Undergrad The Sleeping Beauty Problem: Any halfers here?

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The Sleeping Beauty problem raises a debate between "thirders," who argue the probability of the coin landing heads is 1/3, and "halfers," who believe it is 1/2. Proponents of 1/3 argue that the princess's amnesia prevents her from gaining new information, thus her a posteriori probability remains unchanged. Conversely, halfers contend that since she learns nothing new upon waking, her initial probability of 1/2 should hold. The discussion also explores various scenarios and thought experiments to illustrate the implications of the problem, emphasizing the importance of conditional probabilities. Ultimately, the debate hinges on interpreting the information available to the princess at the moment she is awakened.

What is Sleeping Beauty's credence now for the proposition that the coin landed heads?

  • 1/3

    Votes: 12 33.3%
  • 1/2

    Votes: 11 30.6%
  • It depends on the precise formulation of the problem

    Votes: 13 36.1%

  • Total voters
    36
  • #301
Boing3000 said:
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?
 
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  • #302
Boing3000 said:
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?
 
  • #303
PeterDonis said:
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.
 
  • #304
Dale said:
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.

Dale said:
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.

Dale said:
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.
 
  • #305
Edit: I tried to come up with an equivalent situation but I don't think it worked.
 
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  • #306
stevendaryl said:
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.
 
  • #307
Stephen Tashi said:
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.
 
  • #308
Dale said:
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.
 
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  • #309
PeterDonis said:
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.

PeterDonis said:
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?
 
  • #310
Stephen Tashi said:
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.
 
  • #311
JeffJo said:
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.
 
  • #312
JeffJo said:
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.
 
  • #314
stevendaryl said:
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".
 
  • #315
Stephen Tashi said:
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.
 
  • #316
stevendaryl said:
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?
 
  • #318
Stephen Tashi said:
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.
 
  • #319
Dale said:
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.
 
  • #320
JeffJo said:
But it does not differ
There are 4 Beautys in yours, but 1 in the OP.
 
  • #321
Dale said:
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.
 
  • #322
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.
 
  • #323
Stephen Tashi said:
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).
 
  • #324
Dale said:
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.
 
  • #325
Dale said:
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.
 
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  • #326
Stephen Tashi said:
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.
 
  • #327
Stephen Tashi said:
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.
 
  • #328
stevendaryl said:
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.
 
  • #329
Stephen Tashi said:
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?
 
  • #330
Stephen Tashi said:
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.
 

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