JeffJo

I have liked this scenario ever since I found out that there are three Michelin 3-star restaurants in San Francisco. And that their names (almost) line up in a convenient order. They are Atelier Crenn (which I will call "A"), Benu ("B"), and Quince ("C" to fit in the order). (All of this is irrelevant, except it helps the narrative.)

I will use them to distinguish H&Mon, T&Mon, and T&Tue in Beauty's world, without affecting her credence in any way. Each will be randomly assigned one of these combinations at the same time the coin is flipped. Whe... (read more)

You are right, I didn't look closely at your betting argument, because (as I have said) betting arguments are invalid. They can be manipulated - as you do - to get the answer you want.

I pick a random day to approach Beauty with an offer of a bet. If she guesses right, on Wednesday I’ll give her $1.5, otherwise she’ll give me $2. If she truly believes that the coin flipped Tails with a probability of 2⁄3, then she’ll take this bet.

You are saying the odds (easier to work with for betting) you offer are 1.5:2 for Tails (representing probability 4/7), and ... (read more)

Why couldn’t SB use her credence of the coin for the bet I described?

First, I'm still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?

But to answer your question, she can. On the inside of the experiment, where there are three equivalently-placed choices. It's a common justification for 1/3. You can't, on the outside, where there is no way to make a random choice that is equivalent. It's a different experiment than the one where she is asked about her credence.

Do you mean that she goes shopping on Tuesd

I pick a random day ...

You say you pick a random day, but you do not specify the set from which you pick. Is it {Mon, Tue}, {H+Mon, T+Mon, T+Tue}, or {H+Mon, T+Mon, H+Tue, T+Tue}? In each case, it is possible that the bet will not be offered, and possible that SB is asked for a credence that is never "tested" with a bet. (And please, don't say "Monday if Heads, and Monday or Tuesday if Tails." That's biasing the method to what the method is supposed to test.)

That is not how the problem works. You are manipulating the odds to make your answer look correc... (read more)

What observation are you trying to update on?

There are three die rolls where SB would be awake at thevmoment.

There are three where she would be asleep.

She is awake.

What is so difficult to see about this?

the Thirder logic results in credences of expected value that cannot be used without leading to losing bets

That depends entirely on how you evaluate the"bet." If it is evaluated once, over the two days of the experiment, "1/2" produces balanced bets and "1/3" produces unbalanced bets. If it is evaluated individually each time SB is asked the question... (read more)

I’m not sure what you’re trying to say.

With all due respect, I don't sense that you are trying to understand it. Your position seems to be firmly that SB has gained no information, so she can't use conditional probability. Mine is that she does, and all I'm trying to do is make it hard to ignore why it is.

As you point out, there are six possible die rolls. But on any one day, only three remain possible. This is what is known, in probability theory, as a "condition." It is what allows one to "update" a prior probability, which is what you calculated, to ... (read more)

Certainly, since it is wrong.

There is much obfuscation surrounding the SB problem, that appear in this thread. One is that when you ask about a probability in the present tense about an occurrence described in the past tense, you are asking about the conditional probability of that event given the present information. What this means is that there is only one question that is relevant: "Given SB's knowledge when she is awake, what is the probability that the coin result was Heads?" Not "What was the probability, ignoring any information obtained since, tha... (read more)

The correct model for the Sleeping Beauty Problem has to recognize that in a probability experiment, the prior sample space and its related prior probabilities are based on the possible outcomes without considering any evidence that might result from any outcomes.

The issue with the Sleeping Beauty Problem, that makes it unique, is that each running of the experiment produces two outcomes. One on Monday, and another on Tuesday. These are different, and independent, outcomes because an outcome is defined by SB's ability to experience it. Notice I didn't say... (read more)

Q1: Is the answer the same if SB is left asleep on H+Mon, but wakened on H+Tue?

Q2: Is the answer the same if the day SB is left asleep, H+Mon or H+Tue, is determined by a second coin flip?

Q3:  On Sunday Night, flip two coins. Call them C1 and C2. Coin C1  is the one SB is asked about, and C2 is the second on in Q2. So, on Monday wake SB if either coin is showing Tails. On Monday Night, turn C2 over to show the opposite side. And on Tuesday, again wake SB if either coin is showing Tails. Whenever awakened, what should SB answer about coin C1? But before ans... (read more)

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

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 ... (read more)

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

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

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... (read more)

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 bef... (read more)

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 tw

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 h... (read more)

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 awakene... (read more)

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 b... (read more)