I propose the bold conjecture that it is the geometric average of all transcendental numbers that appear in fundamental physics with the first and last digits swapped

In this comment, I will disprove the 1/3 answer in two different ways. Because I don't want my comments to be too long, I will prove that the 1/2 answer is correct in a later comment. Moreover, I will sell my shares and send a gift of at least 500 mana to anyone who can successfully refute my arguments. I don't care about profiting anyway; I created a Manifold account solely to bet in this market.

Proof 1: Suppose thirders were right, and P(Heads and Monday) = P(Tails and Monday) = P(Tails and Tuesday) = 1/3. Also, they will definitely agree that a random Beauty awakening has a 2/3 probability of being on Monday. However, since the event "random Beauty awakening being on Monday" does not depend on when it is mentioned, this probability remains 2/3 even before the coin flip. In Beauty's world before the experiment starts (by "world" I mean "epistemic state"), we think about what makes up the 2/3 probability. Since "Heads" and "Tails" are disjoint sets whose union is the universe, we get that P(Monday) = P(Heads)×P(Monday|Heads) + P(Tails)×P(Monday|Tails) = 2/3. Substituting P(Heads) = 1/2, P(Tails) = 1/2, and P(Monday|Heads) = 1, we get a unique solution, where P(Monday|Tails) = 1/3. It is unreasonable to assume that if the coin lands tails, then a random Beauty awakening has a higher probability of being on Tuesday than on Monday (2/3 > 1/3). Therefore, assumptions made by thirders lead to an unreasonable conclusion (i.e. a contradiction).

Proof 2: This proof is based on an example from this unfinished article: https://perso.lpsm.paris/~oadelman/Sleeping%20Beauty%20is%20a%20Double%20Halfer.pdf

I attached a photo of the article’s version of this proof, but you might have to read the introduction of the article first to understand the notation. Let’s start off with some preliminaries. Consider the modified Sleeping Beauty problem where Beauty wakes up 3 times instead of 2 if the coin lands tails. In this case, almost all thirders will agree that in Beauty's world just after she wakes up, the probability that the coin landed heads is 1/4. Now consider another modified Beauty problem, where two coins are flipped, and Beauty wakes up once if both of them land heads but wakes up twice otherwise. Again, almost all thirders will agree that when Beauty wakes up, the probability that it is her only wake-up is 1/7. These “facts” are all tied to the faulty belief that in Beauty's world just after she wakes up, the probability that she wakes up n times in total during the experiment is proportional to n×p, where p is the probability before the experiment starts that she will be chosen to wake up n times. The article calls this the "proportionality credo". Now, the real example: Suppose a coin is flipped until it lands tails, and Beauty wakes up 2^h times, where h is the number of heads. Since the coin is mathematically guaranteed to eventually land tails, and no matter what happens, Beauty will always wake up a finite number of times, this process is well-defined. The proportionality credo clearly implies that in Beauty's world just after she wakes up, each of the events "the coin landed heads n times" for nonnegative integers n has equal probability; let's call this common probability x. If x > 0, then since there are an infinite number of possibilities, the sum of all the probabilities is ∞ and 1 = ∞. However, if x = 0, then the sum of all the probabilities is 0 and 1 = 0. This concludes the proof.

In the original Sleeping Beauty problem, Lewisian halfers are also wrong in thinking that P(Monday) exists and is 3/4. Simply put, the sets "Monday" and "Tuesday" cannot exist in the Kolmogorov probability space (or σ-algebra equipped with a probability measure) that represents Beauty's world just after she wakes up. This means that if, after she wakes up, Sleeping Beauty learns that it is Monday, then she is in Knightian uncertainty. So what is right? Double halfism, which asserts that P(Heads) = P(Heads|Monday*) = 1/2, where "Monday*" is the event that Beauty will or did wake up on Monday at some point during the experiment (which is, of course, guaranteed). This should clear up a lot of confusion, because people tend to think that P(Heads) = P(Heads|Monday) = 1/2 is a belief of double halfers and also think that it leads to a contradiction. It does, and in fact, I'll derive that contradiction myself. Exercise: Prove that if 0 < P(A) = P(A|B) < 1 and A → B, then P(B) = 1. Since Heads → Monday, the assumption that P(Heads) = P(Heads|Monday) = 1/2 is sufficient to prove that P(Monday) = 1, but clearly this cannot be the case as it is possible for it to be Tuesday.

I expect that if humans are sensible, this market will be <10% in a short period of time.

Jan 3: Edited once for better wording and clarity. Will not edit again.

@luvkprovider Proof 1 is wrong. The entire argument for 1/3 is that, during her awakenings, Beauty updates her credence to account for the fact that she's awake right now, which is more likely if the coin landed heads. If Beauty is awake before the coin is flipped, she should have a 1/2 credence in both heads and tails. There's no sense in which a random beauty awakening has a 2/3 probability of being on Monday given her epistemic state at that time. You have just assumed that Beauty's credence before the coin is flipped is the same as her credence while being awakened after the flip, which is a circular argument.

For Proof 2, you're assuming that there is a well-defined, non-infinitesimal, normalizable probability for each number n of awakenings, given Beauty's situation. The thirder just denies that this is the case. This isn't that weird - there are relatively uncontroversial examples of non-measurable events such as the probability of a uniformly selected number in [0,1] being in a given Vitali set. Also, any theory of anthropics will lead to cases where there is either an undefined probability, an infinitesimal probability, or a non-normalizable probability distribution in cases where the number of potential observers (or observer-moments) is infinite, so the problem presented by Proof 2 gives no evidence against the thirder position. Any position has the same problem, including halferism.

@luvkprovider Regarding proof 1, why are you assuming that P(Heads) = P(Tails) = 1/2? Why can't P(Heads) = 1/3 and P(Tails) = 2/3? You're assuming the very thing you're trying to prove.

@AndrewHebb I clearly stated that I was considering Beauty’s world before the coin was flipped. I think Plasma has a higher level of intelligence than you and is thus more worth talking to.

@PlasmaBallin Both of your arguments that my proofs are wrong can easily be refuted. I will formally refute each, but first I need you to agree on some facts.

σ-algebra and probability measure: A "σ-algebra" on a set X is a set of subsets of X that is closed under complements and countable unions. A "probability measure" P on a σ-algebra is a measure under the usual definition but with the additional restraint that P(X) = 1.

Basis: A σ-algebra on X is said to be "generated" by a set of subsets of X, denoted B, iff it is a superset of B and is a subset of every other σ-algebra on X that is a superset of B. B is said to be a "basis" of the σ-algebra that it generates.

Some theorems follow, which I named myself:

Theorem of Deduction: Given a σ-algebra Σ on X with probability measure P, and any measurable subset S of X with nonzero measure, the set {A∩S : A∈Σ} is itself a σ-algebra, with a probability measure Q satisfying Q(A∩S) = P(A∩S)/P(S).

Theorem of Induction: Given a countable collection of σ-algebras Σ_i on X_i with probability measures P_i and assigned positive real numbers r_i for i ∈ I, where I is a countable indexing set, the X_i are pairwise disjoint, and the sum of all the r_i is 1, there exists a σ-algebra Σ on the union of all the X_i such that Σ is generated by the basis B = {A : ∃i∈I(A∈Σ_i)} with the unique probability measure Q that satisfies Q(A) = P_i(A)×r_i if A∈Σ_i.

I call each of the following "principles" because they are more abstract, and without them I would have no way to relate any of my theorems to actual situations:

Principle of Universalism: Any epistemic state with events that have well-defined probabilities can be represented by a σ-algebra with a probability measure, whose elements enumerate all possible combinations of events that could occur.

In each of the following, assume that the current epistemic state of the observer is represented by the σ-algebra Σ on X with probability measure P:

Principle of Deduction: Probabilities can be updated over time as the observer receives new "information". Every piece of well-defined information can be represented by a measurable subset S of X, where if the observer is told that piece of information, they enter a new epistemic state represented by the σ-algebra on S that is constructed in the Theorem of Deduction.

Principle of Induction: If the observer's epistemic state can be partitioned into a countable number of sub-possibilites, each represented by Σ_i for some i in a countable set I, and the observer is guaranteed to be in exactly one of those sub-possibilities, Σ is equal to the construction in the Theorem of Induction (generated from the elements of the Σ_i).

You shouldn't disagree with any of the above definitions/theorems/principles because they are the standard assumptions when dealing with probability, but if you do, please specify exactly what you disagree with and why. After you state your agreement with all of the above, I will formally prove that your arguments are contradictory.

@luvkprovider There is no new information gained after the coin is flipped from the perspective of Sleeping Beauty. Why would her probabilities change? I don't see why the probability of P(Heads) can't equal 1/3 before the coin is flipped.

@AndrewHebb Because the coin is said to be fair. If the Beauty believes that P(Heads) = 1/3 before the experiment starts then she will be making bad betting decisions.

In such case she doesn't have any clever rationalization to twist her utilities to compensate for her incorrect credence.

@luvkprovider Congratulations on figuring out that Double Halfism is right. Sadly humans are not sensible enough. I've posted links to my own careful analysis of the problem long ago, where I came to the similar conclusions, and the market stayed the same. I think you'll enjoy reading my posts if you haven't seen them already:

https://www.lesswrong.com/posts/SjoPCwmNKtFvQ3f2J/lessons-from-failed-attempts-to-model-sleeping-beauty

https://www.lesswrong.com/posts/gwfgFwrrYnDpcF4JP/the-solution-to-sleeping-beauty

https://www.lesswrong.com/posts/cvCQgFFmELuyord7a/beauty-and-the-bets

https://www.lesswrong.com/posts/Gf4WtPfrELwRtfaM9/semantic-disagreement-of-sleeping-beauty-problem

I don't think Proof 2 formally works, because statement "the coin is mathematically guaranteed to eventually land tails" is incorrect. There is an infinitesimal chance for an infinite-long chain of Heads. That said the situation is very exploitable, nevertheless. A thirder Beauty's credence that any coin toss happened not to be Heads is approaching zero, therefore you can money-pump her proposing some 1000000:1 odds on every awakening that k-th coin toss is not Heads, then that k+1-st coin toss is not Heads and so on till the Beauty runs out of money.

For Proof 1, I feel confused about what is meant by "Random Beauty awakens on Monday" before the experiment even started, as being awake implies that experiment is going on. When thirders try to describe the update on awakening they use P(Monday) = 1/2, P(Monday|Awake) = 2/3. This model is coherent, even though not actually applicable to the Sleeping Beauty problem, because her awakenings are not random. I'm not sure whether you have managed to successfully corner thirders in their own lies or just added an extra level of sleight of hand of your own.

What I believe does catch thirders in their lies is a situation where Beauty is told that the coin is Tails. Then, non-Double Halfers would assume that she is supposed to reason as if her two awakenings happen on random, where Monday and Tuesday are mutually exclusive and collectively exhaustive equiprobable events: P(Monday)=P(Tuesday)=1/2. But then she is supposed to believe that probability to experience at least one Monday out of two awakening is only 3/4, which is clearly wrong.

@a07c Even if the coin is fair, the fact that she is twice as likely to experience the world in which it comes up tails means P(Heads) = 1/3 before it is tossed. Her probability doesn't have to be the same as it is for others.

@AndrewHebb this whole line of reasoning is incorrect to begin with, as she has 100% chance to experience the world when the coin is Heads and probabilities are capped at 100%.

But it's a whole extra level of wrong when we are talking about a priori probability of a future coin toss. If her probability estimate is not 1/2 on Sunday (before the experiment starts) or on Wednesday (after it has ended), she can be systematically pumped for money. Even thirders will tell you that.

@a07c How can she be systemically pumped for money? The payoff would be double if it were tails because she would be woken up twice and paid twice.

@AndrewHebb The rationalization with double payoff somewhat works if we are talking about credence on awakening. It absolutely doesn't work for credence before the experiment.

Consider:

1) On every awakening during the experiment Beauty signs a contract: if the coin in this iteration of experiment is Tails, she is payed 100$ but if the it's Heads she pays 150$. This is a good deal for her. On Tails she will sign 2 contracts, therefore winning 200$, while on Heads she signs only one contract and loses 150$. She has 1:2 betting odds in this situation.

2) Before the experiment Beauty signs a contract: if the coin in the experiment will be Tails, she is payed 100$ but if the it's Heads she pays 150$. This is not a good deal for her. Whether the coin is Tails or Heads only one contract will be signed and the losses are greater than wins. She has only 1:1 betting odds in this situation.

In the 1) one can incorrectly factorize expected utility function and assume that betting odds 1:2 imply that Beauty has 1/3 credence. It's wrong, but at least it's an understandable mistake. In 2) there is no such excuse.

@luvkprovider

This part of Proof 2 is leaky (I call it leaky because it's not iron-clad and could definitely be wrong, but I'm not exactly sure how to prove it must necessarily be wrong):

"The proportionality credo clearly implies that in Beauty's world just after she wakes up, each of the events "the coin landed heads n times" for nonnegative integers n has equal probability; let's call this common probability x. If x > 0, then since there are an infinite number of possibilities, the sum of all the probabilities is ∞ and 1 = ∞. However, if x = 0, then the sum of all the probabilities is 0 and 1 = 0. This concludes the proof."

Refutation: Can't x be 1/∞? Just like how integration adds an infinite number of infinitely small pieces, can't the infinite sum of infinitely small but nonzero probability also add to one?

Proof 1 is also wrong, but in a more complicated way. It is true that if you plug in P(Monday|Tails) = 1/2 into the given P(Monday) = P(Heads)×P(Monday|Heads) + P(Tails)×P(Monday|Tails) equation, you get P(Monday) = 3/4, and not 2/3. This is because you are weighting each awakening differently.

The simple, intutive explanation to why 1/2 can't be right is: Sleeping Beauty doesn't know which awakening she is on, and only knows that for each experiment, Heads and Tails are equally likely but the Tails case awakens her twice while Heads awakens her once, for a total of 3 possible awakenings per experiment. Remember that the Tails case has 2 awakenings, which means the probability that the coin has flipped tails when Sleeping Beauty wakes up should be weighted heigher. To make this even more intuitive, consider a modified scenario where everything else is the same, but now if the coin flips tails Sleeping Beauty is woken up 10^100 times, or any other arbitrarily high number. Now, it is obvious that every time Sleeping Beauty wakes up, she should believe that P(Heads) << P(Tails), because while P(Heads) = P(Tails) = 1/2 for each experiment, P(Heads) << P(Tails) for each awakening.

A more rigid, mathematical proof to the original problem: in Sleeping Beauty's frame, P(Heads) is actually 1/3 and P(Tails) is 2/3, because what she considers is not P(Heads) and P(Tails), but actually P(Heads|Awakened) and P(Tails|Awakened). Then plugging in the obviously correct P(Monday|Heads) = 1 and P(Monday|Tails) = 1/2 gives P(Monday) = 2/3, as expected. This is because while P(Monday) = 3/4, the 2/3 is from P(Monday and Awakened) = P(Heads|Awakened) * P(Monday|Heads) + P(Tails|Awakened) * P(Monday|Tails) = 2/3. The self-contradiction expressed in Proof 1 is created by using P(Monday|Awakened) = 2/3, but using P(Heads) and P(Tails) and NOT P(Heads|Awakened) and P(Tails|Awakened). Naturally this will generate an incorrect result, because you've equaled apples and oranges.

Extra edit, to push my point even further: Yet another way to think about it is by taking direct probabilities. Since P(Heads) = P(Tails) = 1/2, we can naturally take a sample of two experiments, one Heads and one Tails, and check the probability ourselves. In the Heads experiment, there's one awakening, and the correct answer would be Heads. In the Tails experiment, there's two awakenings, and the correct answer there would be Tails, both times. Then the probability that the coin flipped Heads when Sleeping Beauty awakens is 1/3.

Another one: Imagine you replace the problem. Instead of the whole sleeping beauty + awakening thing, it's:
Flip a bunch of coins. For each Heads, put a green ball in a box. For each Tails, put two red balls in the same box. What is the chance you pull out a green ball?

^^^This question is equivalent to the Sleeping Beauty problem. Naturally, you will answer 1/3.

@AndrewHebb Surprisingly, your initial idea that got immediately shut down by @luvkprovider was - albeit informal and not rigorous - correct! This was because instead of considering objective P(Heads) and P(Tails), you considered P(Heads|Awakened) and P(Tails|Awakened), which were indeed 1/3 and 2/3, respectively.

@Gameknight Waking up doesn't reveal any information so the probability of heads is the same as the probability of heads given that she wakes up.

@AndrewHebb Waking up in itself means that tails is more likely, because there's more wake-ups if it's tails.

@a07c

You said: "The rationalization with double payoff somewhat works if we are talking about credence on awakening. It absolutely doesn't work for credence before the experiment."

But that's exactly what the question is asking. We are not asking about the credence before the experiment (Which would, indeed, be 1/2). We are asking about the credence on awakening. That's what the question is about.

This section in wikipedia is more clear and concise than anything I could type up:

@Gameknight But she already knows that she will wake up. Things that you know will happen cannot change probabilities.

@AndrewHebb That is correct, she knows she will wake up. But she will wake up more if the coin flip was tails. Therefore, when she wakes up, it is more likely that she is in an experiment where the coinflip was tails.

@Gameknight @AndrewHebb I agree. Here's a related thought experiment. Suppose there are infinite universes with all different Planck's constants. Suppose only one Planck's constant supports life -- 6.6262 x 10^-34 Joule⋅seconds -- and that constant is very rare among all universes. You "wake up" in a universe. What is the probability that Planck's constant is 6.6262 x 10^-34 Joule⋅seconds in your universe?

@DavidPennock Oh hey, I was just reading about the Anthropic principle! Yeah, this is basically my point - you know that the universe has a Planck Constant that can or can't support life - so when you check it, what's the chance it doesn't support life? Obviously 0, because you wouldn't be there to observe it if it didn't support life. Similarly, when Sleeping Beauty wakes up, what's the chance that it's Heads? Obviously lower than the chance that it's Tails, because she only wakes up once if it's Heads but wakes up multiple times if it's Tails, therefore P(Heads|Awakened) = 1/3 and P(Tails|Awakened) = 2/3

@luvkprovider I'm aware that the usual axioms of probability imply countable additivity and only use real-valued probability. That's why I said that if you assume the probability is well-defined in the infinite case you mentioned, you just can't use a real-valued normalizable probability (by normalizable, I meant that if you add up all the probabilities you get one, so it's equivalent to countable additivity in this case) - you either have to modify probability theory to allow for infinitesimal values or not to be countably additive. Note that this is the case no matter what answer you give to the Sleeping Beauty Problem - you will end up with some scenarios involving countably many people where your probabilities are either undefined, infinitesimal, or zero but not countably additive. So this can't be used as an argument against 1/3 in particular.

@Gameknight

when she wakes up, it is more likely that she is in an experiment where the coinflip was tails.

This is literally false.

If an event happens in every iteration of experiment, observation of such event doesn't reveal you any information about which iteration of experiment you are in.

P(Awake|Heads) = P(Awake|Tails) = P(Awake) = 1

P(Heads|Awake) = P(Heads) = 1/2

P(Tails|Awake) = P(Tails) = 1/2

If Beauty could observe event "I'm awaken twice" - she would be be able to update in favor of Tails. But due to the amnesia, such event is unobservable to her.

@a07c The sample space is literally larger. Again, you're ignoring the fact that she wakes up twice. As noted in my previous long comment: "Since P(Heads) = P(Tails) = 1/2, we can naturally take a sample of two experiments, one Heads and one Tails, and check the probability ourselves. In the Heads experiment, there's one awakening, and the correct answer would be Heads. In the Tails experiment, there's two awakenings, and the correct answer there would be Tails, both times. Then the probability that the coin flipped Heads when Sleeping Beauty awakens is 1/3."

The key here is that the two awakenings are not mutually exclusive - they are mutually inclusive, meaning that if one happens the other must NECESSARILY happen, and thus one Tails Awakening actually implies TWO Tails Awakenings, while one Heads Awakening is just one heads awakening.

@luvkprovider, I disproved both points. If you can't find the comment, here's a link: https://manifold.markets/MartinRandall/is-the-answer-to-the-sleeping-beaut#antjq6ra7kg

So. Where's my bounty money, hm?

@Gameknight Of course you did not disprove my points. You just wrote a bunch of nonsense.

Refutation to your refutation: 1/∞ does not exist.

Refutation to your second argument that my Proof 1 is wrong: You basically made the same argument that Plasma made. I already said I was going to formally defend my Proof 1 in a reply to Plasma (which I am still working on).

I expect you have nothing to say now.

@luvkprovider "1/∞ does not exist."

Oooookay... So the entirety of calculus is, I suppose, thoroughly worthless. Since infinitesimal values cannot exist, derivatives (and by extension integrals) are made-up concepts. Thanks for the heads-up. I guess we'll have to go dig up Newton and Leibniz from their graves to be posthumously shamed.

You also can't just ignore my refutation with "you just wrote a bunch of nonsense", that's like being told 1 + 1 = 2 and going "nuh-uh", as if that'll change the fact that 1 + 1 = 2. I'll be waiting for your defense to Proof 1, as my argument also elaborated on Plasma's proof with more calculation and more methods of proving 1/3.

@Gameknight

The key here is that the two awakenings are not mutually exclusive

Exactly. And as sample space has to consists of mutually exclusive and collectively exhaustive outcomes, the mathematical model that assumes that Tails&Monday and Tails&Tuesday are different outcomes is inapplicable.

In the Heads experiment, there's one awakening, and the correct answer would be Heads. In the Tails experiment, there's two awakenings, and the correct answer there would be Tails, both times.

Yes. Both Tails awakenings are part of the same iteration of the experiment. Tails&Monday and Tails&Tuesday are the same outcome that we can call Tails&Monday&Tuesday. Meaning that in this iteration of the experiment the coin is Tails, Monday awakening happens and Tuesday wakening happens.

Then the probability that the coin flipped Heads when Sleeping Beauty awakens is 1/3."

This doesn't follow.

you're ignoring the fact that she wakes up twice

This fact is irrelevant to the question of probability, as both awakenings are part of the same outcome of the probability experiment.

@Gameknight You don’t understand calculus at all. Calculus does not imply that 1/∞ exists. Also, your refutation is equivalent to 1 + 1 = 3, not 1 + 1 = 2.

@a07c

You said: "

Then the probability that the coin flipped Heads when Sleeping Beauty awakens is 1/3."

This doesn't follow."

what? You can't just say "this doesn't follow", that's a childish "nuh-uh" to basic logical reasoning. If she wakes up 3 times, and two of her awakenings were awakenings where the coin flipped landed on tails, then naturally 2/3 of her awakenings should be tails, and 1/3 should be heads. It obviously naturally follows.

You also said: "

you're ignoring the fact that she wakes up twice

This fact is irrelevant to the question of probability, as both awakenings are part of the same outcome of the probability experiment."

You're missing the point here. Yes, both awakenings are part of the same outcome (tails) of the probability experiment, but Sleeping Beauty doesn't have a f*cking clue whether the coin flipped heads or tails and can't tell whether her current awakening is part of the singular heads awakening or the double tails awakening. She has to consider that tails will make her awaken more, and thus the chances that she woke up on a tails setup is more likely.

I'd like it if you'd engage with my whole argument, because both you and @luvkprovider have been thoroughly ignoring the obvious answer to the Sleeping Beauty variant of "she wakes up an indefinitely large number of times if the coin flips tails". Halfers will still say that Sleeping Beauty should answer 1/2 upon waking up, but if anyone has any logical reasoning skill it should obviously be almost 0, because over a large number of experiments the total number of times that Sleeping Beauty wakes up in a Tails setup is much more than the number of times that she wakes up in a Heads setup.

@luvkprovider "Calculus does not imply that 1/∞ exists." Oh boy, now you're pulling semantics. You know full well what I'm trying to say here, as calculus uses the concept of "the limit of 1/n as n goes to infinity" and other related variations of it. Similarly, we can have n different outcomes, each with probability 1/n, where n -> ∞, so that the total probability adds to 1.

@Gameknight Yes, we can have "probability 1/n, where n -> ∞, so that the total probability adds to 1." But you completely ignored my point. My point was that the proabilities of each scenario "coin flipped heads n times" are equal (according to thirders), so we can’t have "probability 1/n, where n -> ∞" because that would make the probabilities unequal.

@luvkprovider

"My point was that the proabilities of each scenario "coin flipped heads n times" are equal (according to thirders)"

Yes. Let's denote the common probability of each of those "coin flipped heads n times" scenarios as x, just like you did in your original longpost.

My point is that x can equal 1/m, m -> ∞. That was my whole refutation to your Proof 2.

@Gameknight I though you meant that the probabilities could equal 1/2, 1/4, 1/8, etc. so they sum to 1 (because the limit of the denominators goes to infinity). But now I see you mean that each x is individually 1/n, n -> ∞. Well, you know full well that the limit of 1/n as n -> ∞ is 0. And the sum of a countable number of 0's is 0.

@luvkprovider Okay now I feel like you're just ignoring my point by continually misrepresenting it, so I'm going to restate it fully here to resolve any ambiguity:

Let the probability of "the coin landing heads n times" be x(n). x(n) = x(n+1), according to the thirder position - I'm sure we can agree on that. Let us denote this common x(n) as just x, for simplicity. My refutation to your proof 2 is x = 1/m, m -> ∞. Since there are also m -> ∞ possible values of n, then x(0) + x(1) + x(2) + ... = m * x = m * 1/m = 1.

If you distort my argument again by pointing out some "flaw" in a strawman you made, I'm going to assume that you are arguing in bad faith.

@Gameknight You say that "x = 1/m, m -> ∞". But that means x = 0.

@luvkprovider We literally just talked about calculus and how they use the exact same idea I just used: by integrating the sum of 1/m, m -> ∞, m times, to get a finite, nonzero* result. Dear god, I'm starting to really think you're arguing in bad faith here.

*footnote: of course, some integrations add to 0, but you know what i mean - integration is not forced to add to zero just because the dx or d[insert variable here] is 1/m for m -> ∞.

@Gameknight If you don’t agree that the limit of 1/m as m -> ∞ is 0, go search it up.

@luvkprovider Okay, so once again you refuse to acknowledge the concept of a nonzero derivative and a nonzero integral function. Good work. I'm not gonna put up with your bullsh*t.

@Gameknight Ok then, you tell me what the limit of 1/m as m -> ∞ is.

@luvkprovider Oh boy, are you really such a jack*ss that you are going to argue that (1/m) * m = 1, for m -> ∞? Really?

I do recognize that the limit of 1/m = 0 as m -> ∞. However, the limit of m* (1/m) as m -> ∞ is 1. That's been my whole point here. x(n), the common probability that the coin landed heads n times, can be a nonzero but infinitesimally small value (1/m), such that the sum of x(0) + x(1) + x(2) + ... = m * (1/m) = 1.

I agree that the limit of (1/m) * m as m -> ∞ is 1. But you said that x = 1/m as m -> ∞, which means that x = 0. What don't you get.

@Gameknight Omg, stop editing your comments after I already replied. Basically what you are saying that x(n) = 1/100 when you evaluate 100 terms of the partial sum, and it changes to x(n) = 1/101 when you evalutate 101 terms of the partial sum. But the value of x(n) is constant, it doesn’t change continuously.

@luvkprovider "Basically what you are saying that x(n) = 1/100 when you evaluate 100 terms of the partial sum, and it changes to x(n) = 1/101 when you evalutate 101 terms of the partial sum. But the value of x(n) is constant, it doesn’t change continuously."

AND ONCE AGAIN THE MOTHERF*CKER STRAWMANS MY ARGUMENT BY SPECIFICALLY TAKING A CERTAIN SECTION OUT OF CONTEXT OKAYYYY LETS DO THIS

YES, I am saying x(n) = 1/101 when you evaluate 101 terms of the partial sum, etc etc. HOWEVER I AM NOT ACTUALLY SAYING x(n) = 1/101, BECAUSE I AM NOT EVALUATING 101 TERMS OF THE PARTIAL SUM, I AM EVALUATING m -> ∞ TERMS OF THE PARTIAL SUM, WHICH WOULD THEN BE THE WHOLE SUM. THEREFORE x(n) = 1/m, m -> ∞ IS CONSTANT, AND THEREFORE DOES NOT POSE AN ISSUE, AND THE SUM OF THE PROBABILITIES OF ALL EVENTS HAPPENING IS ALSO CONSTANT, WHICH SHOULD ALSO NOT POSE AN ISSUE. THEREFORE PROOF 2 HAS NO BASIS. JESUS F*CK WHAT DO YOU NOT UNDERSTAND

ohhhhhhhhhhhhhhhh i get it you're trolling

well played, what a long-con, and dangling a reward for refutals of your proof is smart

@Gameknight x(n) cannot be an infinitesimally small number greater than 0. That's equivalent to saying that 0.9999… ≠ 1 because 0.9999… = 1 - x(n), where x(n) is "a nonzero but infinitesimally small value (1/m)".

@luvkprovider No it's not, and you're making a false equivalence here. Again, if you want proof that infinitesimally small numbers can be greater than zero, go look at a derivative. Remember that x -> 0 is different from x = 0.

@Gameknight Let me ask what you are saying here. Are you saying that there is a real number x such that x > 0 and x < 0.5^n for all positive integers n?

@Gameknight

If she wakes up 3 times, and two of her awakenings were awakenings where the coin flipped landed on tails, then naturally 2/3 of her awakenings should be tails, and 1/3 should be heads. It obviously naturally follows.

It would naturally follow if these "3 times" were three different iterations of probability experiment. But as they are not so it doesn't follow.

Yes, both awakenings are part of the same outcome (tails) of the probability experiment, but Sleeping Beauty doesn't have a f*cking clue whether the coin flipped heads or tails and can't tell whether her current awakening is part of the singular heads awakening or the double tails awakening.

Yes, therefore she is indifferent between two possible outcomes of the experiment: Heads&OneAwakening and Tails&TwoAwakenings. She observes that she is awaken at least once. This doesn't give her any new information, so she keeps her prior.

She has to consider that tails will make her awaken more, and thus the chances that she woke up on a tails setup is more likely.

No, that's not how probability function is defined. Probability measures the ratio of events happening in different iterations of the experiment.

You should probably start from here: https://en.wikipedia.org/wiki/Experiment_(probability_theory)

Experiment (probability theory) - Wikipedia

And then follow the hyperlinks to get what it means for an outcome/event to be realized in the particular trial and why the notion of an event happening in the same iteration of the experiment multiple times, which you seem to be confusing yourself with, is not even coherent.

I'd like it if you'd engage with my whole argument, because both you and @luvkprovider have been thoroughly ignoring the obvious answer to the Sleeping Beauty variant of "she wakes up an indefinitely large number of times if the coin flips tails".

This version is not particularly enlightening. Yes, no matter how many awakenings the Beauty has her credence should stay the same as in half the iterations of the experiment when she is awake the coin is Heads.

I think the version which luvkprovider described with 2^n awakenings is more interesting. As it actually leads to different betting decisions by thirders and double halfers and thirders will predictably loose all their money while participating even in a single iteration of such experiment, as I describe here https://manifold.markets/MartinRandall/is-the-answer-to-the-sleeping-beaut#nzzj0y1393

@luvkprovider "Are you saying that there is a real number x such that x > 0 and x < 0.5^n for all positive integers n?"

First of all, this is thoroughly irrelevant to the discussion at hand. I am aware this is an attempt to change the topic. Also, lim (0.5^n) = 0 as n -> ∞, so no, there is no specific value x where x > 0 and x < 0.5^n for every single positive integer n. However, there ARE numbers where, for any given positive integer n, satisfy the condition that x > 0 and x < 0.5^n. For example, 0.5^(n+1) > 0 and 0.5^(n+1) < 0.5^n.

@a07c

You said: "

If she wakes up 3 times, and two of her awakenings were awakenings where the coin flipped landed on tails, then naturally 2/3 of her awakenings should be tails, and 1/3 should be heads. It obviously naturally follows.

It would naturally follow if these "3 times" were three different iterations of probability experiment. But as they are not so it doesn't follow."

She awakens 3 times. Two times, she awaked in a tails-setup. Therefore, the chance that each awakening is on a tails-setup is 2/3, and the chance that each awakening is on a heads-setup is 1/3. If you can't understand simple mathematics, then I'm sorry, but it's not my problem anymore.

@Gameknight Ok, so for the x(n) you described earlier, would you agree that there is a positive integer n such that x(n) > 0.5^n?

@luvkprovider x(n) = 1/m -> 0, for m -> ∞, while 0.5^n -> 0 for n -> ∞. It's not really a comparable thing since both are limits.

@Gameknight so you are saying that x(n) is not a real number?

@luvkprovider x(n) is undeniably a real number. If it were not a real number, then calculus itself would unravel entirely. The exact explanation for why and how is beyond me, and I am not qualified to speak about it. But undeniably, as calculus itself is based upon the concept, 1/m as m -> ∞ is a very real and mathematically valid idea. All I'm trying to say is that it is possible that x(n) = 1/m.

@Gameknight If x(n) is a real number greater than 0, then just as you said, there exists a positive integer m such that x(n) > 0.5^m. Thus, adding x(n) to itself infinitely many times will be equal to ∞.

@luvkprovider I see your point, but I feel like the whole "there exists a positive integer m such that x(n) > 0.5^m" doesn't really apply to limits? Because indeed, x -> 0, but x =/= 0.

@Gameknight Would you say that it applies to real numbers?

@Gameknight

If you can't understand simple mathematics

Ironic, considering that I specifically pinpointed the place where your reasoning fails, explained why and provided link to further reading material of conventional wisdom which would allow you to understand it in full details.

Usually I would consider further discussion useless, but you do show a lot more hope than an average thirder. It's funny, really, you seem to have done all the right reasoning steps and then suddenly jump to a completely wrong conclusion.

So I'll give it the last shot.

She awakens 3 times. Two times, she awaked in a tails-setup. Therefore, the chance that each awakening is on a tails-setup is 2/3, and the chance that each awakening is on a heads-setup is 1/3.

Here you are implicitly assuming that

P(Tails) = P(Tails&Monday) + P(Tails&Tuesday)

but this assumption only works for mutually exclusive events, which, as you've stated yourself, these awakenings are not.

The correct formula for non-mutually exclusive events is:

P(Tails) = P(Tails&Monday) + P(Tails&Tuesday) - P(Tails&Monday&Tails&Tuesday)

where Tails&Monday&Tails&Tuesday is a non-empty intersection between Tails&Monday and Tails&Tuesday.

As in every iteration of the experiment where Tails&Monday happens, Tails&Tuesday also happens:

P(Tails&Monday) = P(Tails&Tuesday) = P(Tails&Monday&Tails&Tuesday)

Now if we try to put 1/3 here we get:

P(Tails) = 1/3 + 1/3 - 1/3 = 1/3

and then

P(Heads) = 1 - P(Tails) = 2/3

Which is clearly wrong. The correct value is 1/2:

P(Tails) = 1/2 + 1/2 -1/2 = 1/2

P(Heads) = 1 - P(Tails) = 1/2

@a07c

"Here you are implicitly assuming that

P(Tails) = P(Tails&Monday) + P(Tails&Tuesday)"

No I am not, you overcomplicating fool. I count two awakenings where she wakes up tails, and one awakening when she wakes up heads. Let's do a little simple math: How many awakenings does she have in total? It's 1 + 2 = 3, right? Now how many of those are heads? Just one, as previously discussed. So what fraction of awakenings are heads? It's 1 / 3. There's your f*cking answer. I don't need any of your other bullsh*t.

@Gameknight I see I shouldn't have given you the benefit of the doubt.

@a07c I see you don't have a counterargument; thanks for reminding me of how right I am.

@a07c I think that your viewpoint is slightly distorted, but mostly right. Instead of arguing that P(Monday) is 1, you should instead argue that P(Monday) is undefined. It's not that your argument is wrong; it is completely valid to change the definition of an event (you are changing "Beauty wakes up and it is Monday" to "Beauty wakes up on Monday at some point during the experiment") by showing that it is the only way to make it consistent with probability. But although "P(Monday) = 1" can make sense in probability theory, it does not make physical sense, as it is certainly possible that when Beauty wakes up, it is not Monday. Your argument would have been stronger if you had instead said that there is no way to include the event "Monday" in a probability space without contradictions. Also, thank you for recommending me four of your excellent posts, which I’ve read. I will give you personal feedback on those posts in a private message.

@Gameknight You did not provide a counterargument to Ape in the first place. You just kept repeating the faulty arguments from your first reply. I will admit, however, that unlike your usual self, you actually included a single new argument that has not been refuted yet. That argument is: "If you can’t understand simple mathematics, then I’m sorry"

Refutation: You literally thought that the limit of 1/m as m -> ∞ is not 0.

@luvkprovider Man, every time you respond I laugh, because the only thing keeping up your argument is moving the goalposts and strawmanning.

You literally thought that the limit of 1/m as m -> ∞ is not 0.

No, I didn't. Nice try. I said that, for x = 1/m, m -> ∞, the limit of x is 0, but x is not exactly equal to 0. Every time you strawman me, it is just more proof that I am right and you just want to create nonexistent faults for the sake of winning an argument.

Anyway, I've already made my point. You can choose to ignore the basics of fractions and probability, but that's not my problem anymore. I just hope others aren't fooled by your bullsh*t.

@Gameknight From your definition of x, do you agree with these two statements:

1. x is a positive real number

2. x < 1/m for every positive integer m