The bet's positive expected value arises because she is twice as likely to win as she is to lose. This is due to the experimental setup, which on average creates twice as many T-awakenings as H-awakenings. — Pierre-Normand

Which of these are you saying?

1. There are twice as many T-awakenings because tails is twice as likely
2. Tails is twice as likely because there are twice as many T-awakenings

I think both of these are false.

I think there are twice as many T-awakenings but that tails is equally likely.

The bet's positive expected value arises only because there are twice as many T-awakenings.

It's because her appropriately interpreted credence P(T) =def P(T-awakening) = 2/3 that her bet on T yields a positive expected value, not the reverse. If she only had one opportunity to bet per experimental run (and was properly informed), regardless of the number of awakenings in that run, then her bet would break even. This would also be because P(T) =def P(T-run) = 1/2. — Pierre-Normand

I don't think that works.

$\begin{aligned}P(HInterview | HRun) &= {{P(HRun | HInterview) * P(HInterview)} \over P(HRun)}\\\\P(HInterview) &= {{P(HInterview|HRun) * P(HRun)} \over P(HRun|HInterview)}\\\\P(HInterview) &= {{1 * P(HRun)} \over 1}\\\\P(HInterview) &= P(HRun)\end{aligned}$

P(H-awakening) = P(H-run). Therefore either both are 1/2 or both are 1/3. This is expected given the biconditional.

I think it's more rational to say that P(H-awakening) = 1/2 than to say that P(H-run) = 1/3. Therefore I think it's more rational to say that P(H-awakening) = P(H-run) = P(H) = 1/2.

Which of these are you saying?

1. There are twice as many T-awakenings because tails is twice as likely
2. Tails is twice as likely because there are twice as many T-awakenings

I think both of these are false.

I think there are twice as many T-awakenings but that tails is equally likely.

The bet's positive expected value arises only because there are twice as many T-awakenings. — Michael

I am not relying on 1, but it would be a valid inference if we assume that P(T) = 2*P(H). This assumption would hold if we define P(T) as P(T) =def |{T-awakenings}| / |{awakenings}| (and similarly for P(H)).

Your disagreement with 2 appears to stem from an assumption that (at least in the context of the Sleeping Beauty problem) the probability of an outcome must solely be a reflection of the realization of an object's intrinsic properties, such as a fair coin. However, we frequently talk about probabilities of (types of) events that depend on how we interact with objects and that only indirectly depend (if at all) on the propensities of those objects had to actualize their properties. For instance, if there are twice as many Italians as Tunisians in my city (and no other nationalities), but for some reason, Tunisians go out four times more often than Italians, then when I go out, the first person I meet is twice as likely to be a Tunisian.

Here's another example. The management of the Sleeping Beauty Experimental Facility organizes a cocktail party for the staff. The caterers circulate among the guests serving drinks and sandwiches. Occasionally, they flip a coin. If it lands heads, they ask a random guest to guess the result. If it lands tails, they ask two random guests. The guests are informed of this protocol (and they don't track the caterers' movements). When a caterer approaches you, what are the odds that the coin they flipped landed heads? This situation is similar to the case of encountering a Tunisian in my previous example. Upon meeting someone, you could rationally say, "The probability that this person is a Tunisian is 2/3," even though people don't inherently possess a propensity to be born a different nationality than the one they were actually born into.

In the cocktail party scenario, the probability that "this" coin landed tails is a function of both the coin's propensities (its fairness) and the method used to engineer interactions between guests and coins based on the outcomes of the tosses.

However, we frequently talk about probabilities of (types of) events that depend on how we interact with objects and that only indirectly depend (if at all) on the propensities of those objects had to actualize their properties. For instance, if there are twice as many Italians as Tunisians in my city (and no other nationalities), but for some reason, Tunisians go out four times more often than Italians, then when I go out, the first person I meet is twice as likely to be a Tunisian. — Pierre-Normand

In this case:

1. there are twice as many Tunisian-meetings because Tunisian-meetings are twice as likely
2. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often

This makes sense.

So:

1. there are twice as many T-awakenings because T-awakenings are twice as likely
2. T-awakenings are twice as likely because ...

How do you finish 2? It's circular reasoning to finish it with "there are twice as many T-awakenings".

The management of the Sleeping Beauty Experimental Facility organizes a cocktail party for the staff. The caterers circulate among the guests serving drinks and sandwiches. Occasionally, they flip a coin. If it lands heads, they ask a random guest to guess the result. If it lands tails, they ask two random guests. The guests are informed of this protocol (and they don't track the caterers' movements). When a caterer approaches you, what are the odds that the coin they flipped landed heads? — Pierre-Normand

To make this comparable to the Sleeping Beauty problem; there are two Sleeping Beauties, one will be woken if heads, two will be woken if tails. When woken, what is their credence in heads? In such a situation the answer would be 1/3. Bayes' theorem for this is:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over2}*{1\over2}}\over{3\over4}}\\&={1\over3}\end{aligned}$

This isn't comparable to the traditional probem.

Incidentally, what is your version of Bayes' theorem for this where P(Heads) = 1/3?

What wouldn't make sense is just to say that Tunisian-meetings are twice as likely because there are twice as many Tunisian-meetings. That is a non sequitur. — Michael

But why wouldn't it make sense? For example, if you're an immigration lawyer and your secretary has arranged for you to meet with twice as many Tunisians as Italians in the upcoming week, when you walk into a meeting without knowing the client's nationality, isn't it logical to say that it's twice as likely to be with a Tunisian?

To make this comparable to the Sleeping Beauty problem; there are two Sleeping Beauties, one will be woken if heads, two will be woken if tails. When woken, what is their credence in heads? In such a situation the answer would be 1/3. Bayes' theorem for this is:

P(Heads|Awake)=P(Awake|Heads)∗P(Heads)/P(Awake)

=(1/2)∗(1/2) / (3/4)=1/3

=1/3

This isn't comparable to the traditional problem. — Michael

Why isn't it comparable? I had proposed an identical version earlier. One effective way to erase Sleeping Beauty's memory without any side effects from an amnesia-inducing drug might be to switch her with her identical twin for the second awakening. They would each only experience one awakening at most as part of a team. Their epistemic perspectives regarding the coin toss would remain the same, and therefore so should their rational credences.

Incidentally, what is your version of Bayes' theorem for this where P(Heads) = 1/3?

I am unsure what it is that you are asking here.

But why wouldn't it make sense? For example, if you're an immigration lawyer and your secretary has arranged for you to meet with twice as many Tunisians as Italians in the upcoming week, when you walk into a meeting without knowing the client's nationality, isn't it logical to say that it's twice as likely to be with a Tunisian? — Pierre-Normand

I've since edited my post to make my point clearer. To repeat:

In this case:

1. there are twice as many Tunisian-meetings because Tunisian-meetings are twice as likely
2. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often

This makes sense.

So:

1. there are twice as many T-awakenings because T-awakenings are twice as likely
2. T-awakenings are twice as likely because ...

How do you finish 2? It's circular reasoning to finish it with "there are twice as many T-awakenings".

I am unsure what it is that you are asking here. — Pierre-Normand

Starting here you argued that P(Heads) = 1/3.

So, what do you fill in here for the example of one person woken if heads, two if tails?

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\\\&={{P(Awake | Heads)*{1\over3}}\over{P(Awake)}}\end{aligned}$

Starting here you argued that P(Heads) = 1/3.

So, what do you fill in here for the example of one person woken if heads, two if tails? — Michael

I see. I was filling up a template that you had provided where P(Monday) = 2/3, thus making it clear that we were quantifying awakening episodes.

In that case P(Monday|Heads) = 1, and P(Heads) = 1/3 since one third of the awakenings are H-awakenings.

Therefore P(Heads|Monday) = P(Monday|Heads)∗P(Heads)/P(Monday) = (1)*(1/3)/(2/3) = 1/2.

Likewise, P(Heads|Awake) = P(Awake|Heads)∗P(Heads)/P(Awake) = (1)*(1/3)/(1) = 1/3

Note that when we quantify awakening episodes, P(Awake|Heads) = 1 since all H-awakenings are awakenings.

In this case:

1. there are twice as many Tunisian-meetings because Tunisian-meetings are twice as likely
2. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often

This makes sense.

So:

1. there are twice as many T-awakenings because T-awakenings are twice as likely
2. T-awakenings are twice as likely because ...

How do you finish 2? It's circular reasoning to finish it with "there are twice as many T-awakenings". — Michael

To fill in your number 2 with no circularity, we can draw a parallel to the first example:

2a. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often (and thus, Tunisian-meetings are twice as frequent as Italian-meetings)

Likewise:

2b. T-awakenings are twice as likely because, based on the experiment's design, Sleeping Beauty is awakened twice as often when the coin lands tails (and thus, T-awakenings are twice as frequent as H-awakenings)

T-awakenings are twice as likely because, based on the experiment's design, Sleeping Beauty is awakened twice as often when the coin lands tails — Pierre-Normand

This is just repeating the same thing in a different way. That there are twice as many T-awakenings just is that Sleeping Beauty is awakened twice as often if tails. So your reasoning is circular.

This is just repeating the same thing in a different way. That there are twice as many T-awakenings just is that Sleeping Bauty is awaked twice as often if tails — Michael

You accepted the validity of the reasoning when probability was deduced from frequencies in the Tunisian-meetings scenario. Why is this reasoning acceptable for people who were born Tunisian but questionable for coins that landed tails?

These are two different sets of claims:

A1. there are twice as many Tunisian-meetings because Tunisian-meetings are twice as likely
A2. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often

B1. there are twice as many T-awakenings because T-awakenings are twice as likely
B2. T-awakenings are twice as likely because Sleeping Beauty is woken twice as often if tails

"there are twice as many T-awakenings" is biconditional with "Sleeping Beauty is woken twice as often if tails" and so B uses circular reasoning.

"there are twice as many Tunisian-meetings" isn't biconditional with "there are half as many Tunisians and Tunisians go out four times more often" and so A doesn't use circular reasoning.

We have two different experiments:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, both A and B once each if tails

Given that I'm guaranteed to wake up if heads in the first experiment but not guaranteed to wake up if heads in the second experiment (and guaranteed to wake up if tails in both experiments) I think it only reasonable to conclude that P(Heads|Awake) in the first experiment is greater then P(Heads|Awake) in the second experiment.

And given that P(Heads|Awake) = 1/3 in the second experiment I think it only reasonable to conclude that P(Heads|Awake) > 1/3 (i.e. 1/2) in the first experiment.

"there are twice as many Tunisian-meetings" isn't biconditional with "there are half as many Tunisians and Tunisians go out four times more often" and so A doesn't use circular reasoning. — Michael

However, you seem to agree that in this scenario, one is twice as likely to encounter a Tunisian. The conclusion that there are twice as many Tunisian-meetings emerges from the premises: (1) there are half as many Tunisians and (2) Tunisians venture out four times more often. This inference is simply an intermediate step in the argumentation, providing an explanation for why there are twice as many Tunisian-meetings. Analogously, the Sleeping Beauty setup explains why there are twice as many T-awakenings. If the reason for twice as many Tunisian-meetings is that Tunisians venture out twice as often (assuming there are an equal number of Tunisians and Italians), then the analogy with the Sleeping Beauty scenario is precise. The attribute of being Tunisian can be compared to a coin landing tails, and encountering them on the street can be paralleled to Sleeping Beauty encountering such coins upon awakening. In the Sleeping Beauty setup, coins that land tails are 'venturing out' more often.

However, you seem to agree that in this scenario, one is twice as likely to encounter a Tunisian. The conclusion that there are twice as many Tunisian-meetings emerges from the premises: (1) there are half as many Tunisians and (2) Tunisians venture out four times more often. This inference is simply an intermediate step in the argumentation, providing an explanation for why there are twice as many Tunisian-meetings. Analogously, the Sleeping Beauty setup explains why there are twice as many T-awakenings. If the reason for twice as many Tunisian-meetings is that Tunisians venture out twice as often (assuming there are an equal number of Tunisians and Italians), then the analogy with the Sleeping Beauty scenario is precise. The attribute of being Tunisian can be compared to a coin landing tails, and encountering them on the street can be paralleled to Sleeping Beauty encountering such coins upon awakening. In the Sleeping Beauty setup, coins that land tails are 'venturing out' more often. — Pierre-Normand

This goes back to my distinction between:

1. One should reason as if one is randomly selected from the set of all participants
2. One should reason as if one's interview is randomly selected from the set of all interviews

In the case where I go out and meet someone on the street it is certainly comparable to 2, and this is why when we consider the sitters it is correct to say that the probability that they are assigned a heads interview is 1/3.

But Sleeping Beauty isn't assigned an interview in the same way. It's not the case that there is one heads interview, two tails interviews, and she "meets" one of the interviews at random (such that P(T interview) = 2/3); instead it's the case that there is one heads interview, two tails interviews, and first she is assigned one of the interview sets at random (such that P(T interviews) = 1/2) and then she "meets" one of the interviews in her set at random.

If we were to use the meetings example then:

1. A coin is tossed
2. If heads then 1 Italian walks the streets
3. If tails then 2 Tunisians walk the streets
4. Sleeping Beauty is sent out into the streets

What is the probability that she will meet a Tunisian? That there are twice as many Tunisians isn't that her meeting a Tunisian is twice as likely.

Your argument is that: if 1) there are twice as many T-awakenings and if 2) I randomly select one of the awakenings then 3) it is twice as likely to be a T-awakening.

This is correct. But the manner in which the experiment is conducted is such that (2) is false. (3) doesn't follow from (1) alone.

(2) is true for the sitter assigned an interview but not for the participant.

For the participant it is the case that 1) there are twice as many T-awakenings, 2) I randomly select one of the awakening sets, 3) it is equally likely to be a T-awakening set, and so 4) it is equally likely to be a T-awakening. (1) it turns out is irrelevant.

You can't just ignore (or change) the manner in which Sleeping Beauty participates in the experiment, which is what your various analogies do.

If we were to use the meetings example then:

1. A coin is tossed
2. If heads then 1 Italian walks the streets
3. If tails then 2 Tunisians walk the streets
4. Sleeping Beauty is sent out into the streets

What is the probability that she will meet a Tunisian? — Michael

In the Sleeping Beauty problem, she isn't asked to estimate the probability of being awakened in the future with the coin having landed heads. Instead, she's awakened and then questioned about her current belief regarding the coin's outcome. To maintain this structure in the street encounter example, we should consider Sleeping Beauty meeting a wanderer and then being asked to consider the probability that this wanderer is an Italian. If, over time, the setup leads to twice as many Tunisian encounters (perhaps because Tunisians wander about twice as long as Italians), then Sleeping Beauty's rational credence should be P(Italian) = 1/3.

If, over time, the setup leads to twice as many Tunisian encounters (perhaps because Tunisians wander about twice as long as Italians), then Sleeping Beauty's rational credence should be P(Italian) = 1/3. — Pierre-Normand

I believe this credence is based on fallacious reasoning as explained here.

Her reasoning is: if 1) there are twice as many Tunisian walkers and if 2) I randomly meet one of the walkers then 3) it is twice as likely to be a Tunisian walker.

Given the manner in which the experiment is conducted (2) is false and so this isn't the correct reasoning with which to determine one's credence.

Your argument is that: if 1) there are twice as many T-awakenings and if 2) I randomly select one of the awakenings then 3) it is twice as likely to be a T-awakening.

This is correct. But the manner in which the experiment is conducted is such that 2) is false. — Michael

Your point (2) doesn't factor into my argument. I've consistently held to the premise, as dictated by the problem statement, that Sleeping Beauty awakens once when the coin lands heads and twice when it lands tails. There's no necessity for an external agent to select an awakening, just as there's no need for someone to choose a street encounter. Instead, Sleeping Beauty, upon each awakening (or encounter), should consider the long-term distribution of these awakenings (or encounters) to formulate a rational belief about the current situation.

We start with the mutually agreeable premise:

P1) there are twice as many T-awakenings

Your conclusion is:

C) T-awakenings are twice as likely

Obviously this is a non sequitur. We need the second premise:

P2) if there are twice as many T-awakenings then T-awakenings are twice as likely

This is something that I disagree with and that you need to prove.

In the case of the meetings we have:

*P1) there are twice as many Tunisian walkers
*P2) if I meet a walker at random then I am twice as likely to meet a Tunisian walker (from *P1)
*P3) I meet a walker at random
*C) I am twice as likely to have met a Tunisian walker (from *P2 and *P3)

In Sleeping Beauty's case we have:

P1) there are twice as many tails interviews
P2) ?
P3) I am in an interview
C) I am twice as likely to be in a tails interview

What is your (P2) that allows you to derive (C)? It doesn't follow from (P1) and (P3) alone.

In the case of the meetings we have:

*P1) there are twice as many Tunisian walkers
*P2) if I meet a walker at random then I am twice as likely to meet a Tunisian walker (from *P1)
*P3) I meet a walker at random
*C) I am twice as likely to have met a Tunisian walker (from *P2 and *P3)

In Sleeping Beauty's case we have:

P1) there are twice as many tails interviews
P2) ?
P3) I am in an interview
C) I am twice as likely to be in a tails interview

What is your (P2) that allows you to derive (C)? It doesn't follow from (P1) and (P3) alone. — Michael

You are introducing premises *P2 and *P3 in an attempt to emphasize a perceived disanalogy between the cosmopolitan meeting scenario and the Sleeping Beauty problem. Both *P1 and *P2 seem to imply that there exists a pre-determined set of potential encounters (many Tunisians and half as many Italians strolling around), from which a random selection process subsequently generates an encounter. There indeed is no analogous situation in the Sleeping Beauty problem, as there isn't a pre-determined set of pre-tossed coins from which Sleeping Beauty randomly encounters one upon awakening. However, I would argue that this misrepresents the cosmopolitan meeting scenario.

In the cosmopolitan situation, the probability of meeting a Tunisian doubles because Tunisians are around twice as often. There is no need for a random selection from a pre-existing set of potential encounters. Similarly, in the Sleeping Beauty problem, coins that have landed on tails "hang around" twice as long (i.e., for two sequential awakenings instead of one), which makes it twice as likely for Sleeping Beauty to encounter this outcome each time she is awakened and interviewed throughout the experiment.

The random process is fully specified by the equal distribution of coin toss outcomes (over the long run) and the longer "hang around" times of tails outcomes. Likewise, in the cosmopolitan meeting case, the process is fully specified by the equal distribution of Italians and Tunisians in the city and the increased frequency of encounters generated by Tunisians due to their longer "hang around" times. In neither case are additional random selection processes from a pre-determined set of possible encounters necessary.

I introduce the additional premise(s) because this is a non sequitur:

A1. there are twice as many Tunisian walkers as Italian walkers
A2. I am twice as likely to meet a Tunisian walker

If all the Tunisian walkers are in one area of the town but I'm walking in another then the conclusion is false. If, unknown to me, all the Tunisian walkers are wearing red and all the Italians wearing blue and I'm told to meet someone wearing blue if a coin lands heads or red if tails then the conclusion is false. If I don't go out to meet anyone then the conclusion is false. The experiment needs to be set up in such a way that the walkers are randomly distributed throughout the town and that I meet at random any one of the walkers. Only when this setup is established as a premise will the conclusion follow:

A1. there are twice as many Tunisian walkers as Italian walkers
A2. if I meet a walker at random from a random distribution of all walkers then I am twice as likely to meet a Tunisian walker

Similarly, this is a non sequitur:

B1. there are twice as many T-interviews as H-interviews
B2. my interview is twice as likely to be a T-interview

You would instead need something like:

C1. there are twice as many T-interviews as H-interviews
C2. if my interview is randomly assigned from the set of all interviews then my interview is twice as likely to be a T-interview

But this reasoning doesn't apply to Sleeping Beauty because her interview isn't randomly assigned from the set of all interviews.

For Sleeping Beauty the correct argument is the one that properly sets out the manner in which the experiment is conducted:

D1. there are an equal number of T-interview sets as H-interview sets
D2. If I am assigned at random either a T-interview set or a H-interview set then my interview set is equally likely to be a T-interview set
D3. I am assigned at random either a T-interview set or a H-interview set
D4. my interview is a T-interview iff my interview set is a T-interview set
D5. my interview is equally likely to be a T-interview

B is fallacious, C is inapplicable, and D is sound, hence why P(Heads|Awake) = 1/2 is the only rational conclusion. The fact that there are twice as many T-interviews as H-interviews is irrelevant. It's a premise from which no relevant conclusion regarding credence can be derived. It's only use is to explain betting outcomes.

In the cosmopolitan situation, the probability of meeting a Tunisian doubles because Tunisians are around twice as often. — Pierre-Normand

This is an ambiguous claim. If there are half as many Tunisians but they go out four times as often but are only out for 10 mins, whereas Italians are out for 20 mins, then it would be that Tunisians are around equally as often as measured by time out. The only way you could get this to work is if the argument is set out exactly as I have done above:

A1. there are twice as many Tunisian walkers as Italian walkers (out right now)
A2. if (right now) I meet a walker at random from a random distribution of all walkers (out right now) then I am twice as likely to meet a Tunisian walker

But there's nothing comparable to "if (right now) I meet a walker at random from a random distribution of all walkers (out right now)" that has as a consequent "then my interview is twice as likely to be a T-interview".

P1. If I am assigned at random either a H-interview set or a T-interview set then my interview set is equally likely to be a H-interview set
P2. I am assigned at random either a H-interview set or a T-interview set
P3. My interview is a H-interview iff my interview set is a H-interview set
C1. My interview is equally likely to be a H-interview

The premises are true and the conclusion follows, therefore the conclusion is true.

However, consider:

P4. If my sitter is assigned at random either a H-interview or a T-interview then his interview is half as likely to be a H-interview
P5. My sitter is assigned at random either a H-interview or a T-interview
P6. My interview is a H-interview iff my sitter's interview is a H-interview
C2. My interview is half as likely to be a H-interview

Prima facie the premises are true and the conclusion follows, therefore prima facie the conclusion is true. However, C1 and C2 are contradictory, therefore one of the arguments must be unsound.

Let's say that my sitter happens to be John:

P7. If John is assigned at random either a H-interview or a T-interview then his interview is half as likely to be a H-interview
P8. John is assigned at random either a H-interview or a T-interview
P9. My interview is a H-interview iff John's interview is a H-interview
C3. My interview is half as likely to be a H-interview

The issue is with P9. My interview is not biconditional with John's interview given that he is not guaranteed to be my sitter. That second argument commits a fallacy. P4 and P5 are true only under a de re interpretation of "my sitter" and P6 is true only under a de dicto interpretation.

This is why the participant shouldn't update his credence to match his sitter's.

We have two different experiments:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails

Your version of the experiment is comparable to the second experiment, not the first. Your case of HH is equivalent to the case that the coin landed heads and I am B. The very fact that I'm being asked my credence allows me to rule out one of the non-zero prior probabilities, i.e. P(HH) = 1/4 and P(Heads and B) = 1/4.

The second experiment is not equivalent to the first experiment. In the first experiment there is no non-zero prior probability that I can rule out when being asked my credence.

Given that I'm guaranteed to be woken up if heads in the first experiment but not the second (and guaranteed to be woken up if tails in both experiments) it makes sense that my credence in heads is greater in the first experiment than in the second. If I'm less likely to be woken up if heads then it's less likely to be heads if I'm woken up.

P1. If I am assigned at random either a T-interview set or a H-interview set then my interview set is equally likely to be a T-interview set
P2. I am assigned at random either a T-interview set or a H-interview set
P3. My interview is a T-interview iff my interview set is a T-interview set
C1. My interview is equally likely to be a T-interview

The premises are true and the conclusion follows, therefore the conclusion is true. — Michael

The conclusion doesn't follow because, while the biconditional expressed in P3 is true, this biconditional does not guarantee a one-to-one correspondence between the set of T-interviews and the set of T-runs (or "T-interview sets"). Instead, the correspondence is two-to-one, as each T-run includes two T-interviews. This is a central defining feature of the Sleeping Beauty problem that your premises fail to account for.

The random process is fully specified by the equal distribution of coin toss outcomes (over the long run) and the longer "hang around" times of tails outcomes — Pierre-Normand

Haven't read all the recent back and forth here, but I think the usual examples of conditional probability do not apply.

If you want a closer analogy with pedestrians, it's Tunisians walking around in pairs. If the chances of meeting an Italian or a pair of Tunisians are equal, then the chances of meeting *a* Tunisian are either nil, since you can't meet just one, or the same as meeting a pair.

Look at how hang-around times affect the pedestrian-encountering odds. Roughly, if you miss a short walker, you've missed him, but if you miss a long walker you get another chance. That's not how Sleeping Beauty works at all. There's no way to miss your first tatils interview but still catch the second one.

This is an ambiguous claim. If there are half as many Tunisians but they go out four times as often but are only out for 10 mins, whereas Italians are out for 20 mins, then it would be that Tunisians are around equally as often as measured by time out. The only way you could get this to work is if the argument is set out exactly as I have done above:

A1. there are twice as many Tunisian walkers as Italian walkers (out right now)
A2. if (right now) I meet a walker at random from a random distribution of all walkers (out right now) then I am twice as likely to meet a Tunisian walker

But there's nothing comparable to "if (right now) I meet a walker at random from a random distribution of all walkers (out right now)" that has as a consequent "then my interview is twice as likely to be a T-interview". — Michael

To fine-tune the analogy, let's assume that there are an equal number of Tunisians and Italians, that they are out for the same duration, and that Tunisians go out twice as frequently. Importantly, there's no need for an extraneous process of random selection to generate an encounter with a citizen, or a tossed coin, in either example. In the cosmopolitan encounter case, the random distributions of citizens in the street at any given time (with, on average, twice as many Tunisians out) directly result in twice as many encounters with Tunisians. It is a straightforward causal relationship between the distribution of wanderers and the distribution of encounters. Similarly, in the Sleeping Beauty case, the setup guarantees that Sleeping Beauty will encounter twice as many coins having landed tails when she awakens, simply by ensuring that she is awakened twice as often when the coins land tails.

If you want a closer analogy with pedestrians, it's Tunisians walking around in pairs. If the chances of meeting an Italian or a pair of Tunisians are equal, then the chances of meeting *a* Tunisian are either nil, since you can't meet just one, or the same as meeting a pair.

Look at how hang-around times affect the pedestrian-encountering odds. Roughly, if you miss a short walker, you've missed him, but if you miss a long walker you get another chance. That's not how Sleeping Beauty works at all. There's no way to miss your first tatils interview but still catch the second one. — Srap Tasmaner

If I were to adjust the analogy, suppose that meeting a Tunisian pedestrian guarantees that you have met or will meet their sibling either in the previous or next encounter. In this scenario, would your credence that the pedestrian you're encountering is a Tunisian change? As long as you meet Tunisians twice as often as Italians, your credence P(Tunisian encounter) should remain 2/3 at the time of each individual encounter, regardless of the pairing situation.

But haven't you lost Sleeping Beauty's other constraint, that the chances of encountering one Italian or two Tunisians are equal?

The conclusion doesn't follow because, while the biconditional expressed in P3 is true, this biconditional does not guarantee a one-to-one correspondence between the set of T-interviews and the set of T-runs (or "T-interview sets"). Instead, the correspondence is two-to-one, as each T-run includes two T-interviews. This is a central defining feature of the Sleeping Beauty problem that your premises fail to account for. — Pierre-Normand

That doesn't mean that the credence isn’t transitive. My premises "fail" to account for it because it's irrelevant.

A iff B
P(B) = 1/2
Therefore, P(A) = 1/2

The conclusion has to follow.

See also what I said before:

$\begin{aligned}P(TInterview | TRun) &= {{P(TRun | TInterview) * P(TInterview)} \over P(TRun)}\\\\P(TInterview) &= {{P(TInterview|TRun) * P(TRun)} \over P(TRun|TInterview)}\\\\P(TInterview) &= {{1 * P(TRun)} \over 1}\\\\P(TInterview) &= P(TRun)\end{aligned}$

In the cosmopolitan encounter case, the random distributions of citizens in the street at any given time (with, on average, twice as many Tunisians out) directly result in twice as many encounters with Tunisians. — Pierre-Normand

Because of what I said before:

A1. there are twice as many Tunisian walkers as Italian walkers
A2. therefore, if I go out and meet at random one of the walkers then I am twice as likely to meet a Tunisian walker
A3. therefore, there will be twice as many Tunisian-walker meetings

In Sleeping Beauty's case:

B1. Sleeping Beauty wakes up twice as often if the coin lands tails
B2. the coin is equally likely to land tails
B3. therefore, there will be twice as many tails awakenings

This argument is sound and fully explains the betting outcome. Your conclusion that "therefore, if Sleeping Beauty wakes up then the coin is twice as likely to have landed tails" just doesn't follow.

What would follow is "therefore, if I pick at random one of Sleeping Beauty's awakenings then it is twice as likely to be a tails awakening," but given that the experiment isn't conducted this way it doesn't make sense for Sleeping Beauty to reason this way to determine her credence.