All of JeffJo's Comments + Replies

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)

1Ape in the coat9d
I understand that they are different, that's the whole point. They are different in such a way that we can agree that the answer to my problem is clearly 1/2, while we can't agree to the answer to your problem.  But none of their differences actually affect the mathematical argument you have constructed, so the way you arrive to an answer 1/3 in your problem, would arrive to the same answer in mine. What amnesia does in Sleeping Beauty is ensuring that the Beauty can't order the outcomes. So when she is awaken she doesn't know whether it's the first awakening or the second. She is unable to observe the event "I've been awaken twice in this experiment". The similar effect is achieved by the fact that she is given a random ball from the box. She doesn't know whether it's the first ball or the second. And she can't directly observe whether there are two balls in the box or only one. Which is completely irrelevant to your mathematical argument about four equiprobable states because you've constructed it in such a manner, that the same probabilities are assigned to all of them regardless of whether the Beauty is awake or not. All your argument is based on "There are four equiprobable states and one of them is incompatible with the observations", it is not dependent on the number of observations. Now, there is a different argument that you could've constructed that would take advantage of two awakening. You could've said that when the first coin is Tails there are twice as many awakenings as when it's Heads and claim that we should interpret it as P(Awake|T1)=2P(Awake|H1) but it's very much not the argument you were talking about. In a couple of days, in my next post I'm explicitly exploring both of them. This is wrong. And it's very easy to check. You may simulate your experiment, a large number of times, writing down the states of the coins on every awakening and notice that there is a clear way to predict the next token beter than chance:  if i-th token == TH and

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)

2Ape in the coat11d
You keep repeating the same points and they are all based on faulty assumptions. Which you would have already seen if you properly evaluated my example with balls in a box. Let me explicitly do it for you: Two coins are tossed. Then if it's not Heads Heads one ball is put into a box. Then the second coin is turned to the other side and again if it's not Heads Heads a ball is put into a box. After this procedure is done, you are given a random ball from the box. What is the probability that the first coin is Heads after you've got the ball?  The correct answer here is unambiguosly 1/2, which we can check by running the experiment multiple times. On every iteration you get only one ball and on 1/2 of them the first coin is Heads. Getting a ball is not evidence in favor of anything because you get it regardless of the outcome of the coin toss. But if we reason about this problem the same way you try to reason about Sleeping Beauty we inevitably arrive to the conclusion that it has to be 1/3. After all, there are four equiprobable possible states {HH, TT, HT, TH}. The ball you've just got couldn't be put in the box on HH, so we have to update to three equiprobable states {HT, TH, TT} and the only one of them where the first coin is Heads is HT. P(HT) = 1/3. This show that such reasoning method can't generally produce correct answers. So when you applied it to Two-Coin-Toss version of Sleeping Beauty you didn't actually show that 1/3 is the correct answer.

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
... (read more)
1JeffJo11d
There are several, valid solutions that do not always introduce the details that are misinterpreted as ambiguities. The two-coin version is one, which says the answer is 1/3. Here's another, that I think also proves the answer is 1/3, but I'm sure halfers will disagree with that. But it does prove that 1/2 can't be right. * Instead of leaving SB asleep on Tuesday, after Heads, we wake her but do not interview her. We do something entirely different, like take her on a $5000 shopping spree on Rodeo Drive. (She can get maybe one nice dress.) This way, when she is first wakened - which can only mean before she learns if it is for an interview or a shopping spree, since she can't know about any prior/subsequent waking - she is certain that the probability of Heads and Tails are each 50%. But when she is interviewed, she knows that something that only happens after a Heads has been "eliminated." So the probability of Heads must be reduced, and the probability of Tails must be increased. I think that she must add "Heads and it is Tuesday" to the sample space Elga used, and each observation has a probability of 25%. Which makes the conditional probability of Heads, given that she is interviewed, 1/3. BUT IT DOES NOT MATTER WHAT HAPPENS ON "HEADS AND IT IS TUESDAY." The "ambiguity" is created by ignoring that "HEADS and it is Tuesday" happens even if SB sleeps through it. OR, we could use four volunteers but only one coin. Let each on sleep through a different combination of "COIN and it is DAY." Ask each for the probability that the coin landed on the side where she might sleep through a day. On each day, three will be wakened. For two of them, the coin landed on the side that means waking twice. For one, it is the side for waking once. All three will be brought into a room where they can discuss the answer, but not share their combination. Each has the same information that defines the correct answer. Each must give the same answer, but only one matches the conditi

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?
1Ape in the coat13d
Your Two Coin Toss version is isomorphic to classical Sleeping Beauty problem with everything this entails.  The problem Elga solved in his paper isn't actually Sleeping Beauty problem - more on it in my next post. Likewise, the solution you propose to your Two Coin Toss problem is actually solving a different problem: Here your reasoning is correct. There are four equiprobable possible outcomes and awakening illiminates one of them. Person who participates in the experiment couldn't be certain to experience an awakening and that's why it is evidence in favor of Tails. 1/3 is unambiguously correct answer. But in Two Coin Toss version of Sleeping Beauty this logic doesn't apply. It would proove too much. And to see why it's the case, you may investigate my example with balls being put in the box, instead of awakenings and memory erasure.

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
... (read more)
1Radford Neal14d
Interesting.  I hadn't heard of the Child Born on Tuesday Problem.  I think it's actually quite relevant to Sleeping Beauty, but I won't go into that here... Both problems (your 1 and 2) aren't well-defined, however. The problem is that in real life we do not magically acquire knowledge that the world is in some subset of states, with the single exception of the state of our direct sense perceptions. One could decide to assume a uniform distribution over possible ways in which the information we are supposedly given actually arrives by way of sense perceptions, but uniform distributions are rather arbitrary (and will often depend on arbitrary aspects of how the problem is formulated). Here's a boys/girls puzzle I came up with to illustrate the issue:  The symmetrical summaries of what is learned are intentionally misleading (it's supposed to be a puzzle, after all).  The way in which you learned they have at least one girl is not the same as the way you learned that they have at least one boy. And that matters.

"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.

1Radford Neal18d
You may think the difference between "the card is an Ace" and "JeffJo says the card is an Ace" is just a quibble.  But this is actually a very common source of error.   Consider the infamous "Linda" problem, in which researchers claim that most people are irrational because they think "Linda is a bank teller" is less likely than "Linda is a bank teller and active in the feminist movement".  When you think most people are this blatantly wrong, you maybe need to consider that you might be the one who's confused...

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.

1Ape in the coat20d
I mean that there are no sleeping or awakenings, instead there are balls in a box that follow the same logic: Two coins are tossed, if both are Heads, nothing happens, otherwise a ball is put into a box. Then the second coin is placed the other side and once again, the ball is placed into the box unless both coins are Heads. Then you are randomly given a ball from the box. Should you reason that there is another ball in a box with probability 2/3? After all, there are four equiprobable combinations: HH, TT, HT, TH. Since the ball, you were given, was put into the box, it couldn't happen when the outcome was HH, so we are left with HT, TH and TT.

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.

2Radford Neal19d
Actually, there is no answer to the problem as stated. The reason is that the evidence I (who drew the card) have is not "the card is an Ace", but rather "JeffJo said the card is an Ace". Even if I believe that JeffJo never lies, this is not enough to produce a probability for the card being the Ace of Spades. 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. However, I agree that a "reward structure" is not required, unless possible rewards are somehow related to my beliefs about what JeffJo might do. For example, I can assess my probability that the store down the street has ice cream sundaes for sale when I want one, and decide that the probability is 3/4. If I then change my mind and decide that I don't want an ice cream sundae after all, that should not change my probability that one is available.
2Said Achmiz20d
If there is no reward structure, then neither answer is meaningfully more “correct” than the other. Beliefs are for actions.

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)

1Ape in the coat20d
Suppose we have the same two coin setting but instead of steps 1, 2, 3 a ball is put into the box. Then, after the procedure is done and there are either one or two balls in the box, you are to be given random balls from it as long as there any. You've just gotten a random ball. Should you, by the same logic, assume that probability to get a second ball is 2/3?
1Ape in the coat20d
So, could you answer the initial question? Were you always a thirder? Or is this two coin version of Sleeping Beauty what changed your mind to become one? Would you change your mind if the two coin case was found to be flawed?

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)

2Ben20d
Echoing Said's comment, what does it mean to be "correct" in this context? If we ask Beauty to pick between heads or tails, and she picks heads, then sometimes this will be correct, and sometimes not. In order for Beauty to give a (correct) probabilistic answer to the question (1/3 or 1/2) we need to introduce the idea of some proportion of trials. We need to at least imagine running the situation many times, and talk about some proportion of those imagined repeats. These imagined trials don't need to actually happen, they are imaginary. But they are an indispensable fiction. Now, we imagine 100 repeats. 50 heads, 50 tails. Beauty is awoken a total 150 times. For 100 awakenings it was a head that was flicked, for 50 awakenings a tail. >For 1/3rd of the awakenings the coin was tails. For 1/2 of the trials the coin was tails. I don't think anyone (halfer or thirder) disputes the line directly above (with the >). There is agreement on what proportion of awakenings tails was tossed, and on what proportion of trials a tails was tossed. We can all see that one of the two proportions is 1/3 and the other is 1/2. Which of the two proportions is picked out by the word "probability" is the entire argument. The rewards structure @Said Achmiz is talking about is a nice way of making people either aim to be right in as many guesses as possible or in as many trials as possible, which demand different strategies. 
4Said Achmiz20d
Correct answer depends on the reward structure. Absent a reward structure, there is no such thing as a correct answer. See this post. In your card-drawing scenario, there is only one plausible reward structure (reward given for each correct answer). In the Sleeping Beauty problem, there are two plausible reward structures. Of those two reward structures, one results in the correct answer being one-third, the other results in the correct answer being one-half.

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

... (read more)

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)

1Ape in the coat7mo
No lies are necessary. I didn't have to tell her beforehand that she will be asked any questions at all. Or I could have told her that she will be asked on Monday and Tuesday (if she is awake) without knowing which day it is and then she will be asked on Wednesday, knowing that it is indeed Wednesday. None of it changes the experiment. And even if I didn't ask her about her credence that the coin is Heads on Wednesday, she still has to have some probability estimate doesn't she? The point of the Wednesday question is to highlight, that, what you mean by "credence", isn't actually a probability estimate that the coin is Heads. What you are talking about, is the probability that the coin is Heads, weighted by the number of times this question is asked. Which is a meaningful category. But confusing it with the probability that the coin is Heads can be extremely misleading 

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.

1Ape in the coat7mo
  Well, when you explicitly acknowledge that "2/3 credence for Tails" doesn't correspond to Tails being the result of the coin toss in 2/3 of experiments where the beauty is thinking this way, we have cleared the possible misunderstanding. You are talking about the motte not the bailey. We can still argue about whose definitions are better, and I'd like to make a few points about it. But essentially the crux is resolved. In such mind experiments the possibility of lies is usually stated explicitly. Like in the riddle of three idols. And people do not usually claim that if the person trusted the lying idol their belief is "correct" and it's not their fault they got lied to. When the possibility of a lie wasn't explicitly stated but the lie happened, then your point is valid. We can say that the credence corresponded to a different experiment setting, thus in this sense can still be correct, even if it's not correct in the new experimental setting.  However, this is absolutely not the case here. No one is lying to the beauty. The experiment is going exactly is stated. So if beauty's knowledge doesn't represent the state of the system, it means she made some incorrect inference from her evidence which is her failure as a rational agent. When you declare that map representing the territory is irrelevant it's not clear how "correct" is a meaningful term anymore. I suppose you mean that it corresponds to your mathematical model. This is true but you can always find some mathematical model that your answer corresponds to. The interesting question is whether this model corresponds to reality.

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)

1Ape in the coat7mo
  Your knowledge about the state of the system can either correctly represent the state of the system or not. If you believe that the probability that the coin is Tails is 2/3, while on a repeated experiment about 50% times the coin is Heads that means that your probability estimate is not correct. Of course you can declare that your credence is answering some other question and this is fine. But there is a potential for misunderstanding - people can mistakenly think that they can actually guess the result of the coin toss better than chance as if they received some relevant evidence.  I explore this dynamics in my recent post.

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)

2Dagon7mo
Much of the debate gets obscured in the question of outside (of SB's perception) information.  That's really the purpose of the thought experiment - if the memory wipe is complete and SB has literally no way of knowing whether it's monday or tuesday, why are they considered (by her) to be different experiences?  For purposes of sampling, there is only one waking, the one she's experiencing. Alternate formulations (especially frequentist models where there is some "truth" to probability, as opposed to it being purely subjective) can be different, but only by introducing some distinction of experience to make them into two wakings, somehow. My view is that to an outside observer, the probability is 1/2 before the flop, 1 or 0 after the flip.  To Beauty, it remains 1/2 until she receives evidence, in the form of something she can observe that would be different with heads than with tails.  She should WAGER 1/3 in some formulations (for instance, if the wagers for Monday and Tuesday are summed, rather than being one evaluation). I'm sure you'd agree that, if the room has a calendar, Beauty should assign 1/2 on Sunday and Monday, and 0 on Tuesday.  Never 1/3, right?
Answer by JeffJoJul 17, 202330

I can't tell you whether it is considered to be sound. In my opinion, it is. But I do know that the issues causing the controversy to continue for 23 years are not a part of the actual problem. And so are unnecessary.

If anyone thinks that sounds odd, they should go back and read the question that Elga posed. It does not mention Monday, Tuesday, or that a Tails-only waking follows a mandatory waking. Those were elements he introduced into the problem for his thirder solution.

In the problem as posed, the subject (I'll call her SB, even though in the posed pr... (read more)

1Ape in the coat7mo
I originally upvoted your answer as it presented an interesting version of the SB which I didn't see before. However, it has similar problems to the original and it's even more convoluted, you may notice that there is no need to throw the second coin other than to confuse everyone. You can just put it tails and then turn over to get the exact same results. HH, HT, TH and TT are not four elementary outcomes of the experiment, as there is causal connection between them, even if such formulation makes it less obvious and harder to talk about it. 

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. 

1[anonymous]3y
She gets paid once because that's how I choose to demonstrate my point, supported by your reply, that arguments concerning the "correctness" of halving/thirding are impotent.

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)

4dadadarren3y
I fail to see how this variation is going to settle the debate. Thirders will agree with your solution but halfers would disagree with it the same way as in the original sleeping beauty problem. Halfers will ask why should beauty regard the four outcomes (HH, HT, TH, TT) equal probable? Yes they are equal probables if this is a simple tossing of two coins. Yet the experiment is far from that simple: my awakenings depend on it, the dime is being manipulated half way, my memory is erased after the first awakening....Halfers will say HH is never in the sample space to begin with, and there is no good reason to believe HT, TH and TT are equal probables. Beauty should just examine the information she has once waken up. She knew that she would definitely find herself awake in the experiment since the dime is manipulated, so right now being awake gives no new information about the Quarter, the probability ought to remain at 1/2. The same old dispute as in the original sleeping beauty. Without trying to figure out the correct way to interpret today or this awakening the debate is not going to be settled. Some halfers (SSA) think this awakening shall be interpreted as a random awakening. Some thirders (SIA) think this awakening should be regarded as a random sample from all potential awakenings (thus being actually awake gives new information as the case of your argument). There are others (FNC) who think we should ignore indexicals such as today or this awakening all together but consider all objective information available. And I'm suggesting treating indexicals like fundamentals: they are primitively understood from the first-person perspective and irreducible. These are all attempts to solve the anthropic mystery. I don't think this debate can magically go away just by using a different experiment setup.
1[anonymous]3y
Let's say Beauty is paid (once, at the end) if & only if she guesses the Quarter correctly on every wake-up.  The reward will be £111 if she correctly guessed Heads, £100 if she correctly guessed Tails.  So she should guess Heads. You would still say that her credence for Heads was 13, but you'd argue that she adapts her bet to take account of the experimental protocol, betting in defiance of her greater credence for Tails.  Right? Now if the single non-HH Monday morning wake-up was replaced with some undisclosed number of wake-ups, I imagine you'd say that her credence was undefined.  Yet she's still able to take the bet, absent of any credence.  How is this? The answer is that she does not need to resort to credence in making the decision.  So it is vacuous to argue that Halfing or Thirding is the "correct" approach.

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)

1dadadarren5y
Hello Jeff. Good seeing you again. Happy new year. A typical thirder argument follows the Self-Indication Assumption. That this awakening should be regarded as being randomly selected from one of the three possible awakenings. One awakening for Monday-Heads, one for Monday-Tails and another for Tuesday-Tails. The way I understand what your experiment does is, very cleverly, using actually existing participants to represent these possibilities. I.e. today, out of the three awaking participants two of witch would be awake on both days and one would be awake on one day only. So my argument against you is the same as my argument against SIA (or SSA as a matter of fact): in the same logical framework it is wrong to use terms such as [today] or [myself] as self-explanatory concepts while also treat them as in the same reference class of all days and all people. Because [now] or [I] is only meaningful if reasoning from a first-person perspective for they are inherently unique to oneself. If they are treated as unique then it is logically inconsistent to treat them as ordinary time or people. Even though they might be ordinary by objective measures, i.e. in the same reference class as other time and people from a third-person perspective/uncentered reasoning. More specifically applying to your argument, I disagree with assertion C. Just because there are three people awake today does not mean I should regard myself as randomly selected among those three or as if today is randomly selected among Monday or Tuesday. In this question Monday and Tuesday are not indexical. They are defined by a calendar which could be interpreted as defined by the relative positions of planetary bodies. In this sense the dates are defined by objective events. They can be treated as ordinary compare to one another from a third-person perspective. What is indexical is rather the concept of [today]. Which require a perspective center (first-person) to define. My argument is that first-person and t

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)

1Chris_Leong6y
Wait for my next post on this topic. Unfortunately, I chose a narrow scope for this post (only explaining the halfer response to the specific Beauty and the Prince objection, not justifying this approach in general) and everyone is posting objections that would require a whole post to answer. But basically, I will argue that there are valid reasons for adopting this formalism that aren't merely trivial.

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)

1Chris_Leong6y
"Determining whether "today counts" still depends on it" - No, you just ask about the (second) coin which determines what day counts and whether it shows heads or tails (for consistency assume that we flip a heads-only coin if the first coin comes up heads). So questions becomes, "What is the chance of the (first) coin being heads given Sleeping Beauty's non-indexical state of knowledge on Monday if the second coin is heads or Sleeping Beauty's non-indexical state of knowledge on Tuesday if the second coin is tails?"

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)

0Zaq11y
You can have a credence of 1/2 for heads in the absence of which-day knowledge, but for consistency you will also need P(Heads | Monday) = 2/3 and P(Monday) = 3/4. Neither of these match frequentist notions unless you count each awakening after a Tails result as half a result (in which case they both match frequentist notions).

The random process a frequentist should repeat is flipping a random biased coin, and getting a random bias b and either heads or tails. You are assuming it is flipping the same* biased coin with fixed bias B, and getting heads or tails.

The probability a random biased coins lands heads is 1/2, from either point of view. And for nshepperd, the point is that a Frequentist doesn't need to know what the bias is. As long as we can't assume it is different for b1 and 1-b1, when you integrate over the unknown distribution (yes, you can do that in this case) the answer is 1/2.

My first post, so be gentle. :)

I disagree that there is a difference between "Bayesian" and "Frequentist;" or at least, that it has anything to do with what is mentioned in this article. The field of Probability has the unfortunate property of appearing to be a very simple, well defined topic. But it actually is complex enough to be indefinable. Those labels are used by people who want to argue in favor of one definition - of the indefinable - over another. The only difference I see is where they fail to completely address a problem.

Tak... (read more)

3[anonymous]12y
I can't speak for the rest of your post, but is pretty clearly wrong. (In fact, it looks a lot like you're establishing a prior distribution, and that's uniquely a Bayesian feature.) The probability of an event (the result of the flip is surely an event, though I can't tell if you're claiming to the contrary or not) to a frequentist is the limit of the proportion of times the event occurred in independent trials as the number of trials tends to infinity. The probability the coin landed on heads is the one thing in the problem statement that can't be 1/2, because we know that the coin is biased. Your calculation above seems mostly ad hoc, as is your introduction of additional random variables elsewhere. However, I'm not a statistician.