Here’s a new problem that requires the same solution methodology as Sleeping Beauty.
It uses the same sleep and amnesia drugs. After SB is put to sleep on Sunday Night, a card is drawn at random from a standard deck of 52 playing cards.
On Monday, SB is awakened, interviewed, and put back to sleep with amnesia.
On Tuesday, if the card is a Spade, a Heart, or a Diamond—but not if it is a Club—SB is awakened, interviewed, and put back to sleep with amnesia.
On Wednesday, if the card is a Spade or a Heart—but not if it is a Diamond or a Club—SB is awakened, interviewed, and put back to sleep with amnesia.
On Thursday, only if the card is a Spade, SB is awakened, interviewed, and put back to sleep with amnesia.
On Friday, SB is awakened and the experiment ends.
In each interview, SB is asked for the probability that the drawn card is the Ace of Spades.
By Halfer logic, it is 1⁄52. In fact, the probability for each card in the deck is 1⁄52. Yet it is possible that the card cannot be a Club when it can be a spade, so they cannot have the same probability. This contradiction disproves the halfer logic.
The correct probabilities are:
For each Spade, [(1/52)+(1/39)+(1/26)+(1/13)]/4 = 25⁄624
For each Heart, [(1/52)+(1/39)+(1/26)]/4 = 1⁄48
For each Diamond, [(1/52)+(1/39)]/4 = 7⁄624
For each Club, (1/52)/4 = 1⁄208
The correct solution to Elga’s Sleeping Beauty problem is that SB’s prior sample space, for what the researchers saw this morning, is {H&Mon, T&Mon, H&Tue T&Tue}. And yes, H&Tue is a member since the prior sample space does not depend on observation. These may be “sequential,” whatever than means, to the researchers. But they are independent outcomes to SB because her memory loss prevents any association of the current day with another. Each has a prior probability of 1⁄4.
The solution to the problem is that H&Tue is eliminated by SB’s observation that she is awake. The other probabilities update to 1⁄3.
By Halfer logic, it is 1⁄52. In fact, the probability for each card in the deck is 1⁄52.
Yes, seems right. Whatever card is picked the Sleeping Beauty is awakened at least once and this is the event that she observes. In about 1⁄52 iterations of the probability experiment, when the SB is awakened at all, the card is indeed Ace of Spades.
Yet it is possible that the card cannot be a Club when it can be a spade, so they cannot have the same probability.
I don’t think I understand what you’ve written here. It’s indeed possible that the card is not Club when it’s Spade. As a matter of fact, it’s the only possibility, because the card can’t be both Spade and a Club.
Probably you are talking about the fact that there are more awakenings when the card is Spade then when its Club. This is indeed the case, but doesn’t affect the probability to be awaken in the experiment. It matters when the Beauty is proposed a specific betting scheme which is repeated on every awakening. The next post explains how to use the correct probability estimate to deal with every betting scheme in Sleeping Beauty. The same logic applies to this version of the problem.
The correct probabilities are:
For each Spade, [(1/52)+(1/39)+(1/26)+(1/13)]/4 = 25⁄624
For each Heart, [(1/52)+(1/39)+(1/26)]/4 = 1⁄48
For each Diamond, [(1/52)+(1/39)]/4 = 7⁄624
For each Club, (1/52)/4 = 1⁄208
These are not probabilities but weighted probabilities, where the measure function is re-normalized by the number of awakenings, even though the awakenings are not mutually exclusive.
The correct solution to Elga’s Sleeping Beauty problem is that SB’s prior sample space, for what the researchers saw this morning, is {H&Mon, T&Mon, H&Tue T&Tue}. And yes, H&Tue is a member since the prior sample space does not depend on observation. These may be “sequential,” whatever than means, to the researchers. But they are independent outcomes to SB because her memory loss prevents any association of the current day with another. Each has a prior probability of 1⁄4.
Sigh. We’ve been through it a couple of times already.
The prior sample space depends on which observations is possible to make in the experiment at all, according to the prior knowledge state of a particular person.
{H&Mon, T&Mon, H&Tue T&Tue} is a sample space for observer problem, not for SB. For a researcher working on a random day, observing outcome T&Tue is a priori possible, for the Beauty participating on every day, it’s not.
Amnesia doesn’t make Monday and Tuesday awakenings mutually exclusive according to the Beauty’s knowledge state. It simply returns her to the initial knowledge state, in which she already knows the whole setting of the experiment, including the fact that there are such iterations of experiment where both Monday and Tuesday awakening happen and therefore they are not mutually exclusive.
Look, we were actually making some real progress when we were discussing two-coin-toss problem, and agreed on its sample space. I feel that if, instead of starting yet another thread where we cycle over the same arguments, we continue this one, we are much more likely to resolve our disagreement.
I don’t think I understand what you’ve written here. It’s indeed possible that the card is not Club when it’s Spade. As a matter of fact, it’s the only possibility, because the card can’t be both Spade and a Club.
There are four different days when SB could be awakened. On three of them, she would not have been awakened if the card was a club. This makes it more likely that the card is a club. This really is very simple probability. If you have difficulty with it, wake her every day. But in the situations where she was left asleep before, wake her and ask her for the probability that the card is the Ace of Clubs.
My points are that (A) the event where SB is left asleep is still an event in SB’s sample space, (B) her “new information” is that she is observing the event in her sample space that match certain conditions, not that the negation of that event is not a part of her sample space, and (C) the amnesia drug disassociates the current day from all others.
This … doesn’t affect the probability to be awaken in the experiment.
Yes, it does. We can argue back and forth about who is correct, but all you have provided is word salad to support a conclusion you reached before choosing the logic. I keep providing examples that contradict it, and your counter-argument is that it contradicts your word salad.
Try another example; and this is actually closer to Zuboff’s original problem than Elga’s version of it, or the version he solved which has become the canonical form. (It’s problem is that the Monday:Tuesday difference obfuscates the probability.)
2N players agree to the following game, and are fully informed of all details. Before the game starts, each is put to sleep. Over the next N days, each will be wakened on either every one of the N days, or on a single, randomly-selected day in the interval, based on the result of a single coin flip that applies to all 2N players and all days. With 2 possible flip results, and N days, each player is randomly assigned a different combination of a coin result and a day.
On each day, each of the N+1 players who are wakened is asked for her probability/credence/confidence/whatever for the proposition that she will be wakened only once. After answering, each will be put back to sleep with amnesia.
In Zuboff’s version, N=one trillion, and the day is random, the coin result is unspecified, but this is the question the player is asked; in other words, it does not matter that the coin result is unspecified.In the problem Elga posed, N=2 but the days are not specified, and the question is effectively the same since Heads is specified. In the problem Elga solved, N=2, the day is day #1, and Heads is specified.
Nothing about these variations change the solution methodology.
The halfer solution is that each of the N+1 players who are awake should say the probability that this is her only waking is 1⁄2. And that each of the other N awake payers should answer the same. This is despite the fact that she knows, for a fact, that only one of the N+1 awake players satisfies the proposition. Additionally, they can assign a probability of 1⁄2 to an asleep player, despite knowing that this player does satisfy the proposition.
These are not probabilities but weighted probabilities, where the measure function is re-normalized by the number of awakenings, even though the awakenings are not mutually exclusive.
These are not weighted probabilities, they are probabilities using all of the 4*52=208, equally-likely members of the prior sample space that can apply to a randomly-selected day in the experiment. They are mutually exclusive, to SB, because the amnesia drug isolates each day from the others that may be “sequential” with it, whatever that is supposed to mean to SB since she sees only one day.
Sigh. We’ve been through it a couple of times already.
Yes, we have, and you ignore every point I make.
The prior sample space depends on which observations is possible to make in the experiment at all, according to the prior knowledge state of a particular person.
A prior sample space depends ONLY on how the circumstances occur. Observation has nothing to do with it.
On HEADS+TUESDAY, wake SB but take her on a shopping spree instead of interviewing her. When interviewed, should she think that Pr(H)=1/2, or Pr(H)=1/3? Of course it is 1⁄3, because her set of possible observations includes four (not two) mutually exclusive outcomes, with one ruled out by observation. My point is that she knows she is in NOT(H&Tue) regardless of what observation is possible in H&Tue.
{H&Mon, T&Mon, H&Tue T&Tue} is a sample space for observer problem, not for SB. For a researcher working on a random day, observing outcome T&Tue is a priori possible, for the Beauty participating on every day, it’s not.
{H&Mon, T&Mon, H&Tue T&Tue} is the set of possible circumstances for a single day in the experiment. The “observer problem” you describe includes only two outcomes, Heads and Tails. Both Monday and Tuesday exist in each observation, since they are not isolated by amnesia. What you listed is the sample space when the random experiment can include only one day—AMNESIA!! H&Tue is in SB’s prior = unaffected by observation sample space.
My generalized version of Zuboff’s problem has 2N members in its sample space. When a player is awake, N+1 of these members are consistent with observation. One matches the proposition that the player will be wakened once. Making the probability of the proposition 1/(N+1).
In the canonical SB problem, N=2 and the probability is 1/(N+1)=1/3.
We can argue back and forth about who is correct, but all you have provided is word salad to support a conclusion you reached before choosing the logic.
Ironic.
If you are confused about some term that I’m using, I’m ready to formally specify it. For example “sequential events”, that you seem to have troubles with means that on a repetition of probability experiment, knowledge that first of such event is realized in i-th iteration means that you can be absolutely certain that the second of such events also is realized in i-th iteration, for every value of i. Or, if instead of iterations of probability experiment you want to talk about individual awakenings, knowledge that first sequential event from a pair is realized in j-th awakenings, means that the second sequential event from a pair is realized in j+1-st awakening for every j.
As you are familiar with the idea of probability experiment there shouldn’t be any more misunderstandings. Or do I need to explain to you what “realization of event” is?
Now would you be able to provide the same courtesy? Could you formally explain the terms you are using in the motivated world salad of yours?
There are four different days when SB could be awakened. On three of them, she would not have been awakened if the card was a club. This makes it more likely that the card is a club.
What makes it more likely that the card is club? What is this event you are conditioning on? Not “Being awaken in the experiment at all” because there is always at least one awakening in every iteration of probability experiment, as you’ve described it.
If it’s “Being awakened today” please formally specify what is this “today” variable you are talking about.
For example in No-Coin-Toss Problem, we can formally specify today as Monday xor Tuesday. We can see that in every iteration of probability experiment this variable has one value. And so statement “Today is Monday” has a coherent truth value, from the perspective of the Beauty, in every iteration of probability experiment.
Please, do the same thing here. Define “today” in your card picking version of the problem so that statement “Today is Monday” was coherent for every iteration of the experiment.
My points are that (A) the event where SB is left asleep is still an event in SB’s sample space, (B) her “new information” is that she is observing the event in her sample space that match certain conditions, not that the negation of that event is not a part of her sample space, and (C) the amnesia drug disassociates the current day from all others.
Try formulating (C) as a statement of probability theory. What does “disassociates the current day from all others” means to the knowledge state of the Beauty and therefore to the sample space of the problem?
It can’t mean that the Beauty is under the assumption that probability experiment consist of only one day, because amnesia doesn’t erase her knowledge of the setting of experiment. So what does it mean, then? And how does it formally imply that {H&Mon, T&Mon, H&Tue T&Tue} is the correct sample space for the problem?
I keep providing examples that contradict it
You keep providing alternative formulations of our disagreement without bringing any progress into adressing it. Let me demonstrate how unhelpful it is. I also can provide related examples to the initial problem. For instance, consider Unbounded Sleeping Beauty.
A fair coin is tossed until the first Heads outcome. Beauty is awakened 2^n times where n is the number of Tails outcomes the coin produced.
H: One awakening TH: Two awakenings TTH: Four awakenings TTTH: Eight awakenings TTTTH: Sixteen awakenings
and so on.
Now, what credence Beauty should have that k-th coin toss came up Tails when she is awoken in the experiment? According to thirder logic, her confidence approaches 1 for any k, which is clearly absurd.
There. Has it changed your mind? I really don’t think so. The same faulty reasoning that makes you accept thirdism in SB makes you likewise accept it in the alternative problems and you do not even see any problem with it, on the contrary, now the correct answer to the problem appears absurd to you, which you demonstrate again and again while coming up with your alternative problems.
Of course, from your perspective, the same can be said about me. Yes, I do indeed claim that probability that the card is Ace of Spades, conditionally on being awakened in the card picking experiment is the same as for Ace of Clubs, or any other card. This problem doesn’t make thirdism any more persuasive for me.
Therefore, I think we should abandon the strategy of going through alternative problems and engage with the reasoning itself while talking about one particular problem. But in case I’m wrong, and you feel that Unbounded Sleeping Beauty has shaken your confidence, please tell me—I have a bunch of other examples of absurd consequences of thirdism.
A prior sample space depends ONLY on how the circumstances occur. Observation has nothing to do with it.
Which observation can occur is, naturally, part of the circumstances.
If a red ball is given when the coin is Heads and blue ball when the coin is Tails, a color-blind person and non-color-blind-person have very different circumstances and therefore possible observations, credences and optimal betting odds.
On HEADS+TUESDAY, wake SB but take her on a shopping spree instead of interviewing her. When interviewed, should she think that Pr(H)=1/2, or Pr(H)=1/3?
Still 1⁄2 for all the same reasons. But this problem is more convoluted, so its really not going to add any more clarity to this conversation.
Of course it is 1⁄3, because her set of possible observations includes four (not two) mutually exclusive outcomes
Here set of possible observations may include whatever you want—it’s neither preciesly defined, nor relevant for probability theory. What is relevant is set of possible outcomes of probability experiment. And there are exactly two outcomes: {Heads&MondayExperiment&TuesdayShopping, Tails&MondayExperiment&TuesdayExperiment}. Her observation of being awaken during experiment doesn’t mean that !TuesdayShoping outcome is realized—it may still happen tomorrow.
{H&Mon, T&Mon, H&Tue T&Tue} is the set of possible circumstances for a single day in the experiment
Outcomes are defined for probability experiment as a whole. There is no such thing as outcomes for a single day of probability experiment if probability experiment goes for multiple days. Once again I recommend to continue the thread about two-coin-toss problem where we’ve already cleared out this confusion.
The “observer problem” you describe includes only two outcomes, Heads and Tails. Both Monday and Tuesday exist in each observation, since they are not isolated by amnesia.
No it doesn’t, which you would’ve noticed if you carefully read the description of the problem. In observer problem you work in the laboratory only for one day, which according to your knowledge state can be either the first day of the experiment or the second. This is what allows to treat them as random pick from two options, like in every other probability theory problem where you get one outcome from several—you may notice that absolute majority of them don’t include experiencing amnesia.
Once again, feel free to formally justify how experiencing amnesia allows us to treat getting multiple options in a particular order as a random sample, but until you or someone else actually does so, this idea is completely ungrounded.
Here’s a new problem that requires the same solution methodology as Sleeping Beauty.
It uses the same sleep and amnesia drugs. After SB is put to sleep on Sunday Night, a card is drawn at random from a standard deck of 52 playing cards.
On Monday, SB is awakened, interviewed, and put back to sleep with amnesia.
On Tuesday, if the card is a Spade, a Heart, or a Diamond—but not if it is a Club—SB is awakened, interviewed, and put back to sleep with amnesia.
On Wednesday, if the card is a Spade or a Heart—but not if it is a Diamond or a Club—SB is awakened, interviewed, and put back to sleep with amnesia.
On Thursday, only if the card is a Spade, SB is awakened, interviewed, and put back to sleep with amnesia.
On Friday, SB is awakened and the experiment ends.
In each interview, SB is asked for the probability that the drawn card is the Ace of Spades.
By Halfer logic, it is 1⁄52. In fact, the probability for each card in the deck is 1⁄52. Yet it is possible that the card cannot be a Club when it can be a spade, so they cannot have the same probability. This contradiction disproves the halfer logic.
The correct probabilities are:
For each Spade, [(1/52)+(1/39)+(1/26)+(1/13)]/4 = 25⁄624
For each Heart, [(1/52)+(1/39)+(1/26)]/4 = 1⁄48
For each Diamond, [(1/52)+(1/39)]/4 = 7⁄624
For each Club, (1/52)/4 = 1⁄208
The correct solution to Elga’s Sleeping Beauty problem is that SB’s prior sample space, for what the researchers saw this morning, is {H&Mon, T&Mon, H&Tue T&Tue}. And yes, H&Tue is a member since the prior sample space does not depend on observation. These may be “sequential,” whatever than means, to the researchers. But they are independent outcomes to SB because her memory loss prevents any association of the current day with another. Each has a prior probability of 1⁄4.
The solution to the problem is that H&Tue is eliminated by SB’s observation that she is awake. The other probabilities update to 1⁄3.
Yes, seems right. Whatever card is picked the Sleeping Beauty is awakened at least once and this is the event that she observes. In about 1⁄52 iterations of the probability experiment, when the SB is awakened at all, the card is indeed Ace of Spades.
I don’t think I understand what you’ve written here. It’s indeed possible that the card is not Club when it’s Spade. As a matter of fact, it’s the only possibility, because the card can’t be both Spade and a Club.
Probably you are talking about the fact that there are more awakenings when the card is Spade then when its Club. This is indeed the case, but doesn’t affect the probability to be awaken in the experiment. It matters when the Beauty is proposed a specific betting scheme which is repeated on every awakening. The next post explains how to use the correct probability estimate to deal with every betting scheme in Sleeping Beauty. The same logic applies to this version of the problem.
These are not probabilities but weighted probabilities, where the measure function is re-normalized by the number of awakenings, even though the awakenings are not mutually exclusive.
Sigh. We’ve been through it a couple of times already.
The prior sample space depends on which observations is possible to make in the experiment at all, according to the prior knowledge state of a particular person.
{H&Mon, T&Mon, H&Tue T&Tue} is a sample space for observer problem, not for SB. For a researcher working on a random day, observing outcome T&Tue is a priori possible, for the Beauty participating on every day, it’s not.
Amnesia doesn’t make Monday and Tuesday awakenings mutually exclusive according to the Beauty’s knowledge state. It simply returns her to the initial knowledge state, in which she already knows the whole setting of the experiment, including the fact that there are such iterations of experiment where both Monday and Tuesday awakening happen and therefore they are not mutually exclusive.
Look, we were actually making some real progress when we were discussing two-coin-toss problem, and agreed on its sample space. I feel that if, instead of starting yet another thread where we cycle over the same arguments, we continue this one, we are much more likely to resolve our disagreement.
There are four different days when SB could be awakened. On three of them, she would not have been awakened if the card was a club. This makes it more likely that the card is a club. This really is very simple probability. If you have difficulty with it, wake her every day. But in the situations where she was left asleep before, wake her and ask her for the probability that the card is the Ace of Clubs.
My points are that (A) the event where SB is left asleep is still an event in SB’s sample space, (B) her “new information” is that she is observing the event in her sample space that match certain conditions, not that the negation of that event is not a part of her sample space, and (C) the amnesia drug disassociates the current day from all others.
Yes, it does. We can argue back and forth about who is correct, but all you have provided is word salad to support a conclusion you reached before choosing the logic. I keep providing examples that contradict it, and your counter-argument is that it contradicts your word salad.
Try another example; and this is actually closer to Zuboff’s original problem than Elga’s version of it, or the version he solved which has become the canonical form. (It’s problem is that the Monday:Tuesday difference obfuscates the probability.)
2N players agree to the following game, and are fully informed of all details. Before the game starts, each is put to sleep. Over the next N days, each will be wakened on either every one of the N days, or on a single, randomly-selected day in the interval, based on the result of a single coin flip that applies to all 2N players and all days. With 2 possible flip results, and N days, each player is randomly assigned a different combination of a coin result and a day.
On each day, each of the N+1 players who are wakened is asked for her probability/credence/confidence/whatever for the proposition that she will be wakened only once. After answering, each will be put back to sleep with amnesia.
In Zuboff’s version, N=one trillion, and the day is random, the coin result is unspecified, but this is the question the player is asked; in other words, it does not matter that the coin result is unspecified.In the problem Elga posed, N=2 but the days are not specified, and the question is effectively the same since Heads is specified. In the problem Elga solved, N=2, the day is day #1, and Heads is specified.
Nothing about these variations change the solution methodology.
The halfer solution is that each of the N+1 players who are awake should say the probability that this is her only waking is 1⁄2. And that each of the other N awake payers should answer the same. This is despite the fact that she knows, for a fact, that only one of the N+1 awake players satisfies the proposition. Additionally, they can assign a probability of 1⁄2 to an asleep player, despite knowing that this player does satisfy the proposition.
These are not weighted probabilities, they are probabilities using all of the 4*52=208, equally-likely members of the prior sample space that can apply to a randomly-selected day in the experiment. They are mutually exclusive, to SB, because the amnesia drug isolates each day from the others that may be “sequential” with it, whatever that is supposed to mean to SB since she sees only one day.
Yes, we have, and you ignore every point I make.
A prior sample space depends ONLY on how the circumstances occur. Observation has nothing to do with it.
On HEADS+TUESDAY, wake SB but take her on a shopping spree instead of interviewing her. When interviewed, should she think that Pr(H)=1/2, or Pr(H)=1/3? Of course it is 1⁄3, because her set of possible observations includes four (not two) mutually exclusive outcomes, with one ruled out by observation. My point is that she knows she is in NOT(H&Tue) regardless of what observation is possible in H&Tue.
{H&Mon, T&Mon, H&Tue T&Tue} is the set of possible circumstances for a single day in the experiment. The “observer problem” you describe includes only two outcomes, Heads and Tails. Both Monday and Tuesday exist in each observation, since they are not isolated by amnesia. What you listed is the sample space when the random experiment can include only one day—AMNESIA!! H&Tue is in SB’s prior = unaffected by observation sample space.
My generalized version of Zuboff’s problem has 2N members in its sample space. When a player is awake, N+1 of these members are consistent with observation. One matches the proposition that the player will be wakened once. Making the probability of the proposition 1/(N+1).
In the canonical SB problem, N=2 and the probability is 1/(N+1)=1/3.
Ironic.
If you are confused about some term that I’m using, I’m ready to formally specify it. For example “sequential events”, that you seem to have troubles with means that on a repetition of probability experiment, knowledge that first of such event is realized in i-th iteration means that you can be absolutely certain that the second of such events also is realized in i-th iteration, for every value of i. Or, if instead of iterations of probability experiment you want to talk about individual awakenings, knowledge that first sequential event from a pair is realized in j-th awakenings, means that the second sequential event from a pair is realized in j+1-st awakening for every j.
As you are familiar with the idea of probability experiment there shouldn’t be any more misunderstandings. Or do I need to explain to you what “realization of event” is?
Now would you be able to provide the same courtesy? Could you formally explain the terms you are using in the motivated world salad of yours?
What makes it more likely that the card is club? What is this event you are conditioning on? Not “Being awaken in the experiment at all” because there is always at least one awakening in every iteration of probability experiment, as you’ve described it.
If it’s “Being awakened today” please formally specify what is this “today” variable you are talking about.
For example in No-Coin-Toss Problem, we can formally specify today as Monday xor Tuesday. We can see that in every iteration of probability experiment this variable has one value. And so statement “Today is Monday” has a coherent truth value, from the perspective of the Beauty, in every iteration of probability experiment.
Please, do the same thing here. Define “today” in your card picking version of the problem so that statement “Today is Monday” was coherent for every iteration of the experiment.
Try formulating (C) as a statement of probability theory. What does “disassociates the current day from all others” means to the knowledge state of the Beauty and therefore to the sample space of the problem?
It can’t mean that the Beauty is under the assumption that probability experiment consist of only one day, because amnesia doesn’t erase her knowledge of the setting of experiment. So what does it mean, then? And how does it formally imply that {H&Mon, T&Mon, H&Tue T&Tue} is the correct sample space for the problem?
You keep providing alternative formulations of our disagreement without bringing any progress into adressing it. Let me demonstrate how unhelpful it is. I also can provide related examples to the initial problem. For instance, consider Unbounded Sleeping Beauty.
A fair coin is tossed until the first Heads outcome. Beauty is awakened 2^n times where n is the number of Tails outcomes the coin produced.
H: One awakening
TH: Two awakenings
TTH: Four awakenings
TTTH: Eight awakenings
TTTTH: Sixteen awakenings
and so on.
Now, what credence Beauty should have that k-th coin toss came up Tails when
she is awoken in the experiment? According to thirder logic, her
confidence approaches 1 for any k, which is clearly absurd.
There. Has it changed your mind? I really don’t think so. The same faulty reasoning that makes you accept thirdism in SB makes you likewise accept it in the alternative problems and you do not even see any problem with it, on the contrary, now the correct answer to the problem appears absurd to you, which you demonstrate again and again while coming up with your alternative problems.
Of course, from your perspective, the same can be said about me. Yes, I do indeed claim that probability that the card is Ace of Spades, conditionally on being awakened in the card picking experiment is the same as for Ace of Clubs, or any other card. This problem doesn’t make thirdism any more persuasive for me.
Therefore, I think we should abandon the strategy of going through alternative problems and engage with the reasoning itself while talking about one particular problem. But in case I’m wrong, and you feel that Unbounded Sleeping Beauty has shaken your confidence, please tell me—I have a
bunch of other examples of absurd consequences of thirdism.
Which observation can occur is, naturally, part of the circumstances.
If a red ball is given when the coin is Heads and blue ball when the coin is Tails, a color-blind person and non-color-blind-person have very different circumstances and therefore possible observations, credences and optimal betting odds.
Still 1⁄2 for all the same reasons. But this problem is more convoluted, so its really not going to add any more clarity to this conversation.
Here set of possible observations may include whatever you want—it’s neither preciesly defined, nor relevant for probability theory. What is relevant is set of possible outcomes of probability experiment. And there are exactly two outcomes: {Heads&MondayExperiment&TuesdayShopping, Tails&MondayExperiment&TuesdayExperiment}. Her observation of being awaken during experiment doesn’t mean that !TuesdayShoping outcome is realized—it may still happen tomorrow.
Outcomes are defined for probability experiment as a whole. There is no such thing as outcomes for a single day of probability experiment if probability experiment goes for multiple days. Once again I recommend to continue the thread about two-coin-toss problem where we’ve already cleared out this confusion.
No it doesn’t, which you would’ve noticed if you carefully read the description of the problem. In observer problem you work in the laboratory only for one day, which according to your knowledge state can be either the first day of the experiment or the second. This is what allows to treat them as random pick from two options, like in every other probability theory problem where you get one outcome from several—you may notice that absolute majority of them don’t include experiencing amnesia.
Once again, feel free to formally justify how experiencing amnesia allows us to treat getting multiple options in a particular order as a random sample, but until you or someone else actually does so, this idea is completely ungrounded.