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
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
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
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
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
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
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
Incidentally, what is your version of Bayes' theorem for this where P(Heads) = 1/3?
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 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? — Michael
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
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 Bauty is awaked twice as often if tails — Michael
"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. — Pierre-Normand
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
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
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
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
In the cosmopolitan situation, the probability of meeting a Tunisian doubles because Tunisians are around twice as often. — Pierre-Normand
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 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
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
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
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
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
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