I The Sleeping Beauty Problem: Any halfers here?

[stevendaryl]

That's not my approach. That's the one explicitly described in the article as the only reason to change her credence between Sunday evening and awakening. I am still waiting your other alternative explanation.

Once again, could you give me the answer to the following modified Sleeping Beauty experiment:

Suppose that in the case of tails, she is awakened twice, but in the case of heads, she is not awakened at all. Then do you agree that when she is awakened, she will know for certain that the coin is heads? Yes or no?

 

[PeterDonis]

She certainly can distinguish Monday from Tuesday with the explicit apparatus called a brain, with her explicit memory erase (which is explicitly identical to no having those memory yet)

In other words, she can distinguish Monday from Tuesday, except when she can't? So which of the two applies to this experiment, in your view?

 

[Dale]

However, I'm wondering how you would compute the expected payoff for bets in this way in a scenario like the second one in post #67 (where only one bet is paid off even if Beauty is awakened twice).

Going back to this, I don't think that Beauty can formulate a winning strategy if the wager is malicious, meaning she is unaware of the possibility of a bet not being honored. But if she is made aware of the possibility then she can probably marginalize over the possibility of a malicious wager.

However, since she is being asked for her credence "now", I think that the relevant wager would be an immediate wager offered, resolved, and honored "now". Since she knows in advance that she will be asked the question on each awakening the wager would need to be offered at each awakening in order to be an indication of the probability asked.

 

[PeterDonis]

I don't think that Beauty can formulate a winning strategy if the wager is malicious, meaning she is unaware of the possibility of a bet not being honored.

As I understand that scenario in post #67, it isn't that a bet isn't honored. It's that Beauty is told in advance that, if it should happen that she is awakened multiple times and asked to bet, only the last bet will count. Of course, while the experiment is in progress, she won't know whether the bet she is being asked to make is one of multiple bets she will make or not. But she will know that it is possible that she will be asked to bet multiple times, and if so, only the last one counts, so she can take that into account in deciding how to bet.

if she is made aware of the possibility then she can probably marginalize over the possibility of a malicious wager

That is my understanding of what she is expected to do in the scenario in question. And given that she knows in advance how the bets will be processed, I don't think the wager is "malicious"; unusual, perhaps, but not malicious.

Since she knows in advance that she will be asked the question on each awakening the wager would need to be offered at each awakening

In this scenario, as I understand it, Beauty is indeed asked the question on each awakening. But you might be putting a tighter meaning on the word "wager" than was intended in that scenario, since the scenario explicitly says that the "wagers" are resolved at the end of the experiment, and that if Beauty is awakened multiple times, only the last "wager" will actually be paid off (any others will be "resolved" by being discarded). This might be an unusual use of the word "wager", but that can be fixed by changing the word if it's thought necessary. The actual conditions of the experiment are, as far as I can see, the way I have described them.

 

[Marana]

Edit: I tried to come up with an equivalent situation but I don't think it worked.

 

[Stephen Tashi]

That's the challenge: to figure that out.

I agree.

It's interesting that students in introductory probability courses are admonished that the probability of an event is undefined until the probability space containing the event is stated - and they are assigned exercises where they must state the probability space for situations that are verbally described; yet when experts tackle verbal problems they often charge in and offer solutions without saying what probability space they are talking about.

1. (tails, monday, awake)
2. (tails, tuesday, awake)
3. (heads, monday, awake)
4. (heads, tuesday, asleep)

Those are possible states for Sleeping Beauty and the coin. To define a probability space also requires saying what probability is assigned to each event and that is the subject of debate.

The probability space you gave implies that in order to realize an event, we must pick the day to be Monday or Tuesday. The statement of the problem does not explicitly say anything about an experiment where we pick a day. So to compute the probability of an event involving picking a day, we need to use a different probability space where an event like "It is Monday" can be defined in terms of the events described in the problem.

A straightforward translation of the events described in the problem doesn't give a probability space where an event like "It is Monday" is defined.

For example, we could omit the state of the coin. The definition of a probability space requires specifying both the events and the probability assigned to those events. The coin is what implements the probability measure.

I'll use the word "treated" to mean "awakened and interviewed".

Event---------- Probability
A: SB treated on Monday and SB treated on Tuesday, 1/2
B: SB treated on Monday and SB not treated on Tuesday, 1/2
C: SB not treated on Monday and SB treated on Tuesday, 0
D: SB not treated on Monday and SB not treated on Tuesday, 0

In that probability space, how shall we interpret the verbal information "Sleeping beauty is awakened and interviewed"? when no day for the treatment is specified? I would interpret that event to be the event $A \cup B \cup C$.

In asking for the conditional probability that "It is Monday given that SB is awakened and interviewed", how do we define the event "It is Monday"? I don't see how to define such an event in the sample space above.

 

[stevendaryl]

It's interesting that students in introductory probability courses are admonished that the probability of an event is undefined until the probability space containing the event is stated - and they are assigned exercises where they must state the probability space for situations that are verbally described; yet when experts tackle verbal problems they often charge in and offer solutions without saying what probability space they are talking about.

I would say that in this case, it's pretty clear that there are exactly 4 events, or situations, or whatever. The question is how to assign probabilities to them.

The probability space you gave implies that in order to realize an event, we must pick the day to be Monday or Tuesday.

No. We're talking subjective Bayesian probability. The probability is the measure of uncertainty about the truth of a statement. There is no implication that we're picking anything at random. Sleeping Beauty is uncertain about whether today is Monday or Tuesday. That doesn't mean that she is randomly picking between those two possibilities, it just means that there are two possibilities consistent with her knowledge.

She's also uncertain about whether the coin was heads or tails. She quantifies her uncertainty by assigning likelihoods to them. The rules of the experiment imply that if she is awake, then she knows for certain that the combination (Tuesday, Heads) is ruled out. So the question is: how to sensibly assign the other three probabilities.

You can certainly reject the idea of subjective probability, but that's the whole basis of the Sleeping Beauty problem is to ask what is a sensible assignment of subjective probability.

 

[JeffJo]

One of the problems in this thread is the wide variety of alternative scenarios proposed, which seem to be having the opposite effect as intended regarding clarifying the original scenario.

Obviously, I disagree. And I think the length of this stalemate, which is more about what Beauty's point of view should be than probability, proves that trying to decide the proper point of view first is a futile exercise.

If the correct answer can be determined without having to decide point of view first - as I feel I have done - then you can use it to help settle your debate. That won't happen without an independently justified answer. Existing probability theories just don't support including the number of trails as a random variable.

So, counter to your intent, here is as another version of my variation. Use four volunteers, and the four cards I described before (with (H,Mon), (H,Tue), (T,Mon), and (T,Tue) written on them). Deal the cards to the four, and put them in separate rooms. Using one coin flip, and waken three of them on Monday, and Tuesday. Leave the one whose dealt card matches both the day, and the coin flip, asleep. Ask each for her confidence that the coin matches her card.

Obviously, if you show each Beauty her card, her answer has to be the same as the original Beauty's. Since it is the same regardless of what card is dealt, you don't have to show it to any of them. If you don't show it to any of them, you can put all three awake Beauties in a room together to discuss their answers. All have the same information, so all answers have to be the same. Since exactly one of the three has a card that matches the coin flip, that answer must be 1/3.

If you can find a flaw in my logic, I'd be happy to discuss why it is to correct to disregard the day and why Beauty has no new information to provide a probability update. But I feel my approach shows, quite trivially, that the answer is 1/3. So the proper discussion is why there is new information.

 

[Dale]

That is my understanding of what she is expected to do in the scenario in question. And given that she knows in advance how the bets will be processed, I don't think the wager is "malicious"; unusual, perhaps, but not malicious.

I agree. She would have to marginalize over the different options correctly, but it could be done. It would be complicated but, as you say, not malicious.

However, I don't think that this wager would be the one that corresponds to the credence of heads. It would be the credence of something more convoluted, like the credence of heads and amnesia drug positive.

since the scenario explicitly says that the "wagers" are resolved at the end of the experiment,

I didn't see that. Are you talking about the Wikipedia article? If so, where does it say that?

 

[Dale]

In asking for the conditional probability that "It is Monday given that SB is awakened and interviewed", how do we define the event "It is Monday"?

You don't need to do that in order to calculate the requested probability. There is no need to bring Monday or Tuesday into the calculations at all.

 

[stevendaryl]

So, counter to your intent, here is as another version of my variation. Use four volunteers, and the four cards I described before (with (H,Mon), (H,Tue), (T,Mon), and (T,Tue) written on them). Deal the cards to the four, and put them in separate rooms. Using one coin flip, and waken three of them on Monday, and Tuesday. Leave the one whose dealt card matches both the day, and the coin flip, asleep. Ask each for her confidence that the coin matches her card.

Obviously, if you show each Beauty her card, her answer has to be the same as the original Beauty's. Since it is the same regardless of what card is dealt, you don't have to show it to any of them. If you don't show it to any of them, you can put all three awake Beauties in a room together to discuss their answers. All have the same information, so all answers have to be the same. Since exactly one of the three has a card that matches the coin flip, that answer must be 1/3.

If you can find a flaw in my logic, I'd be happy to discuss why it is to correct to disregard the day and why Beauty has no new information to provide a probability update. But I fell my approach shows, quite trivially, that the answer is 1/3. So the proper discussion is why there is new information.

I think it's a little difficult to see the connection with the original problem, so the fact that the results are the same are a little less than compelling...

Oh, now I get it! (I think). The original Sleeping Beauty problem singled out a single event, (H, Tues) as the day that Sleeping Beauty sleeps through the interview. You're restoring the symmetry by having 4 different beauties, each with a different event singled out. So no matter what the day is, and no matter what the coin flip result, 1 of the beauties is sleeping, 1 has a matching coin flip result, and 2 have a non-matching coin-flip result. So the odds of a match among those who are awake is 1/3.

Very clever.

 

[Dale]

Existing probability theories just don't support including the number of trails as a random variable.

That simply isn't true. In fact, it is one of the recognized advantages of Bayesian statistics.

That said, I have no objection to your alternative scenario other than just the fact that it differs from the scenario in the OP.

 

[PeterDonis]

Are you talking about the Wikipedia article?

No, I'm talking about post #67 by @Demystifier . Link here:

https://www.physicsforums.com/threa...m-any-halfers-here.916459/page-4#post-5777796

 

[Stephen Tashi]

No. We're talking subjective Bayesian probability.

The fact that a probability is called a "Bayesian" probability does not exempt it from being defined on a probability space. In the probability space that you gave, you may choose to subjectively assign probabilities to each of the events, but, naturally, that subjective assignment is not mathematical demonstration that your assignments are correct.

As you know, the usual scenario in applying Bayesian probability is to have a "prior" distribution on some probability space and then to compute a "posterior distribution" by conditioning on some event that can be defined in that space. So if we wish to apply the Bayesian method to compute "The probability that the coin landed Heads" given that "Sleeping Beauty is awakened and interviewed", we need to have a probability space where those events can be defined. And to have a convincing argument that our calculation is correct, we need to show that the information given in the problem implies the probabilities we assign in our probability space.

The probability is the measure of uncertainty about the truth of a statement. There is no implication that we're picking anything at random. Sleeping Beauty is uncertain about whether today is Monday or Tuesday. That doesn't mean that she is randomly picking between those two possibilities, it just means that there are two possibilities consistent with her knowledge.

As far as I can see, those statements have no mathematical interpretation in the theory of probability. The usual measure of "uncertainty" is the standard deviation of a random variable, which is not a probability. People do say informally that a probability is a "measure of uncertainty", but what is the mathematical definition for a "measure of uncertainty"?

I agree that the definition of a probability distribution (of Bayesian origin or otherwise) has no notion of picking something at random. It's the application of probability theory to specific problems that makes the connection between a probability distribution and an experiment where something is picked at random. If we are trying to apply probability theory to a problem where the object is to quantify a persons credence and we say credence has nothing to do with picking something at random then we'd better define "credence" and state some axioms that give its properties.

You can certainly reject the idea of subjective probability, but that's the whole basis of the Sleeping Beauty problem is to ask what is a sensible assignment of subjective probability.

Then what the problem asks is undefined until we define the meaning of "a sensible assignment". I don't reject the idea of assigning prior distributions and I don't reject assigning them according to certain criteria such as maximum entropy or symmetry. I do object to the verbal description of an event without specifying what probability space contains that event - for example "P( Monday)" or "Given that Sleeping beauty is awakened".

The following problem is similar to the Sleeping Beauty problem except that it poses some well defined mathematical questions.

Experiment A outputs two possible strings. String "h, M" has probability 1/2 of being the output. String "t,M,T" has probability 1/2 of being the output. Experiment B consists of running Experiment A N times (where N is some given number) and concatenating the outputs of these repeated experiments. For example, a possible result of Experiment B when N = 4 is "h,M,t,M,T,t,M,T,h,M". Experiment C consists of performing experiment B and then selecting a lower case letter from the output of experiment B according the following procedure. First pick an occurrence of a capital letter from the output of experiment B at random, given each occurrence of a capital letter an equal probability of being selected. Then pick the lower case letter to be the first lower case letter preceding the occurrence of the capital letter that was selected.

Question 1: What is probability the lower case letter selected in experiment C is "h" when N = 1?
Question 2: What is the limiting value of the the probability that the lower case letter selected in experiment C is "h" as N approaches infinity?

It seems to me the 1/3 vs 1/2 debate amounts to asking which of those two questions defines Sleeping Beauty's "creedence".

 

[stevendaryl]

The fact that a probability is called a "Bayesian" probability does not exempt it from being defined on a probability space.

But it means that any question that your uncertain about, such as "Is today Monday?" can have an associated probability. The issue is then just to assign probabilities to the exclusive and exhaustive cases: Monday & Heads, Monday & Tails, Tuesday & Heads, Tuesday & Tails. So your claim that Monday isn't an event is not true for Bayesian probability. Any statement with uncertainty can be an event.

As you know, the usual scenario in applying Bayesian probability is to have a "prior" distribution on some probability space and then to compute a "posterior distribution" by conditioning on some event that can be defined in that space.

Yes, the obvious prior for the coin toss is P(H) = P(T) = 1/2. The difficulty is to calculate priors for P(Monday) and P(Tuesday). However, there is really only one choice that makes sense: P(Monday) = P(Tuesday) = 1/2. (Argument at the end)

Given these priors, we compute:

P(Awake) = P(Monday) + P(Tuesday) P(T)

(she's always awake on Monday, and she's awake on Tuesday only if it's tails).

So we conclude: P(Awake) = 3/4

Now finally we compute:

P(H | Awake) = P(H & Awake)/P(Awake)

Since P(H & Awake) only happens if it's Monday, then we can write: P(H & Awake) = P(H & Monday) = P(H) P(Monday) = 1/4

So we conclude:

P(H | Awake) = (1/4)/(3/4) = 1/3

Here's an argument for why P(Monday = 1/2):

First, note that if you told Sleeping Beauty that today is Monday, then she would have no reason to think heads more likely than tails, since the difference between them only shows up on Tuesday. So P(H | Monday) = 1/2.

Second, note that if you told Sleeping Beauty that the coin result was tails, then she would have no reason to think Monday more likely than Tuesday, since they are only different in the case of heads. So P(Monday | T) = 1/2.

Now, compute the conditional probability: P(Monday | H)

We can use Bayes' theorem to write that as:

P(Monday | H) = P(H | Monday) P(Monday)/P(H)

But we already have P(H | Monday) = P(H) = 1/2. So we get:

P(Monday | H) = P(Monday)

Finally, we compute:

P(Monday) = P(Monday | H) P(H) + P(Monday | T) P(T) = 1/2 P(Monday) + 1/4

which implies that P(Monday) = 1/2

As far as I can see, those statements have no mathematical interpretation in the theory of probability. The usual measure of "uncertainty" is the standard deviation of a random variable, which is not a probability. People do say informally that a probability is a "measure of uncertainty", but what is the mathematical definition for a "measure of uncertainty"?

It's subjective. There isn't a right or wrong answer, except that it has to obey the laws of conditional probability. However, in many cases of interest (such as this one), we can appeal to symmetry: If there are two possibilities, and there is no reason to think one more likely than the other, then the probability of each should be 1/2. That principle alone is enough to get unique probabilities in this case.

 

[Stephen Tashi]

So your claim that Monday isn't an event is not true for Bayesian probability. Any statement with uncertainty can be an event.

I'm not claiming that a person can't make up some probability space where "Today is Monday" is a defined event. I'm saying that a direct interpretation of the experiment described in the statement of the problem (e.g. the Wikipedia's version) does not describe a probability space where "Today is Monday" is a defined event.

Yes, the obvious prior for the coin toss is P(H) = P(T) = 1/2. The difficulty is to calculate priors for P(Monday) and P(Tuesday). However, there is really only one choice that makes sense: P(Monday) = P(Tuesday) = 1/2. (Argument at the end)

You stated that the events in your probability space are:

1. (tails, monday, awake)
2. (tails, tuesday, awake)
3. (heads, monday, awake)
4. (heads, tuesday, asleep)

Instead of assigning a prior distribution on those events. It looks like you are assigning two distributions, neither of which , by itself, defines a probability distribution on that space of events. Then you deduce the prior from those two distributions.

Given these priors, we compute:

P(Awake) = P(Monday) + P(Tuesday) P(T)

How are the events you denote by "awake" , "Monday" and "Tuesday" defined in terms of the events in the outcomes of the experiment (as described in the Wikipedia version of the Sleeping Beauty problem) ? The only way that I see to define an event like "Monday" is to conduct the experiment described in the Sleeping Beauty problem and then do another experiment that involves selecting a day from the first experiment. The first experiment produces one of two possible sequences of the events in your sample space. One sequence is (heads, Monday, awake), (heads, Tuesday, asleep). The other sequence is (tails, Monday, awake), (tails, Tuesday, awake). How is the event "Monday" defined in terms of those outcomes?

 

[Dale]

No, I'm talking about post #67 by @Demystifier . Link here:

https://www.physicsforums.com/threa...m-any-halfers-here.916459/page-4#post-5777796

Ah, ok. I think that only wager A corresponds to the requested credulity as described in the Wikipedia article. The interviewer would have to ask a more complicated question or have a different experimental protocol to describe wager B.

 

[Dale]

the statement of the problem (e.g. the Wikipedia's version) does not describe a probability space where "Today is Monday" is a defined event

As I showed earlier, it is not necessary to include "today is Monday" in order to solve the problem. However, it is perfectly acceptable to introduce new unmeasured or unmeasurable variables that you then sum over, which is what is done with the day of the week.

 

[JeffJo]

That simply isn't true. In fact, it is one of the recognized advantages of Bayesian statistics.

Maybe I didn't describe what I meant well. I believe (not my area) that what you mean is that the number of trails isn't predetermined. I meant that it is determined by the event you are trying to find a probability for. The fact that you are arguing about whether it is important to model the day as an event shows that it is not part of the paradigm.

That said, I have no objection to your alternative scenario other than just the fact that it differs from the scenario in the OP.

But it does not differ. That's the point. All I did was remove the distractions that cannot be resolved by your stalemated discussion. Each Beauty in my "alternative scenario" is undergoing the same scenario as that in the OP; one exactly and three with a change in the names applied to the specifics. By removing the names from the solution, the distractions go away.

 

[Dale]

But it does not differ

There are 4 Beautys in yours, but 1 in the OP.

 

[stevendaryl]

There are 4 Beautys in yours, but 1 in the OP.

It took me a while to understand his scenario, but the way I understand it, his 4 beauties amount to 4 isomorphic copies of the original problem. The original sleeping beauty is one of them. For her, things proceed exactly as they did for the original problem, so whatever probabilities she comes up should be the same as for the original problem.

 

[Dale]

I agree, but it is still a clear difference. Someone who is not able to calculate the probability on the original scenario is unlikely to agree that the different scenario is equivalent.

 

[stevendaryl]

I'm not claiming that a person can't make up some probability space where "Today is Monday" is a defined event. I'm saying that a direct interpretation of the experiment described in the statement of the problem (e.g. the Wikipedia's version) does not describe a probability space where "Today is Monday" is a defined event.

The problem statement doesn't ask about probability spaces, it doesn't ask about events. It asks about Sleeping Beauty's subjective probability that the coin flip result was heads. To me, it's a matter of:

Sleeping Beauty is uncertain about a number of things: She's uncertain about what day it is (Monday or Tuesday). She's uncertain about what the coin flip result was (heads or tails). We're asked to quantify the second uncertainty. My approach involves quantifying both uncertainties, which might be overkill, but it produces the asked-for subjective probability of Heads.

Instead of assigning a prior distribution on those events. It looks like you are assigning two distributions, neither of which , by itself, defines a probability distribution on that space of events. Then you deduce the prior from those two distributions.

Yes. You're being asked to deduce the probabilities (from some plausible set of assumptions), not to make them up. I did that. The assumptions can be summarized by:

1. The prior probability of heads and tails are equal: P(Heads) = P(Tails) = 1/2
2. The conditional probability of heads given Monday is equal to the conditional probability of tails given Monday: P(Heads | Monday) = P(Tails | Monday) = 1/2
3. The conditional probability of Monday given tails is equal to the conditional probability of Tuesday given tails: P(Monday|Tails) = P(Tuesday | Tails) = 1/2
4. Awake is equivalent to Monday or Tails

How are the events you denote by "awake" , "Monday" and "Tuesday" defined in terms of the events in the outcomes of the experiment (as described in the Wikipedia version of the Sleeping Beauty problem) ?

Who says I need to do that? What we're asked for is to compute P(Heads | Awake).

 

[PeterDonis]

I think that only wager A corresponds to the requested credulity as described in the Wikipedia article. The interviewer would have to ask a more complicated question or have a different experimental protocol to describe wager B.

The protocol, in terms of what is done to Beauty and what questions are asked of her, is the same in both A and B. The only difference is the payoffs attached to the answers she gives. The original description in the article does not attach payoffs to the answers at all; that is a key reason why I think that description is not precise enough, and why I voted for the third option in the poll attached to this thread.

Of course if you don't think the term "subjective credence" implies a specification of payoffs, your opinion will differ as to whether the original description in the article is or is not precise enough. But I don't think that term has a unique meaning that is that specific; I've seen discussions of it (including one in one of the Wikipedia articles linked to in the article that describes this problem) that talk about subjective credence in terms of bets and payoffs. So I don't think the term "subjective credence" itself is precise enough to point at a unique specification of the problem; both of our interpretations of that term could well be valid.

 

[Stephen Tashi]

As I showed earlier, it is not necessary to include "today is Monday" in order to solve the problem. However, it is perfectly acceptable to introduce new unmeasured or unmeasurable variables that you then sum over, which is what is done with the day of the week.

I can't tell whether you assert that the events used in your solution can be defined in terms of the events in the probability space of the experiment in the Sleeping Beauty problem.

If the events "Awake", "Monday", "Tuesday" in your probability space cannot be defined in terms of events in the probability space of the experiment described in the Sleeping Beauty problem then probability theory cannot be used to deduce the probability of those events from the information that specifies that experiment.

The Sleeping Beauty problem says to consider one state in the sequence of states that the experiment produces. (i.e. we are told the coin has been flipped and that Sleeping Beauty has been awakened on a particular day). The problem does not say how this state has been selected. In particular, no stochastic process is given for picking the situation we are considering.Arguments such as:

Second, note that if you told Sleeping Beauty that the coin result was tails, then she would have no reason to think Monday more likely than Tuesday, since they are only different in the case of heads. So P(Monday | T) = 1/2.

are talking about a particular day being selected from the results of the experiment. What your argument considers can be modeled by a second experiment that picks a particular day from the results of the first experiment by some stochastic process (e.g. when coin lands tails, Monday is said to be "no more likely" than Tuesday, so we can model this by assigning each day a probability of 1/2 of being selected when the coin lands tails).A way to visualize the "halfer" argument is to say that we pick the situation we are considering by first flipping the coin. Then, if there are two days on which Sleeping Beauty will be awakened, we pick the day to consider at random, giving each day a probability of 1/2 of being selected. (This method contradicts the concept that "if you told Sleeping Beauty that today is Monday, then she would have no reason to think heads more likely than tails". )

A way to visualize the "thirder" argument is that we have 4 ordered triples: (heads, Monday, awake), (heads, Tuesday, asleep), (tails, Monday, awake), (tails, Tuesday, asleep) and we pick the situation to consider by given each choice an prior probability of 1/4 and then we pick one of the choices at random enforcing the condition that it must be one of those that contains "awake".

Since the Sleeping Beauty problem does not say what process is used to pick the situation we are considering, it is an ill-posed mathematical problem.

 

[stevendaryl]

If the events "Awake", "Monday", "Tuesday" in your probability space cannot be defined in terms of events in the probability space

The statement of the problem doesn't prescribe a probability space. It's just asking what should Sleeping Beauty's subjective likelihood of Heads be. You're free to formalize the question in terms of probability spaces however you like.

 

[stevendaryl]

Since the Sleeping Beauty problem does not say what process is used to pick the situation we are considering, it is an ill-posed mathematical problem.

That's in the nature of a thought-experiment or puzzle. Solving a problem like this typically has three parts: (1) A translation of the informal description into a mathematical question, (2) an informal argument that the translation captures the essence of the problem (or is at least a plausible way to capture it), and (3) the mathematical solution to the translated question. Calling the problem "ill-posed" just means that the real challenge is in parts 1 & 2; part 3 is usually trivial in comparison.

 

[Stephen Tashi]

The statement of the problem doesn't prescribe a probability space.

The experiment described in the Sleeping Beauty problem could be assigned as an exercise in an introductory probability class with requirement that the students define a probability space for the experiment. Yes, the students answers could take different forms, but there do exist probability spaces that describe the experiment.

It is reasonable to ask people who are doing calculations with events (like "Monday") whether the events they are using can be defined using the events in some probability space that describes the experiment. If the events a post uses cannot be defined in terms of events in a probability space for the experiment then the post cannot be deducing an answer by applying probability theory to the description of the experiment.

I agree that people can invent their own probability spaces and do calculations that have nothing to do with the probability space of the experiment. I'm just asking for clarity about whether this sort of invention is taking place. The tone of some posts is that merely applying probability theory to the description of the experiment is sufficient to compute the answer that the post advocates.

 

[stevendaryl]

A way to visualize the "halfer" argument is to say that we pick the situation we are considering by first flipping the coin. Then, if there are two days on which Sleeping Beauty will be awakened, we pick the day to consider at random, given each day a probability of 1/2 of being selected. (This method contradicts the concept that "if you told Sleeping Beauty that today is Monday, then she would have no reason to think heads more likely than tails". )

Which makes it an incorrect solution, in my opinion. As @PeroK pointed out, since the coin toss need not be consulted until Tuesday morning, there is no reason to flip it until then. So if you take the variant where you flip on Tuesday morning, then the question becomes:

On Monday, you tell Sleeping Beauty that you're going to flip a coin tomorrow. You ask her: What is the likelihood that the result will be heads?

Why would the details that tomorrow she may be asleep, and may have amnesia if she is awake affect her answer to that question?

 

[stevendaryl]

It is reasonable to ask people who are doing calculations with events (like "Monday") whether the events they are using can be defined using the events in some probability space that describes the experiment

I don't think it's reasonable. I don't see why it is necessary, or even helpful.

 

[Dale]

The protocol, in terms of what is done to Beauty and what questions are asked of her, is the same in both A and B. The only difference is the payoffs attached to the answers she gives.

No, A and B don't just change the payoff, they change the wager.

It is like two different wagers on the same horse race. The race is the same, but "Lucky Strike to win" is a different wager than "Lucky Strike to place", with a correspondingly different probability.

Here Beauty's credence "now" that the toss was heads gives the indifferent payout for your wager A. Your wager B would give something like the credence "now" that the toss was heads and that it is Monday.

 

[Stephen Tashi]

Which makes it an incorrect solution, in my opinion.

On Monday, you tell Sleeping Beauty that you're going to flip a coin tomorrow. You ask her: What is the likelihood that the result will be heads?

I don't understand your argument. You say the "halfer" model for selecting the situation we consider is incorrect. Then you consider a situation that definitely takes place on a Monday, without specifying the "correct" way for selecting the situation to be considered in the problem - which presumably would include the possibility that the day is Tuesday. What is your opinion of the correct way to pick the situation we consider?

Are you saying that if we tell Sleeping Beauty we are going to flip the coin tomorrow, that she will say the probability of heads is not 1/2? Or are you agreeing with the answer given by the "halfer" model and disagreeing with the way the model implements picking the day we consider?

As I understand the statement of the Sleeping Beauty problem, we are to consider a situations where the coin has been tossed and Sleeping Beauty has been awakened. No information about the day is specified. No information is specified that would allow the day to be deduced. So we would not tell Sleeping Beauty that the coin will tossed tomorrow if this allows her to deduce that the day is Monday.

 

[Dale]

The fact that you are arguing about whether it is important to model the day as an event shows ...

The fact that there is an argument about the topic says little if anything about the topic. There are still arguments about whether or not the Earth is flat.

 

[Dale]

I can't tell whether you assert that the events used in your solution can be defined in terms of the events in the probability space of the experiment in the Sleeping Beauty problem

I assert that P(Heads|Awake) can be calculated without any reference to P(Monday) or anything similar. Indeed, the whole day of the week thing is a "red herring" for this problem.

However, please note that, in general, it is perfectly acceptable to model unobserved or unobservable events or parameters and then marginalize over them to get the probability of observable outcomes. This is commonly done, for example where a range of possible unobservable population means is assumed and used to calculate the distribution of the observable sample mean.

So it is not necessary for Beauty to be able to observe Monday in order for her to use it to compute probabilities for observable events. What the others are doing is not wrong even though it is not necessary.

 

[stevendaryl]

I don't understand your argument. You say the "halfer" model for selecting the situation we consider is incorrect. Then you consider a situation that definitely takes place on a Monday, without specifying the "correct" way for selecting the situation to be considered in the problem - which presumably would include the possibility that the day is Tuesday. What is your opinion of the correct way to pick the situation we consider?

It's a matter of the meaning of conditional probability. Sleeping Beauty finds herself in the situation of not knowing what day it is, and not knowing what the coin flip result was (or will be). She can reason as follows:

• If today were Monday, then the coin flip result hasn't been determined yet (or at least, it's value has no effect on anything I see). Therefore, I knew that it were Monday, I would say the probability of heads is 50/50. I write down this conclusion as: P(heads | Monday) = P(tails | Monday) = 1/2.

She can also reason as follows:

• If the coin flip result was tails, then it has no impact on me at all, because I always wake up in the case of tails. Therefore, there is no difference between Monday and Tuesday in the case of tails. So if I knew that the coin flip were tails, then the probability of it being Monday would be 50/50. I write this down as: P(Monday | Heads) = P(Tuesday|Heads) = 1/2.

Are you saying that if we tell Sleeping Beauty we are going to flip the coin tomorrow, that she will say the probability of heads is not 1/2?

I'm saying the opposite. If the coin flip is definitely in the future, then she should assume a 50/50 chance of heads and tails. If the coin flip is definitely in the past, and she knows by the rules that she is only awake in the case of tails, then she knows that the chance of the result being heads is 0. If she's uncertain whether today is Monday or Tuesday, then she should use a number between 1/2 and 0, a weighted average of the two.

Or are you agreeing with the answer given by the "halfer" model and disagreeing with the way the model implements picking the day we consider?

No. I said that if Sleeping Beauty knows that the coin toss is in the future, she should use the likelihood 1/2. If she knows that the coin toss is in the past, she should use the likelihood 0. If she doesn't know whether it's in the future or in the past, she should use some number between those two. The halfer position is inconsistent.

 

[stevendaryl]

I assert that P(Heads|Awake) can be calculated without any reference to P(Monday) or anything similar. Indeed, the whole day of the week thing is a "red herring" for this problem.

I didn't see how that is done. You know that there are three possibilities, given that Sleeping Beauty is awake:

(Monday, Heads)
(Monday, Tails)
(Tuesday, Tails)

But how do you know that all three are equally likely?

 

[Dale]

I didn't see how that is done.

That was in post 255 using Bayes rule in odds form:

the best formulation would be Bayes rule stated in odds form:
$$O(H:T|A) = O(H:T) \frac{P(A|H)}{P(A|T)}$$...
The conditional probability P(A|H) is strange, but it is actually not important. What is important is the ratio of P(A|H)/P(A|T), which is clearly 1/2. So then O(H:T|A) = 1/2 (2:1 odds against H), which is a conditional probability of 1/3 for H.

 

[stevendaryl]

That was in post 255 using Bayes rule in odds form:

But the claim that P(A|H)/P(A|T) = 1/2 seems to be weighting Monday and Tuesday equally, which is implicitly saying P(Monday) = P(Tuesday) = 1/2.

Otherwise, how do you get that ratio? I would say that what we know is:

P(A | Monday & H) = P(A | Monday & T) = 1
P(A | Tuesday & H) = 0
P(A | Tuesday & T) = 1

But how do you combine the Monday and Tuesday numbers to come up with a day-independent value for P(A|H)/P(A|T)?

 

[JeffJo]

There are 4 Beautys in yours, but 1 in the OP.

And each is answering one question, based on a scenario that is either identical to the OP, or completely equivalent. In other words, the presence of two other awake Beauties does not alter the circumstances pertaining to an answer, it just facilitates the calculation.

I agree, but it is still a clear difference. Someone who is not able to calculate the probability on the original scenario is unlikely to agree that the different scenario is equivalent.

To both statements, only if they have pre-determined that they don't want to accept the easily calculated answer it provides. But that's why my first "alternate version" used only one Beauty, so this invalid objection couldn't be raised.

To refresh your memory, it didn't let the three awake Beauties interact. The answer is still easily seen to be 1/3. But maybe I need to go slow again. Do you, or do you not, agree that someone who is "not able to calculate the probability on the original scenario" will understand that the answer is trivially the same if she is given the schedule (H,Tue) and asked about Heads, or if she is dealt a random schedule and asked about a match?

 

[Dale]

But the claim that P(A|H)/P(A|T) = 1/2 seems to be weighting Monday and Tuesday equally, which is implicitly saying P(Monday) = P(Tuesday) = 1/2.

It has nothing to do with Monday or Tuesday. The T in that expression is Tails, not Tuesday.

Otherwise, how do you get that ratio?

Beauty is awoken twice if tails and once if heads. The days don't matter, only the total number of times she is awoken in the event of heads or tails.

 

[Stephen Tashi]

The halfer position is inconsistent.

Do you mean that the halfer position is inconsistent with your assumptions? It isn't inconsistent with the statement of the problem.

 

[Dale]

It isn't inconsistent with the statement of the problem.

I think that it is. I think that the problem is well specified, and it has a single correct answer of 1/3.

 

[Stephen Tashi]

I think that it is. I think that the problem is well specified, and it has a single correct answer of 1/3.

The problem is a not well specified problem of computing a probability because there is no information about how the situation to be considered is selected from among the 3 compatible situations that can arise when the experiment is performed.

In fact, if we imagine we are considering a specific situation that arises in the experiment, the "probability" the coin landed heads is either 0 or 1. It is never 1/2 or 1/3. If we wish to pose a question and claim the that the probability the coin landed heads is 1/3 or 1/2, we must say that the situation we are considering ( Sleeping beauty awake and being interviewed) has probabilities of being any of the 3 compatible situations that can arise in the experiment. The problem does not specify any procedure for selecting the situation to be considered from those 3 situations.

I agree that there may be ways to define "credence" so the answer for "credence" is well specified. But no definition for "credence" has yet been offered in this thread.

 

[PeterDonis]

A and B don't just change the payoff, they change the wager.

I don't want to quibble over the meanings of words. My point was that both A and B are consistent with the description of the scenario as given in the Wikipedia article linked to in the OP of this thread; they take the same scenario, the one described in that article, and add different things to it. There is no wager described in that article, any more than there are payoffs described, so where we draw the line between those two words is, IMO, irrelevant. (But see the end of this post for a further note about my interpretation of the Wikipedia article's description.)

Here Beauty's credence "now" that the toss was heads gives the indifferent payout for your wager A. Your wager B would give something like the credence "now" that the toss was heads and that it is Monday.

I understand that this is how you are defining "credence". I am just pointing out that this is a definition. At least, that is the view I am taking, but perhaps I should expand on it some.

I don't think "credence", or probability for that matter, is an intrinsic property. It depends on what purpose you are going to use it for. If you were to put me in Beauty's place in this experiment and ask me my "credence" that the coin would turn up heads, my question in return would be, in effect: "What are the consequences of my answer going to be?" Probabilities, credences, etc. are tools we use to guide our actions based on expected consequences of those actions. You can't compute them if you don't know the consequences of the different possibilities, and that is the crucial information that the scenario on the Wikipedia page linked to in the OP of this thread leaves out. And post #67 shows that, by adding different consequences to the same scenario, you can get different answers to the "credence" question.

To put this another way: "credence" is a word. "Probability" is a word. Words can refer to different concepts for different people. You are basically arguing that the concept "credence" should refer to (or more specifically the concept that the phrase "credence now that the coin came up heads" should refer to) is the one that leads to the answer 1/3 in the Sleeping Beauty scenario. But that, in itself, is an argument about what words should mean, not about how Beauty should answer when she's in the experiment. The latter question depends on what the consequences of her answer are going to be, and if the consequences are such that the answer that maximizes her expected return is 1/2, not 1/3 (e.g., if they are as in the B version in post #67), then saying that her answer isn't properly labeled as her "credence that the coin landed heads" doesn't change the consequences at all. It just changes what words we use to label things.

My understanding from reading the Wikipedia article and other links in the OP, and what other information I have been able to find online, is that the word "credence" does not have a sufficiently precise meaning to support the claim that, as the Sleeping Beauty scenario is described in the Wikipedia article, only the assignment of consequences in the A version in post #67 is consistent with it. That is why I said, above, that both the A and B versions in post #67 are consistent with the scenario as described in the Wikipedia article. If I'm mistaken in that belief, if in fact the term "credence" is in fact a more precise technical term in probability theory than I understand it to be, I would be interested in a reference, since my background in the probability theory literature is not very extensive.

 

[Dale]

The problem is a not well specified problem of computing a probability because there is no information about how the situation to be considered is selected from among the 3 compatible situations that can arise when the experiment is performed

I showed how you can compute the probability. There is no need to "select scenarios".

In fact, if we imagine we are considering a specific situation that arises in the experiment, the "probability" of the coin landed heads is either 0 or 1. It is never 1/2 or 1/3.
...
I agree that there may be ways to define "credence" so the answer for "credence" is well specified. But no definition for "credence" has yet been offered in this thread.

You are thinking of frequentist probability. Credence is closer to Bayesian probability. It is basically the odds at which a person would become ambivalent about accepting a wager. I would recommend reading the Wikipedia article.

 

[Marana]

Consider "time dilation beauty." On heads, beauty is given a drug that makes her experience monday and tuesday as if it were one day.

Then we can do the whole thing with no sleeping. Heads: wake monday, stay awake until tuesday night. Tails: Wake monday, memory loss on monday night that gives the impression of waking up again on tuesday morning with no memory of monday.

Now we are forced to answer P(H)=1/2 purely by symmetry, no calculations required. You are never asleep, and every experience (waking in particular) you have can be mapped by the dilation, occupying the same "proportion" of the time.

This shows that frequency and betting strategy alone are not sufficient to make the answer 1/3. Those are unchanged from the original.

If we are to answer 1/3, it will have to be based specifically on how we experience time. 1/3 is based on knowing "I am awake" for double the length of time on tails (technically by this logic the probability of "I am awake" or "it is monday" is 0 since we are almost always dead and far in the future). This seems similar to sampling bias since you can never learn "I am asleep".

Use four volunteers, and the four cards I described before (with (H,Mon), (H,Tue), (T,Mon), and (T,Tue) written on them). Deal the cards to the four, and put them in separate rooms. Using one coin flip, and waken three of them on Monday, and Tuesday. Leave the one whose dealt card matches both the day, and the coin flip, asleep. Ask each for her confidence that the coin matches her card.

Consider "the 1001 beauties."

On sunday, 1 beauty is randomly selected as the winner. The other 1000 lose. The winner wakes up every day for 1000 days. The losers each wake up on one day in random order.

Using your argument, when they wake up each beauty should have 1/2 credence that they are the winner. And that seems plausible since they wake up next to another beauty who seemed to begin with the same information, and therefore should divide the probability evenly between them.

But there is first-person information which cannot be shared. If we were to select a random beauty, then randomly select someone who wakes up next to them, the chances would be 1000/1001 that the person waking up next to them is the winner. If we were to select a random day, and then ask the probability of each beauty on that day being the winner, it would be 1/2. But I believe that each beauty can begin by reasoning that "I am randomly selected from the beauties", "I am awake today as a result of a process of being randomly selected from the beauties and then having the day(s) of awakening randomly chosen", information which is first-person, related to the process of discovering information, and can't be shared. If I believe that I am randomly selected from the beauties I can't believe that a beauty waking up next to me is also randomly selected from all the beauties. After all, I was almost certain I'd be waking up next to the winner.

 

[PeterDonis]

Credence is closer to Bayesian probability. It is basically the odds at which a person would become ambivalent about accepting a wager.

This is my understanding, and as I expounded in my last post, it means you have to know the terms of the wager--or, as I put it in that post, you have to know the consequences of your answer. Otherwise you don't know what your "credence" is. And the description of the scenario in the Wikipedia article on the Sleeping Beauty problem doesn't describe any consequences at all. So the information given there is insufficient to determine an answer.

 

[PeterDonis]

every experience (waking in particular) you have can be mapped by the dilation

But the mapping is not one-to-one. Your argument that we must have P(H) = 1/2 by symmetry is only valid if the mapping is one-to-one. Otherwise the mapping itself breaks the symmetry.

 

[stevendaryl]

This is my understanding, and as I expounded in my last post, it means you have to know the terms of the wager--or, as I put it in that post, you have to know the consequences of your answer. Otherwise you don't know what your "credence" is. And the description of the scenario in the Wikipedia article on the Sleeping Beauty problem doesn't describe any consequences at all. So the information given there is insufficient to determine an answer.

As you and @Demystifier pointed out, there is a payoff scheme that supports the halfer position, but I have to say it's a little strange: Make the rule that only the last bet counts. That puts the first bet (in the case of two) in a strange position: You want to define probability in terms of consequences, but the first bet has no consequences. Of course, the person making the bet doesn't know that it has no consequences...

 

[stevendaryl]

Consider "time dilation beauty." On heads, beauty is given a drug that makes her experience monday and tuesday as if it were one day.

Then we can do the whole thing with no sleeping. Heads: wake monday, stay awake until tuesday night. Tails: Wake monday, memory loss on monday night that gives the impression of waking up again on tuesday morning with no memory of monday.

Now we are forced to answer P(H)=1/2 purely by symmetry, no calculations required. You are never asleep, and every experience (waking in particular) you have can be mapped by the dilation, occupying the same "proportion" of the time.

I don't see much symmetry between the heads and tails situations. How do you get 1/2 from this situation?

Sleeping Beauty could reason, as in the original situation:

• Under the assumption that today is Monday, I would conclude that there is an equal probability of heads and tails. So P(Heads & Monday) = P(Tails & Monday)
• Under the assumption that the coin toss result was tails, I would conclude that there is an equal probability of it being Monday or Tuesday. So P(Tails & Monday) = P(Tails & Tuesday)

Those two assumptions imply that the three probabilities are equal: P(Heads & Monday) = P(Tails & Monday) = P(Tails & Tuesday)

(By "it is monday", I mean that according to nondilated clocks.)

I'm not sure I understand how the time dilation affects the answer.

If we are to answer 1/3, it will have to be based specifically on how we experience time. 1/3 is based on knowing "I am awake" for double the length of time on tails (technically by this logic the probability of "I am awake" or "it is monday" is 0 since we are almost always dead and far in the future). This seems similar to sampling bias since you can never learn "I am asleep".

I am not sure I agree that the thirder answer is based on knowing "I am awake" is experienced for double the length of time. Instead, it's based on the question being asked double the number of times.