Which context is relevant depends on the perspective one is taking - the perspective of the independent observer or the awakened Beauty. — Andrew M

The relevant context is the awakened Beauty after the interview, as she is asked the question when awakened; it is right in the OP, which would mean you don't consider periods of sleep in the sample space. They effectively do not exist to Beauty. I mean you are arguing against yourself here.

If we are to argue temporal location as relevant to update the probability model, and our task is to do so as if we are Beauty, then sleeping periods are irrelevant, Beauty has no temporal awareness while asleep. Also the moment she wakes up is also irrelevant sure she might be thinking about the probability already, but she could also be thinking about apple pie; we'd be speculating on her thoughts and we can't ground a reliable probability model in speculation.

Our task is to formulate Beauty's response to the interview; that means the relevant temporal locations, are not when she is awakened, but immediately after the interview. By the nature of the experiment the interview itself conveys information, which eliminates all days not contained in the three awakened days of the experiment. This means by the time Beauty is ready to response she already knows the relevant possible awakened states.

Beauty can either consider the probability based on the three possible awakenings that include an interview, or she could consider the coin flip, because regardless of any other considerations, a coin flip is still just a coin flip. Those are the only two relevant sample spaces.

One more point I want to press.

You can slice up probability up whichever way you like, and argue all types of samples space that pay no mind to relevant temporal location or Occam's razor; however, when you get right down to it: it is still just a coin flip. Pushing numbers around on a page will not change that.

when you get right down to it: it is still just a coin flip. — Jeremiah

And when I see the word "just" used as it is here, I assume someone is trying to manipulate my intuitions. If ever a coin flip wasn't "just" a coin a flip, surely it's Sleeping Beauty.

I'm for trying to tackle the problem as posed. I don't think we should assume, for instance, that Beauty is informed by being awakened. But I'm also for examining the problem statement carefully to see if information is being smuggled to Beauty.

Assume you need information to update your prior; either Beauty doesn't update (and the problem has manipulated our intuitions to suggest she has) or she has received information (and the problem makes it appear she hasn't) or some third option (like a really complicated credence function).

We don't know which. Maybe the problem is just underdetermined or ambiguous (but is pretending it can be answered). We have to look. Eliminate the blind alleys. Maybe when we're done there will be a choice we can't eliminate, maybe not.

And when I see the word "just" used as it is here — Srap Tasmaner

Or perhaps that is just innocent word usage. Just is kind of a hard word to avoid, there are not a lot of options that fill its role so concisely.

If ever a coin flip wasn't "just" a coin a flip, surely it's Sleeping Beauty. — Srap Tasmaner

Why? There was one coin, it was flipped once, it was fair and only had two sides. Seems to have all the characteristics of being just a coin flip.

This is a common failing of human perception, people tend to overthink things until they can no longer see the simple. It is still just a coin flip; let's not think our way out of that practical aspect. Probability is not about pushing numbers around on a page, it is about making a reliable model.

I don't think we should assume, for instance, that Beauty is informed by being awakened. — Srap Tasmaner

I didn't say she was being informed by being awakened, I said she was being informed by the interview.

Here's the picture halfers actually use:



And I think they use the same sort of weighted expectation I keep posting, only theirs looks like this:

$\small P(HEADS)=\cfrac{3}{4}(67\%)+\cfrac{1}{4}(0\%)$

$\small \phantom {P(HEADS)}=50\%$

There are a couple of curiosities here:

• in the first (and perhaps only) interview, heads is twice as likely as tails;
• there aren't half as many second interviews as firsts, but one third.

I find these proportions strange. Lewis ends up here and shrugs. I'm not sure what to make of it, but this is a far cry from the way I think halfers want to think of their position, that it's just a coin flip with some meaningless frosting on it. Whichever position we take, something about it is counterintuitive.

There is another sample space you are not even considering.

{H,T}

Meaning?

Meaning? — Srap Tasmaner

Really?


Also make sure we don't confuse proportion with the sample space.

I know what you meant by "{H,T}". I was asking what your point is.

You know the math better than I do, so if you have something to say, I'm going to listen.

I'm willing to do the work to figure things out on my own, but what you have in mind is not one of those things.

Can we just get back to SB now?

Can we just get back to SB now? — Srap Tasmaner

I have been talking about SB this entire time.

First a few terms:

You have a sample space, that is the total possible outcomes, then you have the event, that is the subset randomly selected from that sample space, then you have frequency of occurrences, which is just the proportion of the outcomes.

Lewis is very much arguing it is a simple coin flip with a sample space of {H,T}. Elga does not disagree with that but rather that when Beauty is awakened and interviewed then her temporal location becomes relevant. Elga then argues, that due to this the sample space then becomes {H1, T1, T2}; the periods in which she is awakened and interviewed.

When you see Lewis assign 1/4 to T1 and 1/4 to T2 then what are looking at is the frequency in which is he assigning to those two possible outcomes from the 1/2 in the event of tails. He saying that since being awakened on Monday and Tails and Tuesday and Tails is under the same event then you are equally likely to be in either one upon being awakened and interviewed. However, that does not mean they are pulled from a separate sample space. You have to understand that Monday Tails and Tuesday Tails are pulled from the same event of the coin landing on tails, which is 1/2.

To justify the 1/3 argument you need to give a good reason why the sample space should be from the periods of her being awakened and interviewed over the sample space of two sides of the coin.

If you want you could break it down like this:

This is your sample space {H,T}

In the event of T, which is 50% then you have a new sample space {M,Tu} where M and Tu have an equal chance of occurring. In the event of H your new sample space is just {M}.

However, that never removes the 50% of T or the 50% of H, those sill remain regardless.

You can follow conditional probability from there, but whatever you do, no amount of number pushing will change the 50% T and the 50% H. The only way to do that is to redefine the sample space, which is Elga's argument.

I think some of you beleive that Lewis and Elga are pushing numbers around to support their argument, but they are actually arguing between two different sample spaces.

Demonstration of concept #1: Beauty is put to sleep on Sunday, as in the original problem. Then a single, fair six-sided die is rolled. Based on its result, she is wakened once during the ensuing week: 1=Monday, 2=Tuesday, ... 6=Saturday. While awake, she is asked for two beliefs: That a 3 rolled, and that it is Wednesday.

I don't think anybody will argue that her belief that a 3 rolled should be 1/6. She has no evidence that could make it anything else. But "3 rolled" and "it is Wednesday" represent the same concept: if a 3 rolled, it must be Wednesday, and if anything except a 3 rolled, it can't be Wednesday.

The point is, that "it is Wednesday" is a perfectly valid proposition to evaluate. Like "a 3 rolled," its probability is 1/6.

Demonstration of concept #2: Same as #1, but two dice are rolled until the result is not doubles. Beauty is wakened twice, and the same amnesia drug is used in between wakings. She is asked about the same two beliefs.

Note that "a 3 rolled" can refer to either die, so her belief on Sunday is 10/30=1/3 (remember, the six possible ways doubles could roll are eliminated). When awake during the week, she gains no information that can affect that, so her belief remains 1/3.

"It is Wednesday" is still a valid proposition. Jeremiah, this does not mean that Beauty has "temporal awareness while asleep." Only the period of time from when she woke, until when is put to sleep again, exists in her awareness. But that entire period exists within a single day; a day that has a constant name, even if she does not know it. So she can represent that awareness with a probability for each possible name. And since she has no evidence that any name is more or less likely than "Wednesday," her belief must be 1/6 in Wednesday.

It is is not surprising that this is half of her belief in a 3, since she wakes twice. In the combinations where a 3 rolled, she is awake on Wednesday half of the days.

Demonstration of concept #3: Same as #2, but roll only once and accept doubles. In that case, she will be wakened only once.

Her temporal awareness is still that it can be only one day, within a set of six equally-likely days. Her belief that it is Wednesday is still 1/6. (Whether a 3 rolled is not quite the same issue as "Heads" in the original problem, so I won't obfuscate the point by discussing it.

+++++

The halfer argument is based entirely on treating Monday and Tuesday as the same day in Beauty's awareness. They are not; Beauty cannot distinguish them through her senses, but she knows that one name has applied since she woke up, and the other name has not. She can treat that name with probability.

Each of the four combinations "Heads+Monday, Heads+Tuesday, Tails+Monday, Tails+Tuesday" is equally likely to represent a random point in time during the experiment. The probability is 1/4 each.

Yes, Tails+Tuesday" still can happen, even if Beauty has no awareness of it when it does. The point is that it can exist, she knows it can exist, and she knows she won;y be awake to see it. From that, she can update her belief in Heads to 1/3.

Beauty can either consider the probability based on the three possible awakenings that include an interview, or she could consider the coin flip, because regardless of any other considerations, a coin flip is still just a coin flip. Those are the only two relevant sample spaces. — Jeremiah

I'm suggesting that we should start with a background sample space that includes all possible combinations of days and coin toss outcomes and then assign probabilities according to a principle of indifference. That reflects our prior intuitions that we should be indifferent to an unknown coin toss outcome for a fair coin and also indifferent to an unknown day of the week (or any two instants in time) ceteris paribus. This is the probability distribution for the resulting background sample space:


The next step is to transform it to a second sample space that reflects the Sleeping Beauty scenario by conditionalizing on when Beauty is awake. That excludes the Tuesday/Heads state and gives us the thirder distribution, as follows:


So {H1, T1, T2} is the relevant sample space for Beauty. But it has been derived from a background sample space that distributes all possible coin toss and time interval outcomes according to an indifference principle.

And that's it. That is an independent motivation for the thirder distribution based on the consistent application of a plausible principle.

The halfers use the same principle of indifference, but apply it preferentially to the coin toss outcomes over the day outcomes. Which skews their resulting probability distribution for Beauty.

There are a couple of curiosities here:
in the first (and perhaps only) interview, heads is twice as likely as tails;
there aren't half as many second interviews as firsts, but one third.
I find these proportions strange. Lewis ends up here and shrugs. I'm not sure what to make of it, but this is a far cry from the way I think halfers want to think of their position, that it's just a coin flip with some meaningless frosting on it. — Srap Tasmaner

Yes, it would be good to hear some halfer reasoning for the P(Heads|Monday) = 2/3 consequence.

Whichever position we take, something about it is counterintuitive. — Srap Tasmaner

I agree. That's why I think we should seek to derive the distribution from assumptions that are independently plausible. As I argue, the halfer is treating indifference to coin toss outcomes preferentially to indifference over day outcomes. Whereas the thirder is treating them on an equal footing (which is a further application of the indifference principle). (The hypothetical quarterer is the symmetric complement to the halfer, preferring indifference to day outcomes to indifference over coin toss outcomes.)

When you see Lewis assign 1/4 to T1 and 1/4 to T2 then what are looking at is the frequency in which is he assigning to those two possible outcomes from the 1/2 in the event of tails. He saying that since being awakened on Monday and Tails and Tuesday and Tails is under the same event then you are equally likely to be in either one upon being awakened and interviewed. However, that does not mean they are pulled from a separate sample space. You have to understand that Monday Tails and Tuesday Tails are pulled from the same event of the coin landing on tails, which is 1/2. — Jeremiah

The double interview is not a single event, for the simple reason that Beauty makes two decisions.

UNIMPORTANT ASIDE:

I think this problem arose out of earlier problems and chitchat about decision-making given imperfect memory.

I keep thinking that if it had developed on its own, it would be a time travel puzzle. "Tuesday" appears here essentially as "another Monday". You have no way of knowing for sure it's the first Monday or the last you will experience, etc. etc.

You are misunderstanding, it is an event inside of an event. That is conditional probability, and that is also the reason you can never get away from T 50% and H 50%.

I'm suggesting that we should start with a background sample space that includes all possible combinations of days and coin toss outcomes and then assign probabilities according to a principle of indifference. — Andrew M

You forgot Heads and Wednesday, Heads and Thursday, Heads and Friday. . . . . and on forever.

Then do the same with Tails.

And in the process don't forget the terms on the days Beauty will be awakened AND interviewed (AKA the sample space), was defined before the experiment started.

You forgot Heads and Wednesday, Heads and Thursday, Heads and Friday. . . . . and on forever.

Then do the same with Tails. — Jeremiah

I was assuming the two possible interview days in the experiment. But since the probabilities are all equal, it doesn't matter how large the background sample space is (at least in a finite universe). On conditionalization, all the probability will still be distributed equally between H1, T1 and T2.

And in the process don't forget the terms on the days Beauty will be awakened AND interviewed (AKA the sample space), was defined before the experiment started. — Jeremiah

Yes. So do you see any problem with my approach in principle?

. But since the probabilities are all equal, it doesn't matter how large the background sample space is (at least in a finite universe). — Andrew M

Oh really? So you think they are the same thing.

If your distribution is the same as Elga's then they should have the same center.

Let's find out, by comparing the means.

Egla's distribution: (1/3+1/3+1/3)/3=1/3

Your distribution: (1/3+1/3+1/3+0)/4= 1/4

So in your distribution, on average Beauty will get one out of every four attempts correct, because oddly enough you are actually suggesting zero is in that distribution of possible outcomes. When working with sets zero is not the same thing as null.

Let's look at this from the ground up.

A sample space is the set of all possible outcomes of a random process.

An event is a subset of that sample space.

Let E be the event and let S be the sample space.

Then the Equally Likely Probability Formula is:

P(E) = the number of outcomes in E/ the number of outcomes in S or P(E) = N(E)/N(S)

(N(R) is just the number of elements in R)

Now just to clear it up, in set theory {A,B,C} is not equal to {A,B,C,0}. Let {A,B,C} bet set 1, and let {A,B,C,0} bet set 2.

Consider,

We randomly select one element from each sample space, then our possible outcomes are:

For set 1: A or B or C

For set 2: A or B or C or 0

That means for set 2 there is a one in four chance of 0 being selected. Put that in the context of our problem and that really does not make any sense.

Now conditional probability is the the probability of event K given that event L has already occurred.

---


Consider this argument:


Our sample space is {H,T} with P(H)=N(H)/N(S) or P(H) = 1/2.

Let H equal the set {M1} and let T equal the set {M2, Tu}

Where,

M1 equals Monday and Heads
M2 equals Monday and Tails
Tu equals Tails and Tuesday.

So now our possible events are sets. Set H has one element and set T has two elements, each with a 50% chance of being selected. Then P(M2) or P(Tu), given the event T, by our Equally Likely Probability Formula is P(M2) = 1/4 and P(Tu) = 1/4. Given Tails she could be in P(M2) or P(Tu), so 1/4 + 1/4 = 1/2 therefore P(M2) + P(Tu) = 1/2 = P(T) = P(H).

Oh really? So you think they are the same thing. — Jeremiah

I was referring to the background sample space in my earlier post that included all combinations of day outcomes and coin toss outcomes and that are assigned the same probability. I'm saying that it doesn't matter what size the background sample space is, as long as it is finite and the elements have equal probabilities.

I'm then conditionalizing on Beauty being awake (and interviewed) to produce a second sample space with only three elements {H1, T1, T2} with 1/3 probability each. That second sample space is the relevant sample space and matches Elgas.

The second probability distribution from my earlier post should have been:

Thanks. I think I finally understand the halfer position. (The one thing I'm not completely clear on is how the Monday interview is retroactively determined to be a single or half of a double in the variant where the coin is tossed after the first interview.)

What puzzles me is why Beauty would reason this way.

My Beauty reasons this way, as I've said before:
(1) If I knew it was Monday, I'd know it could be heads or tails, even chance.
(2) If I knew it was Tuesday, I'd know it was tails.
(3) I know I'll be interviewed on Monday, but interviewed on Tuesday only half the time.
(4) Therefore my weighted expectation of heads is 2/3(1/2) + 1/3(0) = 1/3

The halfer Beauty reasons this way:
(1) If I knew I was in the single interview track, I'd know it was heads.
(2) If I knew I was in the double interview track, I'd know it was tails.
(3) I'm in the first track half the time and in the second half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(0) = 1/2

But this is just pretend reasoning.

It's like "working out" your expectation of heads in a simple coin toss this way:
(1) If I knew it was heads, I'd know it was heads.
(2) If I knew it was tails, I'd know it was tails.
(3) It's heads half the time and tails half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(0) = 1/2

What's the point of that?

And indeed, Lewis's "proof" has but a single step.

(No argument in this post, just clearing my head.)

Continuing:

How many times does Beauty expect to be asked for her credence?
(1) If I knew it was heads, I'd know I'll be asked once.
(2) If I knew it was tails, I'd know I'll be asked twice.
(3) It's heads half the time and tails half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(2) = 3/2.

If I'm in the single interview track, and I am half the time, I get 2/3 of my expected interviews.
If I'm in the double interview track, and I am half the time, I get 4/3 of my expected interviews.

My expectation for getting to say "heads" and be right, because I'm in the single interview track, is 2/3(50%) = 1/3.
My expectation for getting to say "tails" and be right, because I'm in the double interview track, is 4/3(50%) = 2/3.

H has one element and set T has two elements, each with a 50% chance of being selected. — Jeremiah

When does anyone ever make a random selection from among only the tails interviews?

The 50% each applies to H and T; not the elements in the sets, but the sets themselves.

I misunderstood.

But you are going to say exactly this about M2 and Tu (or T1 and T2), so the question stands.

In the event of Tails, Beauty will be awakened on Monday and Tuesday, but due to the nature of the experiment she will not be able to tell the difference, either one is equally likely when interviewed. Hence P(M2) = P(Tu) =1/4. It is 1/4 as only a total of 50% was allotted to T.

This temporal uncertainty, is actually where the 1/3 argument is placed. If the uncertainty is about her location in time, and probability is the measure of uncertainty then shouldn't her sample space be {M, T1, T2}?

It all depended on which uncertainty Beauty decides to consider.