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.
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.