JeffJo

Karma: 6

JeffJo 7 Jun 2026 16:54 UTC
1 point
0
on: Just Use Bayes: Sleeping Beauty and Monty Hall
The following is what no Halfer has ever responded to, except to say “Gee, that might be interesting to discuss. We’ll get back to it.” They never do.

The experiment:
- Beauty will be put to sleep on Sunday Night, and a fair coin will be flipped. On each of the next two days (Monday and Tuesday), she will participate in one of two different procedures, named A and B. There are two requirements for them: #1: During either procedure, Beauty will be unaware of what may have, or will have, happened on the “other” day. #2: In procedure A she will be asked a probability (or credence, if you prefer) question about the coin flip. But no such question will be asked in procedure B.
- On Monday, Beauty will participate in procedure A. On Tuesday, she will participate in procedure B if the coin landed on Heads, and in procedure A is it landed on Tails.
- The question, that is only asked in procedure A, is “What do you believe is the probability that the coin landed on Heads?”
This is a simple conditional probability question. There are four situations during the experiment when Beauty could participate in a procedure. Before she is—or, if she won’t be—asked the question, Requirement #1 makes her current situation an independent sampling into that set of four possibilities. If she is asked the question, Beauty knows that she is participating in procedure A, and so it is either Monday after Heads, Monday after Tails, or Tuesday after Tails. The probability that the coin landed on Heads is ¹⁄₃. If she is not (yet) asked, well, does it really matter?

The popular version of the Sleeping Beauty Problem fits this outline. Halfers don’t want to acknowledge two facts about that implementation of my outline. First, that “new Information” does not mean that something happened that was not guaranteed to happen, or is all that Beauty can perceive. It means that something happened that eliminates a possibility from the sample space. Second, that the possibility of something happening on Tuesday makes everything that could happen on Tuesday a member of that sample space. EVEN IF SHE WOULD SLEEP THROUGH IT, Beauty knows that the events of Tuesday, after Heads, must be represented is the sample space.

There is no probability issue in the Sleeping Beauty Problem, just obfuscation about what belongs in a sample space.

JeffJo 27 May 2026 13:37 UTC
1 point
0
in reply to: Steff’s comment on: Just Use Bayes: Sleeping Beauty and Monty Hall
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. When Beauty is awakened, she will be taken to the restaurant assigned to that day and coin result, and asked for her credence that the coin landed Heads. The point of all this is that there is no useful information in whether she dines at “A”, “B”, or “C”

There are also two important times in this process: T1 is after Beauty is woken but before the limo leaves to take her to a restaurant. And T2 is after dinner. So far, this is identical to the canonical problem. Since Beauty has no information about the assignments, it cannot matter if she is asked the question at T1 or T2. This IS the canonical Sleeping Beauty Problem regardless of the timing.

But I will now propose two variations. In Variation 1, Beauty is not woken on H&Tue, as in the canonical version. In Variation 2, she is told in advance that she will be woken both days. The same randomization applies to those three combinations, and Denny’s (“D”) is always assigned to H&Tue.

At V1T1 (that is, time T1 in Variation 1) or V1T2, there seems to be controversy about what credences Beauty should assign to each member of the sample space {H&Mon, T&Mon, T&Tue}. So I’ll delay discussing it.

At V2T1, there can be no controversy about the credences for the sample space {H&Mon, T&Mon, H&Tue, T&Tue}. Each has a credence of ¹⁄₄. At V2T2, Beauty has received “new, centered evidence” about the sample space. If she is dining at “D”, she knows the coin landed on Heads. If she is at “A”, “B”, or “C”, she can use that “new, centered evidence” to eliminate H&Tue. She can update her credences for H&Mon, T&Mon, or T&Tue from ¹⁄₄ each to ¹⁄₃ each.

My point in this discussion is that the only difference between the two variations is when Beauty receives this information. The information itself is the same. In Variation 1, she receives it at time T1, when she is woken up. The equiprobable ¹⁄₄ credences apply to the time T0, when the experimenters are deciding if she should be woken. BEAUTY KNOWS THIS TIME WILL HAPPEN, AND WHAT INFORMATION APPLIES TO IT, EVEN IF SHE DOES NOT EXPERIENCE IT. In Variation 2, she receives it at time T2. Regardless, it is not her ability to perceive the state that determines her credence. It is the knowledge she gains—between T0 and T1 in Variation 1, or between T1 and T2 in Variation 2 - that constitutes “new, centered evidence” for the purpose of the update.

The answer to canonical Sleeping Beauty Problem is ¹⁄₃. Halfers confuse the perception of the information with the knowledge of when it is made evident.

JeffJo 9 May 2026 20:53 UTC
1 point
0
in reply to: Steff’s comment on: Just Use Bayes: Sleeping Beauty and Monty Hall
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 so should be 2.67:2 (representing ³⁄₇) for Heads. If Beauty thinks they are 1:2 and 4:2 (2/3 and ¹⁄₃), Beauty should indeed take this bet, on Tails.

But if this is what you mean, you did not do what claimed to do. You did not puck a random day, you choose a random timeline and inserted a bet into a waking moment in that timeline. She may as well have chosen the bet on Sunday.

By flipping the coin on Sunday and resolving the bet, once, on Wednesday, you have manipulated the wager. What happens in between—all the sleeping, waking, and amnesia—is completely irrelevant. So what is the point of what happens in between? Isn’t the entire point of the thought experiment to evaluate how, and yes, if, it matters? Yet you designed your question to eliminate any issue, not to resolve it.

Try this: actually choose a random day for the offer, but use her “thirder” odds. Note that now there is a chance that no bet will be offered, if the random day is Tuesday and the coin landed on Heads.

If the coin landed Tails, Beauty will be offered a wager. She will win $1 if she bet on Tails (at 1:2), and lose $2 if she bet on Heads. But if the coin landed Heads, she will lose an expected $1 if she chooses Tails when the bet is offered, and win an expected $2 (at 4:2) if she chooses Heads when the bet is offered.

In other words, if we remove the bias you built into your challenge, Beauty expects to wager $2 and a either choice wins $1 if correct.

What if you have Beauty put $2 into each of two identical envelopes? Each time she is awake, we give her one and she chooses either Tails (at 1:2 odds) or Heads (at 4:2). We again evaluate on Wednesday, but this time the number of bets evaluated can vary.

If the coin lands Tails, she is up $2 if she chooses Tails (two wins of $1 since we assume she is consistent over the two days) and down $2 if she chooses Heads. If it lands Heads, she is down $4 if she chooses Tails (two losses of $4) and up $4 if she chooses Heads. This is a balanced wager even though the balance is differs depending on the result.

But it is the differing balance that halfers object to. It is an invalid objection, but again it is why I try to avoid betting arguments altogether.

+++++

It’s a random choice among the days she’ll be awoken. If she’s to awake just Mon, then she’ll be offered on Mon. If she’s to awake on Mon and Tues, then a coin is flipped.

Sorry, I somehow confused your similar response to another comment with this one. This models a very different experiment. One where Beauty is woken exactly once regardless of the coin. It happens on Monday if the coin landed Heads. Or on Monday or Tuesday, according to the second coin, if the first coin landed on Tails.

JeffJo 8 May 2026 1:25 UTC
2 points
1
in reply to: Steff’s comment on: Just Use Bayes: Sleeping Beauty and Monty Hall

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 Tuesday and is told on Tuesday?

On Monday, and on Tuesday after Tails, she is asked about her credence. On Tuesday after Heads, she goes shopping. She doesn’t have to be told anything in this case, since her credences that it is Tuesday, and that the coin landed Heads, will both be 100%. This takes what some want to call “probability mass” away from her Sunday-Night credence in Heads. That was ¹⁄₂, and removing probability mass is what reduces it to ¹⁄₃.

And the point is that IT DOES NOT MATTER what would happen on H+Tue. That is still a day in the experiment, just as valid a result as T+Mon. If you do this, the proper odds for the bet if she is offered one are 2:1.

I’m not denying that those are the possibilities, but that they share equal probabilities.

Please justify how, without using the circular argument that it is the only way to get the answer you want.

But you gave me an idea, to alter the “shopping” version to the red/green die version.

Wake her on all three days. If the result was a green number matching the current day, or a red number not matching it, ask her for her credence in green. If it was a red number matching the current day, or a green one not matching it, ask her for her credence in red.

Before she is asked, there are six-equally likely faces (which contradicts your claim about them being unequal). If she is asked about green, it is reduced to three (still equally likely) faces, where one is green. There is absolutely no difference between these “ask about green” cases and the “awake” cases from before.

I’m not following this part.

And it is what I keep trying to describe. You, on the outside, see a two-day experiment where SB is wakened once, or twice. To SB, it is a one-day sample of that experiment, but limited to the three equally-likely situations where she will be awake. When the sun rose this morning (remember that it is a one-day sample), there were four possible combinations. Three of them, still equally likely, result in waking her. Her evidence is that it is one of these three, and not H+Tue.

Pr(Awake) vs. Pr(Awake on Tues) is awkward because you are confusing two-day and one-day probabilities. So I’ll use the shopping variant. Instead of “Awake” I’ll use “Ask” and “Shop.”

In the two-day experiment—that is, before the coin flip on Sunday and anticipating the next two days—Pr(Ask)=1 and Pr(Shop)=1/2. Note that these are not independent events, so these don’t add up to 1. When SB awakes, but before she learns what will happen today, Pr(Ask)=3/4 and Pr(Shop)=1/4. These are independent events that represent all possibilities, so they do add up to 1.

BOTH OF THESE ARE “PRIOR” PROBABILITIES, meaning they describe the possibilities with no evidence. What you are “not following,” is that the probability space for SB’s credence is the second one, not the first.

Your Bayesian analysis is wrong because it is based on the first. Pr(Awake)=1 means she will be awake at some time over two days. This is why you think you can choose one day at random from what I suspect you mean the two “days” Monday and Tuesday_If_Tails.

SB’s credence is about one day only. Its prior sample space (remember, no evidence) is {H+Mon,T+Mon, H+Tue,T+Tue} with a ¹⁄₄ probability for each. Her evidence is that H+Tue is eliminated, either because she is awake of because she didn’t go shopping.

Your Bayesian analysis “folds” H+Tue into H+Monday because that is the only way she is awake after Heads.

JeffJo 6 May 2026 20:04 UTC
3 points
1
in reply to: Steff’s comment on: Just Use Bayes: Sleeping Beauty and Monty Hall

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 correct. This is why it does not represent the canonical version of the problem, where she is asked on each waking day, not a random one.

The bet must be offered on each waking day, and evaluated each time SB is asked fro a credence, to correctly represent her credence. If you do this, the experiment is a zero-sum game if and only if 2:1 odds are used. But I suspect you will disagree with me. Which is why I do not use betting arguments. Even though I know yours can’t be right based on the issues mentioned above, I know I won’t convince you with my betting scheme, either

You’re not making yourself clear.

I am. You are doing everything you can to avoid seeing it.

Would you like to share some Bayesian equations of your own?

I have described it well enough. But, since you insist:

In the canonical version, on any single day (and this includes Tuesday, after Heads), the sample space for that day includes four outcomes: {H+Mon, T+Mon, H+Tue, T+Tue}. Each has a prior probability—in Bayesian probability, this means a result of the experiment prior to evidence being considered—of ¹⁄₄. SB’s evidence is that it can’t be H+Tue, so the “new data” is the event {H+Mon, T+Mon, T+Tue}.

Bayesian probability says:

Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = Pr({H+Mon, T+Mon, T+Tue}|{H+Mon}) * Pr({H+Mon}) / Pr({H+Mon, T+Mon, T+Tue}}

Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = (1)*(1/4)/(3/4) Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = ¹⁄₃

And again, the mistake you make in your equations is ignoring a valid outcome of the experiment because it goes unobserved by SB.

You have ignored how I point out that the need to consider it is demonstrated by letting SB observe it in a way that distinguishes it from the others: Wake her on Tuesday, after Heads, but take her shopping instead of asking about the coin. The above equation clearly gives her credence, in the situations where she is asked.

And the point is that the correct Bayesian solution does not depend, in any way, on what would happen on Tuesday, after Heads. Only that she makes he same observation for any outcome in {H+Mon, T+Mon, T+Tue}, and uses Bayesian probability based on that observation.

Did you read my post about the Day 1 Objection, which I strongly suspect is very relevant?

Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.

So Pr(Awake) is not 1. “Awake” will not happen on Tuesday, after Heads, which is a possible result.

Please, please, please try to see that I am not saying she might not be awakened over the course of the experiment. I am saying that there are circumstances, consistent with how the question is posed on a single day that is made independent of the other day by amnesia, where she will not be wakened.

And the point of the six-sided die version was to isolate the problem to just the circumstance where she is awake. With no ambiguity about one or two wakings, and how to handle them. That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.

JeffJo 4 May 2026 23:29 UTC
2 points
1
in reply to: Steff’s comment on: Just Use Bayes: Sleeping Beauty and Monty Hall

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, “1/2” produces unbalances bets and “1/3” produces balanced bets.

Please tell me why you think the question is not about the moment the question is asked, but instead includes another moment that may, or may not, happen.

JeffJo 3 May 2026 21:33 UTC
3 points
0
in reply to: Steff’s comment on: Just Use Bayes: Sleeping Beauty and Monty Hall

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 a “posterior” or “conditional” probability, which is what the Sleeping Beauty Problem asks for. This is true even if she gains no information. The condition will be the entire sample space, and the “updated” probability will be the same as the prior probability. So there is no reason to not explore what the condition is.

It’s easier if we number the days 0 (for “Monday”), 1 and 2. The set of possible die rolls—can we agree that this is the sample space? - is {G_0, G_1, G_2, R_0, R_1, R_2}. When she is awake, she does not know which day it is, but she does know that it is only one day. Call it day D. The condition is the event {G_D, R_MOD(D+1,3), R_MOD(D+2,3)}. The condition is the same regardless of what D is. Each outcome in this set has a prior probability of ¹⁄₆. The posterior (or conditional) probability is of G_D is (1/6)/[(1/6)+(1/6)+(1/6)]=1/3.

JeffJo 2 May 2026 17:55 UTC
2 points
0
in reply to: Steff’s comment on: Just Use Bayes: Sleeping Beauty and Monty Hall
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, that a coin flip result is Heads?” (For comparison, say I draw a card at random from a standard 52-card deck, and then tell you that it is a spade. The probability, that I drew the Ace of Spades, is ¹⁄₁₃. Not ¹⁄₅₂, the probability when I drew it.)

Then, and this will become important later, the question Adam Elga asked is equivalent to, but not identical to, the one he answered. In the one he asked, SB is woken once, or twice, based on a coin flip. In the one he answered, SB is always woken on Monday, and then again on Tuesday, if the coin result was Tails.

But in the one he asked, he added two details that presaged his solution. Neither detail has any impact on what was asked. The first is that he mentioned that the experiment occurs over two days. But days are irrelevant in what was asked. The only significance of these two days are in his solution. And he asked “when you are first awakened...”. His solution supposed that SB could be told that it is Monday, or that the coin landed Tails, and evaluated that probability after this information was revealed. The point of “when you are first awakened...” was simply to evaluate the probability before either revelation.

+++++

My point is that these timing details are completely irrelevant, and are used to confuse the probability by establishing different kinds of conditions for each waking. A trivial solution can be obtained by making these conditions equivalent for each. Here is one way:
1. The experiment proceeds over three days, not two. Monday, Tuesday, and Wednesday.
2. Only one set of conditions produce a waking on each day.
3. Replace the coin with a six-sided die. Three faces will be green, representing “Heads,” and three will be red, representing “Tails.”
4. The words “Monday,” “Tuesday” and “Wednesday” are each written on two sides, once in each color.
5. The die is rolled on Sunday Night, and preserved throughout the experiment. If it [landed on/is currently showing] a green side, SB is woken only on the day indicated on that side. If red, SB is woken twice, on the two days not indicated on that side.
6. SB is asked for the probability that the die [landed on/is showing] a green side.
BY HALFER LOGIC, the probability that the die “landed on” each of the six sides is ¹⁄₆. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green.

“New information” does not mean learning something happened that was not guaranteed to happen. It means that the information eliminates at least one result that was possible without the information. My version does this transparently, by ruling out three die results.

In the classic version (the one in Elga’s solution), the information is that the combination of coin and day allows SB to be awake. Before that information was obtained, Heads+Tuesday is possible. By being awake, SB can eliminate one of the four equally-likely possibilities, leaving three. The obfuscation in that version happens because the two combinations that can occur on Monday are treated as equivalent to only one that can occur on Tuesday.

JeffJo 25 Mar 2026 22:40 UTC
1 point
0
on: The Solution to Sleeping Beauty
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 observe it. All she needs to know is that she can exist in four different states. When she is in, say, state T&Mon, she doesn’t know that. But she does know that she is in only one state, and that there are others that she is not experiencing. Four, in all.
This is where the definition of “prior” comes in. Yes, she knows that she cannot be awake in state H&Tue, but that is evidence that results from that particular outcome. It is not considered when constructing the prior sample space and probability distribution. The prior sample space is {H&Mon, T&Mon, H&Tue, T&Tue}. The prior probability distribution is {1/4, ¹⁄₄, ¹⁄₄, ¹⁄₄}.
When SB is awake, she has “new information.” The does not mean something she didn’t know would happen ahead of time; in fact, it isn’t defined in probability theory at all. But it is used, informally, to mean information about a specific outcome that eliminates portions of the prior sample space. The actual defined term is of a condition, which is the event that remains possible given the evidence. Like how being awake eliminates H&Tue and leaves he condition AWAKE={H&Mon, T&Mon, T&Tue}
So Pr(H&Mon|AWAKE) = Pr(H&Mon)/Pr(Awake) = (1/4)/(3/4) = ¹⁄₃.
=====
An equivalent problem is to wake SB both days, but do something other than the interview on H&Tue. So when she is interviewed, she has the same information, defining the same condition, as in the popular version.
And the method can be demonstrated by expanding the problem. Use Monday thru Saturday, a six-sided die, and a 6x6 calendar where columns are days and rows are die results. Randomly choose one of six activities for each “date” in the calendar. When SB participates in activity A, her confidence that the die rolled a “3″ is the number of times activity A appears in row, divided by the number of times it appears in the calendar. Even if one of the activities is “don’t wake her on this day.”

JeffJo 9 Mar 2026 23:52 UTC
1 point
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on: Lessons from Failed Attempts to Model Sleeping Beauty Problem
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 answering that....
Q3A: At any point in time, what is the probability, to an outside observer, that the coins are showing HH? HT? TH? TT?
Q3B: Does SB know and agree with these probabilities?
Q3C: Does an awake SB have “new information” about this set of possibilities?
Q3D: Is the conditional probability that the coins are showing HT, the same as the conditional probability that coin C1 landed on Heads?
+-+-+-+-+
A1: Yes.
A2: Yes.
Q3A: Each has a prior probability of ¹⁄₄.
Q3B: Yes.
Q3C: Yes, the combination HH is eliminated.
Q3D: Yes.
Q3: The conditional probability of {HT}, given {HT, TH, TT} is ¹⁄₃.

JeffJo 20 Jul 2025 22:31 UTC
1 point
0
in reply to: Ape in the coat’s comment on: What’s the Deal with Logical Uncertainty?
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.

JeffJo 19 Jul 2025 19:11 UTC
1 point
0
in reply to: TAG’s comment on: What’s the Deal with Logical Uncertainty?
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?

JeffJo 19 Jul 2025 15:55 UTC
−1 points
0
in reply to: Ape in the coat’s comment on: What’s the Deal with Logical Uncertainty?
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.

JeffJo 12 Jul 2025 19:44 UTC
1 point
0
in reply to: JeffJo’s comment on: Sleeping Beauty and the Forever Muffin
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).

JeffJo 12 Jul 2025 16:58 UTC
−1 points
0
in reply to: Ape in the coat’s comment on: What’s the Deal with Logical Uncertainty?
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.

JeffJo 12 Jul 2025 15:34 UTC
1 point
0
on: Sleeping Beauty and the Forever Muffin
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?”

JeffJo 8 Jan 2025 19:48 UTC
2 points
1
on: Practicing Bayesian Epistemology with “Two Boys” Probability Puzzles
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.

JeffJo 10 Oct 2024 17:30 UTC
−1 points
−4
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.