What kind of misrepresentation you are talking about? You have literally just claimed that there is no way to know whether a particular probabilistic model correctly represents the system.

Since what I said was that probability theory makes no restrictions on how to assign values, but that you have to make assignments that are reasonable based on things like the Principle of Indifference, this would be an example of misrepresentation.

If what you are suggesting were true, then the correct answer to "What is a probability of a fair coin toss, about which you know nothing else, to come Heads?", would be: "It is either 0 or 1, but we can't tell.

What? No, absolutely not! It would be: we lack any reason to expect any outcome more than another, so we are indifferent between both outcomes so 1/2 for both of them. The exact same kind of reasoning for both logical and empirical uncertainty.

You claim that "the probability that the nth digit of pi, in base 10, is 1/10 for each possible digit," is assigning a non-zero probability to nine false statements, and probability that is less than 1 to one true statement. I am saying that it such probabilities apply only if we do not apply the formula that will determine that digit.

I claim that the equivalent statement, when I flip a coin but place my hand over it before you see the result, is "the probability that the coin landed Heads is 1/2, as is the probability that it landed Tails." And that the truth of these two statements is just as deterministic, if I lift my hand. Or if I looked before I hid the coin. Or if someone has a x-ray that can see it. That the probability in question is about the uncertainty when no method is applied that determine the truth of the statements, even when we know such methods exist.

I'm saying that this is not a question in epistemology, not that epistemology is invalid.

And the reason the SB problem is pertinent, is because it does not matter if H+Tue can be observed. It is one of the outcomes that is possible.

My criteria are what is required for the Laws of Probability to apply. There are no criteria for being a "true probability statement," whatever it is that you think that means.

There is only one criterion - it is more of a guideline than a rule - for how to assign values to probabilities. It's called he Principle of Indifference. If you have a set of events, and no reason to think that any one of them is more, or less, likely than any of the other events in the set (this is what "indifferent" means), then you should assign the same probability to each. You don't have to, of course, but that makes for unreasonable (not "false") probabilities if you don't.

This is why, in a simple coin flip, both "Heads" and "Tails" are assigned a 50% probability. Even if I tell you that I have determined experimentally, with 99% confidence, that the coin is biased. Unless I tell you which face is favored, you should assign 50% to each until you learn enough to change that (it's called Baysean Updating). I suppose that is what you would call a "false probability statement," but it is the correct thing to do. and you will not find a "true probability statement," just one that you have more confidence in.

+++++

Now, in the Sleeping Beauty Problem, there are ways to test it. And they get met with what I consider to be unfounded, and unreasonable, objections because some don't like the answer. Start with four volunteers instead of one. Assign each a different combination from the set {H+Mon, H+Tue, T+Mon, T+Tue}. Using the same coin and setup as in the popular version, wake three of them on each day of the experiment. Exclude the one assigned the current day and coin result.

Ask each for the probability that this is the only time she will be wakened. This was actually the question in the original version of the SB problem, but it is equivalent to "the probability of Heads" for the first two combinations in that set, and "the probability of Tails" for the last two. This problem is identical to the popular one for the volunteer assigned H+Tue, and equivalent for the others.

After each of the three who are awake has answered, bring them together to discuss the same question. The only taboo topic is their assignments. Now, each of them knows that exactly one of the three will be woken only once, that each has the same information about whether she is the one, and that the same information applies to each. In other words, the Principle of Indifference applies, and the answers for all three are 1/3.

The "check" here, is what could possibly have changed in between the individual question, and the group question?

Were it true, then the correct answer to "What is a probability of a fair coin toss, about which you know nothing else, to come Heads?", would be: "Any real number between zero and one".

Yes, you can make a valid probability space where the probability of Heads is any such number. What you can't do, is say that probability space accurately represents the system.

Probability Theory only determines the requirements for such a space. It makes no mention of how you should satisfy them. And I assumed you would know the difference, and not try to misrepresent it.

I have no idea how you managed to come to the conclusion that I don't understand that probability is about lack of knowledge.

Because of responses like that last one, and this:

It is not about what what can be shown, deterministically, to be true when we have perfect knowledge of a system. It is about the different outcomes that could be true when your knowledge is imperfect.

Makes little sense.

Okay, I misspoke. Because I also assumed you would at least try to understand. You need complete knowledge of an assumed (Baysean) or actual (Frequentist) system that can produce different results. But lack at least some knowledge of the one instance of the system in question; that is, one particular result.

The system in the SB problem is that it is a two-day experiment, where the individual (single) days are observed separately. There are four combinations, that depend on the separation, and the four are equally likely to apply to any single day in the system. An "instance" is not one experiment, it one day's observation by SB. Because it is independent of any other instance, which includes a different day in the same experiment, due to the amnesia. The partial information that she gains is that it is not Tuesday, after Heads. What I keep trying to convince you of, is that that combination is part of the system and has a probability in the probability space.

And how is it different from the initial example with googolth digit of pi? Why can't we say that we are not assigning non-zero probability to a false statement.

Because the system is "a digit in {0,1,2,3,4,5,6,7,8,9} is selected in a manner that is completely hidden from us at the moment, even if it could be obtained. The probability refers to what the result of calculating the nth digit could be when we do not apply n to the formula for pi, not when we do.

If what you are suggesting were true, then the correct answer to "What is a probability of a fair coin toss, about which you know nothing else, to come Heads?", would be: "It is either 0 or 1, but we can't tell.

And reason I "[concluded] that [you] don't understand that probability is about lack of knowledge", is because you are trying to incorporate what happens in an instance into the probability for system elements. It is because fail to recognize, in that last quote, that the probability is a property of what you could know about an instance but don't due to ignorance, not what is true when you don't know.

Here is a  similar problem based on the same Mathematics. I call it Camp Sleeping Beauty. The same sleep and Amnesia details will be used, but at a week-long camp. On the arrival day, Sunday, you will be told all these details.

1. There are six "camp" activities, and you will partake of a random one on each day: Archery, Boating, Canoeing, Dodge Ball, Exercising, and Free Time.
2. The random part is that, after you are put to sleep on Sunday, a six-sided die will be rolled. But before you are put to sleep, you will help to randomly populate a six-by-six Camp Calendar with these activities. The board has one column for each of the days, Monday through Saturday, and one row for each die roll.
   1. If you think it necessary, assume the non-random assignment of Monday&1=A, Tuesday&2=B, etc., so that every activity is guaranteed to happen.
3. When you awake each day, you will not be told the day, but you will be taken to that day's assigned activity using Sunday's die roll.
4. At lunch, you will be interviewed and asked to assign probabilities to each of the six die rolls, based on the knowledge you gained from the activity. You will have access to the calendar you helped to make.

Like I said, this is a Math problem. The probability for each die roll is the number of times today's activity appears in that row, divided by the number of times it appears on the board. That is, if the activity appears only once on the board - or if every time it appears it is in the same row - there is a 100% chance that that was the die roll. If it appears twice on different rows, there is a 1/2 chance for each of those two rolls. Etc.

The important part here is that the arrangement of the other activities, or the actual activities, is completely irrelevant to this answer. "Today" is not an enigmatic value that you need a "universe model" to access, it is a random sampling of this six-by-six array.

Now, change the "Exercise" activity to "Extra sleep," where you get to sleep all day. That is, no interview. Nothing changes on the days where you do wake up. The answers do not change to the "Halfer" solution of 1/6 for each just because you know you would not wake up those days. The information you have about them is exactly the same - they are inconsistent with what happened today.

The popular Sleeping Beauty Problem is just a two-by-two version of this, with three of the day's activities being "wake up" and one "sleep." The answer, when you do wake up, is the number of times "wake" appears in the Heads row (one) divided by the number of times it appears on the board (three).

Well, it is true, and your confusion demonstrates what you don't want to understand about probability. It is not about what what can be shown, deterministically, to be true when we have perfect knowledge of a system. It is about the different outcomes that could be true when your knowledge is imperfect.

Consider your "opaque box" problem. You are not "assigning non-zero probability to a false statement," you are assigning it to a possibility that could be true based on the incomplete knowledge you have. This is what probability is supposed to be - a measure of your lack of knowledge about a result, not a measure of absolute truth. Even if there are others who possess more knowledge.

So yes, it does provide an answer to the question you asked. That question was just not what you thought you were asking. Like "If a plane traveling from New York to Montreal, carrying 50 Americans and 50 Canadians, crashes exactly on the border, where do we bury the survivors?" Answer: you don't bury the survivors.

You hit one point on the head: "it’s really just a math problem." But what that means is that it is not a philosophy problem. It was originally posed as a problem in philosophy, but the only thing such considerations do is obfuscate the math problem. Without justification. Just the need to put it in the realm of philosophy to match the original intent.

Consider a variation of the problem, that isn't a variation at all, but I'm sure will get a response as to why it doesn't fit in with a desired philosophical solution. I know this, because it has happened before. Use the same details with four volunteers, but one coin, and different schedules. One volunteer will be left asleep on Tuesday, after Heads, as in the popular[1] version. Each of the other three will be left asleep on a different one of the other three possible combinations of the day (Monday or Tuesday) and the coin result (Heads or Tails).

So on each day in the experiment, three of the volunteers will awakened. Each will be asked, individually, for the probability[2] that this is her only waking[3]. Then, the three will be brought together and asked to discuss the same question. The only taboo topic is the combinations where each would be left asleep.

In the individual interview, each is solving an problem that is identical to the popular version. In the combined one, each knows that she has a 1/3 chance to have the same assigned coin result as the missing volunteer. The only possible issue, is why should the answer change between these interviews.

--

[1] It isn't the original, or even the version posed in the paper that created the popular version.

[2] "Odds" means a ratio, not a probability. Odds of X:Y mean a probability of X/(X+Y). Halfers say the odds are 1:1, or a probability of 1/2. Thirders say the odds are 1:2, or a probability or 1/3.

[3] This is the question as it was in the original version. It is equivalent to "What is the probability that the coin result means you will be wakened only once?"

These problems are actually variations of one of the oldest "probability paradoxes" ever. And I put that in quotes, because in 1889 when Joseph Bertrand published it, "Bertrand's Box Paradox" meant how he proved that one proposed answer could not be right, because it produced a contradiction.

Here is that paradox, applied to Problem #1. With a slight modification that changes nothing except using a complimentary probability. In each question, what is the probability that I have a boy and a girl?

• I have two children, at least one of whom is a boy.
• I have two children, at least one of whom is a girl.
• I have two children, and have written the gender of at least one in this sealed envelope.

The answers to #1 and #2 have to be the same. It is tempting to say that answer is 2/3.

BUT, in #3, there are only two possibilities for what is written in the envelope. Once opened, it reduces to either #1, or #2. So the Law of Total Probability tells us that #3, even before the envelope is opened, also has that same answer.

But with no information given, the answer to #3 is clearly 1/2. So it can't be 2/3. This is the paradox in Bertrand's Box Problem. If you add a fourth box to that one, with a gold and a silver coin, it is identical to this Boy or Girl problem. I'm not saying that this directly proves that 1/2 is right. Just that no other answer can be right, and 1/2 can. Much like how proof by contradiction works.

Martin Gardner popularized this problem in the May, 1959 issue of Scientific American. But his wording was "Mr. Smith has two children. At least one is a boy." At first he said the answer was 2/3, but he withdrew that answer the next October, in a column titled "Probability and Ambiguity." His problem statement was ambiguous because it did not specify how the information came to be known. As others have pointed out, some methods lead to the answer 2/3, and others to the answer 1/2.

This wording is also ambiguous, but for a different reason. Whoever "I" is, knows the genders of the two children, and picked one of them. What we don't know, is how. But a Bayesian approach says that if he has a boy and a girl, we should assume he picked at random. This makes the answer 1/2.

To demonstrate this, Gardner published the Three Prisoners Problem in that October 1959 column. It is identical to the more modern puzzle, the Monty Hall Problem. With one exception: Gardner specified that a coin flip was used to choose what information is revealed, when both kinds of information are possible. That is usually not made explicit in Monty Hall, yet it is a necessary assumption to get that switching doors has a 2/3 probability of winning.

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.

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.