I take SB's expected value to be a function of both her credence and of the payout structure. — Pierre-Normand

So then what is this calculation using pounds and pence?

I, however, don't take her credence to be a well defined value in the original Sleeping Beauty problem due to an inherent ambiguity in resolving what kind of "event" is implicitly being made reference to in defining her "current epistemic situation" whenever she awakens. — Pierre-Normand

I disagree that there's any ambiguity.

This is an "$n$-sided die rolled a 1" awakening if and only if this is an "$n$-sided die rolled a 1" experiment.

Therefore, a rational person's credence that this is an "$n$-sided die rolled a 1" awakening must equal their credence that this is an "$n$-sided die rolled a 1" experiment, because anyone who claims both of these isn't rational:

1. I believe it most likely that I am in a "die rolled a 1" awakening
2. I believe it most likely that I am not in a "die rolled a 1" experiment

And anyone who claims both of these isn't rational:

1. If I get to place a new bet each time I wake up then I believe it most likely that the die rolled a 1
2. If I only get to change my original bet each time I wake up then I believe it most likely that the die did not roll a 1

(on this point: what is your credence if you don't know the payout structure or if you don't get to place a bet at all?)

I really just think that there's only one correct answer, regardless of "interpretation" or payout structure (whether or not there even is is one, and whether or not it is known), and that answer is most obvious when we consider the 2¹⁰⁰-sided die; a rational person's credence is $1\over{2^{100}}$, never $2\over3$.

Therefore, a rational person's credence that this is an "n

-sided die rolled a 1" awakening must equal their credence that this is an "n

-sided die rolled a 1" experiment, because anyone who claims both of these isn't rational:

1. I believe it most likely that I am in a "die rolled a 1" awakening
2. I believe it most likely that I am not in a "die rolled a 1" experiment

And anyone who claims both of these isn't rational — Michael

It's precisely due to the usual one-to-one mapping I mentioned in my previous post that believing both those propositions at the same time usually is irrational. It is precisely what makes the Sleeping Beauty problem so interesting (and makes it contested between smart people in the literature to this day) that this usual one-to-one mapping doesn't hold in the proposed scenario. In order to display how the first proposition, as asserted by Thirders is consistent with the second one, as asserted by a Halfers, it is necessary to explicitly disambiguate the two different uses of the word "likely" that they are employing:

(1) I believe that awakening episodes such as the one I am currently experiencing turn out to be (i.e. are expected by me to be) 1-awakenings (i.e. awakening episodes that have been spawned by a die landing on "1") two thirds of the time on typical experimental runs.

(2) I believe that experimental runs such as the one I am currently being awakened into typically turn out not to be (i.e. are expected by me not to be) 1-runs (i.e. runs that have not been triggered by a die landing on "1") ((2^(100)-1)/2^100)% of the time.

Even though the biconditional proposition "this is a 1-awakening if and only if this is a 1-run" holds true, the many-to-one mapping of 1-awakenings to 1-runs accounts for the divergence of the two different ratios expressed above.

(1) I believe that awakening episodes such a the one I am currently experiencing turn out to be (i.e. are expected by me to be) 1-awakenings (i.e. awakening episodes that have been spawned by a die landing on "1") two thirds of the time on typical experimental runs. — Pierre-Normand

Which is true.

But I believe the step from this to "therefore, my credence that this is a 1-awakening is $2\over3$" is a non sequitur.

This extreme example is a reductio ad absurdum that shows that given the peculiarities of the Sleeping Beauty experiment it is irrational for one's credence to be determined by the long-term average ratio of awakenings.

So it doesn't matter that (1) is true. Any rational person being subject to this experiment should be almost certain that a) the die didn't roll a 1, and so that b) this is not a 1-awakening.

I don't know if it's quite comparable but it reminds me of the Boltzmann brain paradox. Although the probability of a Boltzmann brain forming is vanishingly small, given sufficient time they would dwarf the number of "real" brains.

Should I then believe that I am most likely a Boltzmann brain?

Although perhaps this is best comparable to the variation where both a) the experiment is repeated 2¹⁰¹ times and b) I am made to forget which experiment I'm on. I think only then could the long-term average ratio of awakenings factor into my credence.

It doesn't matter that (1) is true. Any rational person being subject to this experiment should be almost certain that a) the die didn't roll a 1, and so that b) this is not a 1-awakening. — Michael

That's because the "rational person" that you are picturing is thinking about the Halfer ratio expressed in the second proposition as the definition of their credence. And it would indeed be a non-sequitur to argue from the truth of (1) that (2) is false. But I am not doing that.

Since (1), just like (2), merely is a definition of SB's credence, as such, it is neither rational nor irrational to believe in them. What it is that Halfers or Thirders go on to do or infer on the basis of such claims (given their definitions) can be qualified as being rational or irrational. But, given any payout structure, both (1) and (2), in addition to being logically compatible, recommend the exact same betting strategies in all cases.

I think your intuition that the credence expressed by (1) is irrational in the extreme case comes from you being focussed on the prior unlikelihood of 1-runs to the expense of considering the compensating effect of there being so many 1-awakenings in one single 1-run.

Imagine a lottery where 1,000,000 tickets are printed, numbered from 1 to 1,000,000. A fair coin is tossed and if the result is Heads, one random ticket is designated to win $1,000,000, or else all tickets win $1. After the draw, the ticket validation process has two steps: one revealing if the ticket is a winner and the second one revealing the amount. Suppose you are a member of a team of 1,000,000 friends who all bought a ticket. You all bring your tickets together to the store and the clerk validates them all. You then ask the clerk to hand a winning ticket (irrespective of how many there are) to its initial buyer. A Halfer would reasonably infer that since no information was gained regarding the coin toss result, the winning amount is equally likely to be $1 or $1,000,000. I think this is the epistemic perspective you are focused on, where the winning/awakening event is being considered merely as part of a run.

Consider the different epistemic perspective of a ticket buyer who independently validates their own ticket (first validation step only). You find out that your ticket is a winner. Are the chances of your prize being $1,000,000 still 1/2? Obviously not. One million out of two million winning tickets sold in such a lottery win $1 while only one of them wins $1,000,000, and your winning ticket just is one such ticket out of 1,000,001.

In a scenario similar to one @JeffJo suggested, the 1,000,000 tickets could be distributed to Sleeping Beauty, with one of them being assigned to her on each one of 1,000,000 consecutive days. On each day, she is awakened and enabled to find out if her ticket is winning. If not, she is put back to sleep. If it is winning, before being put back to sleep, she is asked about her credence in Heads (or equivalently, that the ticket she currently holds is worth $1,000,000). A Thirder would argue, quite sensibly, that her credence is 1/1,000,001 since she is now in the exact same epistemic situation with respect to the ticket and coin as a buyer of a single ticket who has just validated it as a winning ticket.

These are clearly different experiments:

1. If a D3 rolls a 1 then she is woken once on Tuesday, else she is woken once on Monday.

2. If a D2¹⁰⁰ rolls a 1 then she is woken 2¹⁰¹ times on Monday, else she is woken once on Tuesday.

And these are clearly different questions:

a. What fraction of interviews does she expect to be Monday interviews if the experiment were to be repeated indefinitely?

b. How confident is she that her current interview is a Monday interview if she knows that the experiment is only performed once?

Everyone agrees that the answers to (1a) and (2a) are the same: $2\over3$.

The paradox concerns (1b) and (2b). Intuitively, the answer to (1b) should be much greater than the answer to (2b) — but certain lines of reasoning suggest that they should be the same. This counter-intuitive conclusion can't be justified simply by reinterpreting the question in such a way that (b) means the same thing as (a). Although in many cases the answer to (a) determines the answer to (b) it is a mistake to conflate the two, and I believe the paradox is peculiar precisely because this usual determination does not hold.

I think given the common-sense interpretation of (b) she ought be fairly confident that her current interview in (1) is a Monday interview and almost certain that her current interview in (2) is a Tuesday interview — and if expressed as fractions her answer to (1b) should be $2\over3$ and her answer to (2b) should be $1\over{2^{100}}$.

1. If a D3 rolls a 1 then she is woken once on Tuesday, else she is woken once on Monday. — Michael

I'm not sure what you mean this scenario to be. It's possible you mistyped something.

I'm not sure what you mean this scenario to be. It's possible you mistyped something. — Pierre-Normand

A 3-sided die is rolled.
If it rolls a 1 then she is woken once on Tuesday.
If it rolls a 2 then she is woken once on Monday.
If it rolls a 3 then she is woken once on Monday.

Sorry if this was said before but this seems impossible. In other words, why can she remember the experiment, but not one day prior?

Sorry if this was said before but this seems impossible. In other words, why can she remember the experiment, but not one day prior? — NotAristotle

It's a thought experiment, so you have to imagine that the drug erases only short term memory. You can also imagine Sleeping Beauty to be someone who suffers from anterograde amnesia like Leonard Shelby in Nolan's Memento movie although her short term memories would last for an hour or so. She could have the protocol of the experiment written down in her notebook and read it each time she is awakened.

Okay, fair point. Supposing it is possible, the answer would seem to be that the probability of the coin being Heads is 50% or (1/2). The number of times the participant awakens has no influence on whether the coin is Heads or Tails. It is true that if the participant guesses Heads both days when the coin has instead landed Tails, that she will be incorrect two times. However, she will still be correct when she states that there is a 50% probability that the coin has landed heads after one flip of the coin, regardless of how many times she awakens, as one coin flip always carries a 50% probability that either side will land facing up.

Something related that perplexes: if a coin lands Heads and is flipped a second time, is it more likely to land Tails on the second flip? It seems to me that it is more likely, the likelihood being greater than 50% but less than 100%. And, the probability of landing Tails seems to go up after many consecutive Heads flips. However, this would seem to contradict that there is always a 50% chance of landing Heads or Tails. Thoughts?

if a coin lands Heads and is flipped a second time, is it more likely to land Tails on the second flip? — NotAristotle

No, that’s the gambler’s fallacy: https://en.wikipedia.org/wiki/Gambler's_fallacy

:up: Okay, thanks.

No, that’s the gambler’s fallacy: https://en.wikipedia.org/wiki/Gambler's_fallacy — Michael

If your life depended on a person flipping a coin a million times in a row. Would you rather be killed if the person ended up flipping heads or tails one time OR only if the person flipped the same result each of those millions times?

Why is that? :chin:

I don't understand the point of your question, but to explain why NotAristotle is wrong, let's consider all possible permutations of two coin tosses and their prior probabilities of occurring, which are equal:

HH = 0.25
HT = 0.25
TH = 0.25
TT = 0.25

If the first coin lands heads then we just rule out two of the four permutations, with each of the other two remaining equally likely:

HH = 0.5
HT = 0.5
TH = 0
TT = 0