Among other things, there I address Elga's model and explain why it's valid but not sound. His proof is about a different problem which I call "No-Coin-Toss problem":
You are put to sleep and then awaken according to the outcome of a random number generator. If the generator outputs 0, you are awaken on Monday and a coin is put Heads. If 1 - you are also awaken on Monday but the coin is put Tails. If 2 - you are awakened on Tuesday and the coin is also put Tails.
It's kind of obvious in retrospect, because according to Elga's model:
P(Heads)=P(Heads&Monday)=P(Heads|Awake)=1/3 that is, non-conditional probability of allegedly fair coin being Heads is 1/3, which can only happen when there was no coin toss in the first place.
> Therefore, the odds of each event is equal, and there are 3 events, so your credence in the coin having come up heads and it being the first day should be 1/3.
This is true when the events are mutually exclusive so that sum of their probabilities adds up to 1. Like, once again, in No-Coin-Toss problem. In Sleeping Beauty, however, this is not the case because Tails&Monday and Tails&Tuesday are sequential. You can't treat them as individual outcomes and formally construct a probability space.
Ironically enough, this is exactly what Elga's paper shows - that you can't lawfully use probability theory with the framework of "centred possible works" which David Lewis came up with no mathematical justifications. Because if you do, it inevitably leads to fair coin having arbitrary probability.
//P(Heads)=P(Heads&Monday)=P(Heads|Awake)=1/3 that is, non-conditional probability of allegedly fair coin being Heads is 1/3, which can only happen when there was no coin toss in the first place.//
No...sometimes your credence in a past fair event turning out some way should be above .5. For instance, if a fair coin was flipped and if it came up heads I'd be instantly killed, I should think with credence greater than .5 that it came up tails.
//How has Elga managed to accidentally model a different problem? Well, he was coming from the base thirder's assumption that "this awakening" is randomly sampled from three possible outcomes. Plus the belief that the Beauty doesn't get any new information on awakening. So he modeled a problem where exactly that happens, which, as it turned out, isn't a Sleeping Beauty problem.//
That is neither premise of his argument. Elga argues only that your credence in it being day 1 and heads = day 1 and tails, and day 1 and tails = day and tails, so 1/3 is right.
//Either Thirdism is wrong or there is an update on awakening//
As far as I can tell, your second essay doesn't have any argument--it just sort of asserts this very weird model on which you don't update without explaining where Elga goes wrong.
//How come our interest in something else is affecting the statistical properties of events? Why do we think that Beauty has to consider the problem from a "temporally located viewpoint"? //
The evidence beauty has that she's updating on is de se evidence. The facts are about which one happens to be the current moment.
//Don't get me wrong, that's a fair question, that deserves a fair answer. Which at this point should be clear - by doing so you may end up contradicting probability theory.//
No probability theory is contradicted by SIA. You're just assuming that you learn nothing new from your waking up--but that isn't true.
> No...sometimes your credence in a past fair event turning out some way should be above .5. For instance, if a fair coin was flipped and if it came up heads I'd be instantly killed, I should think with credence greater than .5 that it came up tails.
This is not an unconditional probability, though. This is probability conditional on survival. In such situation you can't be sure that you survive and therefore your survival is evidence in favor of Tails and thus you update. Unlike in Sleeping Beauty where you are certain to be awaken and therefore you do not get to update your probability estimate where you indeed awake.
Moreover, as soon as we start to talk about update on awakenings due to some new evidence that the Beauty allegedly receives, we already discard Elga's model because, as he explicitly states in his paper the Beauty doesn't receive any new information:
"At the start of the experiment, you had credence 1/2 in H. But you were also certain
that upon being awakened on Monday you would have credence 1/3 in H — even though
you were certain that you would receive no new information and suffer no cognitive
mishaps during the intervening time. "
> Elga argues only that your credence in it being day 1 and heads = day 1 and tails, and day 1 and tails = day and tails, so 1/3 is right.
I've already addressed it in the comment to which you are answering:
"This is true when the events are mutually exclusive so that sum of their probabilities adds up to 1. Like, once again, in No-Coin-Toss problem. In Sleeping Beauty, however, this is not the case because Tails&Monday and Tails&Tuesday are sequential. You can't treat them as individual outcomes and formally construct a probability space."
The statement P(Heads&Monday) = P(Tails&Monday) = P(Tails&Tuesday) is correct,
it's just that their sum isn't 1 and so they are equal to 1/2 not to 1/3, because Tails&Monday is the same event as Tails&Tuesday.
> There should be an update on wakening
Then, once again, we are discarding Elga's model in favor of Updating one, which I explore as well. I'll also leave a comment under the post you link to, highlighting all the places where you are mistaken.
> As far as I can tell, your second essay doesn't have any argument
There are several arguments made in the post, none of which I've ever seen addressed.
First, that Elga's model just like Lewis's and Updating model fail the statistical test, as they are based on the faulty assumptions that three awakenings can be treated as mutually exclusive events when according to the experimental settings they are not.
Second, that amnesia doesn't make the awakenings random from the Beauty's perspective because she is still aware about the setting of the experiment.
Third, that the whole framework of "de se evidence" isn't based on anything. It occasionally tries to treat sequential events as mutually exclusive in direct contradiction with the core principles of probability theory and therefore leads to paradoxes, which Sleeping Beauty is an example of.
Forth, is that as soon as we stop using this framework and engage with the problems strictly according to the rules of probability theory - it's easily resolved without any issues. No probability updates out of nowhere, no presumptuousness, no suddenly unfair coins.. I demonstrate it by explicitly presenting the correct model in extreme details
> The evidence beauty has that she's updating on is de se evidence. The facts are about which one happens to be the current moment.
And as I explain in details, "de se evidence" is utter non-sense with no mathematical justification and, therefore, we are not allowed to use probability theory in such a manner. Probability theory can *sometimes* be used to reason about uncertainty in time - when we are confident that the events in question are truly mutually exclusive like in No-Coin-Toss or Single-Awakening problems. But not when we are dealing with sequential events. I'm formally proving a lemma that sequential events can't be mutually exclusive in the comments, even though it's completely obvious.
> No probability theory is contradicted by SIA. You're just assuming that you learn nothing new from your waking up--but that isn't true.
SIA and SSA both contradicts probability theory every time they attempts to construct a sample space from non-mutually exclusive events. The fact that they both just assume that a particular sort of sampling necessary happens every time without any justifications should be telling enough.
The question you quoted is the one Elga posed, but not quite the one he solved. You summarized his solution, but without introducing the differences that explain it. The two should be equivalent, but those differences opened the door to the controversy over his answer.
The one he posed does not differentiate the "two days" in any way. The one he solved assumed that you will always be wakened on the "first day," which he called Monday, and that you will be wakened on Tuesday only if the coin came up Tails. And what does "first awakened" mean, since you cannot determine if it is "the first day?" Elga meant before you might be told either that the coin came up Tails (premise #1), or that it is the first day (premise #2).
The interesting part of his setup, is that it creates the situation that obfuscates the solution. But he also uses it to avoid addressing that situation. Which is, what is the status of Tuesday, after Heads, in the solution? This obfuscates the solution since halfers will try to rationalize a relationship between H&Tue and T&Tue. Specifically, since any effect due to H&Tue is ignored, they want to ignore T&Tue the same way. I'll return to this.
But there is a better setup. Once you are fully asleep, the researchers flip two coins; call them C1 and C2. Then this procedure is performed:
Procedure: If both coins are showing Heads, the procedure is ended without waking you. But if either coin is showing Tails, you are wakened and asked "To what degree ought you believe that coin C1 is showing Heads?" After you answer, you are put back to sleep with amnesia.
Once this procedure is ended, and you are again fully asleep, coin C2 is turned over and the procedure is performed again.
When you are awake, which will happen once or twice, you know that at the beginning of the procedure there were four equally-likely combinations for C1C2. They were HH, HT, TH, and TT. But since you were awakened in this iteration of the procedure, you know that the two coins are not showing HH. Of the three remaining possibilities, which must still be equally likely, only one has C1=H. The answer is 1/3.
Returning to Elga's solution, in it there are four possible situations where something happens in the experiment: {H&Mon, T&Mon, H&Tue, T&Tue}. What confuses halfers is that being left asleep counts as "something." The day still occurs, you just won't observe it. But in fact, it doesn't matter if you are awakened, as long as something different happens when you are..
And the "somethings" that might happen on different days count as independent outcomes to you, since you cannot connect it to another day due to the amnesia drug. The "new information" that you have is that the sample space is reduced to {H&Mon, T&Mon, T&Tue}, each with 1/3 probability. And I can't emphasize this enough: By saying you have no "new information," Halfers are implicitly claiming that Tuesday CEASES TO EXIST if you sleep through it. Because if it does exist, and you know it is not your current situation, that is "new information."
What Elga did, in your first two premises, was reduce the sample space further by eliminated both H&Tue, which is already eliminated, and one additional situation. So he IGNORED H&Tue, while Halfers imply it cannot exist.
> halfers will try to rationalize a relationship between H&Tue and T&Tue.
The relationship is between T&Mon and T&Tue, more specifically they are sequential, not mutually exclusive events and therefore can't be used to construct a sample space the way thirders try to.
> But there is a better setup.
As I've already told you, your version with the two coins has all the same issues. The pairs TH and TT; HH and HT are also sequential events and therefore can't be used to construct a sample space. Granted, it may appear a bit more confusing, but the correct answer to your version of Sleeping Beauty is also 1/2.
> What confuses halfers is that being left asleep counts as "something." The day still occurs, you just won't observe it
If you are certain that you don't observe something, you do not get to update your credence while you indeed do not observe it. See conservation of expected evidence.
> And the "somethings" that might happen on different days count as independent outcomes to you, since you cannot connect it to another day due to the amnesia drug.
Amnesia doesn't change Beauty's knowledge state about the experiment to the point where she can rationally believe that the awakenings are exclusionary and happen at random. She still remembers the setting of the experiment, and the order between awakenings, even though she doesn't remember whether she was awakened before or not.
> By saying you have no "new information," Halfers are implicitly claiming that Tuesday CEASES TO EXIST if you sleep through it. Because if it does exist, and you know it is not your current situation, that is "new information."
What Halfers should be claiming is that you are not allowed to use probability theory in such a way, where you treat sequential events Monday and Tuesday as mutually exclusive. Sadly, due to historical reasons, the same person who came up with the "bright idea" to construct sample space from sequential events is the same person who came up with first (and obviously wrong) Halfer model. This created a convenient weakman for Thirders to defeat, while keeping the correct Halfer reasoning obscured.
> The relationship is between T&Mon and T&Tue, more specifically they are sequential, not mutually exclusive events and therefore can't be used to construct a sample space the way thirders try to.
That's the argument halfers try to use. It's not true. You provide no justification for how it affects SB's belief. But it is the point of my variation, where it no longer applies. Everything is based on what happens on a single day.
> As I've already told you, your version with the two coins has all the same issues.
And when you do, you still fail to justify how "sequential" affects the beliefs of a subject who cannot see it, and you ignore that it is not true for the two-coin version.
> The pairs TH and TT; HH and HT are also sequential events and therefore can't be used to construct a sample space.
Why? But aside from that, SB's sample space can be based on the four possible combinations - and they are "outcomes" - that exist on a single day.
> Granted, it may appear a bit more confusing, but the correct answer to your version of Sleeping Beauty is also 1/2.
When the researchers look at the coins, there are four possible combinations. Each has a 14 probability of existing, and this does not depend in any way on whatever "sequential" means. One of these combinations is eliminated, in SB's knowledge. This is what "new information" means. The remaining three all have an updated probability of 1/3.
> If you are certain that you don't observe something, you do not get to update your credence while you indeed do not observe it. See conservation of expected evidence.
Say I am put to sleep, and I will wake up in a green room if a die roll is odd, or a red room if it is even. If what happens is that I wake up in a green room, I am observing that the die roll was odd and was not even. I "get to" update on this information.
If a theory of "the conservation of expected evidence" says otherwise, it is bogus.
> Amnesia doesn't change Beauty's knowledge state about the experiment to the point where she can rationally believe that the awakenings are exclusionary and happen at random.
????
Amnesia separates her knowledge in one pass from anything having to do with the other. The issue wilt Elga's version is that she may (depending on who you listen to) need to factor what could happen on the other day into her assessment. In mine, her assessment can be based entirely on one day. "Sequential" does not change that fact that at the start of this say, there were four equally-likely combinations. If she is awake, one is no longer possible.
> She still remembers the setting of the experiment, and the order between awakenings,
There is no ordering of awakenings. Each looks exactly the same as the other., and functions as a stand-alone experiment.
> What Halfers should be claiming is that you are not allowed to use probability theory in such a way, where you treat sequential events Monday and Tuesday as mutually exclusive.
> You provide no justification for how it affects SB's belief.
Beauty literally knows that. Do I need to justify that knowing things affect your beliefs?
> But it is the point of my variation, where it no longer applies. Everything is based on what happens on a single day.
Still sequential events. All the same problems.
> Why?
> Why? And please explain why "sequential" matters.
Sure. Let's prove a simple lemma.
Let A1 and A2 be two sequential events so that A2 happens always and only before or after A1.
Now, let's suppose they are also mutually exclusive. Then we can construct a sample space Ω={A1, A2} and a probability space (Ω,F,P), satisfying the axioms of probability theory.
By Kolmogorov's third axiom:
P(A1) + P(A2) = 1
There are two possibilities, either P(A1) = 0 or P(A1) > 0
if P(A1) = 0 then P(A2) =1, but this contradicts the initial premise that A2 happens only before or after A1.
so P(A1)> 0 and P(A2) < 1
But then with 1-P(A2) probability A2 will not happen before or after A1 has happened. This contradicts the initial premise that A2 always happens before or after A1.
Q.E.D. Sequential events are not mutually exclusive.
> SB's sample space can be based on the four possible combinations
Unless they are mutually exclusive, it can't. See Conditions of Sample Space
Researchers perspective is irrelevant. We are talking about the Beauty.
> Say I am put to sleep, and I will wake up in a green room if a die roll is odd, or a red room if it is even. If what happens is that I wake up in a green room, I am observing that the die roll was odd and was not even. I "get to" update on this information.
Yes! Absolutely! You were not sure which color of room you'll observe beforehand, and so you update. But if you were meant to be put in a green room regardless of the outcome of a die and, therefore, could be certain not to observe that the room is red, you do not update. It's very straightforward and intuitive.
Likewise, Sleeping Beauty awakens in the experiment regardless of the outcome of the coin, so she doesn't update. She was sure not to observe the outcome HH and so she doesn't update. She would've updated, if she could observe the event "I'm awakened the second time in the same experiment", but she can't do it due to the memory loss.
> The issue wilt Elga's version is that she may (depending on who you listen to) need to factor what could happen on the other day into her assessment.
I suppose some people may find this to be the case, but I don't think I've ever claimed that. So there is no point in arguing about it with me.
> There is no ordering of awakenings.
There is. It's in the definition of the experiment that you've yourself described. TH and TT are sequential events. And Beauty gets told about it.
> Each looks exactly the same as the other.
Beauty knowledge isn't limited to her observation of the awakenings. She also was explicitly told about the order of the awakenings. She has to ignore this knowledge in order to reason about awakenings as happening at random.
Here is a simpler case, which, I believe, can capture our disagreement. Suppose there is a rigged dice that shows 6 always and only if it showed 5 on a previous roll. You are told that is the way the dice works and this it the only thing you know about this dice. Are you justified to treat it as a fair one? If no, why not?
Let's try to make some sense of your “proof.” I’ll start with some definitions:
An outcome is a measurable result of a random experiment. A sample space is a set of all possible distinct outcomes for an experiment. An event is any subset of a sample space.
What confuses many non-mathematicians, is that there is no single “correct” sample space. Roll two six-sided dice, and the sample space could be 36 pairs of values, 21 ordered pairs, or eleven sums. The only requirements are that the members be mutually exclusive, and the set be exhaustive. But the choice of your sample space usually is driven by how you can exhaustively discriminate the results.
You seem to confuse “outcome” and “event” when you think you are applying Kolmogorov’s theory.
> Let A1 and A2 be two sequential events so that A2 happens always and only before or after A1.
If the words “happens after” are involved, you are describing two different random experiments. One may depend on the other, but they are different. In all variations of the SB problem, there is just one experiment with two opportunities to observe it.
So, what you describe, is two sequential observations of the same random experiment. That is, flip the coin(s) once, observe the outcome one way, and then observe it again in another way. The point is that you are observing the same result, not two different results.
If the sample space is well constructed, each observation divides the sample space into two events; one event where every outcome in the set is 100% consistent with the observation, and its complement where every outcome in it contradicts the observation.
>> Now, let's suppose they are also mutually exclusive.
Two different observations of a single outcome cannot be “mutually exclusive.” Each observation defines an event that includes the member of the sample space that describes the actual result. This seems to be the conclusion you reach by faulty logic, but it is irrelevant. It doesn’t affect probabilities the way you think it does.
>> Then we can construct a sample space Ω={A1, A2} and a probability space (Ω,F,P), satisfying the axioms of probability theory.
No, we actually can’t. Ω is the set that contains all distinct outcomes that are possible. F is Kolmogorov's set that contains events. There are some restrictions on what events go into it, but the set of all possible events satisfies them.
But you don’t use these terms correctly, or recognize the differences between an outcome and an observation of an outcome. Your statement here makes no sense in the context of Kolmogorov’s theory. NOT(QED).
+++++
But it can be done, with a well constructed sample space that is sufficient to represent bit observations. Like my two-coin version.
My sample space has four equally-likely outcomes: {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. The first observation recognizes the consistent event {HT1_HH2, TH1_TT2, TT1_TH2} and the inconsistent event {HH1_HT2}. Note that how HH1_HT2 might be recognized, if it were consistent, is irrelevant since it is not consistent. The second observation recognizes the consistent event {HH1_HT2, TH1_TT2, TT1_TH2} and the inconsistent event {HT1_HH2}.
You call these observations “sequential events.” They aren’t events, they are observations. Because of the amnesia drug, they are unrelated observations of the same probability experiment.
In each one, the conditional probability of the event where coin C1 is showing Heads during that observation is the prior probability of that event, divided by the prior probability of the “consistent” event. The result is 1/3.
> Ω is the set that contains all distinct outcomes that are possible. F is Kolmogorov's set that contains events.
Yep, I'm well aware of the definitions.
> No, we actually can’t.
You failed to explain why.
> There are some restrictions on what events go into it, but the set of all possible events satisfies them.
So? How does it make us unable to construct a probability space? What restrictions do I cross?
> You seem to confuse “outcome” and “event”
What you seem to be implying, is that we should,'t be able to make the transition from A1 to {A1} or vice versa. Dare to explain why?
> If the words “happens after” are involved, you are describing two different random experiments.
I'm describing two runs of the same experiment.
> recognize the differences between an outcome and an observation of an outcome
Oh! Maybe I don't. I know only about outcomes from Ω and events from F and a measure function P which domain is F. What is this "observation of outcome" thing your are talking about. Please specify it formally.
> My sample space has four equally-likely outcomes: {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}
I'm glad that you finally stopped claiming that HH TT HT and TH are themselves individual outcomes. We are making progress here! We would've saved each other a lot of time if you were rigorous to begin with.
> The first observation recognizes the consistent event {HT1_HH2, TH1_TT2, TT1_TH2} and the inconsistent event {HH1_HT2}.
> The second observation recognizes the consistent event {HH1_HT2, TH1_TT2, TT1_TH2} and the inconsistent event {HT1_HH2}
By first observation you mean the event "Beauty is awakened before the second coin was turned". And by the second - "the Beauty is awakened after the second coin was turned". Indeed, if the Beauty would be able to to make such observations, she should conclude that the first coin is Heads only with 1/3 probability.
There is only a tiny problem here. She can't. She is not told whether she is experiencing the awakening before the turn of a coin or after it. And so her two observations does not separate the outcome space, each time she observes the event "Any outcome of the sample space is realized".
> You call these observations “sequential events.”
No, I call TT and TH sequential events, but as you stopped committing the error, of treating them this way, we can move past it.
You don’t use them correctly. You use the word “event” for both how a result is observed, and a member of a sample space.
> You failed to explain why [we can’t construct a sample space Ω={A1, A2}, where A1 and A2 are events that you imply are the observations of the same result]
See the definitions of “outcome”, “sample space”, and “event” that I used to explain why you can’t do that.
> So? How does it make us unable to construct a probability space?
…. With a sample space comprising two non-distinct non-events that you call events? When the members of a sample space are, by the definitions you claim to understand, supposed to be outcomes and not events?
> What you seem to be implying, is that we shouldn’t be able to make the transition from A1 to {A1} or vice versa. Dare to explain why?
If A1 is an outcome, I do it all the time.
> I'm describing two runs of the same experiment.
No, you are describing the result of the same experiment two different times, using different criteria. "Sequential" means that (1) the result is determined, (2) you observe it one way and call it one "event," (3) maybe you forget that obsevation?, and (4) you observe it again and call it a different "event," even tho nothing has changed in the random state.
>> My sample space has four equally-likely outcomes: {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}
> I'm glad that you finally stopped claiming that HH TT HT and TH are themselves individual outcomes.
They are. In the first observation, XY1_XZ2 is the same as XY. In the second, it is XZ. This is what you call “sequential events.”
> She is not told whether she is experiencing the awakening before the turn of a coin or after it.
And I never said she was. But she does know that she is experiencing only one awakening. Her treatment of the sample space is different only in that she uses different names for equivalent outcomes. The shorter name is how she makes these equivalences to the longer names. Since the equivalences work the same regardless of which observation it is, this works.
But this is the “Halfer’s Fallacy.” They think that, by not knowing which observation, it works like Schrodinger’s Cat: she is "in" both observations at the same time.
> No, I call TT and TH sequential events
No, you call TT1_TH2 observed the first time, and then the second time, “sequential events.”
Different "simpler case": I tell you that I have a biased coin, but not how it is biased. I ask you to guess what the result will be if I flip it. If you are right, I'll give you $1; if wrong, you give me $1. Do you choose Heads to Tails?
The point is that your rational belief - that is, *your* probability based on *your* knowledge that includes it being biased - can only be 50:50. While you do have the information that "sequential" flips of the coin will trend to one face or the other, you do not know what that trend is. You cannot base your belief on long-term results, only on what you know. I believe I once saw an article on lesswrong about this, you can look it up.
To answer your "simpler case," over many rolls this die will roll each result about the same number of times. A single roll, with no knowledge of any effect of ordering, is "fair" the same way a single flip of my coin is "fair."
More appropriate "simpler case": Suppose there is a rigged die with a counter inside it, for every time it has been rolled. For odd-numbered rolls, it rolls with a uniform probability distribution. For even-numbered rolls, the result is the previous roll, plus one (cycling back to 6-->1). This is what, I believe, you would call "sequential."
If you roll this die, knowing that it is an odd-roll, the probability distribution you would use for the sample space {1,2,3,4,5,6} is uniform. If you roll it without knowing the previous outcome, the probability distribution is uniform regardless of whether it is an odd-roll or an even-roll. It is only if you know it is an even-roll, AND you know something about the previous result, that the distribution changes.
Said another way, while you know there is an effect due to order, if you do not know the position in that order, or anything about another sample in that order that is related to this one, THERE IS NO EFFECT DUE TO ORDER. So there essentially IS NO ORDER to the version of the random experiment you are observing.
In my two-coin version of the SB problem, to an outside observer there are four outcomes (not "event;," you do not seem to know what that word means in probability) of the experiment that comprise a sample space. Named fully, they are {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. Each has a probability of 1/4.
What you seem to mean by "sequential" is that, for example, the outcome HH(n) is the same outcome as HT(3-n). That's why I used combined names. Nobody doubts this. But SB only sees the "n" side of the combined outcome. The fact that she knows the relationships between {HH(n), HT(n), TH(n), TT(n)} and {HH(3-n), HT(3-n), TH(3-n), TT(3-n)} can only affect her belief if she has some information about what happened in the "3-n" side. Just like my biased coin example, the existence of some influencing factor can only influence her belief if she also knows something about a result that shares that factor.
So, SB knows the sample space is {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}, and the probability distribution is {1/4,1/4,1/4,1/4}. She knows with certainty (that is, probability 1) that EXACTLY one of these outcomes is eliminated, either HH1_HT2 or HT1_HH2. So her belief that Coin C1 is currently showing Heads is either:
Please note how this allows for the effect of being "sequential", by being based on the combined outcomes instead of the one-day outcomes. The significant difference between this, and every other halfer-thirder debate, is that the two calculations are the same. So we do not need to consider the position in any order. Even if that should have an effect, which it does not.
Her prior should be 1/2, because that should be one's credence in the flip of a fair coin. I obviously disagree that she should maintain her prior for reasons I give in this article.
https://www.lesswrong.com/posts/SjoPCwmNKtFvQ3f2J/lessons-from-failed-attempts-to-model-sleeping-beauty
Among other things, there I address Elga's model and explain why it's valid but not sound. His proof is about a different problem which I call "No-Coin-Toss problem":
You are put to sleep and then awaken according to the outcome of a random number generator. If the generator outputs 0, you are awaken on Monday and a coin is put Heads. If 1 - you are also awaken on Monday but the coin is put Tails. If 2 - you are awakened on Tuesday and the coin is also put Tails.
It's kind of obvious in retrospect, because according to Elga's model:
P(Heads)=P(Heads&Monday)=P(Heads|Awake)=1/3 that is, non-conditional probability of allegedly fair coin being Heads is 1/3, which can only happen when there was no coin toss in the first place.
> Therefore, the odds of each event is equal, and there are 3 events, so your credence in the coin having come up heads and it being the first day should be 1/3.
This is true when the events are mutually exclusive so that sum of their probabilities adds up to 1. Like, once again, in No-Coin-Toss problem. In Sleeping Beauty, however, this is not the case because Tails&Monday and Tails&Tuesday are sequential. You can't treat them as individual outcomes and formally construct a probability space.
Ironically enough, this is exactly what Elga's paper shows - that you can't lawfully use probability theory with the framework of "centred possible works" which David Lewis came up with no mathematical justifications. Because if you do, it inevitably leads to fair coin having arbitrary probability.
I comprehensively talk about it here:
https://www.lesswrong.com/posts/gwfgFwrrYnDpcF4JP/the-solution-to-sleeping-beauty
//P(Heads)=P(Heads&Monday)=P(Heads|Awake)=1/3 that is, non-conditional probability of allegedly fair coin being Heads is 1/3, which can only happen when there was no coin toss in the first place.//
No...sometimes your credence in a past fair event turning out some way should be above .5. For instance, if a fair coin was flipped and if it came up heads I'd be instantly killed, I should think with credence greater than .5 that it came up tails.
//How has Elga managed to accidentally model a different problem? Well, he was coming from the base thirder's assumption that "this awakening" is randomly sampled from three possible outcomes. Plus the belief that the Beauty doesn't get any new information on awakening. So he modeled a problem where exactly that happens, which, as it turned out, isn't a Sleeping Beauty problem.//
That is neither premise of his argument. Elga argues only that your credence in it being day 1 and heads = day 1 and tails, and day 1 and tails = day and tails, so 1/3 is right.
//Either Thirdism is wrong or there is an update on awakening//
There should be an update on wakening https://benthams.substack.com/p/contra-philosophy-bear-and-aaron.
As far as I can tell, your second essay doesn't have any argument--it just sort of asserts this very weird model on which you don't update without explaining where Elga goes wrong.
//How come our interest in something else is affecting the statistical properties of events? Why do we think that Beauty has to consider the problem from a "temporally located viewpoint"? //
The evidence beauty has that she's updating on is de se evidence. The facts are about which one happens to be the current moment.
//Don't get me wrong, that's a fair question, that deserves a fair answer. Which at this point should be clear - by doing so you may end up contradicting probability theory.//
No probability theory is contradicted by SIA. You're just assuming that you learn nothing new from your waking up--but that isn't true.
> No...sometimes your credence in a past fair event turning out some way should be above .5. For instance, if a fair coin was flipped and if it came up heads I'd be instantly killed, I should think with credence greater than .5 that it came up tails.
This is not an unconditional probability, though. This is probability conditional on survival. In such situation you can't be sure that you survive and therefore your survival is evidence in favor of Tails and thus you update. Unlike in Sleeping Beauty where you are certain to be awaken and therefore you do not get to update your probability estimate where you indeed awake.
Moreover, as soon as we start to talk about update on awakenings due to some new evidence that the Beauty allegedly receives, we already discard Elga's model because, as he explicitly states in his paper the Beauty doesn't receive any new information:
"At the start of the experiment, you had credence 1/2 in H. But you were also certain
that upon being awakened on Monday you would have credence 1/3 in H — even though
you were certain that you would receive no new information and suffer no cognitive
mishaps during the intervening time. "
> Elga argues only that your credence in it being day 1 and heads = day 1 and tails, and day 1 and tails = day and tails, so 1/3 is right.
I've already addressed it in the comment to which you are answering:
"This is true when the events are mutually exclusive so that sum of their probabilities adds up to 1. Like, once again, in No-Coin-Toss problem. In Sleeping Beauty, however, this is not the case because Tails&Monday and Tails&Tuesday are sequential. You can't treat them as individual outcomes and formally construct a probability space."
The statement P(Heads&Monday) = P(Tails&Monday) = P(Tails&Tuesday) is correct,
it's just that their sum isn't 1 and so they are equal to 1/2 not to 1/3, because Tails&Monday is the same event as Tails&Tuesday.
> There should be an update on wakening
Then, once again, we are discarding Elga's model in favor of Updating one, which I explore as well. I'll also leave a comment under the post you link to, highlighting all the places where you are mistaken.
> As far as I can tell, your second essay doesn't have any argument
There are several arguments made in the post, none of which I've ever seen addressed.
First, that Elga's model just like Lewis's and Updating model fail the statistical test, as they are based on the faulty assumptions that three awakenings can be treated as mutually exclusive events when according to the experimental settings they are not.
Second, that amnesia doesn't make the awakenings random from the Beauty's perspective because she is still aware about the setting of the experiment.
Third, that the whole framework of "de se evidence" isn't based on anything. It occasionally tries to treat sequential events as mutually exclusive in direct contradiction with the core principles of probability theory and therefore leads to paradoxes, which Sleeping Beauty is an example of.
Forth, is that as soon as we stop using this framework and engage with the problems strictly according to the rules of probability theory - it's easily resolved without any issues. No probability updates out of nowhere, no presumptuousness, no suddenly unfair coins.. I demonstrate it by explicitly presenting the correct model in extreme details
> The evidence beauty has that she's updating on is de se evidence. The facts are about which one happens to be the current moment.
And as I explain in details, "de se evidence" is utter non-sense with no mathematical justification and, therefore, we are not allowed to use probability theory in such a manner. Probability theory can *sometimes* be used to reason about uncertainty in time - when we are confident that the events in question are truly mutually exclusive like in No-Coin-Toss or Single-Awakening problems. But not when we are dealing with sequential events. I'm formally proving a lemma that sequential events can't be mutually exclusive in the comments, even though it's completely obvious.
> No probability theory is contradicted by SIA. You're just assuming that you learn nothing new from your waking up--but that isn't true.
SIA and SSA both contradicts probability theory every time they attempts to construct a sample space from non-mutually exclusive events. The fact that they both just assume that a particular sort of sampling necessary happens every time without any justifications should be telling enough.
The question you quoted is the one Elga posed, but not quite the one he solved. You summarized his solution, but without introducing the differences that explain it. The two should be equivalent, but those differences opened the door to the controversy over his answer.
The one he posed does not differentiate the "two days" in any way. The one he solved assumed that you will always be wakened on the "first day," which he called Monday, and that you will be wakened on Tuesday only if the coin came up Tails. And what does "first awakened" mean, since you cannot determine if it is "the first day?" Elga meant before you might be told either that the coin came up Tails (premise #1), or that it is the first day (premise #2).
The interesting part of his setup, is that it creates the situation that obfuscates the solution. But he also uses it to avoid addressing that situation. Which is, what is the status of Tuesday, after Heads, in the solution? This obfuscates the solution since halfers will try to rationalize a relationship between H&Tue and T&Tue. Specifically, since any effect due to H&Tue is ignored, they want to ignore T&Tue the same way. I'll return to this.
But there is a better setup. Once you are fully asleep, the researchers flip two coins; call them C1 and C2. Then this procedure is performed:
Procedure: If both coins are showing Heads, the procedure is ended without waking you. But if either coin is showing Tails, you are wakened and asked "To what degree ought you believe that coin C1 is showing Heads?" After you answer, you are put back to sleep with amnesia.
Once this procedure is ended, and you are again fully asleep, coin C2 is turned over and the procedure is performed again.
When you are awake, which will happen once or twice, you know that at the beginning of the procedure there were four equally-likely combinations for C1C2. They were HH, HT, TH, and TT. But since you were awakened in this iteration of the procedure, you know that the two coins are not showing HH. Of the three remaining possibilities, which must still be equally likely, only one has C1=H. The answer is 1/3.
Returning to Elga's solution, in it there are four possible situations where something happens in the experiment: {H&Mon, T&Mon, H&Tue, T&Tue}. What confuses halfers is that being left asleep counts as "something." The day still occurs, you just won't observe it. But in fact, it doesn't matter if you are awakened, as long as something different happens when you are..
And the "somethings" that might happen on different days count as independent outcomes to you, since you cannot connect it to another day due to the amnesia drug. The "new information" that you have is that the sample space is reduced to {H&Mon, T&Mon, T&Tue}, each with 1/3 probability. And I can't emphasize this enough: By saying you have no "new information," Halfers are implicitly claiming that Tuesday CEASES TO EXIST if you sleep through it. Because if it does exist, and you know it is not your current situation, that is "new information."
What Elga did, in your first two premises, was reduce the sample space further by eliminated both H&Tue, which is already eliminated, and one additional situation. So he IGNORED H&Tue, while Halfers imply it cannot exist.
Sorry, not following.
> halfers will try to rationalize a relationship between H&Tue and T&Tue.
The relationship is between T&Mon and T&Tue, more specifically they are sequential, not mutually exclusive events and therefore can't be used to construct a sample space the way thirders try to.
> But there is a better setup.
As I've already told you, your version with the two coins has all the same issues. The pairs TH and TT; HH and HT are also sequential events and therefore can't be used to construct a sample space. Granted, it may appear a bit more confusing, but the correct answer to your version of Sleeping Beauty is also 1/2.
> What confuses halfers is that being left asleep counts as "something." The day still occurs, you just won't observe it
If you are certain that you don't observe something, you do not get to update your credence while you indeed do not observe it. See conservation of expected evidence.
> And the "somethings" that might happen on different days count as independent outcomes to you, since you cannot connect it to another day due to the amnesia drug.
Amnesia doesn't change Beauty's knowledge state about the experiment to the point where she can rationally believe that the awakenings are exclusionary and happen at random. She still remembers the setting of the experiment, and the order between awakenings, even though she doesn't remember whether she was awakened before or not.
> By saying you have no "new information," Halfers are implicitly claiming that Tuesday CEASES TO EXIST if you sleep through it. Because if it does exist, and you know it is not your current situation, that is "new information."
What Halfers should be claiming is that you are not allowed to use probability theory in such a way, where you treat sequential events Monday and Tuesday as mutually exclusive. Sadly, due to historical reasons, the same person who came up with the "bright idea" to construct sample space from sequential events is the same person who came up with first (and obviously wrong) Halfer model. This created a convenient weakman for Thirders to defeat, while keeping the correct Halfer reasoning obscured.
> The relationship is between T&Mon and T&Tue, more specifically they are sequential, not mutually exclusive events and therefore can't be used to construct a sample space the way thirders try to.
That's the argument halfers try to use. It's not true. You provide no justification for how it affects SB's belief. But it is the point of my variation, where it no longer applies. Everything is based on what happens on a single day.
> As I've already told you, your version with the two coins has all the same issues.
And when you do, you still fail to justify how "sequential" affects the beliefs of a subject who cannot see it, and you ignore that it is not true for the two-coin version.
> The pairs TH and TT; HH and HT are also sequential events and therefore can't be used to construct a sample space.
Why? But aside from that, SB's sample space can be based on the four possible combinations - and they are "outcomes" - that exist on a single day.
> Granted, it may appear a bit more confusing, but the correct answer to your version of Sleeping Beauty is also 1/2.
When the researchers look at the coins, there are four possible combinations. Each has a 14 probability of existing, and this does not depend in any way on whatever "sequential" means. One of these combinations is eliminated, in SB's knowledge. This is what "new information" means. The remaining three all have an updated probability of 1/3.
> If you are certain that you don't observe something, you do not get to update your credence while you indeed do not observe it. See conservation of expected evidence.
Say I am put to sleep, and I will wake up in a green room if a die roll is odd, or a red room if it is even. If what happens is that I wake up in a green room, I am observing that the die roll was odd and was not even. I "get to" update on this information.
If a theory of "the conservation of expected evidence" says otherwise, it is bogus.
> Amnesia doesn't change Beauty's knowledge state about the experiment to the point where she can rationally believe that the awakenings are exclusionary and happen at random.
????
Amnesia separates her knowledge in one pass from anything having to do with the other. The issue wilt Elga's version is that she may (depending on who you listen to) need to factor what could happen on the other day into her assessment. In mine, her assessment can be based entirely on one day. "Sequential" does not change that fact that at the start of this say, there were four equally-likely combinations. If she is awake, one is no longer possible.
> She still remembers the setting of the experiment, and the order between awakenings,
There is no ordering of awakenings. Each looks exactly the same as the other., and functions as a stand-alone experiment.
> What Halfers should be claiming is that you are not allowed to use probability theory in such a way, where you treat sequential events Monday and Tuesday as mutually exclusive.
Why? And please explain why "sequential" matters.
> You provide no justification for how it affects SB's belief.
Beauty literally knows that. Do I need to justify that knowing things affect your beliefs?
> But it is the point of my variation, where it no longer applies. Everything is based on what happens on a single day.
Still sequential events. All the same problems.
> Why?
> Why? And please explain why "sequential" matters.
Sure. Let's prove a simple lemma.
Let A1 and A2 be two sequential events so that A2 happens always and only before or after A1.
Now, let's suppose they are also mutually exclusive. Then we can construct a sample space Ω={A1, A2} and a probability space (Ω,F,P), satisfying the axioms of probability theory.
By Kolmogorov's third axiom:
P(A1) + P(A2) = 1
There are two possibilities, either P(A1) = 0 or P(A1) > 0
if P(A1) = 0 then P(A2) =1, but this contradicts the initial premise that A2 happens only before or after A1.
so P(A1)> 0 and P(A2) < 1
But then with 1-P(A2) probability A2 will not happen before or after A1 has happened. This contradicts the initial premise that A2 always happens before or after A1.
Q.E.D. Sequential events are not mutually exclusive.
> SB's sample space can be based on the four possible combinations
Unless they are mutually exclusive, it can't. See Conditions of Sample Space
https://en.wikipedia.org/wiki/Sample_space
> When the researchers look at the coins
Researchers perspective is irrelevant. We are talking about the Beauty.
> Say I am put to sleep, and I will wake up in a green room if a die roll is odd, or a red room if it is even. If what happens is that I wake up in a green room, I am observing that the die roll was odd and was not even. I "get to" update on this information.
Yes! Absolutely! You were not sure which color of room you'll observe beforehand, and so you update. But if you were meant to be put in a green room regardless of the outcome of a die and, therefore, could be certain not to observe that the room is red, you do not update. It's very straightforward and intuitive.
Likewise, Sleeping Beauty awakens in the experiment regardless of the outcome of the coin, so she doesn't update. She was sure not to observe the outcome HH and so she doesn't update. She would've updated, if she could observe the event "I'm awakened the second time in the same experiment", but she can't do it due to the memory loss.
> The issue wilt Elga's version is that she may (depending on who you listen to) need to factor what could happen on the other day into her assessment.
I suppose some people may find this to be the case, but I don't think I've ever claimed that. So there is no point in arguing about it with me.
> There is no ordering of awakenings.
There is. It's in the definition of the experiment that you've yourself described. TH and TT are sequential events. And Beauty gets told about it.
> Each looks exactly the same as the other.
Beauty knowledge isn't limited to her observation of the awakenings. She also was explicitly told about the order of the awakenings. She has to ignore this knowledge in order to reason about awakenings as happening at random.
Here is a simpler case, which, I believe, can capture our disagreement. Suppose there is a rigged dice that shows 6 always and only if it showed 5 on a previous roll. You are told that is the way the dice works and this it the only thing you know about this dice. Are you justified to treat it as a fair one? If no, why not?
Let's try to make some sense of your “proof.” I’ll start with some definitions:
An outcome is a measurable result of a random experiment. A sample space is a set of all possible distinct outcomes for an experiment. An event is any subset of a sample space.
What confuses many non-mathematicians, is that there is no single “correct” sample space. Roll two six-sided dice, and the sample space could be 36 pairs of values, 21 ordered pairs, or eleven sums. The only requirements are that the members be mutually exclusive, and the set be exhaustive. But the choice of your sample space usually is driven by how you can exhaustively discriminate the results.
You seem to confuse “outcome” and “event” when you think you are applying Kolmogorov’s theory.
> Let A1 and A2 be two sequential events so that A2 happens always and only before or after A1.
If the words “happens after” are involved, you are describing two different random experiments. One may depend on the other, but they are different. In all variations of the SB problem, there is just one experiment with two opportunities to observe it.
So, what you describe, is two sequential observations of the same random experiment. That is, flip the coin(s) once, observe the outcome one way, and then observe it again in another way. The point is that you are observing the same result, not two different results.
If the sample space is well constructed, each observation divides the sample space into two events; one event where every outcome in the set is 100% consistent with the observation, and its complement where every outcome in it contradicts the observation.
>> Now, let's suppose they are also mutually exclusive.
Two different observations of a single outcome cannot be “mutually exclusive.” Each observation defines an event that includes the member of the sample space that describes the actual result. This seems to be the conclusion you reach by faulty logic, but it is irrelevant. It doesn’t affect probabilities the way you think it does.
>> Then we can construct a sample space Ω={A1, A2} and a probability space (Ω,F,P), satisfying the axioms of probability theory.
No, we actually can’t. Ω is the set that contains all distinct outcomes that are possible. F is Kolmogorov's set that contains events. There are some restrictions on what events go into it, but the set of all possible events satisfies them.
But you don’t use these terms correctly, or recognize the differences between an outcome and an observation of an outcome. Your statement here makes no sense in the context of Kolmogorov’s theory. NOT(QED).
+++++
But it can be done, with a well constructed sample space that is sufficient to represent bit observations. Like my two-coin version.
My sample space has four equally-likely outcomes: {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. The first observation recognizes the consistent event {HT1_HH2, TH1_TT2, TT1_TH2} and the inconsistent event {HH1_HT2}. Note that how HH1_HT2 might be recognized, if it were consistent, is irrelevant since it is not consistent. The second observation recognizes the consistent event {HH1_HT2, TH1_TT2, TT1_TH2} and the inconsistent event {HT1_HH2}.
You call these observations “sequential events.” They aren’t events, they are observations. Because of the amnesia drug, they are unrelated observations of the same probability experiment.
In each one, the conditional probability of the event where coin C1 is showing Heads during that observation is the prior probability of that event, divided by the prior probability of the “consistent” event. The result is 1/3.
> Ω is the set that contains all distinct outcomes that are possible. F is Kolmogorov's set that contains events.
Yep, I'm well aware of the definitions.
> No, we actually can’t.
You failed to explain why.
> There are some restrictions on what events go into it, but the set of all possible events satisfies them.
So? How does it make us unable to construct a probability space? What restrictions do I cross?
> You seem to confuse “outcome” and “event”
What you seem to be implying, is that we should,'t be able to make the transition from A1 to {A1} or vice versa. Dare to explain why?
> If the words “happens after” are involved, you are describing two different random experiments.
I'm describing two runs of the same experiment.
> recognize the differences between an outcome and an observation of an outcome
Oh! Maybe I don't. I know only about outcomes from Ω and events from F and a measure function P which domain is F. What is this "observation of outcome" thing your are talking about. Please specify it formally.
> My sample space has four equally-likely outcomes: {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}
I'm glad that you finally stopped claiming that HH TT HT and TH are themselves individual outcomes. We are making progress here! We would've saved each other a lot of time if you were rigorous to begin with.
> The first observation recognizes the consistent event {HT1_HH2, TH1_TT2, TT1_TH2} and the inconsistent event {HH1_HT2}.
> The second observation recognizes the consistent event {HH1_HT2, TH1_TT2, TT1_TH2} and the inconsistent event {HT1_HH2}
By first observation you mean the event "Beauty is awakened before the second coin was turned". And by the second - "the Beauty is awakened after the second coin was turned". Indeed, if the Beauty would be able to to make such observations, she should conclude that the first coin is Heads only with 1/3 probability.
There is only a tiny problem here. She can't. She is not told whether she is experiencing the awakening before the turn of a coin or after it. And so her two observations does not separate the outcome space, each time she observes the event "Any outcome of the sample space is realized".
> You call these observations “sequential events.”
No, I call TT and TH sequential events, but as you stopped committing the error, of treating them this way, we can move past it.
> Yep, I'm well aware of the definitions.
You don’t use them correctly. You use the word “event” for both how a result is observed, and a member of a sample space.
> You failed to explain why [we can’t construct a sample space Ω={A1, A2}, where A1 and A2 are events that you imply are the observations of the same result]
See the definitions of “outcome”, “sample space”, and “event” that I used to explain why you can’t do that.
> So? How does it make us unable to construct a probability space?
…. With a sample space comprising two non-distinct non-events that you call events? When the members of a sample space are, by the definitions you claim to understand, supposed to be outcomes and not events?
> What you seem to be implying, is that we shouldn’t be able to make the transition from A1 to {A1} or vice versa. Dare to explain why?
If A1 is an outcome, I do it all the time.
> I'm describing two runs of the same experiment.
No, you are describing the result of the same experiment two different times, using different criteria. "Sequential" means that (1) the result is determined, (2) you observe it one way and call it one "event," (3) maybe you forget that obsevation?, and (4) you observe it again and call it a different "event," even tho nothing has changed in the random state.
>> My sample space has four equally-likely outcomes: {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}
> I'm glad that you finally stopped claiming that HH TT HT and TH are themselves individual outcomes.
They are. In the first observation, XY1_XZ2 is the same as XY. In the second, it is XZ. This is what you call “sequential events.”
> She is not told whether she is experiencing the awakening before the turn of a coin or after it.
And I never said she was. But she does know that she is experiencing only one awakening. Her treatment of the sample space is different only in that she uses different names for equivalent outcomes. The shorter name is how she makes these equivalences to the longer names. Since the equivalences work the same regardless of which observation it is, this works.
But this is the “Halfer’s Fallacy.” They think that, by not knowing which observation, it works like Schrodinger’s Cat: she is "in" both observations at the same time.
> No, I call TT and TH sequential events
No, you call TT1_TH2 observed the first time, and then the second time, “sequential events.”
Different "simpler case": I tell you that I have a biased coin, but not how it is biased. I ask you to guess what the result will be if I flip it. If you are right, I'll give you $1; if wrong, you give me $1. Do you choose Heads to Tails?
The point is that your rational belief - that is, *your* probability based on *your* knowledge that includes it being biased - can only be 50:50. While you do have the information that "sequential" flips of the coin will trend to one face or the other, you do not know what that trend is. You cannot base your belief on long-term results, only on what you know. I believe I once saw an article on lesswrong about this, you can look it up.
To answer your "simpler case," over many rolls this die will roll each result about the same number of times. A single roll, with no knowledge of any effect of ordering, is "fair" the same way a single flip of my coin is "fair."
More appropriate "simpler case": Suppose there is a rigged die with a counter inside it, for every time it has been rolled. For odd-numbered rolls, it rolls with a uniform probability distribution. For even-numbered rolls, the result is the previous roll, plus one (cycling back to 6-->1). This is what, I believe, you would call "sequential."
If you roll this die, knowing that it is an odd-roll, the probability distribution you would use for the sample space {1,2,3,4,5,6} is uniform. If you roll it without knowing the previous outcome, the probability distribution is uniform regardless of whether it is an odd-roll or an even-roll. It is only if you know it is an even-roll, AND you know something about the previous result, that the distribution changes.
Said another way, while you know there is an effect due to order, if you do not know the position in that order, or anything about another sample in that order that is related to this one, THERE IS NO EFFECT DUE TO ORDER. So there essentially IS NO ORDER to the version of the random experiment you are observing.
In my two-coin version of the SB problem, to an outside observer there are four outcomes (not "event;," you do not seem to know what that word means in probability) of the experiment that comprise a sample space. Named fully, they are {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. Each has a probability of 1/4.
What you seem to mean by "sequential" is that, for example, the outcome HH(n) is the same outcome as HT(3-n). That's why I used combined names. Nobody doubts this. But SB only sees the "n" side of the combined outcome. The fact that she knows the relationships between {HH(n), HT(n), TH(n), TT(n)} and {HH(3-n), HT(3-n), TH(3-n), TT(3-n)} can only affect her belief if she has some information about what happened in the "3-n" side. Just like my biased coin example, the existence of some influencing factor can only influence her belief if she also knows something about a result that shares that factor.
So, SB knows the sample space is {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}, and the probability distribution is {1/4,1/4,1/4,1/4}. She knows with certainty (that is, probability 1) that EXACTLY one of these outcomes is eliminated, either HH1_HT2 or HT1_HH2. So her belief that Coin C1 is currently showing Heads is either:
Pr({HT1_HH2})/Pr({HT1_HH2, TH1_TT2, TT1_TH2} = 1/3
or
Pr({HH1_HT2})/Pr({HH1_HT2, TH1_TT2, TT1_TH2} = 1/3
Please note how this allows for the effect of being "sequential", by being based on the combined outcomes instead of the one-day outcomes. The significant difference between this, and every other halfer-thirder debate, is that the two calculations are the same. So we do not need to consider the position in any order. Even if that should have an effect, which it does not.
Her prior should be 1/2, because that should be one's credence in the flip of a fair coin. I obviously disagree that she should maintain her prior for reasons I give in this article.
I argued against your “proof” here https://benthams.substack.com/p/contra-philosophy-bear-and-aaron