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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:
- Every probability is greater than, or equal to, zero.
- The probability of the union of two distinct outcomes A and B is Pr(A)+Pr(B).
- 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.
I don't think I understand what you've written here. It's indeed possible that the card is not Club when it's Spade. As a matter of fact, it's the only possibility, because the card can't be both Spade and a Club.
There are four different days when SB could be awakened. On three of them, she would not have been awakened if the card was a club. This makes it more likely that the card is a club. This really is very simple probability. If you have difficulty with it, wake her every day. But in the situations where she was left asleep before, wake her and ask her for the probability that the card is the Ace of Clubs.
My points are that (A) the event where SB is left asleep is still an event in SB's sample space, (B) her "new information" is that she is observing the event in her sample space that match certain conditions, not that the negation of that event is not a part of her sample space, and (C) the amnesia drug disassociates the current day from all others.
This ... doesn't affect the probability to be awaken in the experiment.
Yes, it does. We can argue back and forth about who is correct, but all you have provided is word salad to support a conclusion you reached before choosing the logic. I keep providing examples that contradict it, and your counter-argument is that it contradicts your word salad.
Try another example; and this is actually closer to Zuboff's original problem than Elga's version of it, or the version he solved which has become the canonical form. (It's problem is that the Monday:Tuesday difference obfuscates the probability.)
2N players agree to the following game, and are fully informed of all details. Before the game starts, each is put to sleep. Over the next N days, each will be wakened on either every one of the N days, or on a single, randomly-selected day in the interval, based on the result of a single coin flip that applies to all 2N players and all days. With 2 possible flip results, and N days, each player is randomly assigned a different combination of a coin result and a day.
On each day, each of the N+1 players who are wakened is asked for her probability/credence/confidence/whatever for the proposition that she will be wakened only once. After answering, each will be put back to sleep with amnesia.
In Zuboff's version, N=one trillion, and the day is random, the coin result is unspecified, but this is the question the player is asked; in other words, it does not matter that the coin result is unspecified.In the problem Elga posed, N=2 but the days are not specified, and the question is effectively the same since Heads is specified. In the problem Elga solved, N=2, the day is day #1, and Heads is specified.
Nothing about these variations change the solution methodology.
The halfer solution is that each of the N+1 players who are awake should say the probability that this is her only waking is 1/2. And that each of the other N awake payers should answer the same. This is despite the fact that she knows, for a fact, that only one of the N+1 awake players satisfies the proposition. Additionally, they can assign a probability of 1/2 to an asleep player, despite knowing that this player does satisfy the proposition.
These are not probabilities but weighted probabilities, where the measure function is re-normalized by the number of awakenings, even though the awakenings are not mutually exclusive.
These are not weighted probabilities, they are probabilities using all of the 4*52=208, equally-likely members of the prior sample space that can apply to a randomly-selected day in the experiment. They are mutually exclusive, to SB, because the amnesia drug isolates each day from the others that may be "sequential" with it, whatever that is supposed to mean to SB since she sees only one day.
Sigh. We've been through it a couple of times already.
Yes, we have, and you ignore every point I make.
The prior sample space depends on which observations is possible to make in the experiment at all, according to the prior knowledge state of a particular person.
A prior sample space depends ONLY on how the circumstances occur. Observation has nothing to do with it.
On HEADS+TUESDAY, wake SB but take her on a shopping spree instead of interviewing her. When interviewed, should she think that Pr(H)=1/2, or Pr(H)=1/3? Of course it is 1/3, because her set of possible observations includes four (not two) mutually exclusive outcomes, with one ruled out by observation. My point is that she knows she is in NOT(H&Tue) regardless of what observation is possible in H&Tue.
{H&Mon, T&Mon, H&Tue T&Tue} is a sample space for observer problem, not for SB. For a researcher working on a random day, observing outcome T&Tue is a priori possible, for the Beauty participating on every day, it's not.
{H&Mon, T&Mon, H&Tue T&Tue} is the set of possible circumstances for a single day in the experiment. The "observer problem" you describe includes only two outcomes, Heads and Tails. Both Monday and Tuesday exist in each observation, since they are not isolated by amnesia. What you listed is the sample space when the random experiment can include only one day - AMNESIA!! H&Tue is in SB's prior = unaffected by observation sample space.
My generalized version of Zuboff's problem has 2N members in its sample space. When a player is awake, N+1 of these members are consistent with observation. One matches the proposition that the player will be wakened once. Making the probability of the proposition 1/(N+1).
In the canonical SB problem, N=2 and the probability is 1/(N+1)=1/3.
Here's a new problem that requires the same solution methodology as Sleeping Beauty.
It uses the same sleep and amnesia drugs. After SB is put to sleep on Sunday Night, a card is drawn at random from a standard deck of 52 playing cards.
On Monday, SB is awakened, interviewed, and put back to sleep with amnesia.
On Tuesday, if the card is a Spade, a Heart, or a Diamond - but not if it is a Club - SB is awakened, interviewed, and put back to sleep with amnesia.
On Wednesday, if the card is a Spade or a Heart - but not if it is a Diamond or a Club - SB is awakened, interviewed, and put back to sleep with amnesia.
On Thursday, only if the card is a Spade, SB is awakened, interviewed, and put back to sleep with amnesia.
On Friday, SB is awakened and the experiment ends.
In each interview, SB is asked for the probability that the drawn card is the Ace of Spades.
By Halfer logic, it is 1/52. In fact, the probability for each card in the deck is 1/52. Yet it is possible that the card cannot be a Club when it can be a spade, so they cannot have the same probability. This contradiction disproves the halfer logic.
The correct probabilities are:
For each Spade, [(1/52)+(1/39)+(1/26)+(1/13)]/4 = 25/624
For each Heart, [(1/52)+(1/39)+(1/26)]/4 = 1/48
For each Diamond, [(1/52)+(1/39)]/4 = 7/624
For each Club, (1/52)/4 = 1/208
The correct solution to Elga's Sleeping Beauty problem is that SB's prior sample space, for what the researchers saw this morning, is {H&Mon, T&Mon, H&Tue T&Tue}. And yes, H&Tue is a member since the prior sample space does not depend on observation. These may be "sequential," whatever than means, to the researchers. But they are independent outcomes to SB because her memory loss prevents any association of the current day with another. Each has a prior probability of 1/4.
The solution to the problem is that H&Tue is eliminated by SB's observation that she is awake. The other probabilities update to 1/3.
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 before are properties of the experiment; the process. "Outcome" is any potential result of an iteration of that process, not the result itself. We can say that a result belongs to an event, even an event of just one outcome, but the result is not the same thing as that event. THE OBSERVATION IS NOT AN EVENT.
For example, an event for a simple die roll could be EVEN={2,4,6}. If you roll a 2, that result is in this event. But it does not mean you rolled a 2, a 4, and a 6.
So, in ...
By the definition of the experimental setting, when the coin is Tails, what Beauty does on Monday - awakens - always affects what she does on Tuesday - awakens the second time. Sequential events are definitely not mutually exclusive and thus can't be elements of a sample space.
... you are describing one iteration of a process that has an unpredictable result. A coin flip. Then you observe it twice, with amnesia in between. Each observation can have its own sample space - remember, experiments do not have just one sample space. But you can't pick, say, half of the outcomes defined in one observation an half from the other, and use them to construct a sample space. That is what you describe here, by comparing what SB does on Monday, and on Tuesday, as if they are in the same event space.
The correct "effect of amnesia" is that you can't relate either observation to the other. They each need to be assessed by a sample space that applies to that observation, without reference to another.
And BTW, what she observes on Monday may belong to an event, but it is not the same thing as the event.
>That result is observed twice (yes, it is; remaining asleep is an observation of a result that we never make use of, so awareness as it occurs is irrelevant
This is false, but not crucial. We can postpone this for later.
A common way to avoid rebuttal is to cite two statements and make one ambiguous assertion about them, without support or specifying which you mean.
It is true that remaining asleep is a possible result of the experiment - that is, an outcome - since Tuesday exists whether or not SB is awake. What SB observes tells her that outcome is not consistent with her evidence. That's an observation.
It is true that same the result (the coin flip) is observed twice; once on Monday, and once on Tuesday.
Or do you want to claim the calendar flips from Monday to Wednesday? That is, that Tuesday only exists is SB is awake? But if you still doubt this, wake SB on Tuesday but don't ask her for her belief in Heads. Knowing the circumstances where you would not ask, she can then deduce that those circumstances do not exist. This is an observation.
What you do think makes it different than not waking her, since her evidence is the same when she is awake is the same?
>What you call "sequential events" are these two separate observations of the same result.
No, what I call sequential events are pairs HH and HT, TT and TH, corresponding to exact awakening, which can't be treated as individual outcomes.
No, thats how you try to misinterpret my version to fit your incorrect model. You use the term for Elga's one-coin version as well. Strawman arguments are another avoidance technique.
On the other hand, as soon as you connect these pairs and got HH_HT, HT_HH, TT_TH and TH_TT, they totally can create a sample space, which is exactly what I told you in this comment. As soon as you've switched to this sound sample space we are in agreement.
Huh? What does "connect these pairs" mean to pairs that I already connected?
You are describing a situation where the Beauty was told whether she is experiencing an awakening before the second coin was turned or not.
No, I am not. This is another strawman. I am describing how she knows that she is in either the first observation or the second. I am saying that I was able to construct a valid, and useful, sample space that applies symmetrically to both. I am saying that, since it is symmetric, it does not matter which she is in.
I only did this to allow you to include the "sequential" counterpart to each in a sample space that applies regardless of the day. The point is that "sequential" is meaningless.
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 sufficiently for the purposes of the random experiment. An experiment's sample space is a set of all distinct outcomes that are possible for that experiment; it is often represented as Ω.
There is no single way to define outcomes, and so no single sample space, for any experiment. For example, if you roll two dice, the sample space could be 36 unordered pairs of numbers, 21 ordered pairs, or eleven sums. But what "sufficient" means is that every observation you intend to make is its own outcome in the sample space. You could divide the results of rolling a single die into {Odd, Even}, but that won't be helpful if you intend to observe "Prime" and "Composite."
An event is any subset of the sample space, not a specific result. (You, Ape in the Coat, confuse it with either an observation, or an outcome; it isn't clear which.) We could also talk about the event space F and the corresponding probability space P, but those exact details are also not important here. Except that P corresponds to F, not Ω. When I talk about the probability of an outcome, I mean the solitary event containing just that outcome.
Finally, conditional probability is used when an observation divides the sample space into two sets: one where every outcome is 100% consistent with the observation - that is, no un-observed measure makes it inconsistent - and its compliment where no outcome is consistent with the observation. (This is where people go wrong in problems like Monty Hall; the outcome where Monty opens Door 3 to show a goat is not the same as the observation that there is a goat behind Door 3).
With these definitions, we can see that the SB Problem is one random experiment with a single result. That result is observed twice (yes, it is; remaining asleep is an observation of a result that we never make use of, so awareness as it occurs is irrelevant). What you call "sequential events" are these two separate observations of the same result. That's why you want to treat them as dependent, because they are the same result. Just looked at a different way.
And the effect of the amnesia drug is that any information learned in "the other" observation does not, and can not, influence SB's belief in Heads based on the current assessment. Knowing that an observation is made is not the same thing as knowing what that observation was.
Elga's sample space(s) are controversial because he uses different measures for the two observations. So how they relate to each other seems ambiguous to some. He actually did it correctly, by further conditionalizing the result. But since his answer violated many people's intuitions, they invented reasons to argue against how he got it.
My two-coin version does not have this problem. The sample space for the experiment is {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. Each outcome has probability 1/4. The first observation establishes the condition as {HT1_HH2, TH1_TT2, TT1_TH2} and its complement as {HH1_HT2}. Conditional probability says the probability of {HT1_HH2} is 1/3. The second observation establishes the condition as {HH1_HT2, TH1_TT2, TT1_TH2} and its complement as {HT1_HH2}. Conditional probability says the probability of {HH1_HT2} is 1/3.
And the point is that it does not matter which observation corresponds to SB being awake, since the answer is 1/3 regardless.
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 and B are independent events.
The outside world certainly can name the outcomes {HH1_HT2, HT1_HH2, TH1_TT2, TT1_TH2}. But the subject has knowledge of only one pass. So to her, only the current pass exists, because she has no knowledge of the other pass. What happens in that interval can play no part in her belief. The sample space is {HH, HT, TH, TT}.
To her, these four outcomes represent fully independent events, because she has no knowledge of the other pass. To her, the fact that she is awake means the event {HH} has been ruled out. It is still a part of the sample space, but is is one she knows is not happening. That's how conditional probability works; the sample space is divided into two subsets; one is consistent with the observation, and one is not.
What you are doing, is treating HH (or, in Elga's implementation, H&Tuesday) as if it ceases to exist as a valid outcome of the experiment. So HH1_HT2 has to be treated differently than TT1_TH2, since HH1_HT2 only "exists" in one pass, while TT1_TH2 "exists" in both. This is not true. Both exist in both passes, but one is unobserved in one pass.
And this really is the fallacy in any halfer argument. They treat the information in the observation as if it applies to both days. Since H&Tuesday "doesn't exist", H&Monday fully represents the Heads outcome. So to be consistent, T&Monday has to fully represent the Tails outcome. As does T&Tuesday, so they are fully equivalent.
If you are observing state TH it necessary means that either you've already observed or will observe state TT.
You are projecting the result you want onto the process. Say I roll a six-sided die tell you that the result is odd. Then I administer the amnesia drug, and tell you that I previously told you whether th result was even or odd. I then ask you for your degree of belief that the result is a six. Should you say 1/6, because as far as you know the sample space is {1,2,3,4,5,6}? Or should you say 0, because "you are [now] observing a state that you've already observed is only {1,3,5}?
And if you try to claim that this is different because you don't know what I told you, that is exactly the point of the Two Coin Version.
The definition of a sample space [was broken] - it's supposed to be constructed from mutually exclusive elementary outcomes.
It is so constructed.
I've specifically explained how. We write down outcomes when the researcher sees the Beauty awake - when they updated on the fact of Beauty's awakening.
Beauty is doing the updating, not "they." She is in an experiment where there are four possible combinations for what the coins are currently showing. She has no ability to infer/anticipate what the coins were/will be showing on another day.
Her observation is that one combination, of what is in the sample space for today, is eliminated.
No, I'm not complicating this with two lists for each day. There is only one list, which documents all the awakenings of the subject,...
Maybe you missed the part where I said you can look at one, or the other, or bot as long as you don't carry information across.
You are mistaken about what the amnesia acomplishes, Once again I send you to reread the Effects of Amnesia section.
Then you are mistaken in that section.
Her belief can be based only on what she knows. If you create a difference between the two passes, in her knowledge, then maybe you could claim a dependence. I don't think you can in this case, but to do it requires that difference.
The Two Coin Version does not have a difference. Nothing about what she observed about the outcomes HH1_HT2 or HT1_HH2 in another pass can affect her confidence concerning them in the current pass. (And please, recall that these describe the combinations that are showing.)
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, is a fully independent "world" with a mathematically valid sample space that applies to it alone.
“It treats sequential events as mutually exclusive,” No, it treats an observation of a state, when that observation bears no connection to any other, as independent of any other.
“… therefore unlawfully constructs sample space.” What law was broken?
Do you disagree that, on the morning of the observation, there were four equally likely states? Do you think the subject has some information about how the state was observed on another day? That an observer from the outside world has some impact on what is known on the inside? These are the kind of details that produce controversy in the Always-Monday-Tails-Tuesday version. I personally think the inferences about carrying information over between the two days are all invalid, but what I am trying to do is eliminate any basis for doing that.
Yes, each outcome on the first day can be paired with exactly one on the second. But without any information passing to the subject between these two days, she cannot do anything with such pairings. To her, each day is its own, completely independent probability experiment. One where "new information" means she is awakened to see only three of the four possible outcomes.
“Your model treats HH, HT, TH and TT as four individual mutually exclusive outcomes” No, it treats the current state of the coins as four mutually exclusive states.
“However, when you actually do it, you get a different list.”
How so? If your write down the state on the first day that the researchers look at the coins, you will find that {HH, TH, HT, TT} all occur with frequency 1/4. Same on the second day. If you write down the frequencies when the subject is awake, you find that {TH, HT, TT} all have frequency 1/3.
Here is what you are arguing: Say you repeat this many times and make two lists, one for each day. And each entry on the lists includes the attempt number. You will find that every attempt where the first list says TH, the second list will say TT. And that every attempt where the first list says HT, the second list will not have an entry for that attempt.
What I'm saying is that the subject's belief cannot cross-correlate the attempt numbers. She could use just the first list with the attempt numbers. Or the second, since each when viewed in isolation has the same properties as the other and so gets the same answer. She can even use a combined list, but if she does she cannot use the attempt numbers to associate the two observations on that attempt. That is what the amnesia drug accomplishes.
What I am saying is that regardless of which list she uses, the probability of heads is 1/3. And your arguments that this is wrong require associating the attempts, essentially removing the effect of amnesia. That cannot influence the subject's belief.
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 several details. I suppose he thought they would facilitate a solution that would be universally accepted. If he thought that, he was wrong. They introduced more uncertainty than acceptance. And his additions have been treated as if they are part of the problem, which is not true. There is a way to implement it without that uncertainty, but the uncertainty is too ingrained for people to let go of it.
After you are put to sleep, the researchers will flip two coins and set them aside; call then C1 and C2. On the first day of your sleep, they will examine the coins. If both are showing Heads, they will leave you asleep for the rest of the day. Otherwise, they will wake you, ask you to what degree you believe that coin C1 is showing Heads, and then put you back to sleep with that drug. Then they will turn coin C2 over to show its other side.
On the second day, they will perform the exact same steps. Well, I guess they don't need to turn coin C2 over.
This way, the details of the actual problem are implemented exactly. But since the probabilistic details that apply to either day do not change, and do not depend on what you know, or believe, on the other day, the question is easy to answer.
When the researchers examined the two coins, there were four equally likely combinations for what they might be showing: {HH. HT, TH, TT}. This is the initial sample space that applies to either day. Because of the amnesia drug, to you they are independent of the other day. But since you are awake, you know that the examination of the current state of the coins did not find HH. This is classic "new information" that allows you to eliminate one member of the sample space, and update your belief in C2=H from 1/2 to 1/3.
I would certainly accept any argument that shows how this is not a valid implementation of the problem, as stated. Or that my solution is not a valid solution for this implementation. And while this may have implications about how we should treat Elga's, or any other, solution to his implementation? They play no part in evaluating my solution.
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 waking. When you are first awakened[2], to what degree ought you believe that the outcome of the coin toss is Heads?
There are two hints of the details Elga will add, but these hints do not impact the problem as stated. At [1], Elga suggests that the two potential wakings occur on different days; all that is really important is that they happen at different times. At [2], the ambiguous "first awakened" clause is added. It could mean that SB is only asked the first time she is awakened; but that renders the controversy moot. With Elga's modifications, only asking on the first awakening is telling SB that it is Monday. He appears to mean "before we reveal some information," which is how Elga eliminates one of the three possible events he uses.
Elga's implementation of this problem was to always wake SB on Monday, and only wake her on Tuesday if the coin result was Tails. After she answers the question, Elga then reveals either that it is Monday, or that the coin landed on Tails. Elga also included DAY=Monday or DAY=Tuesday as a random variable, which creates the underlying controversy. If that is proper, the answer is 1/3. If, as Neal argues, it is indexical information, it cannot be used this way and the answer is 1/2.
So the controversy was created by Elga's implementation. And it was unnecessary. There is another implementation of the same problem that does not rely on indexicals.
Once SB is told the details of the experiment and put to sleep, we flip two coins: call them C1 and C2. Then we perform this procedure:
- If both coins are showing Heads, we end the procedure now with SB still asleep.
- Otherwise, we wake SB and ask for her degree of belief that coin C1 landed on Heads.
- After she gives an answer, we put her back to sleep with amnesia.
After these steps are concluded, whether that occurred in step 1 or step 3, we turn coin C2 over to show the opposite side. And then repeat the same procedure.
SB will thus be wakened once if coin C1 landed on Heads, and twice if Tails. Either way, she will not recall another waking. But that does no matter. She knows all of the details that apply to the current waking. In step 1, there were four possible, equally-likely combinations of (C1,C2); specifically, (H,H), (H,T), (T,H), and (T,T). But since she is awake, she knows that (H,H) was eliminated in step 1. In only one of the remaining, still equally-likely combinations did coin C1 land on Heads.
The answer is 1/3. No indexical information was used to determine this.
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.
SPECIFICALLY: SB's belief is based entirely one what happens in these steps:
- Two coins are randomly arranged so that each of the four combinations {HH, HT, TH, TT} has a 25% chance to be the outcome.
- If the random combination is HH, one option happens, that does not involve asking for a probability. Otherwise, another option happens, and it does involve asking for a probability.
- SB has full knowledge of these three steps, and knows that the second option was chosen. She can assign a probability based ENTIRELY on these three steps.
This happens twice. What you seem to ignore, is that the method used to arrange the coins is different in the first pass through these three steps, and the second. In the first, it is flipping the coins. In the second, it is a modification of this flips. But BECAUSE OF AMNESIA, this modification does no, in any way, affect SB's assessment that sample space is {HH, HT, TH, TT}, or that each has a 25% chance to be the outcome.
Her answer is unambiguously 1/3 anytime she is asked.
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 condition. That answer is 1/3.
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Yes, these all are just different ways of presenting the same information: that, in the popular version, Tuesday after Heads still happens, but cannot be observed. This is what is wrong in all the debates; they treat it as if Tuesday after Heads does not happen.
In my card example, the "prior" sample space includes all 52 cards. The probability distribution is 1/52 for each card. When I say that one is an Ace, it does not mean that it was impossible for a Seven of Clubs to have been drawn, it means that an observation was made, and in that observation the card wasn't the Seven of Clubs.
In the popular version of the SB problem, there are four possible states that can occur. Three can be observed; and because of the amnesia drug, they are all independent to SB as an observer. Regardless of whether she can know the day, it is part of the observation. Since she is awake, (Heads, Tuesday) is eliminated - as an observation, not from the experiment as a whole - and the updated probability - for this observation, not the experiment as a whole - is 1/3.
Now, you can use this problem to evaluate epistemic probability. It isn't really am epistemic problem, but I supposed you can apply it. The answer is 1/3, and the correct epistemic solution is the one that says so.
It is my contention that:
- The problem, as posed, is not ambiguous and so needs no "disambiguation."
- "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.
- 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 permissible form to a hypothetical one that should have a more obvious solution.
- Any argument that "when you are first awakened" refers to such relative timing, which is impossible for the subject to assess without impermissible information, is obfuscation with the intent to justify a solution that requires such information.
So any comment about first/last/relative awakenings is irrelevant.
Does this help? I know I can't prove that #2 is correct, but it can be. Nothing else can.
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 Mother), or which bedroom each child has. You picked a specific child in an order by looking in a specific room, so the genders of the other two are independent of it and each other. So gBB, gBG, gGB, and gGG are equiprobable at that point in your acquisition of knowledge.
But your second acquisition depends on whether similar help is needed for other sports, and how many gender-specific sports there are. And why there isn't one for girls' sports, since we know there is a girl.
My problems are well-posed for what I intended. You didn't "stumble upon" the information, a source with absolute knowledge told it to you, with no hint of any discrimination between genders. There is an established solution in such cases; it's called Bertrand's Box Paradox. That name did not, originally, refer to a problem; it referred to the following solution. It is best illustrated using a different probability than what I asked for:
- I know Mr. Abbot's two children. At least one is a boy.
- I know Mrs. Baker's two children. At least one is a girl.
- I know the Curry's two children. At least one has the gender that I have written inside this sealed envelope.
In each case, what is the probability that the family has a boy and a girl?
Clearly, the answers A1 and A2 must be the same. This is not using uniform distributions, although that is a valid justification. Probability is not about what is true in a specific instance of this disclosure of information - that's a naive mistake. It is about what we can deduce from the information alone. It is a property of our knowledge of a world where it happens, not the world itself. Since our information is equivalent in Q1 and Q2, that means A1=A2.
But you have no significant information about genders in Q3, so A3 must be 1/2. And that can be used to get A1 and A2. Bertrand argued simply that if the envelope were opened, A3 had to equal A1 and A2 regardless of what it said, so you didn't need to open it. Any change would be a paradox. But there is a more rigorous solution.
If W represents what is written in the envelope, the Law of Total Probability says:
A3 = Pr(W="Boy")*A1 + Pr(W="Girl")*A2
A3 = Pr(W="Boy")*A1 + Pr(W="Girl")*A1
A3 = [Pr(W="Boy") + Pr(W="Girl")]*A1
A3 = A1 = A2 (which all equal 1/2).
This solution is also used for the famous Monty Hall Problem, even if those using it do not realize it. The most common solution uses the assertion that "your original probability of 1/3 can't change." So, since the open door is revealed to not have the car, the closed door that you didn't pick must now have a 2/3 probability.
The assertion is equivalent to my sealed envelope. You see the door that gets opened, which equivalent to naming one gender in Q1 and Q2. Since your answer must be the same regardless of which door that is, it is the same as when you ignore which door is opened.
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.
- Do you think my version accurately implements the problem as posed?
- 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.
- I have exactly two children. At least one is a boy. What are the chances that I have two boys?
- I have exactly two children. At least one is a boy who was born on a Tuesday. What are the chances that I have two boys?
(BTW, both "exactly" and "at least" are necessary. If I had said "I have one" and asked about the possibility of two, it implies that any number I state carries an implicit "at least.")
Far too many "experts" will say that the answers are 1/3 and 13/27, respectively. Of the 4 (or 196) possible combinations of the implied information categories, there are 3 (or 27) that fit the information as specified, and of those 1 (or 13) have two boys.
Paradox: How did the added information change the probability from 1/3 to 13/27?
The resolution of this paradox is that you have to include the choice I made of what to tell you, between what most likely is two sets of equivalent information. If I have a Tuesday Boy and a Thursday Girl, couldn't I have used the girl's information in either question? Since you don't know how this choice is made, a rational belief can only be based on assuming I chose randomly.
So in 2 (or 26) of the 3 (or 27) combinations where the statement I made is true, there is another statement that is also true. And I'd only make this one in half of them. So the answers are 1/(3-2/1)=1/2 and (13-12/2)/(27-26/2)=7/14=1/2. And BTW, this is also how the Monty Hall Problem is solved correctly. That problem originated as Martin Gardner's Three Prisoners Problem, which he introduced in the same article where he explained why 1/3 is not correct for #1 above.
In my card drawing problem, there is only one card rank I can report. If you choose to add information, as done with Linda the Bank Teller, you are not a rational solver.
"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 never asked the question upon waking, and he used the Monday/Tuesday schedule to implement the problem but inadvertently created the unorthodox part.
There is a better implementation, that avoids the unorthodox part.
Before being put to sleep, you are told that two coins will be flipped after you are put to sleep, C1 and C2. And that, at any moment during the experiment, we want to know the degree to which you believe that coin C1 came up Heads. Then, if either coin is showing Tails (but not if both are showing Heads):
- You will be wakened.
- Remember what we wanted to know? Tell us your degree of belief.
- You will be put back to sleep with amnesia.
Once this is either skipped or completed, coin C2 is turned over to show its other side. And the process is repeated.
This implements Elga's problem exactly, and adds less to it than he did. But now you can consider just what has happened between looking at the coins to see if either is showing Tails, and now. When examined, there were four equiprobable combinations of the two coins: HH, HT, TH, and TT. Since you are awake, HH is eliminated. Of the three combinations that remain, C1 landed on Heads in only one.
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 that the coin landed on Tails (and you consider what day it is)? This is how Elga avoided any consideration, given his other additions, of what significance to attach to Tuesday, after Heads. That was never used in his solution, yet could be the crux of the controversy.
I have no definitive proof, but I suspect that Elga was already thinking of his solution. He included two hints to the solution: One was "two days," although days were never mentioned again, and that "when first awakened." Both apply to the solution, not the problem as posed. I think "first awakened" simply meant before you could learn information.
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You point out that, as you are trying to interpret it, SB cannot make the determination whether this is a "first awakening." But the last element that is usually included in the problem, but was not in what Elga actually asked, is that the question is posed to you before you are first put to sleep. So the issue you raise - essentially, whether the question is asked on Tuesday, after Heads - is moot. The question already exists, as you wake up. It applies to that moment, regardless of how many times you are wakened.
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 same as the prior probability. The thirder position is that there are four possible observation opportunities of the coin, all equally likely, and one has been eliminated. To see this better, always wake SB on Tuesday, but instead of asking about the coin, take her on a shopping spree if the coin landed on Heads. If she is asked the question, she knows that one observation opportunity is eliminated, and the answer is clearly 1/3.
The difference between the halfer and thirder, is that the halfer thinks that sleeping thru Tuesday removes Tuesday from the sample space of the full experiment, while the thirder sees it as something that contradicts observation.
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, 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?"
The need was created when Elga created a schedule for waking, that treated two days differently. So that "existence" became an alleged issue. There is a better way. First, consider this "little experiment":
- Two coins, C1 and C2, will be randomly arranged so that the four possible combinations, HH, HT, TH, and TT, are equally likely.
- If either coin is showing Tails, you will be asked for what you believe to be the probability that coin C1 is showing Heads.
- If both are showing Heads, you will not be asked a question.
The answer should be obvious: From the fact that you are in step 2, as established by being asked a question, the combination HH is eliminated. What happens in step 3 is irrelevant, since the question is not asked there. In only one of the three remaining combinations is coin C1 showing Heads, so there is a 1/3 chance that coin C1 is showing heads.
To implement the experiment Elga proposed - not the one he solved - put SB to sleep on Sunday night, and flip the two coins. On Monday, perform the "little experiment" using the result of the flips. You will need to wake SB if step 2 is executed. What you do in step 3 is still irrelevant, but can include leaving her asleep. Afterwards, if she is awake, put her back to sleep with amnesia. AND THEN TURN COIN C2 OVER. On Tuesday, perform the "little experiment" again, using the modified result of the flips.
SB does not need to consider any other observers than herself to answer the question, because she knows every detail of the "little experiment." If she is awake, and asked a question, the coins were arranged as described in step 1 and she is in step 2. The answer is 1/3.
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 waking.2 When you are first awakened[2], to what degree ought you believe that the outcome of the coin toss is Heads?
There are two hints of the details Elga will add, but these hints do not impact the problem as stated. At [1], Elga suggests that the two potential wakings occur on different days; all that is really important is that they happen at different times. At [2], the ambiguous "first awakened" clause is added. It could mean that SB is only asked the first time she is awakened; but that renders the controversy moot. With Elga's modifications, only asking on the first awakening is telling SB that it is Monday. He appears to mean "before we reveal some information," which is how Elga eliminates one of the three possible events he uses.
Elga's implementation of this problem was to always wake SB on Monday, and only wake her on Tuesday if the coin result was Tails. After she answers the question, Elga then reveals either that it is Monday, or that the coin landed on Tails. Elga also included DAY=Monday or DAY=Tuesday as a random variable, which creates the underlying controversy. If that is proper, the answer is 1/3. If, as Neal argues, it is indexical information, it cannot be used this way and the answer is 1/2.
So the controversy was created by Elga's implementation. And it was unnecessary. There is another implementation of the same problem that does not rely on indexicals.
Once SB is told the details of the experiment and put to sleep, we flip two coins: call then C1 and C2. Then we perform this procedure:
- If both coins are showing Heads, we end the procedure now with SB still asleep.
- Otherwise, we wake SB and ask for her degree of belief that coin C1 landed on Heads.
- After she gives an answer, we put her back to sleep with amnesia.
After these steps are concluded, whether it happened in step 1 or step 3, we turn coin C2 over to show the opposite side. And then repeat the same procedure.
SB will thus be wakened once if coin C1 landed on Heads, and twice if Tails. Either way, she will not recall another waking. But that does not matter. She knows all of the details that apply to the current waking. Going into step 1, there were four possible, equally-likely combinations of (C1,C2); specifically, (H,H), (H,T), (T,H), and (T,T). But since she is awake, she knows that (H,H) was eliminated in step 1. In only one of the remaining, still equally-likely combinations, did coin C1 land on Heads.
The answer is 1/3. No indexical information was used to determine this. No reference the other potential waking, whether it occurs before or after this one, is needed. This implements Elga's question exactly; this only possible issue that remains is if Elga's implementation does.
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. "Causal connection," whatever you think that means, has nothing to do with it since we are talking about a fixed state of the coins from the examination to the question.
Which statement here to you disagree with?
- Just before she was awakened, the possible states of the coins were {HH, HT, TH, TT} with known probabilities {1/4, 1/4, 1/4, 1/4}.
- SB knows this with certainty. That is, she did not have to be awake at that moment to know that this was true at that time.
- With no change to the coins, they were examined and the next part of the experiment depends on the actual state.
- SB knows this with certainty.
- Since she is awake, she knows that the possible change is that the state HH is eliminated.
- SB knows this with certainty.
- The probabilities in step 1 constitute prior probabilities, and her credence in the state is the same as the posterior probabilities
- Pr(HT|Awake) = Pr(HT)/[Pr(HT)+Pr(TH)+Pr(TT)] = 1/3.
But there are other ways, that have whatever "causal connection" you think is important. That is, they match the way Elga modified the experiment. That's another point you seem to ignore, that the two-day version differs from the actual question by more than mine does.
Try using four volunteers instead of one, but one coin for all. Each will be wakened on both Monday and Tuesday, except:
- SB1 will be left asleep on Tuesday, if the coin landed on Tails. This is Elga's SB, with the same "causal connection."
- SB2 will be left asleep on Tuesday, if the coin landed on Heads.
- SB3 will be left asleep on Monday, if the coin landed on Tails.
- SB4 will be left asleep on Monday, if the coin landed on Heads.
This way, on each of Monday and Tuesday, exactly three volunteers will be wakened. And unless you think the specific days or coin faces have differing qualities, the same "causal connections" exist for all.
Each volunteer will be asked for her credence that the coin landed on the result where she would be wakened only once. With the knowledge of this procedure, an awake volunteer knows that she is one of exactly three, and that their credence/probability cannot be different due to symmetry. Since the issue "the coin landed on the result where you would be wakened only once" applies to only one of these three, this credence is 1/3.
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 information, not because anything about the coin itself has changed.
But if you think her credence should be based on the actual day, even when you didn't tell her that is was Wednesday, then you have told an implicit lie. You are asking her to formulate a credence based on Monday/Tuesday, but expecting her answer to be consistent with Wednesday.
AINC: 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.
And the point of my answer, is that it is actually a conditional probability based on an unusual state of information.
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 not true yet you want her to believe it is. That is lying to her, in the context of the information you want her to use.
But she can from an opinion. "My credence should be halfer/thirder answer if Wednesday has not yet dawned, or 1/2 if it has. Since the cue I was told would happen - being asked for my credence - has not yet occurred, I am uncertain which and so can't give a more definitive answer." And if you give that cue on Wednesday, you are lying even if you promised you wouldn't.
And yes, my mathematical model corresponds to SB's reality when she is asked for her credence. That is the entire point. If you think otherwise, I'd love to hear an explanation instead of a dissertation that does not apply.
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 of your knowledge about the state of the system. It is not a property of the system itself. You are trying to make it a property of the system.
Halfers are most guilty of not understanding this, since their arguments are based on there being no change to the system despite an obvious change to the knowledge (I’m ignoring how to treat that knowledge). Thirders recognize the change, but keep trying to cast the difference in terms of episiotomy when it really is much simpler.
“However she finds a piece of paper hidden in her pajamas from the past version of herself, experiencing an awakening. This past version managed to cheat and send a message to the future where she claims that her current credence for Heads is 1/3. Should she expect to successfully guess Tails with 2/3 probability?”
No, since her knowledge is different now than it was then.
“Or suppose that SB is given memory loss drug only before she is awaken on Tuesday so that on Wednesday she always remembers her last awakenings though she doesn't know whether it was on Monday or Tuesday. Is her credence for Heads still 1/3?”
If I understand this correctly, she knows that it is Wednesday. Her credence is 1/2. She can, at the same time on Sunday or Wednesday, make an assessment of the state of her knowledge on Monday or Tuesday, and recognize how it is different. In other words, "my credence now is 1/2, but my credence then should be 1/3." The paradox you seem to be fishing for does not exist.
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 one, how can it not be "considered (by her) to be a different experiences."
For the purposes of sampling, there are two parts to the experiment. What you call them is not relevant, only that SB knows that she is in one, and only one, of two parts. In the classic version, as Elga modifided it from the actual problem he proposed, one part must have a waking and one part has either a sleeping or a waking. Are you really claiming that the part where she is left asleep is not a part of the experiment? One that she knows is a possibility that is ruled out by her current state? And so fits the classic definition of "new evidence" that you deny? But if you really believe that, what about my version where she is taken shopping? It is even better as a classic example of evidence.
Or you could look at my answer. There, I explain how the problem you are solving actually is an "alternate formulation." And, while there are still two parts, they are equivalent so we do not need to treat them separately. The answer is unequivocally 1/3.
Or try this: Instead of a coin, roll two six-sided dice. On Monday, ask her for the probability that the resulting sum is 7. On Tuesday, if the sum is odd, ask her the same question. But if it is even, ask her for the probability that the sum is 8.
If the room has a calendar, Beauty should certainly say Pr(S=7)=1/6 on Sunday and Monday. But on Tuesday, in answer to "what is the probability of 7," she should say Pr(S=7)=1/3. But without a calendar, Pr(S=7)=1/6 can't be right. Because it might be Tuesday, and Pr(S=7)=1/3 can't be right. And it might be Monday, when Pr(S=7)=1/6 can't be right. It has to be something in between, and that something is (2/3)*(1/6)+(1/3)*(1/3)=4/18=2/9. Yes, even though this is never the answer if she knows the day. That is how conditional probability works.
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 days is "current." (Yes, I know some debate whether days can be used this way - it is a self-fulfilling argument. If it is assumed to be true, you can show that it is true.) Elga was trying to use the day as a valid descriptor of disjoint outcomes, so that in SB's world the outcomes T&Mon and T&Tue become distinct. The prior probabilities of the three disjoint outcomes {T&Mon, T&Tue, H} are each 1/2, and the probability for H can be updated by the conventional definition:
Pr(H|T&Mon or T&Tue or H) = Pr(H)/[Pr(T&Mon)+Pr(T&Tue)+Pr(H)]
Pr(H|T&Mon or T&Tue or H) = (1/2)/[(1/2)+(1/2)+(1/2)] = 1/3
If you think about it, this is exactly what Elga argued, except that he didn't take the extreme step of creating a probability that was greater than 1 in the denominator. He made two problems, that each eliminated an additional outcome. I suggest, but don't want to get into defending, that this is valid since SB's situation divides the prior outcome T into two that are disjoint. Each still has the prior probability of the parent outcome.
But this is predicated on that split. Which is what I want to show is valid. What if, instead of leaving SB asleep on Tuesday after Heads, wake her but don't interview her? Instead, we do something quite different, like take her on a $5,000 shopping spree at Saks Fifth Avenue? (She does deserve the chance of compensation, after all). This gives her a clear way to distinguish H&Mon&Interview from H&Tue&Saks as disjoint outcomes. The use of the day as a descriptor is valid, since this version of the experiment allows for it to be observed.
The actual sample space, on Sunday, for what can happen in an "awake" world for SB is {T&Mon, H&Mon, T&Tue, H&Tue&Saks}. Each has a prior probability of 1/4, for being what will apply to SB on a waking day. When she is actually interviewed, she can eliminate one.
And now I suggest that it does not matter, in an interview, what happens differently on H&Tue. SB is interviewed in the context of an interview day, so the question is placed in that context. Claiming it has the context of Sunday Night ignores how the Saks option can only be addressed in a single-day context, and can affect her answer.
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 problem it is "you") is woken once, or twice, based on the outcome of a fair coin flip (Heads=once, Tails=twice). Elga enacted that description with a mandatory waking on Monday, and an optional one on Tuesday. This way, could create two partial solutions by revealing two different bits of information that removed one of the three possibilities. The unfortunate side effect of this was that the conditions surrounding Monday and Tuesday are different; thirders require Tuesday to be part of a different outcome, and halfers insist it is the same outcome as Monday+Tails.
And that is what is unnecessary. Instead of that Monday/Tuesday schedule, simply flip two coins (call them C1 and C2) after SB is first put to sleep. Then:
- If both coins are showing Heads, skip to step 6.
- Wake SB.
- Ask SB for her credence that coin C1 is showing Heads.
- Wait for her answer.
- Put her back to sleep with amnesia.
- End this stage of the experiment.
Now turn coin C2 over to show its other face, and repeat these steps.
When SB is woken, she knows that: (A) She is in step 2 of a pass thru these six steps. (B) In step 1, there were four equally-likely states for the two coins: HH, HT, TH, and TT. (C) Since she is awake, the state HH is eliminated but the other three remain equally likely. (D) In only one of those states is C1 showing Heads. So she can confidently state that her credence is 1/3.
The difference between this, and how Elga enacted the problem he posed, is that here there is no ambiguity about how she arrived at her current, awake situation.
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], 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 [awakened], to what degree ought you believe that the outcome of the coin toss is Heads?"
Elga introduced the two-day schedule, where SB is always wakened on Monday, and optionally wakened on Tuesday, in order to facilitate his thirder solution. You can argue whether his solution is to the same problem or not. But if it is not, it is the variation problem that is wrong. And it is unnecessary.
First, consider this simplified experiment:
- SB is put to sleep.
- Two coins, C1 and C2, are arranged randomly so that the probability of each of the four possible combinations, {HH, HT, TH, TT} has a probability of 1/4.
- Observe what the coins are showing:
- If either coin is showing Tails:
- Wake SB.
- Ask her for her degree of belief that coin C1 is showing Heads.
- Put SB back to sleep with amnesia.
- If both coins are showing heads:
- Do something else that is obviously different than option 3a.
- Make sure SB is asleep, and can't remember option 3b happening.
- If either coin is showing Tails:
Note that if SB is asked the question in option 3a, she knows that the observation was that at least one coin is showing Tails. It does not matter what would happen - or if anything would happen - in 3b. Her answer can only be 1/3.
We can implement the original problem by flipping these two coins for one possible awakening in the original problem, and then turning coin C2 over for another.
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 the same procedure described for Monday.
In the interview(s), ask SB for her credence that the Quarter is currently showing Heads. Since the Quarter is never changed, it is always showing the same face that was the result of the flip. SB knows that when the locked box was examined, there are four equally probably possibilities for {Dime,Quarter}. They are HH, HT, TH, and TT. Since SB is now awake, she knows that HH is eliminated as a possibility. Her credence for each of TH, HT, and TT are each 1/3.
The only difference between this version, and the original, is that we don't need to say anything about what day it is.
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 will be asked to assess, from your perspective based on these procedures, the probability that you will be wakened only once during the experiment. Surely (call this assertion A) the answer to this question must be the same as in the original Sleeping Beauty Problem.
But when you are awakened, you are told that you are one of four volunteers undergoing the exact same procedures based on the same coin flip; with one exception. The choices for the coin results, and the days, are different for each of the four volunteers. (Since there are four possible combinations, each is used.) You are not told which choices were made for you, or introduced to the others. Surely (Assertion B) this gives you no information about your own situation, so it cannot affect your answer.
But it does allow you to evaluate your perspective with a bit more clarity. Of the four volunteers, you know that one must still be asleep, and it isn't you. That volunteer will be wakened only once. Of the other three, including you, two will be wakened on both days, and one will be wakened only once. Surely (Assertion C) this means that, from your perspective, the probability that you will be wakened once is 1/3.
Unless you can find a flaw with one of my assertions, this means that the answer to the original problem is also 1/3. There are valid, if unorthodox, mathematical proofs of this. But those who trust their intuition over mathematics want the answer to be 1/2. The so-called "double halfer" approach is just a rationalization for ignoring valid mathematics.
The rationalization: an indexical is an identifier for a value relative to some index. If no way to associate the indexical with actual values is provided, as in "the probability it will rain tomorrow," then the indexical cannot be used to assess your perspective. But if such a means is provided, EVEN AS A PROBABILITY DISTRIBUTION, then it can be assessed. For example, say a different volunteer is told she will be wakened once or twice based on the roll of two standard dice; 1=Monday, 2=Tuesday, etc. If both dice results the same, she will be wakened only once. Upon being awake, she can determine that the probability that it is Tuesday is 1/6. "Tuesday" may ne an indexical, but context is provided to index it.
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 eliminate the indexing issue. Its answer is unequivocally 1/3. I think there is an interesting lesson to be learned from the problem, but it can't be approached until people stop trying to make the lesson fit the answer they want.
+++++
The cogent difference between halfers and thirders, is between looking at the experiment from the outside, or the inside.
From the outside, most halfers consider Beauty's awakenings on Mon-T and Tue-T to be the same outcome. They cannot be separated from each other. The justification for this outlook is that, over the course of the experiment, one necessitates the other. The answer from this viewpoint is clearly 1/2.
But it has an obvious flaw. If the plan is to tell Beauty what day it is after she answers, that can't affect her answer but it clearly invalidates the viewpoint. The sample space that considers Mon-T and Tue-T to be the same outcome is inadequate to describe Beauty's situation after she is told that it is Monday, so it can't be adequate before. You want to get around this by saying that one interview "doesn't count." In my four-volunteer proof, this is equivalent to saying that one of the three awake volunteers "doesn't count." Try to convince her of that. Or, ironically, ask her for her confidence that her confidence "doesn't count."
But a sample space that includes Tue-T must also include Tue-H. The fact that Beauty sleeps though it does not make it "unhappen," which is what halfers (and even some thirders) seem to think.
To illustrate, let me propose a slight change to the drugs we assume are being used. Drug A is the "go to sleep" drug, but it lasts only about 12 hours and the subject wakes up groggy. So each morning, Beauty must be administered either drug B that wakes her up and overrides the grogginess, or another dose of drug A. The only point of this change, since it cannot affect Beauty's thought processes, is to make Tue-H a more concrete outcome.
What Beauty sees, from the inside, is a one-day experiment. Not a two-day one. At the start of this one-day experiment, there was a 3/4 chance that drug B was chosen, and a 1/4 chance that it was drug A. Beauty's evidence is that it was drug B, and there is a 1/3 chance of Heads, given drug B.
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 each have a 1/2 *prior* probability of happening to Beauty on a single *indexed* day in the experiment. If you want to get 1/2, you use two. So, that you get 1/2 by your method is not surprising in the slightest. It was pre-ordained.
You need to find a way to justify one or the other that is not a non sequitur, and that isn't possible. You can't justify why "ensuring that only one interview ever 'counts'" solves an issue in the debate. You never tried, you just asserted that it was the thing to do.
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. Another (SB2) will be left asleep on Monday, also if Heads is flipped. This is essentially the same as the OSBP, since in a non-indexical version the day can't matter. The other two (SB3 and SB4) will be left asleep if Tails is flipped, one on Monday and one on Tuesday. And if we ask them about their confidence in Tails instead of Heads, the correct answer should be the same as the OSBP, whatever that turns out to be.
The "de-indexicalization" is accomplished by changing the question to an equivalent one. For SB1 and SB2, the truth value of the statement "I am awake now, and it is my only awakening" is the always the same as "Heads was flipped." For SB3 and SB4, it is the same as "Tails was flipped."
Note that it can't matter if you know which of these volunteers you are, or if you are allowed to discuss the question "Is this my only awakening?" as long as you can't reveal which one you are to the others. One each day of the experiment, exactly three will be awake. For exactly one of those three, it will be her only awakening.
The non-indexical answer is 1/3.
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 mean different things for different people."
But this fractured terminology can be overcome quite simply. Instead of one volunteer, use four.
Each will go through a similar experience where they will be woken at least once and maybe twice, on Monday and/or Tuesday, depending on the result of the same fair coin flip.
All four will be wakened both days with the following exceptions: SB1 will be left asleep on Monday if Heads is flipped. SB2 will be left asleep on Monday if Tails is flipped. SB3 will be left asleep on Tuesday if Heads is flipped. And SB4 will be left asleep on Tuesday if Tails is flipped. Note that SB3's schedule corresponds to the original version of the problem.
This way, three of the volunteers will be wakened on Monday. Two of those will be wakened on again Tuesday, while the third will be left asleep and be replaced by the one who slept through Monday. And each has the same chance to be wakened just once.
Put the three in a room together, and allow them to discuss anything EXCEPT the coin result and day that they would sleep through. Ask each for their confidence in the assertion that she will be wakened just once during the experiment.
No matter what day it is, or how the coin landed, the assertion will be true for one of the three awake volunteers, and false for the other two. So their confidences should sum to 1. No matter what combination of day and result each was assigned to sleep through, each has the same information upon which to base her confidence. So their confidences should be the same.
The only possible solution is that the confidences should all be 1/3. If, instead, SB3 is just told about the other three volunteers, but never meets them, she can still reason the same way and get the answer 1/3. And since "I, SB3, will be wakened only once" is equivalent to "the fair coin landed Heads," our original volunteer can give the same answer.
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 the second coin's result. You now have have six cases that the PoI applies to, counting the "8:00 Monday" case twice, and each has probability 1/6.
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. If this (even-numbered) flip lands heads, wake SB 3*N times and end the experiment; b. If this (even-numbered)flip lands tails, set M=M+1 go to step #2.
In Coscott’s version, we start with N=1 and multiply it by 9 each time we choose a new one; that is, N=9^M. But does the answer depend on N in any way? Halfers don’t think the answer depends on the number of wakings at all, and thirders think it depends only on the ratio of wakings in step 3a to those in step 4a, not the specific values.
So I maintain that my problem is the same as coscott’s, except in scale, no matter how we choose N. We can answer the original question by choosing N=1 every time.
There is a 2/3 chance of ending after an odd number of flips, and a 1/3 chance of ending after an even number. A halfer should claim SB gains no new knowledge by being awake, so P(odd|awake)=2/3 and P(even|awake)=1/3. A thirder should say there are four possible situations that awake SB could be in , and she cannot differentiate between them. Since 3 of them correspond to an even number of flips, P(odd|awake)=1/4 and P(even|awake)=3/4.
But like coscott’s, this variation, by itself, sheds no light on the original problem. We can even change numbers to something else:
1) 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 2. 2) Flip the coin again. a. If this (even-numbered) flip lands heads, wake SB M times and end the experiment; b. If this (even-numbered)flip lands tails, go to step #1.
You can decide for yourself whether you think the answer should depend on M and N, but I suspect most people will decide that based on whether they are halfers (“it can’t depend on M and N!”) or thirders (“it must depend on M and N!”), rather than what makes mathematical sense. (I’m not saying they will ignore mathematical, I’m saying they will define it by getting the answer they prefer.)
But as long as we are accepting the possibility of infinite wakings, what happens if we hold N constant and let M approach infinity? Halfers will still say the answers don’t change, thirders will say P(odd)=N/(M+N) and P(even)=M/(M+N).
But is it, or is it not, the same if we hold M constant, at a very large number, and let N approach 0? Because at N=0, P(odd)=0, which it can’t be if the halfers are right.
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 ambiguity in the situation.
2) Nothing is truly random. If you know the exact shape of a coin, its exact weight distribution, exactly how it is held before flipping, exactly what forces are applied to flip it, the exact properties of the air and air currents it tumbles through, and exactly how long it is in the air before being caught in you open palm, then you can calculate - not predict - whether it will show Heads or Tails. Any lack in this knowledge leaves multiple possibilities open, which is the ambiguity.
3) Saying "the coin is biased" is saying that there is an inherent property, over all of the ambiguous ways you could hold the coin, the ambiguous forces you could use to flip it, the ambiguous air properties, and the ambiguous tumbling times, for it to land one way or another. (Its shape and weight are fixed, so they are unambiguous even if they are not known, and probably the source of this "inherent property.")
4) Your state of mind defines probability only in how you use it to define the ambiguities you are accounting for. Eliezer's frequentist is perfectly correct to say he needs to know the bias of this coin, since in his state of mind the ambiguity is what this biased coin will do. And Eliezer is also perfectly correct to say the actual bias is unimportant. His answer is 50%, since in his mind the ambiguity is what any biased coin do. They are addressing different questions.
5) A simple change to the coin question puts Eliezer in the same "need the environment" situation he claims belongs only to the frequentist: Fli[p his coin twice. What probability are you willing to assign to getting the same result on both flips?
6) The problem with the "B9" question discussed recently, is that there is no framework to place the ambiguity within. No environmental circumstances that you can use to assess the probability.
7) The propensity for some frequentists to want probability to be "in the environment" is just a side effect of practical application. Say you want to evaluate a statistical question, such as the effectiveness of a drug. Drug effectiveness can vary with gender, age, race, and probably many other factors that are easily identified; that is, it is indeed "in the environment." You could ignore those possible differences, and get an answer that applies to a generic person just as Eliezer's answer applies to a generic biased coin. But it behooves you to eliminate whatever sources of ambiguity you easily can.
8) In geometry, "point" and "line" are undefined concepts. But we all have a pretty good idea what they are supposed to mean, and this meaning is fairly universal.
"Length" and "angle" are undefined measurements of what separates two different instances of "point" and "line," respectively. But again, we have a pretty clear idea of what is intended.
In probability, "outcome" is an undefined concept. But unlike geometry, where the presumed meaning is universal, a meaning for "outcome" is different for each ambiguous situation. But an "event" is defined - as a set of outcomes.
"Relative likelihood" is an undefined measurement what separates two different instances of "event." And just like "length," we have a pretty clear idea of what it is supposed to mean. It expresses the relative chances that either event will occur in any expression of the ambiguities we consider.
9) "Probability" is just the likelihood relative to everything. As such, it represents the fractional chances of an event's occurrence. So if we can repeat the same ambiguities exactly, we expect the frequency to approach the probability. But note: this is not a definition of probability, as Bayesians insist frequentists think. It is a side effect of what we want "likelihood" to mean.
10) Eliezer misstated the "classic" two-child problem. The problem he stated is the one that corresponds to the usual solution, but oddly enough the usual solution is wrong for the question that is usually asked. And here I'm referring to, among others, Martin Gardner's version and Marilyn vos Savant's more famous version. The difference is that Eliezer asks the parent if there is a boy, but the classic version simply states that one child is a boy. Gardner changed his answer to 1/2 because, when the reason we have this information is not known, you can't implicitly assume that you will always know about the boy in a boy+girl family.
And the reason I bring this up, is because the "brain-teasing ability" of the problem derives more from effects of this implied assumption, than from any "tendency to think of probabilities as inherent properties of objects." This can be seen by restating the problem as a variation of Bertrand's Box Paradox:
The probability that, in a family of two children, both have the same gender is 1/2. But suppose you learn that one child is in scouts - but you don’t know if it is Boy Scouts or Girl Scouts. If it is Boy Scouts, those who answer the actual "classic" problem as Eliezer answered his variation will say the probability of two boys is 1/3. They'd say the same thing, about two girls, if it is Girl Scouts. So it appears you don’t even need to know what branch of Scouting it is to change the answer to 1/3.
The fallacy in this logic is the same as the reason Eliezer reformulated the problem: the answer is 1/3 only if you ask a question equivalent to "is at least one a boy," not if you merely learn that fact. And the "brain-teaser ability" is because people sense, correctly, that they have no new information in the "classic" version of the problem which would allow the change from 1/2 to 1/3. But they are told, incorrectly, that the answer does change.
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 controversy in the Sleeping Beauty Problem is based on the fallacious conclusion that such "new" information is required to update probabilities this way. This is an example of the logical fallacy called affirming the consequent: concluding that "If A Then B" means "A is required to be true for B to be true" (an equivalent statement is "If B then A").
All that is really needed for updating, is a change in the information. It almost always is an addition, but in the Sleeping Beauty Problem it is a removal. Sunday Sleeping Beauty (SSB) can recognize that "Tails & Awake on Monday" and "Tails & Awake on Tuesday" represent the same future (Manfred's "AND"), both with prior probability 1/2. But Awakened Sleeping Beauty (ASB), who recognizes only the present, must distinguish these two outcomes as being distinct (Manfred's "OR"). This change in information allows Bayes' Rule to be applied in a seemingly unorthodox way: P(H&AonMO|A) = P(H&AonMO)/[P(H&AonMO) + P(T&AonMO) + P(T&AonTU)] = (1/2)/(1/2+1/2+1/2) = 1/3. The denominator in this expression is greater than 1 because the change (not addition) of information separates non-disjoint events into disjoint events.
The philosophical issue about SSA v. SIA (or whatever these people call them; I haven't seen any two who define them agree), can be demonstrated by the "Cloned SB" variation. That's where, if Tails is flipped, an independent copy of SB is created instead of two awakenings happening. Each instance of SB will experience only one "awakening," so the separation of one prior event into two disjoint posterior events, as represented by "OR," does not occur. But neither does "AND." We need a new one called "ONE OF." This way, Bayes' Rule says P(H&Me on Mo) = P(H&Me on MO)/[P(H&Me on MO) + (ONE OF P(T&Me on MO), P(T&Me on TU))] = (1/2)/(1/2+1/2) = 1/2.
The only plausible controversy here is how SB should interpret herself: as one individual who might be awakened twice during the experiment, or as one of the two who might exist in it. The former leads to a credence of 1/3, and he latter leads to a credence of 1/2. But the latter does not follow from the usual problem statement.
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; Pr(observe Heads and Tuesday)=0 •/ + •\ ••\•••• Monday---1---Waken; Pr(observe Tails and Monday)=1/4 ••1/2••••/ ••••\••1/2 •••••\••/ ••• Tails •••••••\ •••••••1/2 ••••••••••\ ••••••• Tuesday--1---Waken; Pr(observe Tails and Tuesday)=1/4
What halfers refuse to recognize, is that whether Beauty is awakened in any specific circumstance is a decision that is part of the process. It is based on the other two random variables, after both – repeat, both – have been determined. The event “Heads and Tuesday” is an event that exists in the sample space, and the decision to not awaken her is made only after that event has occurred. Halfers think they have to force that event into non-existence by making Q=1, when all the experiment requires is that the probability Beauty will observe it is zero. This is the point one thirder argument utilizes, that of Radford Neal’s companion Prince who is always awakened but only asked if Beauty is awakened.
In fact, there is no reason why the probability that it is Monday, given Heads, should be any different than the probability it is Monday, given Tails. So, with Q=1/2, we get that Pr(observe heads)=1/4, Pr(observe anything)=3/4, so Pr(Heads|observe anything)=1/3. QED.
Neq1’s arguments that the thirder positions are wrong are all examples of circular reasoning. He makes some assumption equivalent to saying the answer is 1/2, and from that proves the answer is 1/2. For example, when he uses “Beauty woken up at least once” as a condition, all his terms are also conditioned on the fact that the rules of the experiment were followed. So when he inserts the completely unconditional “Pr(Heads)=1/2” on the right-hand side of the equation, he really should use Pr(heads|rules followed), which is the unknown we are trying to find. It is then unsurprising that he gets the number he inserted, especially if you consider what using a probability-one event as a condition in Bayes’ Rule means.
Where neq1 claims that Nick Bostrom’s argument is wrong in “Disclosure Process 1,” I suggest he go back and use the values from his probability tree. Her credence of heads is (1/2)/(1/2+1/2/1,000,000). In the second process, it is either (1/2)/(1/2+1/2/7,000,000) of (1/2)/(1/2+1/2/1,000,000,000,000), depending on what “specific day” means.
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 prior distribution, an act that is strictly prohibited for a Frequentist. Both claims are absurd. Prior distributions are a fact of the mathematics of probability, and belong to Frequentist and Bayesian alike. The only differences are (1) the Bayesian may use information differently to determine a prior, sometimes in situations where a Frequentist wouldn't see one at all; (2) The Bayesian will prefer solutions based explicitly on that prior, while the Frequentist will prefer solutions based on the how the prior affects repeated experiments; and (3) Some Frequentists might not realize when they have enough information to determine a prior, and/or its effects, that should satisfy them.
If both get answers, and they don't agree, somebody did something wrong.
The answer is 50%. The Bayesian says that, based on available information, neither result can be favored over the other so they must both have probability 50%. The Frequentist says that if you repeat the experiment 100^2 times, including the part where you draw a coin from the bag of 100 coins, you should count on getting each coin 100 times. And you should also count, for each coin, on getting heads in proportion to its probability. That way, you will count 5,000 heads in 10,000 trials, making the answer 50%. Both solutions are based on the same facts and assumptions, just organized differently.
The answer Eliezer_Yudkowsky attributes to Frequentists, for the simpler problem without the bag and stamped coins, is an incorrect Frequentist solution. Or at least, a correct solution to a different problem. One that corresponds to the different question "What proportion of the time will this coin come up heads?" I agree that some who claim to be Frequentists will answer that question. But the true Frequentist will answer the question that was asked: "What proportion of the time will the process of flipping a coin with unknown bias come up heads?" His repetitions must represent the bias for each flip as independent of any other flips, not the same bias each time. The bias B will come up just as often as the bias (1-B), so the number of heads will always be half the number of trials.
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 to theory can provide an answer.
Illuminated by these spirits, once she is fully awake she reasons: "I have no idea whether today is Monday or Tuesday; but it seems that if I did know, I would have no problem answering the question. For example, if I knew it was Monday, my credence that the coin landed heads could only be 1/2. On the other hand, if I knew it was Tuesday, my credence would have to be 0. But on the gripping hand, these two incontrovertible truths can help me answer as my night visitors suggested. There is a theorem in probability, called the Theorem of Total Probability, that says the probability for event A is equal to the probability of the sum of the events (A intersect B(i)), where B(i) partitions the entire event space.
"Today has to be either Monday or Tuesday, and it can't be both, so these two days represent such a partition. Since I want to avoid making any assumptions as long as I can, let me say that the probability that today is Monday is X, and the probability that it is Tuesday is (1-X). Now I can use this Theorem to state, unequivocally, that my credence that the coin landed heads is P(heads)=(1/2)X+0(1-X)=X/2.
"But I know that it is possible that today is Tuesday; even a Bayesian has to admit that X<1. So I know that 1/2 cannot be correct; the answer has to be less than that. A Frequentist would say that X=2/3 because, if this experiment were repeated many times, two out of every three interviews would take place on Monday. And while a Bayesian could, in theory, choose any value that is less than 1, it is a violation of Occam's Razor to assume there is a factor present that would make X different than 2/3. So, it seems my answer must be 1/3.