The Sleeping Beauty Problem: Any halfers here?

[Stephen Tashi]

The "halfer" case hasn't been presented concisely. Here is one version:

Using my previous notation, when we determine the probabilities for which event is selected from $\{A,B,C\}$ we can enforce the constraint that the event selected must be one that actually occurred in the experiment. There is probability of 1/2 that event A = (heads, awake,Monday) is the only event available for selection. If the coin lands tails then either B=(tails, awake,Monday) or C = (tails, awake, Tuesday) can be selected. If desired, one can use the Principle of Indifference to set P(B is selected | tails) = P(C is selected | tails) = 1/2. However, as far as computing the probability of heads, how those probabilities are set doesn't matter.A "halfer" criticism of applying the Principle of Indifference to events $\{A,B,C\}$ is that it assumes that during each specific run of the experiment, all three of those events are available for selection. Another criticism is that a "thirder" answer based on assuming Sleeping Beauty is equally likely to be in each of situations $\{A,B,C\}$ implies the conditional probability of heads is 1/3 independently of whether a fair coin is used in the experiment.

Do you mean my assumptions that Beauty is rational and the experimenters are not malicious?

No, I mean your specific assumptions about the values of probabilities, for which you appear to be using The Principle of Indifference.

 

[Dale]

No, I mean your specific assumptions about the values of probabilities, for which you appear to be using The Principle of Indifference.

Both P(H) and P(A|H)/P(A|T) are given in the problem.

 

[PeterDonis]

I went through and found several sites discussing how credence is mapped to a wager.

Thanks for the references. I was hoping for something more like a standard textbook on probability and statistics, but these are helpful.

It defines the credence as the price at which a reasonable person would buy a $100 wager on X, but since what constitutes X is at dispute here I think that doesn't help.

I agree as far as the original specification in the Wikipedia article linked to in the OP is concerned. But just to clarify, I'm not saying it's impossible, or even very difficult, to get a clear and undisputed definition of X. I thought post #67, scenario A, was fine as a clear definition of X for the thirder position, which is the position you appear to be supporting. It just seems as though I see a difference in the level of precision/clarity between scenario A and the original specification in the Wikipedia article linked to in the OP, and you don't. In any real situation, such a difference of opinion could easily be settled (all you would have to tell me is "I mean scenario A in post #67 as the wager", and I would bet as a thirder).

 

[JeffJo]

Using my previous notation, when we determine the probabilities for which event is selected from $\{A,B,C\}$ we can enforce the constraint that the event selected must be one that actually occurred in the experiment. There is probability of 1/2 that event A = (heads, awake,Monday) is the only event available for selection. If the coin lands tails then either B=(tails, awake,Monday) or C = (tails, awake, Tuesday) can be selected. If desired, one can use the Principle of Indifference to set P(B is selected | tails) = P(C is selected | tails) = 1/2. However, as far as computing the probability of heads, how those probabilities are set doesn't matter.

And in this schema, "awake" is not a valid discriminator of what is, or is not, an event. It always happens, and in fact can happen twice. So your $\{A,B,C\}$ does not represent a partition in any way.

But $\{A1,A2,B,C\}$, where A1 = (Heads, Monday), A2=(Heads, Tuesday), B=(Tails,Monday), C=(Tails,Tuesday), is a partition from Beauty's point of view. There is a prior probability of 1/2 that each of these events will occur.

"Awake" is just a way that Beauty can observe what event has occurred. "Awake" means only that A2 is ruled out, so "A1 or B or C" is left. Pr(A1|A1 or B or C) = Pr(A1 and (A1 or B or C))/[Pr(A1)+Pr(B)+Pr(C)] = (1/2)/(3/2) = 1/3.

 

[Dale]

It just seems as though I see a difference in the level of precision/clarity between scenario A and the original specification in the Wikipedia article linked to in the OP, and you don't.

Yes, I think that sums it up pretty well. I do think your description was more clear, but that the Wikipedia article was sufficiently clear as to be unambiguous.

 

[Stephen Tashi]

Both P(H) and P(A|H)/P(A|T) are given in the problem.

The probability of the event involving "awake" is not given in the problem.

Using the notation
A = (heads, awake,Monday)
B = (tails, awake,Monday)
C = (tails, awake, Tuesday)
D = (heads, asleep, Monday)

and taking the notation to denote events that "do happen" during the experiment,
the problem gives the information
P(H) = 1/2, P(not-H) = 1/2
P(($A \cap D$) | H ) = 1, P(( $B \cup C$) | H) = 0
P(($B \cap C$ | not-H) = 1, P($A \cup D$) | not-H) = 0

You can define the event "awake|tails" to be $B \cup C$ , but B and C are not mutually exclusive events.
The fact that the coin landed tails and Sleeping Beauty is awake on Monday does not exclude the possibility that the coin landed tails and Sleeping Beauty is awake on Tuesday.

If you want to introduce an interpretation of events where B and C become mutually exclusive events then you must imagine some selection process different than the experiment. The question posed by the Sleeping Beauty problem requires that we imagine such a situation. We are asked to imagine a situation where one and only one of the events {A,B,C} is realized. (i.e. We are told Sleeping Beauty is awakened on one particular but unspecified day and with one particular but unspecified state of the coin.)

 

[Stephen Tashi]

Pr(A1|A1 or B or C) = Pr(A1 and (A1 or B or C))/[Pr(A1)+Pr(B)+Pr(C)]

By using Pr(A1) + Pr(B) + Pr(C), you are treating A1,B,C as mutually exclusive events. They are not mutually exclusive events in the experiment.

To claim A1,B,C each have prior probability 1/2, you must refer to the events A1,B,C as they are defined in the experiment. Using that definition, the events B and C are not mutually exclusive.

If we wish to create a scenario where A1,B,C are involved in mutually exclusive events , we can construct a situation where one and only one of A1,B,C is selected. We need a new set of events

a1 = A1 is selected
b = B is selected
c = C is selected

(If people don't like the verb "selected" they can phrase it as "A1 is the situation when Sleeping Beauty awakes", etc.)

A "halfer" objection to applying the Principle of Indifference to a1,b,c is that, if we are restricted to selecting an event from one that actually happened in the experiment then A1,B,C are not simultaneously available to pick from. The selection must either be made from {A1} or from {B,C}.

A "thirder" position is that Sleeping Beauty is "totally ignorant" of both the state of the coin and the day of the week, so , she is permitted to consider a1,b,c equally likely.

The statement of the problem doesn't defined any probability distribution on the set of events {a1,b,c}.

 

[Demystifier]

I am impressed with how many replies such a simple question can evoke.

 

[stevendaryl]

If you want to introduce an interpretation of events where B and C become mutually exclusive events then you must imagine some selection process different than the experiment

I don't think anything else is a reasonable interpretation of the problem. You wake up Sleeping Beauty, explain the situation with the coin and the memory wipe, then you can ask her:

What is the likelihood that today is Monday?
What is the likelihood that today is Tuesday?

But the question: What is the likelihood that today is both Monday and Tuesday? Doesn't make any sense.

Asking about selection processes seems contrary to the statement of the problem. What happens to Sleeping Beauty is explicitly specified. You're asking her what she knows, and if she doesn't know, can she give a numerical level of confidence in the various possibilities.

 

[stevendaryl]

The statement of the problem doesn't defined any probability distribution on the set of events {a1,b,c}.

It seems to me that that's the goal of the problem, to come up with a reasonable distribution.

 

[Dale]

The probability of the event involving "awake" is not given in the problem.

True, but the ratio of P(A|H)/P(A|T) is given.

 

[stevendaryl]

True, but the ratio of P(A|H)/P(A|T) is given.

I would think that you would have to deduce that. That ratio isn't explicitly given, is it?

 

[Dale]

I would think that you would have to deduce that. That ratio isn't explicitly given, is it?

"if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday." So P(A|H)/P(A|T)=1/2. The comments about the day of the week are irrelevant. All that matters is that she is awakened twice as often if tails than if heads.

 

[JeffJo]

Please note that I never include "asleep" or "awake" in the definitions of my events. They do not belong as discriminators of the outcomes, because they are totally determined by the other two factors that are.

By using Pr(A1) + Pr(B) + Pr(C), you are treating A1,B,C as mutually exclusive events. They are not mutually exclusive events in the experiment.

They are not mutually exclusive when you use an external perspective - that is, from Beauty's frame of reference on Sunday or Wednesday; or the lab tech's who is administering the drugs and asking the questions.

They are mutually exclusive events from the point of view of an awake Beauty. And this is the entire point of the problem.

To claim A1,B,C each have prior probability 1/2, you must refer to the events A1,B,C as they are defined in the experiment. Using that definition, the events B and C are not mutually exclusive.

Correct. Each has a probability of 1/2 in the prior - that is, from Beauty's frame of reference on Sunday or Wednesday. And A2 belongs in that prior. To define things "as they are defined in the experiment," you must include A2.

In the posterior - from the point of view of an awake Beauty - we gain "alternate information" that requires an update. This information is that B and C are now mutually exclusive, as are A1 and A2. And that the observation "awake" excludes A2, which is in the prior. I'm sorry if I'm repeating myself, but these facts keep getting ignored.

I also realize that this is an unprovable assertion of how perspective applies. All I ask, is that you recognize that your counterpart assertion, that A2 is not required, is equally unprovable. So, it is only if we can derive an answer that is not based in either perspective, that we can try to address which perspective applies. I think I have done that, and you are avoiding it by trying to discuss perspective first.

If we wish to create a scenario where A1,B,C are involved in mutually exclusive events , we can construct a situation where one and only one of A1,B,C is selected.

You also need to include A2, since from Beauty's frame of reference on Sunday or Wednesday, or the lab tech's who is administering the drugs and asking the question, A2 is a possibility.

Please, please, please try to understand this concept: The occurrence of event A2 has nothing to do with Beauty's ability to observe it. It can occur. Being awake is not a "selection," it is an observation of what has been selected. If you want to examine that more in depth, consider the scenario where Beauty is taken to Disneyworld instead of being interviewed under A2. Ask yourself if it matters how (or if) Beauty can observe A2 when occurs.

A "halfer" objection to applying the Principle of Indifference to a1,b,c is ...

... misguided, because it treats Beauty's inability to observe A2 as a de facto assumption that A2 can't happen. A2 can happen. The PoI is applied to {A1,A2,B,C}, not {A1,B,C}.

Using Elga, the ...

... "thirder" position is that Sleeping Beauty ...

... avoids the fact that A2 is a possibility by applying the PoI separately to {A1, B} (that is, assuming "Monday") and (A1, C} (that is, by assuming "Tails"). It treats each as a subset of the larger set {A1,A2,B,B} where the PoI applies to the entire set.

 

[Stephen Tashi]

"if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday." So P(A|H)/P(A|T)=1/2.

How are you defining the event "A"? If "A" means "Sleeping beauty is awakened" then P(A|H) = P(A|T) = 1 because sleeping beauty is always awakened during the experiment.

It is correct that Sleeping Beauty is awakened on 2 days if the coin lands tails and one day if the coin lands heads. As you know, numbers of outcomes alone don't define probabilities unless each of the outcomes is assumed to have an equal probability and the outcomes are mutually exclusive. To make those numbers relevant to a probability < 1, you must introduce some selection process such as "Pick a situation that happens in the experiment from some probability distribution". Then you can ask "What is the probability that Sleeping Beauty is awake in that (single) situation?". The problem does not specify what distribution to use when you pick a situation in which Sleeping Beauty is awake.

I think your interpretation for "A" refers to an an experiment where a situation from the 4 possible situations is chosen at random, giving each situation an equal probability of being selected. P(A|H) is the probability that the situation selected is one where Sleeping Beauty is awake given the condition that we have selected a situation where the coin landed heads.

The essence of the "halfer" versus "thirder" controversy is whether that experiment is a model appropriate to answering the question posed by the problem.

Is Sleeping Beauty to model the selection of her situation by an experiment where one situation (such as (heads, Monday, asleep) ) is selected at random from 4 possible situations each with probability 1/4? (Using the value 1/4 requires assuming the Principle of Indifference on a set of 4 events, so it isn't information given in the the problem.)

Or is Sleeping Beauty to model the selection of her situation by an experiment where her situation is selected only from those situations where she is awake? (Thus giving the situation (heads, Monday , awake) a probability of 1/2 of being selected because when coin lands heads, it is the only situation that is offered for selection).

The wording of the question asks about what happens "when Sleeping Beauty is awakened". It does specify any probability distribution for picking situations. If the question said "The experiment is run. From those days on which Sleeping Beauty is awakened, a day is picked at random..." this would favor the "halfer" viewpoint. If the question said "Before the experiment is run, one of the situations that arise in the experiment is chosen at random..." this would favor the "thirder position".

 

[Dale]

How are you defining the event "A"? If "A" means "Sleeping beauty is awakened"

Awaken and interviewed, yes.

then P(A|H) = P(A|T) = 1

I never calculate either P(A|H) or P(A|T), but they are certainly not equal.

It is correct that Sleeping Beauty is awakened on 2 days if the coin lands tails and one day if the coin lands heads. As you know, numbers of outcomes alone don't define probabilities ...

To change from frequencies to probabilities does require a normalization constant, but that constant factors out in the ratio. So relative frequencies do define the ratio of probabilities (or Bayes factor). This is a common technique.

All of the rest of the complications you add are unnecessary. I never calculate the actual probabilities.

 

[Stephen Tashi]

They are mutually exclusive events from the point of view of an awake Beauty. And this is the entire point of the problem.

You need to define what probability space you are using in your reasoning and what events that space contains.

Correct. Each has a probability of 1/2 in the prior - that is, from Beauty's frame of reference on Sunday or Wednesday. And A2 belongs in that prior. To define things "as they are defined in the experiment," you must include A2.

The term "frame of reference" has a meaning in physics, but it doesn't have a specific meaning in the theory of probability You seem to be talking about conditional probabilities. If so, it would be clearer just to speak in terms conditional probabilities and the sets of events that define the conditions.

I also realize that this is an unprovable assertion of how perspective applies. All I ask, is that you recognize that your counterpart assertion, that A2 is not required, is equally unprovable. So, it is only if we can derive an answer that is not based in either perspective, that we can try to address which perspective applies. I think I have done that, and you are avoiding it by trying to discuss perspective first.

It's difficult to interpret you remarks. I can interpret them to be in agreement with my last reply to Daryl. That reply says there are are (at least) two distinct probability models for a process satisfying the phrase "Sleeping Beauty is awakened (in some situation)".

One model is the probability space is the set of events S = {A1,A2,B,C} with each event having probabilitiy 1/4. Using that space, the probability space for selecting Beauty's situation is the set of events {A1,B,C} with corresponding probabilities $P(A1| (A1\cup B \cup C)),\ P(B| (A1\cup B \cup C)),\ P(C| (A1\cup B\cup C))$ where the conditional probabilities are computed using the probability distribution define on S.

A different model is the probability space T consisting of the set of events {A1,B,C} with corresponding probabilites {1/2, 1/4, 1/4}. That model uses the concept that the experiment is run and Beauty's situation is selected only from those situations where Beauty was awake in the particular run of the experiment.

You also need to include A2, since from Beauty's frame of reference on Sunday or Wednesday, or the lab tech's who is administering the drugs and asking the question, A2 is a possibility.

Which model to use is a subjective judgment. The "thirder model" It is a reasonable opinion, but it can't be proven that the information given in the problem implies that it is the only model for the process "Sleeping Beauty is awakened in some situation".

 

[Stephen Tashi]

I never calculate either P(A|H) or P(A|T), but they are certainly not equal.

That can't be proven from the information given in the problem.

You have not precisely defined the event denoted by "A". If "A" refers to the union of all events of the form ( coin state = any, day = any, Beauty state = awake) then the given information about the experiment says that P( A | H) = 1 because the set A includes the event (heads, Monday, awake) which always occurs when the coin lands heads.

I think you wish "A" to be an event defined in a probability space where (heads, Monday, awake) and (heads, Tuesday, asleep) are mutually exclusive events. The information in the problem does not assign any probability measure on such a space.

 

[Stephen Tashi]

Asking about selection processes seems contrary to the statement of the problem. What happens to Sleeping Beauty is explicitly specified. You're asking her what she knows, and if she doesn't know, can she give a numerical level of confidence in the various possibilities.

I agree that those things are done. However, the focus of controversy is whether there is a correct answer that Beauty should give if she is "rational". An objectively correct answer would be the numerical value of the probability that the situation that occurs when beauty is awakened is a situation where the coin landed heads. I am using terminology that a situation is "selected". when Beauty is awakened. That seems reasonable terminology for a probability model of what happens. "Beauty is awakened" implicitly says she is awakened in some particular situation. One particular situation is "realized" or "occurs" or "is selected". The choice of verb is a matter of taste.

 

[Dale]

That can't be proven from the information given in the problem.

Yes, it can. Their ratio is 1/2 so they cannot be equal.

You have tied yourself in mental knots here. This is far simpler than you are making it out to be. A happens twice as frequently given T as it does given H, therefore P(A|H)/P(A|T)=1/2. It is not necessary to calculate either probability individually to know their ratio.

 

[Stephen Tashi]

Yes, it can. Their ratio is 1/2 so they cannot be equal.

You are asserting the ratio is 1 to 2 without precisely defining the event A or showing any calculations that derive the result from given information.

You have tied yourself in mental knots here. This is far simpler than you are making it out to be.

If that were true, the problem would no longer be controversial among academics.

A happens twice as frequently given T as it does given H

To demonstrate that, you must define what you mean by "A" and say what probability space you are using.

 

[Dale]

precisely defining the event A

"Awakened and interviewed" as defined in the problem statement.

or showing any calculations that derive the result from given information

The frequency is 2 given T or 1 given H. I don't know how to derive that 1/2=1/2.

 

[Stephen Tashi]

"Awakened and interviewed" as defined in the problem statement.

That is not a specific definition of "A".

Are you saying that (your) "A" is the union of the events:
(my) A = The coin lands heads. Sleeping Beauty is awakened and interviewed on Monday
B = The coin lands tails. Sleeping Beauty is awakened and interviewed on Monday
C = The coin lands tails. Sleeping Beauty is awakened and interviewed on Tuesday

If so, P(your A | the coin lands heads) = 1. That is information given about the experiment.

It seems to me that you wish A to be the union of events like:
a = The coin landed heads. Today is Monday. Today, Sleeping beauty is awakened and interviewed.
b = The coin landed tails. Today is Monday. Today, Sleeping beauty is awakened and interviewed.
c = The coin landed tails. Today is Tuesday. Today, Sleeping beauty is awakened and interviewed.

The description of the experiment gives no information about how the situation "today" is selected.

I agree that one plausible model for selecting the situation "today" is that a situation is selected from 4 possible situations giving each situation an equal probability. Then if the situation selected has Sleeping Beauty alseep, that choice is discarded and we select a different situation at random until we have picked a situation where Sleeping Beauty is awake.

The "halfer" model for selecting the situation "today" is that the situation "today" is selected by first running the experiment. Then the situation "today" is selected at random from the situations that occurred in the experiment , giving each situation an equal probability of being selected. If the situation selected has Sleeping Beauty asleep, that choice is discarded and another situation is selected until a situation where Sleeping Beauty is awake has been chosen.

 

[Dale]

That is not a specific definition of "A".

In the context of this problem it is exactly specific. The problem says "awakened and interviewed" happens twice if "tails" and once if "heads". So "A", "H", and "T" are just shorthand for actual statements in the description.

Are you saying that (your) "A" is the union of the events

I am not saying anything about your definitions and I am not even attempting to map my definitions to yours. The problem statement says that "awakened and interviewed" happens one or more times during the experiment, and I prefer to type "A" rather than "awakened and interviewed". This has nothing to do with your notation or events.

The description of the experiment gives no information about how the situation "today" is selected.

Which is why I chose a form of Bayes rule that does not require that information.

 

[stevendaryl]

"if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday." So P(A|H)/P(A|T)=1/2. The comments about the day of the week are irrelevant. All that matters is that she is awakened twice as often if tails than if heads.

To me, that's a deduction from the facts, and is not obviously equivalent to what is said. We're told:

• If the coin flip was Heads, and today is Monday, then Sleeping Beauty is awake.
• If the coin flip was Heads, and today is Tuesday, then Sleeping Beauty is asleep.

To conclude that "If the coin flip was Heads, then there is for Sleeping Beauty a 50% chance of being awake" seems to me to require some kind of deduction. I would argue that:

P(Awake | Heads) = P(Awake | Heads & Monday) P(Monday | Heads) + P(Awake | Heads & Tuesday) P(Tuesday | Heads) = 1 * P(Monday | Heads) + 0 * P(Tuesday | Heads)

Saying P(Awake | Heads) = 1/2 is equivalent to me to saying P(Monday | Heads) = P(Tuesday | Heads) = 1/2.

Of course, you could do away with the Monday versus Tuesday, and simply consider a single day, where the probability of being awakened is dependent on the coin toss:

• If the coin toss was Heads, then awaken her with 50% probability
• If the coin toss was Tails, then awaken her with 100% probability

With this variant, there is only a single day involved, and you can ask the probability of heads given that she is awake, and your analysis is pretty straight-forward.

 

[Stephen Tashi]

In the context of this problem it is exactly specific. The problem says "awakened and interviewed" happens twice if "tails" and once if "heads". So "A", "H", and "T" are just shorthand for actual statements in the description.

That still doesn't define the event "A" or the probability space that contains it.

The problem statement says that "awakened and interviewed" happens one or more times during the experiment, and I prefer to type "A".

The problem does say that when the coin lands heads there is one situation where SB is awake and when the coin lands tails there are 2 situations where she is awake. But the problem says nothing about how to pick a single situation from the 4 possible situations that can arise in a series of experiments or how to pick one situation from the two possible situations that do arise in one particular experiment.

Are you using "A" to represent the event that "the situation on a particular day is one where SB is awakened and interviewed"? Assigning that event a probability implies a single situation is selected (or "happens" or "is realized) from the possible situations. The problem does not say how to make such a selection.

 

[Stephen Tashi]

One way to see that the Sleeping Beauty problem is ill-posed to imagine conducting an empirical test to see whether the "thirder" or "halfer" or some other viewpoint is correct.

Conduct the Sleeping Beauty experiment N times and record the results of each run of the experiment.

Now implement the process that determines "The situation when Sleeping Beauty is awakened".

How to implement that process is not specified by the problem.

The "thirder method": From all situations in the records of all the experiments pick a situation at random giving each situation the same probability of being selected.

The "halfer" method: From all the experiments in the records, pick an experiment at random giving each experiment the same probability of being selected. Then, from the records of the selected experiment, pick one of the situations in the experiment at random, giving each situation the same probability of being selected.

The bewildering aspect of the Sleeping Beauty problem is that these two "random looking" methods imply different answers for the conditional probability of the event "The situation is one where the coin landed heads given that it is a situation is one where Sleeping Beauty is awake".

For example, the probability of that event by the "halfer" method is: (1/2)(1) + (1/2)(0) because if we select an experiment where the the coin lands heads, the condition "it is a situation Sleeping Beauty is awake" requires that we select 1 event from the single choice (coin landed heads, Today is Monday, Today, Sleeping Beauty is awakened and interviewed).

The probability of the event in question by the "thirder" method is 1/3 -as has been demonstrated in several posts in this thread that tacitly assume the "thirder" method is the one used to determine the situation when Sleeping Beauty awakes.

 

[kith]

The blog post by Allen Downey which is cited in the OP contains an interesting dialogue. It is about a different bet than the ones considered in #67 and #372 and it highlights the role of forgetting.

Experimenter: Ok, SB, it’s Sunday night. After you go to sleep, we’re going to flip this fair coin. What do you believe is the probability that it will come up heads, P(H)?

Sleeping Beauty: I think P(H) is ½.

Ex: Ok. In that case, I wonder if you would be interested in a wager. If you bet on heads and win, I’ll pay 3:2, so if you bet $100, you will either win $150 or lose $100. Since you think P(H) is ½, this bet is in your favor. Do you want to accept it?

SB: Sure, why not?

Ex: Ok, on Wednesday I’ll tell you the outcome of the flip and we’ll settle the bet. Good night.

SB: Zzzzz.

Ex: Good morning!

SB: Hello. Is it Wednesday yet?

Ex: No, it’s not Wednesday, but that’s all I can tell you. At this point, what do you think is the probability that I flipped heads?

SB: Well, my prior was P(H) = ½. I’ve just observed an event (D = waking up before Wednesday) that is twice as likely under the tails scenario, so I’ll update my beliefs and conclude that P(H|D) = ⅓.

Ex: Interesting. Well, if the probability of heads is only ⅓, the bet we made Sunday night is no longer in your favor. Would you like to call it off?

SB: No, thanks.

Ex: But wait, doesn’t that mean that you are being inconsistent? You believe that the probability of heads is ⅓, but you are betting as if it were ½.

SB: On the contrary, my betting is consistent with my beliefs. The bet won’t be settled until Wednesday, so my current beliefs are not important. What matters is what I will believe when the bet is settled.

Ex: I suppose that makes sense. But do you mean to say that you know what you will believe on Wednesday?

SB: Normally I wouldn’t, but this scenario seems to be an unusual case. Not only do I know that I will get more information tomorrow; I even know what it will be.

Ex: How’s that?

SB: When you give me the amnesia drug, I will forget about the update I just made and revert to my prior. Then when I wake up on Wednesday, I will observe an event (E = waking up on Wednesday) that is equally likely under the heads and tails scenarios, so my posterior will equal my prior, I will believe that P(H|E) is ½, and I will conclude that the bet is in my favor.

Ex: So just before I tell you the outcome of the bet, you will believe that the probability of heads is ½?

SB: Right.

Ex: Well, if you know what information is coming in the future, why don’t you do the update now, and start believing that the probability of heads is ½?

SB: Well, I can compute P(H|E) now if you want. It’s ½ -- always has been and always will be. But that’s not what I should believe now, because I have only seen D, and not E yet.

Ex: So right now, do you think you are going to win the bet?

SB: Probably not. If I’m losing, you’ll ask me that question twice. But if I’m winning, you’ll only ask once. So ⅔ of the time you ask that question, I’m losing.

Ex: So you think you are probably losing, but you still want to keep the bet? That seems crazy.

SB: Maybe, but even so, my beliefs are based on the correct analysis of my situation, and my decision is consistent with my beliefs.

Ex: I’ll need to think about that. Well, good night.

SB: Zzzzz.

 

[Dale]

To me, that's a deduction from the facts, and is not obviously equivalent to what is said

Hmm, I am not sure why that is not obvious. If I count warts on my head and warts on my tail and I consistently find twice as many warts on my tail then I can certainly say that the probability of getting a wart on my tail is twice the probability of getting a wart on my head. I can do so without ever determining the probability of getting a wart on my tail.

To conclude that "If the coin flip was Heads, then there is for Sleeping Beauty a 50% chance of being awake" seems to me to require some kind of deduction.

I don't make that conclusion.

Saying P(Awake | Heads) = 1/2 is equivalent to me to saying P(Monday | Heads) = P(Tuesday | Heads) = 1/2

I never say that. I have no idea what P(A|H) is.

 

[Dale]

The problem does say that when the coin lands heads there is one situation where SB is awake and when the coin lands tails there are 2 situations where she is awake.

Hence P(A|H)/P(A|T)=1/2.

But the problem says nothing about how to pick a single situation from the 4 possible situations

Nor does it need to.

Assigning that event a probability implies a single situation is selected (or "happens" or "is realized) from the possible situations.

I never assign that event a probability.

 

[stevendaryl]

Hmm, I am not sure why that is not obvious. If I count warts on my head and warts on my tail and I consistently find twice as many warts on my tail then I can certainly say that the probability of getting a wart on my tail is twice the probability of getting a wart on my head. I can do so without ever determining the probability of getting a wart on my tail.

It is certainly not obvious to me. Why is the ratio of number of days awake the same as the probability of being awake? That seems to assume equal likelihood for the days, which is P(Monday) = P(Tuesday) = 1/2.

I don't make that conclusion.

Well, that's what I don't understand. What you said seems mathematically equivalent to what you're denying.

You said you're assuming P(A|H)/P(A|T) = 1/2. That sounds plausible, but I think it requires proof. But however you derive it, or assume it, it's equivalent to the claim P(A|H) = 1/2, because P(A|T) = 1. (If the coin toss is tails, then she's going to be awake every morning).

To me, the most direct way to the conclusion is this:

There are three possible situations consistent with Sleeping Beauty being awake:

1. Heads & Monday
2. Tails & Monday
3. Tails & Tuesday

To me, it seems that we can reason:

• If the coin toss is tails, then there is nothing observably different about Monday and Tuesday for Sleeping Beauty. So P(Tails & Monday) = P(Tails & Tuesday)
• If the day is Monday, then there is nothing observably different between heads and tails (because the difference only shows up on Tuesday). So P(Tails & Monday) = P(Heads & Monday)

The above assumptions imply that all three are equally likely.

 

[stevendaryl]

You said you're assuming P(A|H)/P(A|T) = 1/2. That sounds plausible, but I think it requires proof. But however you derive it, or assume it, it's equivalent to the claim P(A|H) = 1/2, because P(A|T) = 1. (If the coin toss is tails, then she's going to be awake every morning).

I should clarify that when I talk about being awake, I mean awake during some specified time set aside for conducting interviews. Whether she is awake at other times is not relevant. So P(A|T) = 1.

 

[Stephen Tashi]

But however you derive it, or assume it, it's equivalent to the claim P(A|H) = 1/2, because P(A|T) = 1. (If the coin toss is tails, then she's going to be awake every morning).

It's interesting that both the "halfer" and "thirder" methods described in post #412 agree that P(A|H) = 1/2 and P(A|T) = 1. However, they get different answers for P(H|A). So the assumption that P(A|H)/P(A|T) = 1/2 isn't sufficient to determine P(H|A).

 

[bahamagreen]

I'm in the halfer camp. It looks to me like before Beauty is put to sleep Sunday night, she knows:

- a coin will be tossed p(H)=.5
- H=Monday interview, T=Monday and Tuesday interviews, two possibilities
- a Monday awakening is twice as likely as a Tuesday awakening, irrelevant to p(heads)
- all interviews will be identically empty in that she won't know the day nor gain new information
- all interviews will conclude with the drug so she won't retain any potential new information
- the drug will be administered after each interview, so she will not know upon being awakened on Wednesday what day it is until she is informed or realizes she is not to be interviewed

- there are not three possibilities, only two:
1) (heads) Monday awakening
2) (tails) Monday and Tuesday awakening
- she will be unable to tell which of these two possibilities is the case within any interview

- the idea of three possibilities is incorrect:
Monday awakening (heads)
Monday awakening (tails)
Tuesday second awakening (tails)

This would be splitting Monday and Tuesday interviews for tails into two possibilities, but they are connected by experimental protocol as one. Tails implies and entails both Monday's and Tuesday's interviews. She will know that in the case of tails she would not experience either of these interviews as one of two separate events. She knows both interviews will be experienced as identically and indistinguishably uninformative, then subsequently forgotten, both necessary to comprise the results of tails.

 

[stevendaryl]

I'm in the halfer camp. It looks to me like before Beauty is put to sleep Sunday night, she knows:

- there are not three possibilities, only two:
1) (heads) Monday awakening
2) (tails) Monday and Tuesday awakening
- she will be unable to tell which of these two possibilities is the case within any interview

The fact that she can't tell the difference between Monday, Tails and Tuesday, Tails doesn't mean that they aren't distinct situations. Sleeping Beauty herself, having understood the rules, would say when asked whether today is Monday or Tuesday, would answer: "I don't know". There are two possibilities and she doesn't know which one is correct. It's just like whether the coin is heads or tails; she doesn't know which it is, but she understands that they are different.