Sorry, I meant to say that he can rule out it being the case that the coin landed heads and that this is Day2. — Pierre-Normand

They ruled that out before the experiment begun. You might as well say that they can rule out it being the case that the coin landed heads and that this is day 3.

Do you agree that from the sitter point of view, the probability that the coin landed tails is 2/3?

Yes.

Once John Doe awakens, he can rule out the possibility of it being Day 2 of his participation, and so can you. — Pierre-Normand

Why?

Once you have been assigned to John Doe, your credence (P(H)) in this proposition should be updated from 1/2 to 2/3. This is because you were randomly assigned a room, and there are twice as many rooms with participants who wake up twice as there are rooms with participants waking up once. — Pierre-Normand

Yes, if you are randomly assigned an interview from the set of all interviews then the probability of it being a tails interview is greater than the probability of it being a heads interview.

The question is whether or not it is rational for the participant to reason this way. Given that the experiment doesn't work by randomly assigning an interview to them from the set of all interviews, I don't think it is. The experiment works by randomly assigning an interview set from the set of all interview sets (which is either the head set or the tail set), and so I believe it is more rational to reason in this way.

In this modified scenario, imagine Monty picks one of the unchosen doors at random and reveals a goat by chance. — Pierre-Normand

It's not exactly comparable as in my example he can only put to sleep one of the two people who will be put to sleep; he cannot put to sleep someone who won't be put to sleep.

A more comparable example would be if there are four doors, two containing a goat and two containing a car. You pick a door (say 1) and then Monty opens one of the two doors that contain a goat (say 2). What is the probability that your chosen door contains a car? What is the probability that the car is behind door 3, or door 4?

So there are two ways for the participants to approach the problem:

1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews — Michael

To apply this to the traditional problem: there are two Sleeping Beauties; one will be woken on Monday and one on both Monday and Tuesday, determined by a coin toss.

What is their credence that they have been or will be woken twice?

Do they reason as if they are randomly selected from the set of all participants or do they reason as if their interview is randomly selected from the set of all interviews?

Which is the most rational?

The math can prove that the former reasoning gives an answer of $1\over2$ and the latter an answer of $1\over3$ but I don’t think it can prove which reasoning is the correct to use. It might be that there is no correct answer, only a more compelling answer.

Given the way the experiment is conducted I find the former reasoning more compelling. This is especially so with the example of the coin landing heads 100 times in a row.

From the external point of view of the experimenter, it makes sense that the tree probabilities add up to more than one since the three outcomes are non exclusive. The events (Tails & Monday) and (Tails and Tuesday) can (and indeed must) be realized jointly. The intersection zone in the Venn diagram highlights the fact that those two events aren't mutually exclusive. The perspective (and epistemic position) of Sleeping Beauty is different. When she is awoken, those three possible states are mutually exclusive since it is not possible for her to believe that, in her current state of awakening, the current day is both Monday and Tuesday. Her credences in the three mutually exclusive states that she can be in therefore must add up to 1. — Pierre-Normand

I think this is best addressed by a variation I described here:

There are four people, each assigned a number (unknown to them) between 1 and 4. Two of them are to be put to sleep at random, determined by a single coin toss: if heads then 1 and 2 are put to sleep; if tails then 3 and 4 are put to sleep.

After the coin toss one of the two is put to sleep first. For each of the remaining three, what is the probability that they will be put to sleep? Is it 1/3, because there are three of them and only one is to be put to sleep, or is it 1/2, because that was the probability before the first was put to sleep?

Before anyone was put to sleep, for each of the four participants the probability of being put to sleep was 1/2, which of course doesn't add to 1.

And depending on your answer to this scenario, for each of the remaining three participants the probability of being put to sleep is 1/2.

I agree with the idea that Sleeping Beauty's credence in H is updated to 1/2 after she learns that her current awakening is occurring on a Monday. The question is what was this credence updated from, before she learned that the current day was Monday. At that point she could not rule out that the current day was Tuesday. Does that not imply that her credence in H was lower than 1/2 before she learned that the day was Monday? Before she learned that, all three mutually exclulsive areas of the Venn diagram represent possible states for her to be in. She therefore should have non-zero credences for each one of them. — Pierre-Normand

There are two approaches. The normal halfer approach is:

P(Heads) = 1/2
P(Tails) = 1/2
P(Monday) = 3/4
P(Tuesday) = 1/4

The double halfer approach is:

P(Heads) = 1/2
P(Tails) = 1/2
P(Monday) = 1/2
P(Tuesday) = 1/2

The double halfer approach does entail:

P(Heads & Monday) = P(Tails & Monday) = P(Tails & Tuesday) = 1/2

This reflects what I said before:

The probability that the coin will land heads and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Tuesday is 1/2.

As the Venn diagram shows, there are two (overlapping) probability spaces, hence why the sum of each outcome's probability is greater than 1.

I'm undecided on whether or not I'm a double halfer.

As one of the 2^100 participants who has just been awoken, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, given that half the participant interviews (plus one) are with participants assigned the winning ticket. — Pierre-Normand

I think this is a non sequitur. That most interviews are with a winner isn't that I am mostly likely a winner. Rather, if most participants are a winner then I am most likely a winner.

So there are two ways for the participants to approach the problem:

1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews

I don't think the second is the rational approach. The experiment doesn't work by randomly selecting from the set of all interviews and then dropping each participant into that interview (in such a case the second might be the more rational approach, although how it would be determined which participant(s) are chosen to have more than one interview and the order in which the participants are placed is unclear, and it may be that there is more than one winner). The experiment works by giving each participant a ticket (and in such a case I think the first is the more rational approach).

I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all. — Pierre-Normand

Well, consider the Venn diagram here (which you said you agreed with).

There are two probability spaces. Monday or Tuesday is one consideration and Heads or Tails is a second consideration. Finding out that today is Monday just removes the blue circle.

We must rather imagine that in the unlikely event that the coin lands heads 100 times in a row, 2^101 persons get pulled from the waitlist (or else, only one person does). In that case, if you are one random person who has been pulled from the waitlist, it is twice as likely that your have been so pulled as a result of the coin having landed heads 100 times in a row than not. It is hence rational for you to make this bet and in the long run 2/3 of the people pulled from the waitlist who make that bet with be right. — Pierre-Normand

This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference.

Compare these two scenarios:

1. There is a probability of 1 that I will be a participant if the coin lands heads 100 times and a probability of 1 that I will be a participant if the coin doesn't land heads 100 times

2. There is a probability of 1 that I will be a participant if the coin lands heads 100 times and a probability of $1\over2^{101}$ that I will be a participant if the coin doesn't land heads 100 times

In the first case it is rational to believe that the probability that the coin landed heads 100 times is $1\over2^{100}$

In the second case it is rational to believe that the probability that the coin landed heads 100 times is $2\over3$.

It certainly isn't the case that the two have the same answer. Surely you at least accept that? If so then the question is which of 1 and 2 properly represents the traditional problem. I say 1. There is a probability of 1 that she will be a participant if the coin lands heads and a probability of 1 that she will be a participant if the coin lands tails.

Regarding betting, expected values, and probability:

Rather than one person repeat the experiment 2¹⁰⁰ times, the experiment is done on 2¹⁰⁰ people, with each person betting that the coin will land heads 100 times in a row. 2¹⁰⁰ - 1 people lose, and 1 person wins, with the winner's winnings exceeding the sum of the losers' losses. The expected value is greater than the cost, but the probability of being a winner is still $1\over2^{100}$.

Even though I could win big, it is more rational to believe that I will lose.

It is a non sequitur to claim that a greater EV means a greater probability.

I'm not sure why you think this is absurd. Compare again my lottery study example. Suppose there are one billion people on the waiting list. If a coin lands heads 20 times in a row, then, 100 million people get pulled from the list. Else, one single person gets pulled from the list. I am then informed that I got pulled from the list (but not whether I am alone or one from 100 million). Is it absurd for me to believe that the coin landed heads 20 times in a row? My credence in this proposition should be roughly 99% — Pierre-Normand

The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the Sleeping Beauty case (both the normal and my extreme version), she's guaranteed to be woken up either way.

In Sleeping Beauty's case, your intuition that her credence in the high probability of the sequence of heads ought to be absurd apparently stems from your unwillingness to contemplate the possibility of her being able to updating it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. — Pierre-Normand

She's certainly able to update it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. I'm saying that it's irrational of her to.

The only rational approach, upon waking, is to recognize that it landing heads 100 times in a row is so unlikely that it almost certainly didn't, and that this is her first and only interview.

There is a space of possible awakening/interview events A that are being characterised by the day in which they occur ((M)onday or (T)uesday) and by the state of a coin that has been flipped prior to them occurring ((H)eads or (T)ails). P(H) = P(T) = 0.5. — Pierre-Normand

There's actually two spaces. See here.

One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating. — Pierre-Normand

In my extreme example she wins in the long run (after 2¹⁰⁰ experiments) by betting on the coin landing heads 100 times in a row.

It doesn't then follow that when the experiment is run once that Sleeping Beauty's credence that the coin landed heads 100 times in a row should be greater than her credence that it didn't.

As I have said before, that she can bet more times if one outcome happens just isn't that that outcome is more probable.

I would rather say that the experience works by ensuring that Sleeping Beauty finds herself being awoken in circumstances that she knows to be twice as likely to occur (because twice a frequent) as a result of a coin having landed heads than as a result of this coin having landed tails. — Pierre-Normand

Then you have to say the same about my extreme example; that even when she knows that the experiment is only being run once, Sleeping Beauty's credence that the coin landed heads 100 times in a row should be greater than her credence that it didn't.

And I think that's an absurd conclusion, showing that your reasoning is faulty.

But in that case P(Awake|Heads) should, consistently with this interpretation, refer to your being awakened at all conditioned on the case where the coin landed heads. This is (1/3+2/3)/2 = 0.5 — Pierre-Normand

Me being awaked at all conditioned on the case where the coin lands heads is 1/3, given that if it lands heads then I am only woken up if I was assigned the number 1.

If you mean P(Awake) to refer to the probability of your being awakened at all (on at least one day) then P(Awake) is indeed 0.5. — Pierre-Normand

And it's the same in the normal case. The probability of being awakened at all (on at least one day) is 1. That's what should be used in Bayes' theorem.

I am then using Bayes' theorem to deduce the probability of a random awakening having occurred on a Tuesday. — Pierre-Normand

Then this goes back to what I said above. These are two different questions with, I believe, two different answers:

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?

Thirders answer the second question, which I believe is the wrong answer to the first question. The experiment doesn't work by repeating the experiment several times, randomly selecting an interview from the set of all interviews, and then dropping Sleeping Beauty into it.

This is most clear with my extreme example of 2¹⁰¹ interviews following a coin toss of 100 heads in a row. Any random interview selected from the set of all interviews is most likely to have followed 100 heads in a row. But when we just run the experiment once, it is most likely that the coin didn't land 100 heads in a row, and so Sleeping Beauty's credence should only reflect this fact.

So how would your reasoning work for this situation?

Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

My reasoning is that P(Awake) = 1/2 given that there are 6 possible outcomes and I will be awake if one of these is true:

1. Heads and I am 1
2. Tails and I am 2
3. Tails and I am 3

My reasoning is that P(Awake | Heads) = 1/3 given that if it is heads I will only be awake if I am number 1.

This gives the correct Bayes' theroem:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}$

P(Tuesday|Awoken) = (P(Awoken|Tuesday) / P(Awoken)) * P(Tuesday)

Sleeping Beauty is awoken with probability 3/4 on an average day (Monday or Tuesday). On Tuesdays, she is awoken with P = 1/2. Therefore, P(Awoken|Tuesday) / P(Awoken) = (1/2)/(3/4) = 2/3.

This (2/3) is the Bayesian updating factor. The unconditioned probability of her being awoken on Tuesday is 1/2. The updated probability therefore is P(Tuesday|Awoken) = (2/3)*(1/2) = 1/3, as expected. — Pierre-Normand

I don't think it correct to say P(Awake) = 3/4. P(Awake) is just the probability that she will be woken up, which is 1.

This is clearer if we forget the days. It is just the case that if it is heads then she is woken up once and if it is tails then she is woken up twice. It doesn't make sense to say that she's not awake for her second awakening if heads.

What is the probability of the day being Tuesday? — fdrake

Maybe this Venn diagram helps?

Of course, this is from the experimenter's perspective, not Sleeping Beauty's, but it might help all the same.

I think there are two different questions with two different answers:

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads 100 times in a row?

2. If the experiment is repeated 2¹⁰⁰ times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads 100 times in a row?

I don't think it's rational for Sleeping Beauty to use the answer to the second question to answer the first question. I think it's only rational for Sleeping Beauty's credence that the coin landed heads 100 times in a row to be $1\over2^{100}$.

And so too with the original problem:

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?

Thirders answer the second question, which I believe is the wrong answer to the first question.

This is not a probability. It's a ratio of probabilities. The updated probability P(Heads|Awoken) is 2/3. The quoted ratio being larger than one just reflects the fact that Bayesian updating results in a probability increase in this case. — Pierre-Normand

My mistake. I think your example here is the same as the example I posted at the start?

As I later showed here, it provides a different answer to the original problem.

It makes it twice as likely that individual bets are winning bets. Right? Likewise in Sleeping Beauty's problem, the fact that she is being awoken twice when the coin lands heads makes is more likely that a randomly selected awakening is the result of a coin having landed heads. — Pierre-Normand

This is where I think my extreme example is helpful. Place a bet on each interview whether or not the coin landed heads 100 times in a row. In the long run (e.g. after ~2¹⁰⁰ experiments) you’ll start making money, but it just doesn’t follow that it having landed heads 100 times in a row is more probable.

And also the premise of both the original and my problem is that the experiment is just run once.

So this goes back to what I said above:

After being woken up, which of these is the most rational consideration?

1. The coin almost certainly didn't land heads 100 times, and so this is most certainly my first and only interview, or

2. If this experiment was repeated 2¹⁰⁰ times then the total number of interviews after the coin landed heads 100 times is greater than the total number of interviews after it didn't, and so if I was to pick an interview at random from that set then there is a greater probability that that interview would have followed the coin landing heads 100 times.

I think the first is the most (and only) rational consideration.

[Although] the second is true ... given that the experiment isn't conducted by picking an interview at random from that set and dropping Sleeping Beauty into it, it's also irrelevant.

P(Selected|Heads) / P(Selected) is 2/1.5 = 4/3. — Pierre-Normand

Also this makes no sense. You can't have a probability of 2.

I think you numbers there are wrong. See this.

You're inviting us to imagine ourselves in Sleeping Beauty's shoes to support the halfer position. — Pierre-Normand

Well yes. The very question posed by the problem is “what is Sleeping Beauty’s credence that the coin landed heads?”, or in my version “what is Sleeping Beauty’s credence that the coin landed heads 100 times in a row?”

To make this scenario more directly analogous to the original problem, let's imagine that Sleeping Beauty, upon each awakening, can not only express her belief about the coin toss but also place a bet on it. In the long run, she would profit from taking the bet as a thirder, further reinforcing this position. — Pierre-Normand

Being able to bet twice if it lands tails, and so make more money, doesn’t make it more likely that it landed tails; it just means you get to bet twice.

You might as well just say: you can place a £1 bet on a coin toss. If you correctly guess heads you win £1; if you correctly guess tails you win £2.

Obviously it’s better to bet on tails, but not because tails is more probable.

Perhaps this is more evident with my extreme example. Place a bet on each interview whether or not the coin landed heads 100 times in a row. In the long run (e.g. after ~2¹⁰⁰ experiments) you’ll start making money, but it just doesn’t follow that it having landed heads 100 times in a row is more probable.

And also the premise of the problem is that the experiment is just run once.

Your two definitions of E[z] aren't equivalent. — sime

They are equivalent.

There are two, equally probable, situations that $E(z)$ uses:

1. $z = 2y$ and $y = 10$
2. $z = {y\over2}$ and $y = 20$

So it's doing this:

$E(z) = {1\over2}2y[where\space y=10] + {1\over2}({y\over2})[where\space y=20]$

I'm just making this more explicit:

$E(z) = {1\over2}(2\cdot10) + {1\over2}({20\over2})$

Which leads to:

$E(z) = 15 = {5\over4}12$

Hence the redefinition of $y$. It no longer stands for the value of the chosen envelope given that the value of the chosen envelope isn't 12.

Yes, I see that. So why are you redefining y? — sime

I'm not redefining y, the switching argument is. I'm showing you what it covertly does.

No covert redefinitions of y are happening — sime

There is. I explained it above. I'll do it again.

Assume, for the sake of argument, that one envelope contains £10 and one envelope contains £20, and that I pick an envelope at random.

$P(y = 10) = P(z = 2y) = {1\over2}\\P(y = 20) = P(z = {y\over2}) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y\\E(z) = {1\over2}(2\cdot10) + {1\over2}({20\over2}) = {5\over4}12$

Notice that in $E(z)$ the variable $y$ stands for 3 different values. In the first case it stands for the value of the smaller envelope (10), in the second case it stands for the value of the larger envelope (20), and in the third case it stands for a different value entirely (12).

That third value (12) isn't the value of the chosen envelope, given that the chosen envelope contains either £10 or £20.

This is true for every possible pair of values and is true even when we don't assume the values of the two envelopes (it's just harder to see).

And it should be obvious that the probability assignments are correct. If one envelope contains £10 and one envelope contains £20 and I pick one at random then the probability that I will pick the envelope that contains £10 (the smaller amount) is $1\over2$ and the probability that I will pick the envelope that contains £20 (the larger amount) is $1\over2$.

You may have missed my edit.

It doesn't require anything like that. The only premises are that one envelope contains twice as much as the other and that I pick one at random. The probability that I will pick the envelope with the smaller amount is 1/2.

It's contradictory expectation values don't appeal to faulty reasoning given acceptance of the premises. — sime

I believe it does, as I showed above. It covertly redefines $y$ such that when it concludes $E(z) = {5\over4}y$, $y$ is no longer the value of the chosen envelope.

Rather the switching argument is unsound, for among it's premises is an improper prior distribution over x, the smallest amount of money in an envelope. And this premise isn't possible in a finite universe.

Intuitively, it's contradictory conclusions makes sense; if the smallest amount of money in an envelope could be any amount of money, and if the prior distribution over the smallest amount of money is sufficiently uniform, then whatever value is revealed in your envelope, the value of the other envelope is likelier to be higher. — sime

It doesn't require anything like that. The only premises are that one envelope contains twice as much as the other and that I pick one at random. The probability that I will pick the envelope with the smaller amount is $1\over2$.

This is the reasoning that leads to the switching argument:

$P(y = x) = P(z = 2y) = {1\over2}\\P(y = 2x) = P(z = {y\over2}) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y$

But then by the exact same logic (or by performing the appropriate substitions):

$E(z) = {1\over2}2x + {1\over2}x = {3\over2}x$

Therefore:

$E(z) = {5\over4}y = {3\over2}x$

I don't understand. — SolarWind

This and this explain it quite clearly I think.

You can be sure that the expected value for the other envelope is 5/4 of that of the one you have. — SolarWind

I can be sure it isn't as per the post immediately before yours.

The simplest "experiment" is just to imagine yourself in Sleeping Beauty's shoes. You know that if the coin lands heads 100 times then you will be interviewed 2¹⁰¹ times, otherwise you will be interviewed once, and you know that the experiment will not be repeated.

After being woken up, which of these is the most rational consideration?

1. The coin almost certainly didn't land heads 100 times, and so this is most certainly my first and only interview, or

2. If this experiment was repeated 2¹⁰⁰ times then the total number of interviews after the coin landed heads 100 times is greater than the total number of interviews after it didn't, and so if I was to pick an interview at random from that set then there is a greater probability that that interview would have followed the coin landing heads 100 times.

I think the first is the most (and only) rational consideration.

Your proposed simulation would certainly prove that the second is true (although the math alone is enough to prove it), but given that the experiment isn't conducted by picking an interview at random from that set and dropping Sleeping Beauty into it, it's also irrelevant.

If I understand right, if the coin is heads 100 times, she wakes up on Monday and is not woken up on Tuesday. If the coin is not heads 100 times, she wakes up on Monday and Tuesday? Then the experiment ends. — fdrake

I don't think we need to worry about days. The traditional experiment can be simply stated as: if tails, two interviews, otherwise one interview. In my experiment it is: if 100 heads, 2¹⁰¹ interviews, otherwise one interview.

But if thinking about it in days helps then: if the coin is heads 100 times, she wakes up on Day 1, Day 2, Day 3, ..., and Day 2¹⁰¹. If the coin is not heads 100 times, she wakes up on Day 1. Then the experiment ends.

What I think this variation shows is that it is wrong to determine the probability by imagining that we randomly select from the set of all possible interviews (weighted by their probability), and then "dropping" Sleeping Beauty into that interview. The experiment just doesn't work that way; it works by tossing a coin 100 times and then waking her up. I think it's clear at a glance that these will give different results, and I think that the second is the correct one, even from Sleeping Beauty's perspective.

So I suppose this is a reductio ad absurdum against the self-indication assumption that guides thirder reasoning.

Michael - you ruined my mind again god damnit. — fdrake

You're welcome. ;)

Sleeping Beauty is put to sleep and a coin is tossed 100 times. If it lands heads every time then she is woken up, interviewed, and put back to sleep 2¹⁰¹ times, otherwise she is woken up, interviewed, and put back to sleep once.

When being interviewed, is her credence that the coin landed heads 100 times in a row greater than her credence that it didn’t?

If we accept thirder reasoning then it is, which I think is wrong. It is a mistake to use the number of times that she would wake up were it to land heads 100 times in a row to determine the probability that it did land heads 100 times in a row.

Any reasonable Sleeping Beauty would understand that it almost certainly didn’t land heads 100 times in a row, and so that her current interview is almost certainly her first and only.

Whichever reasoning applies to this experiment must also apply to the traditional experiment.

I'll analyse that case if you can describe it very specifically. Like in the OP. — fdrake

I don't know how to explain it any simpler than the above. It's exactly like the traditional experiment, but rather than two interviews following from a $P = {1\over2}$ coin toss it's 2¹⁰¹ interviews following from a $P = {1\over{2^{100}}}$ coin toss (with just 1 interview otherwise).

Thirder reasoning would entail that, after waking, it is more likely that the coin landed heads 100 times in a row, and I think that's an absurd conclusion.

I would say that it doesn't matter how many times you will wake me if the coin lands heads 100 times in a row. When I wake up the only reasonable conclusion is that it almost certainly didn't land heads 100 times in a row.

What do you make of this?

if it’s heads 100 times in a row I wake you up 2¹⁰¹ times, otherwise I wake you up once

You know the experiment is only being run one time.

When you wake up, do you follow thirder reasoning and argue that it is more probable that the coin landed heads 100 times in a row?

Or do you follow halfer reasoning and argue that the number of times you would wake up were it to land heads 100 times in a row is irrelevant, and that it almost certainly didn't land heads 100 times in a row?

Because SB wakes up more on tails, a given wake up event is more likely to be caused by a tail flip that a head flip. — PhilosophyRunner

And that's the non sequitur.

That I would wake up more often if the coin lands heads 100 times in a row isn't that, upon waking, it is more likely that the coin landed heads 100 times in a row.

the probability of you seeing heads when you wake up is conditional on how often you wake up for heads and how often for tails — PhilosophyRunner

No, the probability of you seeing heads when you wake up is conditional on how likely you wake up for heads and for tails, not on how often you wake up.

In your case, the reason it matters is because the probability of waking on heads is 1 and tails is 0.

In the ordinary case it doesn't matter, because the probability of waking on heads is 1 and tails is 1.

I was talking about frequency not probability. — PhilosophyRunner

And my first comment to you was literally "that it happens more often isn’t that it’s more likely", i.e. that it's more frequent isn't that it's more probable.

Waking on tails is twice as frequent but equally probable.