All of JeffJo's Comments + Replies

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)

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 suffi... (read more)

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."

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 a... (read more)

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... (read more)

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 s... (read more)

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 wak

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.

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.

SPECIFICAL... (read more)

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... (read more)

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 permissib

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... (read more)

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?

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

"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.

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.

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.

I skipped answering the initial question because I've always been a thirder. I'm just trying to comment on the reasons people have given. Mostly how many will try to use fuzzy logic - like "isn't the question just asking about the coin flip?" in order to make the answer that they find intuitive sound more reasonable. I find that people will tend to either not change their answer because they don't want to back down from their intuition, or oscillate back and forth, without recalling why they picked an answer a few weeks later. Many of those will end up with "it depends on what you think the question is."

I try to avoid any discussion of repeated betting, because of the issues you raise. Doing so addresses the unorthodox part of an unorthodox problem, and so can be used to get either solution you prefer.

But that unorthodox part is unnecessary. In my comment to pathos_bot, I pointed out that there are significant differences between the problem as Elga posed it, and the problem as it is used in the controversy. It the posed problem, the probability question is asked before you are put to sleep, and there is no Monday/Tuesday schedule. In his solution, Elga n... (read more)

The same problem statement does not mention Monday, Tuesday, or describe any timing difference between a "mandatory" waking and an "optional" one. (There is another element that is missing, that I will defer talking about until I finish this thought.) It just says you will be wakened once or twice. Elga added these elements as part of his solution. They are not part of the problem he asked us to solve.

But that solution added more than just the schedule of wakings. After you are "first awakened," what would change if you are told that the day is Monday? Or ... (read more)

Say I ask you to draw a card and then, without looking at it, show it to me. I tell you that it is an Ace, and ask you for the probability that you drew the Ace of Spades. Is the answer 1/52, 1/4, or (as you claim about the SB problem) ambiguous?

I think it is clear that I wanted the conditional probability, given the information you have received. Otherwise, what was the point of asking after giving the information?

The "true" halfer position is not that ambiguity; it is that the information SB has received is null, so the conditional probability is the sam... (read more)

The need for distinguishing between SIA and SSA is not needed in the Sleeping Beauty Problem. It was inserted into Adam Elga's problem when he changed it from the one he posed, to the one he solved. I agree that they should have the same answer, which may help in choosing SIA or SSA, but it is not needed. This is what he posed:

"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, ... (read more)

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 wak

The "need to throw the second coin" is to make the circumstances underlying any awakening the same. Using a random method is absolutely necessary, although it doesn't have to be flipped. The director could say that she is choosing her favorite coin face. As long as SB has no reason to think that is more likely to be one result than the other, it works. The reason Elga's version is debated, is because it essentially flips Tails first for the second coin.

What the coins are showing at the moment are elementary outcomes of the experiment-within-the-experiment.... (read more)

AINC: "What's Beauty credence for Heads when she wakes on Wednesday and doesn't remember any of her awakenings on Monday/Tuesday?"

If she has no reason to think this is not one of her awakenings on Monday/Tuesday, then her credence is the same as it would have been then: 1/2 if she is a halfer, and 1/3 if she is a thirder.

AINC: "If it's 1/2 what is the reason for the change from 1/3?"

The only way it could change from 1/3 to 1/2, is if she is a thirder and you tell her that it is Wednesday. And the reason it changes is that you changes the state of her infor... (read more)

My point is that SB must have reason to think that she exists in the "Monday or Tuesday" waking schedule, for her to assign a credence to Heads based on that schedule. If she is awake, but has any reason to think she is not in that situation, her credence must take that into account.

You told her that she would be asked for her credence on Monday and maybe on Tuesday. "What's Beauty credence for Heads when she wakes on Wednesday and doesn't remember any of her awakenings on Monday/Tuesday?" is irrelevant because you are allowing for the case where that is n... (read more)

Whether your knowledge correctly represents the state of the system, or not, is irrelevant. Your credence is based on your knowledge. If they lie to SB and wake her twice after Heads, and three times after Tails? But still tell here it is once or twice? A thirder's credence should still be 1/3.

And those who guess would not be considered the rational probability agent that some versions insist SB must be.

The correct answer is 1/3. See my answer to this question.

Ape in the coat: “What's Beauty credence for Heads when she wakes on Wednesday and doesn't remember any of her awakenings on Monday/Tuesday?”

The snarky answer is “irrelevant.” The assessment of her credence, that she gives on Monday or Tuesday, is based on the information that it is either Monday or Tuesday within the waking schedule described in the experiment. She is not responsible for incorporating false information into her credence.

One thing that seems to get lost in many threads about this problem, is that probability and/or credence is a function o... (read more)

No, much of the debate gets obscured by trying to ignore how SB's information, while it includes all of the information that will be used it separate the experiment into two observations, also is limited to the "inside information" and she is in just one of those parts. That's the purpose of the shopping spree I suggested. Correcting what you wrote, if the memory wipe is complete and SB has literally no way of knowin what information might apply to what is clearly a distinct part of the experiment, but she knows there is a different part than the current o... (read more)

The "valid answers for different questions" position is a red herring, and I'll show why. But first, here is an unorthodox solution. I don't want to get into defending whether its logic is valid; I'm just laying some groundwork.

On Sunday (in Elga's re-framing of the problem he posed - see my answer), the sample space for the experiment seems to be {T,H} with equal probability for each outcome. Both Monday and Tuesday will happen, and so cannot be used to specify outcomes. This is one of your "different questions."

But when SB is awake, only one of those day... (read more)

Betting arguments - including the "expected value of the lottery ticket" I saw when skimming this - are invalid since it is unclear whether there is exactly one collection opportunity, or the possibility of two. You can always get the answer you prefer by rearranging the problem to the one that gets the answer you want.

But the problem is always stated incorrectly. The original problem, as stated by Adam Elga in the 2000 paper "Self-locating belief and the Sleeping Beauty problem," was:

"Some researchers are going to put you to sleep. During [the experiment]... (read more)

Why does she get paid only once, at the end? Why not once for each waking?

This is the problem with all betting arguments. They incorporate an answer to the anthropic question by providing one, or #wakings, payoffs. 

There is a simple way to answer the question without resorting anthropic reasoning. Then you can try to make the anthropic reasoning fit the (correct) answer.

Flip two coins on Sunday Night, a Dime and a Quarter. Lock them in a glass box showing the results. On Monday morning, perform this procedure: Look at the two coins. If either is showing Tails, wake SB, interview her, and put her back to sleep with the amnesia drug. If both are showing Heads, leave her asleep.

On Monday night, open the box, turn the Dime over, and re-lock it. On Tuesday morning, repeat... (read more)

Imagine an only slightly different problem: You volunteer for an experiment where you may be wakened once, or twice, on Monday and/or Tuesday. The administrators of the experiment will flip a coin to decide whether it will be once, or twice; but they do not tell you what coin result determines that the number of wakings. And they do not tell you whether the day you will be left asleep, if only one waking is to occur, will be Monday or Tuesday. Oh, and you will be given the amnesia drug on Monday, if you are wakened on that day.

Whenever you are awake, you w... (read more)

And in my reply I will show how you are addressing the conclusion you want to reach, and not the problem itself. No matter how you convolute choosing the sample point you ignore, you will still be ignoring one. All you will be doing, is creating a complicated algorithm for picking a day that "doesn't count," and it will be probabilisticly equivalent to saying "Tuesday doesn't count" (since you already ignore Tue-H). That isn't the Sleeping Beauty Problem.

But you haven't responded to my proof, which actually does eli... (read more)

If an interview on one day "counts," while an interview on another day doesn't, you are using an indexical to discriminate those days. Adding another coin to help pick which day does not count is just obfuscating how you indexed it.

This is why betting (or frequency) arguments will never work. Essentially, the number of bets (or the number of trials in the frequency experiment) is dependent on the answer, so the argument is circular. If you decide ahead of time that you want to get 1/3, you will use three bets (or "trials") that eac... (read more)

I agree that we need to remove any dependence on the indexical "today," but what you propose doesn't do that. Determining whether "today counts" still depends on it. But there is a way to unequivocally remove this dependence. Use four volunteers, and wake each either once or twice as in the original Sleeping Beauty Problem (OSBP). But change the day and/or coin result that determines the circumstances where each is left asleep.

So one volunteer (call her SB1) will be left asleep on Tuesday, if Heads is flipped, as in the OSBP. Anoth... (read more)

The problem with the Sleeping Beauty Problem, is that probability can be thought of as a rate: #successes per #trials. But this problem makes #trials a function of #successes, introducing what could be called a feedback loop into this rate calculation, and fracturing our concepts of what the terms mean. All of the analyses I've seen struggle to put these fractured meanings back together, without fully acknowledging that they are broken. MrMind comes closer to acknowledging it than most, when he says "'A fair coin will be tossed,' in this context, will... (read more)

these situations are mutually exclusive, indistinguishable and exhaustive.

No, they aren't. "Indistinguishable" in that definition does not mean "can't tell them apart." It means that the cases arise through equivalent processes. That's why the PoI applies to things like dice, whether or not what is printed on each side is visually distinguishable from other sides.

To make your cases equivalent, so that the PoI applies to them, you need to flip the second coin after the first lands on Heads also. But you wake SB at 8:00 regardless of ... (read more)

Coscott’s original problem is unsolvable by standard means because the expected number of wakings is infinite, so you can’t determine a frequency. That doesn’t mean it is unanswerable – we just need an isomorphism. After informing SB of the procedure and putting her to sleep the first time:

1) Set M=0. 2) Select a number N (I’ll discuss how later). 3) Flip a coin. a. If this (odd-numbered) flip lands heads, wake SB N times and end the experiment. b. If this flip (odd-numbered) lands tails, continue to step 4. 4) Flip the coin again. a. ... (read more)

Since this discussion was reopened, I've spent some time - mostly while jogging - pondering and refining my stance on the points expressed. I just got around to writing them down. Since there is no other way to do it, I'll present them boldly, apologizing in advance if I seem overly harsh. There is no such intention.

1) "Accursed Frequentists" and "Self-righteous Bayesians" alike are right, and wrong. Probability is in your knowledge - or rather, the lack thereof - of what is in the environment. Specifically, it is the measure of the amb... (read more)

The problem with the Sleeping Beauty Problem (irony intended), is that it belongs more in the realm of philosophy and/or logic, than mathematics. The irony in that (double-irony intended), is that the supposed paradox is based on a fallacy of logic. So the people who perpetuate it should be best equipped to resolve it. Why they don't, or can't, I won't speculate about.

Mathematicians, Philosophers, and Logicians all recognize how information introduced into a probability problem allows one to update the probabilities based on that information. The controver... (read more)

After tinkering with a solution, and debating with myself how or whether to try it again here, I decided to post a definitive counter-argument to neq1's article as a comment. It starts with the correct probability tree, which has (at least) five outcomes, not three. But I'll use the unknown Q for one probability in it:

••••••• Monday---1---Waken; Pr(observe Heads and Monday)=Q/2 ••••••••••/ ••••••••Q •••••••/ ••• Heads •••••/••\••••••••••••1---Sleep; Pr(sleep thru Heads and Tuesday)=(1-Q)/2 ••••/•••1-Q•••••••/ ••1/2••••\••••••••/ ••/•••• Tuesday--0---Waken;... (read more)

Say a bag contains 100 unique coins that have been carefully tuned to be unfair when flipped. Each is stamped with an integer in the range 0 to 100 (50 is missing) representing its probability, in percent, of landing on heads. A single coin is withdrawn without revealing its number, and flipped. What is the probability that the result will be heads?

You are claiming that anybody who calls himself a Frequentist needs to know the number on the coin to answer this question. And that any attempt to represent the probability of drawing coin N is specifying a pri... (read more)

Sleeping Beauty does not sleep well. She has three dreams before awakening. The Ghost of Mathematicians Past warns her that there are two models of probability, and that adherents to each have little that is good to say about adherents to the other. The Ghost of Mathematicians Present shows her volumes of papers and articles where both 1/2 and 1/3 are "proven" to be the correct answer based on intuitive arguments. The Ghost of Mathematicians Future doesn't speak, but shows her how reliance on intuition alone leads to misery. Only strict adherence... (read more)