JeffJo

Karma: 3

JeffJo 10 Oct 2024 17:30 UTC
1 point
−2
on: What’s the Deal with Logical Uncertainty?
Contrary to what too many want to believe, probability theory does not define what “the probability” is. It only defines these (simplified) rules that the values must adhere to:
1. Every probability is greater than, or equal to, zero.
2. The probability of the union of two distinct outcomes A and B is Pr(A)+Pr(B).
3. The probability of the universal event (all possible outcomes) is 1.
Let A=”googolth digit of pi is odd”, and B=”googolth digit of pi is even.” These required properties only guarantee that Pr(A)+Pr(B)=1, and that each is a non-zero number. We only “intuitively” say that Pr(A)=Pr(B)=0.5 because we have no reason to state otherwise. That is, we can’t assert that Pr(A)>Pr(B) or Pr(A)<Pr(B), so we can only assume that Pr(A)=Pr(B). But given a reason, we can change this.
The point is that there are no “right” or “wrong” statements in probability. Only statements where the probabilities adhere to these requirements. We can never say what a “probability is,” but we can rule out some sets of probabilities that violate these rules.

JeffJo 18 Aug 2024 18:28 UTC
0 points
−1
in reply to: Ape in the coat’s comment on: The Solution to Sleeping Beauty
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.

JeffJo 16 Aug 2024 19:02 UTC
1 point
0
on: The Solution to Sleeping Beauty
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.

JeffJo 1 Apr 2024 21:00 UTC
−2 points
0
in reply to: Ape in the coat’s comment on: The Solution to Sleeping Beauty
Yes! I’m so glad you finally got it! And the fact that you simply needed to remind yourself of the foundations of probability theory validates my suspicion that it’s indeed the solution for the problem.
Too bad you refuse to “get it.” I thought these details were too basic to go into:
A probability experiment is a repeatable process that produces one or more unpredictable result(s). I don’t think we need to go beyond coin flips and die rolls here. But probability experiment refers to the process itself, not an iteration of it. All of those things I defined before are properties of the experiment; the process. “Outcome” is any potential result of an iteration of that process, not the result itself. We can say that a result belongs to an event, even an event of just one outcome, but the result is not the same thing as that event. THE OBSERVATION IS NOT AN EVENT.
For example, an event for a simple die roll could be EVEN={2,4,6}. If you roll a 2, that result is in this event. But it does not mean you rolled a 2, a 4, and a 6.
So, in …
By the definition of the experimental setting, when the coin is Tails, what Beauty does on Monday—awakens—always affects what she does on Tuesday—awakens the second time. Sequential events are definitely not mutually exclusive and thus can’t be elements of a sample space.
… you are describing one iteration of a process that has an unpredictable result. A coin flip. Then you observe it twice, with amnesia in between. Each observation can have its own sample space—remember, experiments do not have just one sample space. But you can’t pick, say, half of the outcomes defined in one observation an half from the other, and use them to construct a sample space. That is what you describe here, by comparing what SB does on Monday, and on Tuesday, as if they are in the same event space.
The correct “effect of amnesia” is that you can’t relate either observation to the other. They each need to be assessed by a sample space that applies to that observation, without reference to another.
And BTW, what she observes on Monday may belong to an event, but it is not the same thing as the event.
>That result is observed twice (yes, it is; remaining asleep is an observation of a result that we never make use of, so awareness as it occurs is irrelevant
This is false, but not crucial. We can postpone this for later.
A common way to avoid rebuttal is to cite two statements and make one ambiguous assertion about them, without support or specifying which you mean.
It is true that remaining asleep is a possible result of the experiment—that is, an outcome—since Tuesday exists whether or not SB is awake. What SB observes tells her that outcome is not consistent with her evidence. That’s an observation.
It is true that same the result (the coin flip) is observed twice; once on Monday, and once on Tuesday.
Or do you want to claim the calendar flips from Monday to Wednesday? That is, that Tuesday only exists is SB is awake? But if you still doubt this, wake SB on Tuesday but don’t ask her for her belief in Heads. Knowing the circumstances where you would not ask, she can then deduce that those circumstances do not exist. This is an observation.
What you do think makes it different than not waking her, since her evidence is the same when she is awake is the same?
>What you call “sequential events” are these two separate observations of the same result.
No, what I call sequential events are pairs HH and HT, TT and TH, corresponding to exact awakening, which can’t be treated as individual outcomes.
No, thats how you try to misinterpret my version to fit your incorrect model. You use the term for Elga’s one-coin version as well. Strawman arguments are another avoidance technique.
On the other hand, as soon as you connect these pairs and got HH_HT, HT_HH, TT_TH and TH_TT, they totally can create a sample space, which is exactly what I told you in this comment. As soon as you’ve switched to this sound sample space we are in agreement.
Huh? What does “connect these pairs” mean to pairs that I already connected?
You are describing a situation where the Beauty was told whether she is experiencing an awakening before the second coin was turned or not.
No, I am not. This is another strawman. I am describing how she knows that she is in either the first observation or the second. I am saying that I was able to construct a valid, and useful, sample space that applies symmetrically to both. I am saying that, since it is symmetric, it does not matter which she is in.
I only did this to allow you to include the “sequential” counterpart to each in a sample space that applies regardless of the day. The point is that “sequential” is meaningless.

JeffJo 31 Mar 2024 19:51 UTC
−3 points
−1
on: The Solution to Sleeping Beauty
A Lesson in Probability for Ape in the Coat
First, some definitions. A measure in Probability is a state property of the result of a probability experiment, where exactly one value applies to each result. Technically, the values should be numbers so that you can do things like calculate expected values. That isn’t so important here; but if you really object, you can assign numbers to other kinds of values, like 1=Red, 2=Orange, etc.
An observation (my term) is a set of one or more measure values. An outcome is an observation that discriminates a result sufficiently for the purposes of the random experiment. An experiment’s sample space is a set of all distinct outcomes that are possible for that experiment; it is often represented as Ω.
There is no single way to define outcomes, and so no single sample space, for any experiment. For example, if you roll two dice, the sample space could be 36 unordered pairs of numbers, 21 ordered pairs, or eleven sums. But what “sufficient” means is that every observation you intend to make is its own outcome in the sample space. You could divide the results of rolling a single die into {Odd, Even}, but that won’t be helpful if you intend to observe “Prime” and “Composite.”
An event is any subset of the sample space, not a specific result. (You, Ape in the Coat, confuse it with either an observation, or an outcome; it isn’t clear which.) We could also talk about the event space F and the corresponding probability space P, but those exact details are also not important here. Except that P corresponds to F, not Ω. When I talk about the probability of an outcome, I mean the solitary event containing just that outcome.
Finally, conditional probability is used when an observation divides the sample space into two sets: one where every outcome is 100% consistent with the observation—that is, no un-observed measure makes it inconsistent—and its compliment where no outcome is consistent with the observation. (This is where people go wrong in problems like Monty Hall; the outcome where Monty opens Door 3 to show a goat is not the same as the observation that there is a goat behind Door 3).
With these definitions, we can see that the SB Problem is one random experiment with a single result. That result is observed twice (yes, it is; remaining asleep is an observation of a result that we never make use of, so awareness as it occurs is irrelevant). What you call “sequential events” are these two separate observations of the same result. That’s why you want to treat them as dependent, because they are the same result. Just looked at a different way.
And the effect of the amnesia drug is that any information learned in “the other” observation does not, and can not, influence SB’s belief in Heads based on the current assessment. Knowing that an observation is made is not the same thing as knowing what that observation was.
Elga’s sample space(s) are controversial because he uses different measures for the two observations. So how they relate to each other seems ambiguous to some. He actually did it correctly, by further conditionalizing the result. But since his answer violated many people’s intuitions, they invented reasons to argue against how he got it.
My two-coin version does not have this problem. The sample space for the experiment is {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. Each outcome has probability ¹⁄₄. The first observation establishes the condition as {HT1_HH2, TH1_TT2, TT1_TH2} and its complement as {HH1_HT2}. Conditional probability says the probability of {HT1_HH2} is ¹⁄₃. The second observation establishes the condition as {HH1_HT2, TH1_TT2, TT1_TH2} and its complement as {HT1_HH2}. Conditional probability says the probability of {HH1_HT2} is ¹⁄₃.
And the point is that it does not matter which observation corresponds to SB being awake, since the answer is ¹⁄₃ regardless.

JeffJo 28 Mar 2024 20:47 UTC
3 points
0
in reply to: JeffJo’s comment on: The Solution to Sleeping Beauty
The link I use to get here only loads the comments, so I didn’t find the “Effects of Amnesia” section until just now. Editing it:
“But in my two-coin case, the subject is well aware about the setting of the experiment. She knows that her awakening was based on the current state of the coins. It is derived from, but not necessarily the same as, the result of flipping them. She only knows that this wakening was based on their current state, not a state that either precedes or follows from another. And her memory loss prevents her from making any connection between the two. As a good Bayesian, she has to use only the relevant available information that can be applied to the current state.”

JeffJo 28 Mar 2024 16:57 UTC
3 points
0
in reply to: Ape in the coat’s comment on: The Solution to Sleeping Beauty
Let it be not two different days but two different half-hour intervals. Or even two milliseconds—this doesn’t change the core of the issue that sequential events are not mutually exclusive.
OUTCOME: A measurable result of a random experiment.
SAMPLE SPACE: a set of exhaustive, mutually exclusive outcomes of a random experiment.
EVENT: Any subset of the sample space of a random experiment.
INDEPENDENT EVENTS: If A and B are events from the same sample space, and the occurrence of event A does not affect the chances of the occurrence of event B, then A and B are independent events.
The outside world certainly can name the outcomes {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. But the subject has knowledge of only one pass. So to her, only the current pass exists, because she has no knowledge of the other pass. What happens in that interval can play no part in her belief. The sample space is {HH, HT, TH, TT}.
To her, these four outcomes represent fully independent events, because she has no knowledge of the other pass. To her, the fact that she is awake means the event {HH} has been ruled out. It is still a part of the sample space, but is is one she knows is not happening. That’s how conditional probability works; the sample space is divided into two subsets; one is consistent with the observation, and one is not.
What you are doing, is treating HH (or, in Elga’s implementation, H&Tuesday) as if it ceases to exist as a valid outcome of the experiment. So HH1_HT2 has to be treated differently than TT1_TH2, since HH1_HT2 only “exists” in one pass, while TT1_TH2 “exists” in both. This is not true. Both exist in both passes, but one is unobserved in one pass.
And this really is the fallacy in any halfer argument. They treat the information in the observation as if it applies to both days. Since H&Tuesday “doesn’t exist”, H&Monday fully represents the Heads outcome. So to be consistent, T&Monday has to fully represent the Tails outcome. As does T&Tuesday, so they are fully equivalent.
If you are observing state TH it necessary means that either you’ve already observed or will observe state TT.
You are projecting the result you want onto the process. Say I roll a six-sided die tell you that the result is odd. Then I administer the amnesia drug, and tell you that I previously told you whether th result was even or odd. I then ask you for your degree of belief that the result is a six. Should you say ¹⁄₆, because as far as you know the sample space is {1,2,3,4,5,6}? Or should you say 0, because “you are [now] observing a state that you’ve already observed is only {1,3,5}?
And if you try to claim that this is different because you don’t know what I told you, that is exactly the point of the Two Coin Version.
The definition of a sample space [was broken] - it’s supposed to be constructed from mutually exclusive elementary outcomes.
It is so constructed.
I’ve specifically explained how. We write down outcomes when the researcher sees the Beauty awake—when they updated on the fact of Beauty’s awakening.
Beauty is doing the updating, not “they.” She is in an experiment where there are four possible combinations for what the coins are currently showing. She has no ability to infer/anticipate what the coins were/will be showing on another day.
Her observation is that one combination, of what is in the sample space for today, is eliminated.
No, I’m not complicating this with two lists for each day. There is only one list, which documents all the awakenings of the subject,...
Maybe you missed the part where I said you can look at one, or the other, or bot as long as you don’t carry information across.
You are mistaken about what the amnesia acomplishes, Once again I send you to reread the Effects of Amnesia section.
Then you are mistaken in that section.
Her belief can be based only on what she knows. If you create a difference between the two passes, in her knowledge, then maybe you could claim a dependence. I don’t think you can in this case, but to do it requires that difference.
The Two Coin Version does not have a difference. Nothing about what she observed about the outcomes HH1_HT2 or HT1_HH2 in another pass can affect her confidence concerning them in the current pass. (And please, recall that these describe the combinations that are showing.)

JeffJo 27 Mar 2024 16:25 UTC
1 point
0
in reply to: Ape in the coat’s comment on: The Solution to Sleeping Beauty
And as I’ve tried to get across, if the two versions are truly isomorphic, and also have faults, one should be able to identify those faults in either one without translating them to the other. But if those faults turn out to depend on a false analysis specific to one, you won’t find them in the other.
The Two Coin version is about what happens on one day. Unlike the Always-Monday-Tails-Tuesday version, the subject can infer no information about coin C1 on another day, which is the mechanism for fault in that version. Each day, in the “world” of the subject, is a fully independent “world” with a mathematically valid sample space that applies to it alone.
“It treats sequential events as mutually exclusive,” No, it treats an observation of a state, when that observation bears no connection to any other, as independent of any other.
“… therefore unlawfully constructs sample space.” What law was broken?
Do you disagree that, on the morning of the observation, there were four equally likely states? Do you think the subject has some information about how the state was observed on another day? That an observer from the outside world has some impact on what is known on the inside? These are the kind of details that produce controversy in the Always-Monday-Tails-Tuesday version. I personally think the inferences about carrying information over between the two days are all invalid, but what I am trying to do is eliminate any basis for doing that.
Yes, each outcome on the first day can be paired with exactly one on the second. But without any information passing to the subject between these two days, she cannot do anything with such pairings. To her, each day is its own, completely independent probability experiment. One where “new information” means she is awakened to see only three of the four possible outcomes.
“Your model treats HH, HT, TH and TT as four individual mutually exclusive outcomes” No, it treats the current state of the coins as four mutually exclusive states.
“However, when you actually do it, you get a different list.”
How so? If your write down the state on the first day that the researchers look at the coins, you will find that {HH, TH, HT, TT} all occur with frequency ¹⁄₄. Same on the second day. If you write down the frequencies when the subject is awake, you find that {TH, HT, TT} all have frequency ¹⁄₃.
Here is what you are arguing: Say you repeat this many times and make two lists, one for each day. And each entry on the lists includes the attempt number. You will find that every attempt where the first list says TH, the second list will say TT. And that every attempt where the first list says HT, the second list will not have an entry for that attempt.
What I’m saying is that the subject’s belief cannot cross-correlate the attempt numbers. She could use just the first list with the attempt numbers. Or the second, since each when viewed in isolation has the same properties as the other and so gets the same answer. She can even use a combined list, but if she does she cannot use the attempt numbers to associate the two observations on that attempt. That is what the amnesia drug accomplishes.
What I am saying is that regardless of which list she uses, the probability of heads is ¹⁄₃. And your arguments that this is wrong require associating the attempts, essentially removing the effect of amnesia. That cannot influence the subject’s belief.

JeffJo 26 Mar 2024 21:13 UTC
0 points
0
on: The Solution to Sleeping Beauty
This is the Sleeping Beauty Problem:
“Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?”
Unfortunately, it doesn’t describe how to implement the wakings. Adam Elga tried to implement by adding several details. I suppose he thought they would facilitate a solution that would be universally accepted. If he thought that, he was wrong. They introduced more uncertainty than acceptance. And his additions have been treated as if they are part of the problem, which is not true. There is a way to implement it without that uncertainty, but the uncertainty is too ingrained for people to let go of it.
After you are put to sleep, the researchers will flip two coins and set them aside; call then C1 and C2. On the first day of your sleep, they will examine the coins. If both are showing Heads, they will leave you asleep for the rest of the day. Otherwise, they will wake you, ask you to what degree you believe that coin C1 is showing Heads, and then put you back to sleep with that drug. Then they will turn coin C2 over to show its other side.
On the second day, they will perform the exact same steps. Well, I guess they don’t need to turn coin C2 over.
This way, the details of the actual problem are implemented exactly. But since the probabilistic details that apply to either day do not change, and do not depend on what you know, or believe, on the other day, the question is easy to answer.
When the researchers examined the two coins, there were four equally likely combinations for what they might be showing: {HH. HT, TH, TT}. This is the initial sample space that applies to either day. Because of the amnesia drug, to you they are independent of the other day. But since you are awake, you know that the examination of the current state of the coins did not find HH. This is classic “new information” that allows you to eliminate one member of the sample space, and update your belief in C2=H from ¹⁄₂ to ¹⁄₃.
I would certainly accept any argument that shows how this is not a valid implementation of the problem, as stated. Or that my solution is not a valid solution for this implementation. And while this may have implications about how we should treat Elga’s, or any other, solution to his implementation? They play no part in evaluating my solution.

JeffJo 10 Mar 2024 20:38 UTC
1 point
0
on: Sleeping Beauty Resolved?
This paper starts out with a misrepresentation. “As a reminder, this is the Sleeping Beauty problem:”… and then it proceeds to describe the problem as Adam Elga modified it to enable his thirder solution. The actual problem that Elga presented was:
Some researchers are going to put you to sleep. During the two days[1] that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened[2], to what degree ought you believe that the outcome of the coin toss is Heads?
There are two hints of the details Elga will add, but these hints do not impact the problem as stated. At [1], Elga suggests that the two potential wakings occur on different days; all that is really important is that they happen at different times. At [2], the ambiguous “first awakened” clause is added. It could mean that SB is only asked the first time she is awakened; but that renders the controversy moot. With Elga’s modifications, only asking on the first awakening is telling SB that it is Monday. He appears to mean “before we reveal some information,” which is how Elga eliminates one of the three possible events he uses.
Elga’s implementation of this problem was to always wake SB on Monday, and only wake her on Tuesday if the coin result was Tails. After she answers the question, Elga then reveals either that it is Monday, or that the coin landed on Tails. Elga also included DAY=Monday or DAY=Tuesday as a random variable, which creates the underlying controversy. If that is proper, the answer is ¹⁄₃. If, as Neal argues, it is indexical information, it cannot be used this way and the answer is ¹⁄₂.
So the controversy was created by Elga’s implementation. And it was unnecessary. There is another implementation of the same problem that does not rely on indexicals.
Once SB is told the details of the experiment and put to sleep, we flip two coins: call them C1 and C2. Then we perform this procedure:
1. If both coins are showing Heads, we end the procedure now with SB still asleep.
2. Otherwise, we wake SB and ask for her degree of belief that coin C1 landed on Heads.
3. After she gives an answer, we put her back to sleep with amnesia.
After these steps are concluded, whether that occurred in step 1 or step 3, we turn coin C2 over to show the opposite side. And then repeat the same procedure.
SB will thus be wakened once if coin C1 landed on Heads, and twice if Tails. Either way, she will not recall another waking. But that does no matter. She knows all of the details that apply to the current waking. In step 1, there were four possible, equally-likely combinations of (C1,C2); specifically, (H,H), (H,T), (T,H), and (T,T). But since she is awake, she knows that (H,H) was eliminated in step 1. In only one of the remaining, still equally-likely combinations did coin C1 land on Heads.
The answer is ¹⁄₃. No indexical information was used to determine this.

JeffJo 16 Feb 2024 21:05 UTC
0 points
0
in reply to: Ape in the coat’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
My problem setup is an exact implementation of the problem Elga asked. Elga’s adds some detail that does not affect the answer, but has created more than two decades of controversy.
The answer of ¹⁄₃.

JeffJo 16 Feb 2024 21:03 UTC
1 point
0
in reply to: Ape in the coat’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
I keep repeating, because you keep misunderstanding how my example is very different than yours.
In yours, there is one “sampling” of the balls (that is, a check on the outcome and a query about it). This one sampling is done only after two opportunities to put a ball into box have occurred. The probability you ask about depends on what happened in both. Amnesia is irrelevant. She is asked just once.
In mine, there are two “samplings.” The probability in each is completely independent of the other. Amnesia is important to maintain the independence.
SPECIFICALLY: SB’s belief is based entirely one what happens in these steps:
1. Two coins are randomly arranged so that each of the four combinations {HH, HT, TH, TT} has a 25% chance to be the outcome.
2. If the random combination is HH, one option happens, that does not involve asking for a probability. Otherwise, another option happens, and it does involve asking for a probability.
3. SB has full knowledge of these three steps, and knows that the second option was chosen. She can assign a probability based ENTIRELY on these three steps.
This happens twice. What you seem to ignore, is that the method used to arrange the coins is different in the first pass through these three steps, and the second. In the first, it is flipping the coins. In the second, it is a modification of this flips. But BECAUSE OF AMNESIA, this modification does no, in any way, affect SB’s assessment that sample space is {HH, HT, TH, TT}, or that each has a 25% chance to be the outcome.
Her answer is unambiguously ¹⁄₃ anytime she is asked.

JeffJo 14 Feb 2024 22:16 UTC
1 point
0
in reply to: JeffJo’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
There are several, valid solutions that do not always introduce the details that are misinterpreted as ambiguities. The two-coin version is one, which says the answer is ¹⁄₃.
Here’s another, that I think also proves the answer is ¹⁄₃, but I’m sure halfers will disagree with that. But it does prove that ¹⁄₂ can’t be right.
- Instead of leaving SB asleep on Tuesday, after Heads, we wake her but do not interview her. We do something entirely different, like take her on a $5000 shopping spree on Rodeo Drive. (She can get maybe one nice dress.)
This way, when she is first wakened—which can only mean before she learns if it is for an interview or a shopping spree, since she can’t know about any prior/subsequent waking—she is certain that the probability of Heads and Tails are each 50%. But when she is interviewed, she knows that something that only happens after a Heads has been “eliminated.” So the probability of Heads must be reduced, and the probability of Tails must be increased. I think that she must add “Heads and it is Tuesday” to the sample space Elga used, and each observation has a probability of 25%. Which makes the conditional probability of Heads, given that she is interviewed, ¹⁄₃.
BUT IT DOES NOT MATTER WHAT HAPPENS ON “HEADS AND IT IS TUESDAY.” The “ambiguity” is created by ignoring that “HEADS and it is Tuesday” happens even if SB sleeps through it.
OR, we could use four volunteers but only one coin. Let each on sleep through a different combination of “COIN and it is DAY.” Ask each for the probability that the coin landed on the side where she might sleep through a day. On each day, three will be wakened. For two of them, the coin landed on the side that means waking twice. For one, it is the side for waking once.
All three will be brought into a room where they can discuss the answer, but not share their combination. Each has the same information that defines the correct answer. Each must give the same answer, but only one matches the condition. That answer is ¹⁄₃.
+++++
Yes, these all are just different ways of presenting the same information: that, in the popular version, Tuesday after Heads still happens, but cannot be observed. This is what is wrong in all the debates; they treat it as if Tuesday after Heads does not happen.
In my card example, the “prior” sample space includes all 52 cards. The probability distribution is ¹⁄₅₂ for each card. When I say that one is an Ace, it does not mean that it was impossible for a Seven of Clubs to have been drawn, it means that an observation was made, and in that observation the card wasn’t the Seven of Clubs.
In the popular version of the SB problem, there are four possible states that can occur. Three can be observed; and because of the amnesia drug, they are all independent to SB as an observer. Regardless of whether she can know the day, it is part of the observation. Since she is awake, (Heads, Tuesday) is eliminated—as an observation, not from the experiment as a whole—and the updated probability—for this observation, not the experiment as a whole—is ¹⁄₃.
Now, you can use this problem to evaluate epistemic probability. It isn’t really am epistemic problem, but I supposed you can apply it. The answer is ¹⁄₃, and the correct epistemic solution is the one that says so.

JeffJo 14 Feb 2024 21:09 UTC
1 point
0
in reply to: Ape in the coat’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
It is my contention that:
1. The problem, as posed, is not ambiguous and so needs no “disambiguation.”
2. “When you are first awakened” refers to the first few moments after you are awakened. That is, before you/SB might learn information that is not provided to you/SB by the actual problem statement. It does not refer to the relative timing of (potentially) two awakenings.
3. Any perceived ambiguity is caused by the misinterpretation of Elga’s solution, which artificially introduces such information for the purpose of updating the probability space from the permissible form to a hypothetical one that should have a more obvious solution.
4. Any argument that “when you are first awakened” refers to such relative timing, which is impossible for the subject to assess without impermissible information, is obfuscation with the intent to justify a solution that requires such information.
So any comment about first/last/relative awakenings is irrelevant.
Does this help? I know I can’t prove that #2 is correct, but it can be. Nothing else can.

JeffJo 12 Feb 2024 15:07 UTC
2 points
1
in reply to: Radford Neal’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
Your problem is both more, and less, well-posed than you think.
The defining feature of the “older child” version of the Two Child Problem has nothing to do with age. It is that you have ordered the children independently of gender, and identified the gender of a child in a position within that order. Age works well here, since it is easy to show why BB, BG, GB, and GG must be equiprobable by examining the event of the second birth.
But any gender-independent ordering works. It could be alphabetizing names, their seats around the dinner table (clockwise from Mother), or which bedroom each child has. You picked a specific child in an order by looking in a specific room, so the genders of the other two are independent of it and each other. So gBB, gBG, gGB, and gGG are equiprobable at that point in your acquisition of knowledge.
But your second acquisition depends on whether similar help is needed for other sports, and how many gender-specific sports there are. And why there isn’t one for girls’ sports, since we know there is a girl.
My problems are well-posed for what I intended. You didn’t “stumble upon” the information, a source with absolute knowledge told it to you, with no hint of any discrimination between genders. There is an established solution in such cases; it’s called Bertrand’s Box Paradox. That name did not, originally, refer to a problem; it referred to the following solution. It is best illustrated using a different probability than what I asked for:
1. I know Mr. Abbot’s two children. At least one is a boy.
2. I know Mrs. Baker’s two children. At least one is a girl.
3. I know the Curry’s two children. At least one has the gender that I have written inside this sealed envelope.
In each case, what is the probability that the family has a boy and a girl?
Clearly, the answers A1 and A2 must be the same. This is not using uniform distributions, although that is a valid justification. Probability is not about what is true in a specific instance of this disclosure of information—that’s a naive mistake. It is about what we can deduce from the information alone. It is a property of our knowledge of a world where it happens, not the world itself. Since our information is equivalent in Q1 and Q2, that means A1=A2.
But you have no significant information about genders in Q3, so A3 must be ¹⁄₂. And that can be used to get A1 and A2. Bertrand argued simply that if the envelope were opened, A3 had to equal A1 and A2 regardless of what it said, so you didn’t need to open it. Any change would be a paradox. But there is a more rigorous solution.
If W represents what is written in the envelope, the Law of Total Probability says:
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A2
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A1
A3 = [Pr(W=”Boy”) + Pr(W=”Girl”)]*A1
A3 = A1 = A2 (which all equal ¹⁄₂).
This solution is also used for the famous Monty Hall Problem, even if those using it do not realize it. The most common solution uses the assertion that “your original probability of ¹⁄₃ can’t change.” So, since the open door is revealed to not have the car, the closed door that you didn’t pick must now have a ²⁄₃ probability.
The assertion is equivalent to my sealed envelope. You see the door that gets opened, which equivalent to naming one gender in Q1 and Q2. Since your answer must be the same regardless of which door that is, it is the same as when you ignore which door is opened.

JeffJo 12 Feb 2024 14:12 UTC
1 point
0
in reply to: Ape in the coat’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
This variation of my two-coin is just converting my version of the problem Elga posed back into the one Elga solved. And if you leave out the amnesia step (you didn’t say), it is doing so incorrectly.
The entire point of the two-coin version was that it eliminated the obfuscating details that Elga added. So why put them back?
So please, before I address this attempt at diversion in more detail, address mine.
1. Do you think my version accurately implements the problem as posed?
2. Do you think my solution, yielding the unambiguous answer ¹⁄₃, is correct? If not, why not?

JeffJo 9 Feb 2024 16:22 UTC
1 point
0
in reply to: Radford Neal’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
Yes, the fact that someone had to chooses the information is an common source of error, but that is not what you describe. I choose a single card and a single value to avoid that very issue. With very deliberate thought. Your example is a common misinterpretation of what probability means, not how to use it correctly according to Mathematics.
A better example, of what you imply, is the infamous Two Child Problem. And its variation, the Child Born on Tuesday Problem.
1. I have exactly two children. At least one is a boy. What are the chances that I have two boys?
2. I have exactly two children. At least one is a boy who was born on a Tuesday. What are the chances that I have two boys?
(BTW, both “exactly” and “at least” are necessary. If I had said “I have one” and asked about the possibility of two, it implies that any number I state carries an implicit “at least.”)
Far too many “experts” will say that the answers are ¹⁄₃ and ¹³⁄₂₇, respectively. Of the 4 (or 196) possible combinations of the implied information categories, there are 3 (or 27) that fit the information as specified, and of those 1 (or 13) have two boys.
Paradox: How did the added information change the probability from ¹⁄₃ to 13/27?
The resolution of this paradox is that you have to include the choice I made of what to tell you, between what most likely is two sets of equivalent information. If I have a Tuesday Boy and a Thursday Girl, couldn’t I have used the girl’s information in either question? Since you don’t know how this choice is made, a rational belief can only be based on assuming I chose randomly.
So in 2 (or 26) of the 3 (or 27) combinations where the statement I made is true, there is another statement that is also true. And I’d only make this one in half of them. So the answers are 1/(3-2/1)=1/2 and (13-12/2)/(27-26/2)=7/14=1/2. And BTW, this is also how the Monty Hall Problem is solved correctly. That problem originated as Martin Gardner’s Three Prisoners Problem, which he introduced in the same article where he explained why ¹⁄₃ is not correct for #1 above.
In my card drawing problem, there is only one card rank I can report. If you choose to add information, as done with Linda the Bank Teller, you are not a rational solver.

JeffJo 7 Feb 2024 20:43 UTC
1 point
0
in reply to: Radford Neal’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
“I would need to also consider my prior probability that JeffJo would say this conditional on it being the Ace of Space, the Ace of Hearts, the Ace of Diamonds, or the Ace of Clubs. Perhaps I believe the JeffJo would never say the card is an Ace if it is a Space. In that case, the right answer is 0.”
And in the SB problem, what if the lab tech is lazy, and doesn’t want a repeat waking? So they keep re-flipping the “fair coin” until it finally lands on Heads? In that case, her answer should be 1.
The fact is that you have no reason to think that such a bias favors any one card value, or suit, or whatever, different than another.

JeffJo 6 Feb 2024 12:31 UTC
1 point
0
in reply to: Ape in the coat’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
You’ll need to describe that better. If you replace (implied by “instead”) step 1, you are never wakened. If you add “2.1 Put a ball into the box” and “2.2 Remove balls from the box. one by one, until there are no more” then there are never two balls in the box.

JeffJo 6 Feb 2024 12:21 UTC
1 point
−2
in reply to: Said Achmiz’s comment on: Has anyone actually changed their mind regarding Sleeping Beauty problem?
If the context of the question includes a reward structure, then the correct solution has to be evaluated within that structure. This one does not. Artificially inserting one does not make it correct for a problem that does not include one.
The actual problem places the probability within a specific context. The competing solutions claim to evaluate that context, not a reward structure. One does so incorrectly. There are simple ways to show this.