she can condition on being interviewed in that scenario and reallocate probability from Tuesday/Heads to the remaining state, then she can also condition on being interviewed in the standard Sleeping Beauty scenario and reallocate probability from Tuesday/Heads to the remaining three states. The condition has obtained just the same whether or not she has learned something new.

Suppose we really were asking Beauty to guess the result of the coin toss, rather than give her credence. We'll do this with 100 Beauties and tally the results.

100 tosses, presumed result of 50 heads and 50 tails, 150 interviews.
If the Beauties all guess tails all the time, they will get 100 right out of their 150 answers.

That looks like a 2/3 success rate, right? But is it?

Out of the 100 tosses, they got 50 of them wrong. Looked at this way, that's a 50% success rate.

There's an element to committed tails-guessing of over-performing in districts you're sure to win, if you see what I mean. That extra interview, the one that tips that the result was tails, it happens in the tails track. If you're guessing tails, you've already guessed it. You're not "extra right" about tails just because you get to be right twice about the same event.

Some of the math puzzles me. Figuring out how to formalize it puzzles me. It's not a situation there's an off-the-shelf model for. Getting the math to work in a satisfying way is a chance to learn. None of that makes me at all uncertain about 1/2 being the right answer though.

If I condition on ~(Tuesday & HEADS), I exclude neither the heads protocol nor the tails protocol, as neither included it.

This is incorrect. What happens on Tuesday&HEADS is a part of the HEADS protocol, so you excluded part of it. And you treat the various possibilities inconsistently.

What you are trying to do is like trying to get a sum of 10 or more on two six-sided dice. If you look at only one, and see that it is not a 5 or a 6, can you conclude that the sum can't be 10 or more? After all, you have to have a 5 or a 6 to get that large of a sum, and you don't see one.

Just like my example excluded the possibility that the unseen die is a 6, you excluded part of the HEADS protocol when you conditioned on ~(Tuesday & HEADS). Specifically, the part where SOTAI happens. Whether or not Beauty is awake to see it, it is still a part of the HEADS protocol and you are treating it as if it is not. You are inconsistent because you insist you can't separate the two parts of the TAILS protocol the same way.

This helps me not at all.

The "help" I am trying to offer, is to get you to see that you have to separate both protocols into individual days. And you are right, it will be of no help to you if you refuse to see this, just like you won't address my "four volunteers" proof that the answer is 1/3.

+++++

In probability, an outcome is a description of a result. A set of such descriptions with the property that every possible result of an experiment fits exactly one of the descriptions is called a sample space of the experiment.

There can be more than one sample space, depending on what you are interested in describing. Possible sample spaces for rolling my two dice include 36 outcomes (if every ordered combination is considered), or 11 (if just the sum is considered), or 2 (if all you care about is whether the sum is, or isn't, 10 or more). But note that it is never wrong to use a larger sample space than the minimum required to distinguish what is of interest to you: the 36-element sample space describes the experiment where you are trying to get 10 or more, and in fact is easier to use. The most common mistake made by beginners in probability may be using the wrong sample space, and assuming the outcomes are equiprobable just because it is a sample space. ("I have two children, and there is a 1/3 chance that I have two boys because the sample space is 0, 1, or 2 boys!")

An event is not the same as an outcome, it is a set of outcomes. The two are easily confused. The difference is that your schema for providing descriptions can, depending on the event, separate it into subsets that are also events. By definition, an outcome can't be separated that way unless you change the schema.

So if your schema is to look at the sum, a 10 is a 10 whether the combination is (4,6) or (5,5). But that schema isn't useful if all you see is one die: "I see a 4" doesn't tell you anything about which "sum" description is appropriate. You have to change to a schema that describes the possible companions of the 4 you see.

In your approach to Sleeping Beauty, you are considering Monday&TAILS and Tuesday&TAILS to be inseperable parts of the same outcome. Probably because they are both part of what you call "the TAILS protocol." That is, you consider Monday&TAILS, Tuesday&TAILS, and just TAILS to be different names for the same outcome. This is a point of view that is only valid from outside the experiment; the lab techs, or Beauty on Sunday night.

What you are ignoring, is that when she is inside the experiment, even though she doesn't know which day it is, she does know that that it is not currently both days. So TAILS (or TAILS protocol) is not an outcome. She can separate it into the distinct outcomes Monday&TAILS and Tuesday&TAILS, and know that only one applies to the current moment. And the fact that you do make this distinction for HEADS requires you to do it with TAILS. (And even if you think it should not be necessary to do so, you can't be wrong by doing it.)

The two-day protocol is irrelevant to Beauty, because she is inside the experiment and so participating in only one day. Her sample space is the set of four possible single-day protocols: {Monday&TAILS, Tuesday&TAILS, Monday&HEADS, Tuesday&HEADS}. Each has a prior probability of 1/4 to apply to a random single day in the experiment, which is all that Beauty knows is happening. But because she kn0ows SOTAI is happening on that single day, she ca rule out one of those outcomes.

Her sample space is the set of four possible single-day protocols: {Monday&TAILS, Tuesday&TAILS, Monday&HEADS, Tuesday&HEADS}. Each has a prior probability of 1/4 to apply to a random single day in the experiment, which is all that Beauty knows is happening. But because she kn0ows SOTAI is happening on that single day, she ca rule out one of those outcomes. — JeffJo

This is why I suggested the alternative where we forget about days and just say that if it's heads then she's woken once and if it's tails then she's woken twice. Our sample space is just {Heads + only awakening, Tails + first awakening, Tails + second awakening}.

What happens on Tuesday&HEADS is a part of the HEADS protocol, so you excluded part of it. — JeffJo

(a) No it isn't. From the OP:

A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.

I personally think it's slightly cleaner to describe the experiment as immediately sending her home at the conclusion of her final interview, whether that's Monday or Tuesday -- it reduces the temptation to argue about Tuesday and Wednesday -- but this is a standard presentation and it works just fine.

There's also little difference if you modify the experiment as @Michael suggested and push the tails interviews back a day to Tuesday & Wednesday, and give Monday entirely to heads. Or do them all at different times of day on Monday, as some internet poster (on one of the LW threads, I think) has suggested.

The only thing that matters is one for heads and two for tails.

(b) If it were part of the heads protocol, by eliminating it, you would be eliminating heads as an outcome. Simply being interviewed would tell you the coin landed tails.

If that seems like a tendentious interpretation, consider what happens as you increase the number of tails interviews: whatever the ratio, that's your odds it was tails. Do a thousand tails interviews, and it's a near certainty -- according to thirders -- that a fair coin lands tails.

The "help" I am trying to offer — JeffJo

I was speaking as Beauty there. I appreciate your input very much -- my point was that Beauty reasoning in this way makes no progress.

you won't address my "four volunteers" proof that the answer is 1/3. — JeffJo

I'll look at it. After 18 days now in this thread, I had grown weary of alternative presentations that require analysis to figure out if they're even equivalent to SB. But I'll look at it.

Probability is about expectations. Success should be measured not by what proportion of your predictions were accurate, but by what proportion of outcomes you successfully predicted. That won't usually make much of a difference, but SB is skewed so that you get twice the credit when you're right about tails but are only singly penalized for being wrong about heads. If you guess heads all the time, in our guessing version of SB, you get exactly the same proportion of outcomes right as guessing tails, but you get doubly penalized for being wrong so that only 1/3 of your predictions were right.

Because the SB scenario doubles tails outcomes, it is difficult, or at least unnatural, to express your confidence about the outcome through wagering. For instance, suppose the coin is biased 2:1 in favor of heads. You have inside information and are thrilled to be given even money odds. Then we get these results:
100 tosses, 67 heads, 33 tails, 133 interviews.
Betting heads consistently on a biased coin at even money, you break even. WTF?
The sucker who was betting tails, who didn't know the coin was biased? He breaks even too.

ADDED:
Here's the spontaneous version of guessing-SB:
Suppose I'm going to teach Andy & Michael a little about probability. I'm going to flip a coin a bunch of times, but before each flip, they each guess. When they're right they get an M&M, and when we're done we'll count the M&M's and stuff. Now suppose before one toss, Michael guesses "Heads! Heads heads heads heads heads!!!!" If the coin lands heads, do I give him 1 M&M or 6?

100 tosses, presumed result of 50 heads and 50 tails, 150 interviews.
If the Beauties all guess tails all the time, they will get 100 right out of their 150 answers.

That looks like a 2/3 success rate, right? But is it?

Out of the 100 tosses, they got 50 of them wrong. Looked at this way, that's a 50% success rate. — Srap Tasmaner

Yes that's the nature of the experiment. There are two ways of looking at it.

In my view, probability is a measure of the state that the agent is in. Unconditionally, there is a 1/2 chance that Beauty will be in a state associated with heads.

However when conditioning on being awake and interviewed, there is a 1/3 chance that Beauty will be in a state associated with heads.

So the issue between double-halfers and thirders is whether conditioning is valid here.

Here's a variation of the experiment. Suppose that for Tuesday and Heads Beauty is also awakened and interviewed. At every interview she is informed whether or not it is a Tuesday and Heads interview. She knows these rules prior to the experiment. Naturally if she is informed that it is Tuesday and Heads at the interview, she can conclude with certainty that she is in a state associated with heads.

However if Beauty is told that it is not Tuesday and Heads at her interview, should she condition on that information or not?

Here's the spontaneous version of guessing-SB:
Suppose I'm going to teach Andy & Michael a little about probability. I'm going to flip a coin a bunch of times, but before each flip, they each guess. When they're right they get an M&M, and when we're done we'll count the M&M's and stuff. Now suppose before one toss, Michael guesses "Heads! Heads heads heads heads heads!!!!" If the coin lands heads, do I give him 1 M&M or 6? — Srap Tasmaner

Normally you would just give 1 M&M. But it is ultimately a question about what sample space and rules are appropriate in the circumstances.

What happens on Tuesday&HEADS is a part of the HEADS protocol, so you excluded part of it. — JeffJo

(a) No it isn't. From the OP:

A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends. — Srap Tasmaner

Yes, it is. The bolded text tells the lab techs what to do - or more accurately, what not to do - on both days. It defines two protocols for TAILS: interview Monday, interview Tuesday. It defines two protocols for HEADS: interview Monday, sleep Tuesday. Even if they send her home that day, that would still be SOTAI.

My point is that it can't matter what SOTAI is. There is a protocol for Tuesday&HEADS, and in an interview Beauty knows that it is not the protocol that is currently in progress.

The only thing that matters is one for heads and two for tails

What matters is that there is a protocol on both days for both HEADS and TAILS. And that one of these four protocols is inconsistent with Beauty being interviewed. You keep treating the fact that she sleeps through a day as if that makes the day nonexistent,or that it is not something the lab techs have to have included in their protocol.

(b) If it were part of the heads protocol, by eliminating it, you would be eliminating heads as an outcome. Simply being interviewed would tell you the coin landed tails.

?????
In general, I dislike the use of the word "eliminate." People forget that it means "An outcome which was possible has been shown to be incompatible with the current information state."

On Sunday Night, Beauty knows that her information state during the experiment will be limited to a single day's experiences. She knows that there are four possible such states:

1. The single day is Monday and HEADS flipped.
2. The single day is Monday and TAILS flipped.
3. The single day is Tuesday and HEADS flipped.
4. The single day is Tuesday and TAILS flipped.

If she is interviewed, she knows that one has been shown to be incompatible with her current information state.

If that seems like a tendentious interpretation, consider what happens as you increase the number of tails interviews: whatever the ratio, that's your odds it was tails. Do a thousand tails interviews, and it's a near certainty -- according to thirders -- that a fair coin lands tails.

Yep. Get two thousand volunteers. Order them randomly from #1 to #2,000. House #1 thru #1,0000 in the HEADS wing of your lab, and #1,001 thru #2,000 in the TAILS wing. Then flip your fair a coin.

On each of the next 1,000 days, wake all of the volunteers in the wing that corresponds to the coin result, and one - the one whose number corresponds to the day - from the other wing. Ask each of the 1,001 awake volunteers for her confidence that her wing is not the one that corresponds to the coin that was flipped.

Each of these women is in an experiment that is identical - except for the labels you put on coins and days - to what you just described. Each knows that 1,000 awake women came from one wing, and only 1 from the other.

Yes, it is a fair coin flip. But Beauty is not asked about its flip in an information vacuum. She knows that there is a 1/N chance that she would have been interviewed today under one result, but a 100% chance under the other. Her confidence in the first result must be 1/(1+N).

Here's the thing: it sure does look like the design of the experiment involves conditioning heads on ~Tuesday, so you get (1/4)/(1/2) = 1/2 for heads -- heads ends up by definition (heads & Monday). I'm tempted to say that since this is baked into the design of the experiment, this bit of conditioning has the status of background knowledge, more or less. At any rate, I consider it an open question whether this bit of restricting the space can or should be treated differently from the conditioning that Beauty might do in considering her personal situation.

Lots more to say, but first I want to ask you two about another quickie alternative experiment, which we might have done before, I've lost track.

I toss fair coin twice. I ask for your credence that the first toss landed heads only on {HH, TH, TT}.

My question is this: do you think this is equivalent to SB? And why or why not?

Sorry, I'm not getting your experiment, or its equivalence to SB.

One thing I'm generally uncertain about is how strongly to lean on "what day today is" being random. There are some things we can say about their equivalence for Beauty, but Elga and Lewis are both pretty cautious about that. I don't think we can just throw a big principle of indifference at "what day today is" and be done.

Here's how I converted from thirderism to halferism. Heads interviews are red marbles, tails interviews are blues.

If you select a marble from an urn with 1 red and 2 blues, sure, chances of getting the red are 1/3. But that is not Beauty's situation. Instead we have two urns, red in one and blue in the other. A coin is tossed to determine which urn to select from. It doesn't even matter how many marbles are in each; your chances of getting red are 1/2. Beauty cannot tell the difference between one interview on heads and any number of interviews on tails, but she knows that each procedure has a 1/2 chance of being followed.

I toss fair coin twice. I ask for your credence that the first toss landed heads only on {HH, TH, TT}.

My question is this: do you think this is equivalent to SB? And why or why not? — Srap Tasmaner

Yes, in the sense that one should condition on being interviewed.

Sorry, I'm not getting your experiment, or its equivalence to SB. — Srap Tasmaner

He's simply saying that if you interview different people for each permutation instead of just one person, then the thirder-style (1/(1+N)) result follows. Which is equivalent to the SB scenario.

In terms of your M&M example, if two people guess tails and they are correct, they both get an M&M. To reflect Sleeping Beauty, the experiment is set up such that only one person gets to guess when the outcome is heads. If the person conditions on the fact that they are getting to guess at all, then they will know that they are more likely to be in the tails track.

Here's a variation of the experiment. Suppose that for Tuesday and Heads Beauty is also awakened and interviewed. At every interview she is informed whether or not it is a Tuesday and Heads interview. She knows these rules prior to the experiment. Naturally if she is informed that it is Tuesday and Heads at the interview, she can conclude with certainty that she is in a state associated with heads.

However if Beauty is told that it is not Tuesday and Heads at her interview, should she condition on that information or not? — Andrew M

I really like this argument. I meant to ask about it myself -- I saw a variation of it on StackExchange a few days ago while I was digging around for other approaches -- but I forgot.

Maybe an even cleaner version is for the additional rule to be: if and only if it's a (Tue & Heads) interview, you will be told at the start that it's a (Tue & Heads) interview; then when Beauty is not told this, she infers ¬(Tue & Heads). Now when you delete the (Tue & Heads) interview, absolutely nothing else changes. You could even run standard Sleeping Beauty by having the experimenter lie, tell her it's Informative-SB, but then never do the (Tue & Heads) interviews. This looks like the perfect way to solve SB by treating it as a special case of something more obviously solvable.

Still some things to puzzle through, but I'm convinced. My sojourn in the land of halferism is at its end.

I do still disagree about how to interpret this thing though. The failure rate of my tails-guessing Beauties is still 1/2, no matter how much they pat themselves on the back. The argument you give here totally justifies conditioning on being interviewed, so the epistemic issue isn't there; it's in this conflict between the two ways of measuring success. Michael (this is my 8 year old, not TPF's Michael) gets 1 M&M and a fatherly lecture on how to measure the success of predictions.

Thanks for hanging so long, @Andrew M (weird, the Andy in my story is my 10 year old, not you). Think I learned some things. Going to take a long break now from Sleeping Beauty.

Still thinking about how to properly score this thing.

The Lewis table is what you get if you try to compensate for SB's structure by treating the coin itself as biased 2:1 heads:tails. You start with this table
drop (H & Tue) and conditionalize on P(H1 ∨ T1 ∨ T2) = 2/3 to get
So it's true that the Lewis does represent an attempt to "discount" the overabundance of tails, but it does it in the wrong place. You can't mess with the coin.

The only thing to do, to get a better measure of success rate, is to score the results at 2:1, so that you can get a payoff table like this:
I'm not sure this is sophisticated enough though. What if instead of going all heads/tails, you use a mixed strategy? The payout events (W and L) have the right ratio, but the payout values are still screwy.

Not sure I even need to worry about mixed strategies here though. The coin being fair gives a lower bound to failure of 50% and an upper bound to success of 50%.

Thought I was done, but I'm going to keep thinking about the scoring problem.

In terms of your M&M example, if two people guess tails and they are correct, they both get an M&M. To reflect Sleeping Beauty, the experiment is set up such that only one person gets to guess when the outcome is heads. If the person conditions on the fact that they are getting to guess at all, then they will know that they are more likely to be in the tails track. — Andrew M

One more point.

If you take a step back, SB looks a bit like a fucked up way of doing two trials of a single experiment. (No worries about the single coin flip -- the trial is asking different subjects for their credence.) But whichever way you split, by toss outcome or by day, it's not two trials: it's one trial each for two different experiments and which experiment is being run is determined by the coin toss, and is thus the source of Beauty's uncertainty.