Undergrad The Sleeping Beauty Problem: Any halfers here?

[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.

 

[Dale]

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;

This is where we disagree. I think that B is not consistent with the description.

I understand that this is how you are defining "credence". I am just pointing out that this is a definition.

This is not a personal definition of mine, this is the standard definition. Credence is used as a synonym of subjective probability. Operationally, when a person is asked for their credence for X they are being asked for the betting odds at which they would be indifferent to a wager for X (then changed from odds form into probability form). Your scenario B is not a wager for X, it is a wager for X and Y.

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.

I think that you are introducing a personal definition here. Credence is used to determine if a person will accept or reject an offered wager at a given price. The confusion here only comes because you want to offer a different wager.

But that, in itself, is an argument about what words should mean,

I recognize that, but you are claiming that the problem is insufficiently specified, and your justification for that is to use a nonstandard meaning for credence and show that with your non standard meaning of credence multiple outcomes are possible.

If the Sleeping Beauty problem were posed as part of a homework assignment then part of the assignment would be to test the student's understanding of the standard meaning of the important terms in the problem. A student giving an answer of 1/2 would be wrong, even if the reason they are wrong is because they misunderstood the meaning of credence.

 

[stevendaryl]

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.

It's a strange situation, definitely. You have two people who are awake. One of them is the winner. One is one of the 1000 losers. By symmetry, they can't come to different conclusions about their chances of being the winner since they have exactly the same information.

The analogy of the thirder argument is:

"One of us is the winner. There is no indication of which one. So I might as well assume that it's a 50/50 chance that it's me."

The analogy of the halfer argument is:

"I only had a 1/1000 chance to start with. I knew ahead of time that I would eventually be awake at the same time as the winner. So there is really no reason to think that I'm more likely to be the winner now."

In the first case, both come to the conclusion that they are likely to be winners. In the second case, both come to the conclusion that it is very unlikely that they are the winner (even though they know that one of them is very wrong).

I really do understand the intuitions behind the halfer position in this case.

I suddenly realized that there is a connection with nonmeasurable sets. Assuming the continuum hypothesis, it is possible to have an ordering \prec on the reals in the range [0,1] such that for every x, there are countably many values of y such that y \prec x, but uncountably many y such that x \prec y.

So suppose Alice and Bob each generate a random number in the range [0,1]. Call Alice's number x and Bob's number y. Now, ask Alice what are her subjective odds that y \prec x. She has two ways of reasoning about this:

1. On the one hand, she knows that the situation is exactly symmetrical. So the odds should be 50/50.
2. On the other hand, when she looks at x, she knows that there are only countably many values of y such that y \prec x. So the odds that Bob picked one of those at random is 0. (Any countable set has measure zero)

So depending on how she looks at it, her odds of having the largest number (according the ordering \prec) is either 50/50 or 0.

 

[Stephen Tashi]

If the Sleeping Beauty problem were posed as part of a homework assignment then part of the assignment would be to test the student's understanding of the standard meaning of the important terms in the problem. A student giving an answer of 1/2 would be wrong,

They'd be wrong if you were the grader.

even if the reason they are wrong is because they misunderstood the meaning of credence.

Are you saying that there is no ambiguity about how credence is defined in the Sleeping Beauty problem?

The Wikipedia, (which might be incorrect), says:

Therefore, the Sleeping Beauty problem is not about mathematical probability theory. Rather, the question is whether subjective probability or credence are well-defined concepts, and how they must be operationalized.

The problem has apparently been the subject of many published articles. Can we appeal to the "weight of authority" and find there is a consensus answer in 2017? The thirders seem to publish the majority of articles, but the problem still appears to be controversial.

 

[PeterDonis]

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.

"No consequences" (zero payoff regardless of the bet) is still "consequences".

Of course, the person making the bet doesn't know that it has no consequences...

Yes, and that would be a key factor in the rationale for that betting scheme if I were the one setting it up. It would be something like: we don't want to "privilege" the situation where Beauty gets two chances to bet instead of one, because both bets must be the same anyway, since Beauty has the same information when making both of them. So we only count the last one.

Yes, this might seem "strange", but so is the whole scenario. Such strange scenarios are typical of philosophical thought experiments, so I don't think "hey, that's strange" is a valid reason to discard an alternative. The point of the strangeness is to push ideas to their limits to see what happens.

 

[PeterDonis]

Your scenario B is not a wager for X, it is a wager for X and Y.

See, here's that vague ordinary language again. I say scenario B is a wager for X, but with different payoffs than scenario A. In both scenarios, Beauty is asked for her credence that the coin came up heads. She is not asked in scenario B for her credence that the coin came up heads and it is Monday. She is asked the same question in both scenarios. Same question, same wager. But prior to the experiment starting, in A and B, she is told different payoff schemes, and that affects her answer to the question when she is asked during the experiment.

you are claiming that the problem is insufficiently specified, and your justification for that is to use a nonstandard meaning for credence

No, I'm using the standard meaning for credence, and pointing out that a wager is insufficiently specified if no payoffs are given. The original scenario does not give payoffs.

A student giving an answer of 1/2 would be wrong, even if the reason they are wrong is because they misunderstood the meaning of credence.

What if the student pointed out to you, the professor, that the standard definition of credence, which you gave in class, involves a wager, and a wager is insufficiently specified if there are no payoffs given, and your exam question gives no payoffs?

 

[stevendaryl]

"No consequences" (zero payoff regardless of the bet) is still "consequences".

Well, it's a degenerate case in which you have no possibility of losing and no possibility of winning. How do you define the probability in a case like that? The betting definition of probability is something like: The probability of winning is the amount (in dollars) I would be willing to bet in order to have a chance to win one dollar. If it's impossible to win or lose any money, then the probability would be undefined. It wouldn't make any difference if I bet 1/100 or 1.

Yes, and that would be a key factor in the rationale for that betting scheme if I were the one setting it up. It would be something like: we don't want to "privilege" the situation where Beauty gets two chances to bet instead of one, because both bets must be the same anyway, since Beauty has the same information when making both of them. So we only count the last one.

Okay, well, that's equivalent to treating the two-wakenings as the same as the one-wakening, which is basically ignoring all the details about wakenings and memory wipes, etc. I guess that's okay, but it seems contrary to the spirit of the puzzle.

 

[PeterDonis]

it's a degenerate case in which you have no possibility of losing and no possibility of winning. How do you define the probability in a case like that?

Ah, ok, I see the issue you're raising, but I think the answer is that this case is an edge case in the ordinary language term "wager". I would say that the actual "wager" is made on Sunday, when Beauty is told the experimental protocol, including the payoff scheme, and has to decide how she will respond when she is awakened and asked what is her subjective credence that the coin came up heads. (She can decide in advance because she knows exactly what information she will have on each awakening.) And what she decides on Sunday will depend on what payoff scheme she is told. So however many awakenings and askings happen during the experiment, they are all part of one "wager", which gets resolved on Wednesday, when the experiment concludes.

Beauty has enough information on Sunday to compute correct expectations for whatever payoff scheme she is told, so in that sense she has all of the "probability" information she needs. If the coin comes up tails, it is true that the actual answer she gives on Monday will not affect her actual winnings on Wednesday at all, but that fact is irrelevant to her decision of what to answer, since her answer is precomputed in advance, and will be the same on both Monday and Tuesday.

I agree this is a somewhat unusual use of the term "wager", but as I said before, the whole scenario is unusual. We don't normally think of it being possible for a person to be offered multiple "wagers" in each of which they have exactly the same information (where "information" here includes all of their memories).

 

[JeffJo]

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 using yours, you seem to think it depends on the number 1001.

What if the number of contestants is unknown to them (except that N>=2); each just finds herself awake in a room with another Beauty, with no recollection of prior days. Just the process that requires one of them to be the winner, and the other to be one of however many losers. Each has the same information, so their credences can't be different. If they must sum to 1, they are each 1/2 regardless of the number. If they don't have to sum to 1, what are they? Do they need to know N? Is there even a mathematics that can tell you what they should sum to?

Or is it simply 1/2? That is an incredibly obvious answer: no matter what circumstances led to each of you being awake at the moment, no matter how different they are, the fact is that your observation is only that one of you was selected by process A, one by process B, and you have no credence that A is preferred over B, or B over A.

Or try this, in the same vein: you win a one-in-a-million lottery for a free cruise. You get to pick 10 friends, one of whom will be selected at random to accompany you. The only stipulation is that the two of you will be amnesia-ed the first night, so that neither knows which won the lottery, or was named as a friend. Does your credence that you won the lottery depend on the number 1,000,000, or 10? Or both? How about your credence that you are the friend? Does either change if you get to go on 5 different cruises, each with a different friend from your list but no memory of the other cruises?

Or is it simply 1/2?

You are the one who is trying to make information, that is available only in the global sense, part of the solution to a local-memory-only question.

 

[Dale]

I say scenario B is a wager for X, but with different payoffs than scenario A.

Scenario B doesn't just change the amount that is paid for a win, it changes the conditions under which Beauty wins. In scenario A the wager can be resolved by revealing the coin to Beauty. In scenario B resolving the wager requires revealing not only the coin but also a clock/calendar.

 

[PeterDonis]

Scenario B doesn't just change the amount that is paid for a win, it changes the conditions under which Beauty wins.

The conditions you refer to are not specified in the original specification of the problem, so even under this interpretation I don't see how scenario B is inconsistent with the original specification of the problem.

At this point it looks to me like we're disagreeing about how to interpret vague ordinary language, which was the original point I was trying to make in this thread: that ordinary language is vague.

In scenario B resolving the wager requires revealing not only the coin but also a clock/calendar.

I'm not sure I agree with this, because (a) the coin result determines the clock/calendar possibilities, and (b) Beauty gives the same answer each time she is asked during the experiment anyway (since she has the same information at each awakening), so the clock/calendar state each time she give the answer doesn't actually affect anything; the coin flip result (plus her precomputation of the answer she will give) is already sufficient to determine everything else.

To put it another way: on Wednesday, when the experiment ends, what information does the experimenter have to give Beauty to resolve all bets? Does he have to show her the clock/calendar states for each of the times she was awakened and asked her credence? No. All he has to show her is the coin.

Again, I think this is just exposing edge cases in ordinary language terms that we usually don't bother digging into in this much detail, because our normal usage of those terms doesn't give rise to those edge cases. We only see the edge cases in specially constructed thought experiments.