To apply this to the traditional problem: there are two participants; one will be woken on Monday, one on both Monday and Tuesday, determined by a coin toss.
I am one of the participants. What is the probability that I am woken twice?
Do I reason as if I am randomly selected from the set of all participants, and so that I am equally likely to be woken twice, or do I reason as if my interview is randomly selected from the set of all interviews, and so that I am more likely to be woken twice?
Halfers do the former, thirders the latter.
Which is the most rational?
Given the way the experiment is conducted I think the former (halfer) reasoning is the most rational. — Michael
A more comparable example would be if there are four doors, two containing a goat and two containing a car. You pick a door (say 1) and then Monty opens one of the two doors that contain a goat (say 2). What is the probability that your chosen door contains a car? What is the probability that the car is behind door 3, or door 4? — Michael
Once John Doe awakens, he can rule out the possibility of it being Day 2 of his participation, and so can you. — Pierre-Normand
Once you have been assigned to John Doe, your credence (P(H)) in this proposition should be updated from 1/2 to 2/3. This is because you were randomly assigned a room, and there are twice as many rooms with participants who wake up twice as there are rooms with participants waking up once. — Pierre-Normand
Why? — Michael
The question is whether or not it is rational for the participant to reason this way. Given that the experiment doesn't work by randomly assigning an interview to them from the set of all interviews, I don't think it is. The experiment works by randomly assigning an interview set from the set of all interview sets (which is either the head set or the tail set), and so I believe it is more rational to reason in this way. — Michael
Sorry, I meant to say that he can rule out it being the case that the coin landed heads and that this is Day2. — Pierre-Normand
Do you agree that from the sitter point of view, the probability that the coin landed tails is 2/3?
Yes. — Michael
They ruled that out before the experiment begun. You might as well say that they can rule out it being the case that the coin landed heads and that this is day 3. — Michael
I would say that both are true, but also contradictory. Which reasoning it is proper to apply depends on the manner in which one is involved.
For the sitter, his involvement is determined by being randomly assigned an interview, and so I think the first reasoning is proper. For the participant, his involvement is determined by tossing a coin 100 times, and so I think the second reasoning is proper. — Michael
I’ve explained the error with betting examples before. Getting to bet twice if it’s tails doesn’t mean that tails is more likely. — Michael
I don’t quite understand this example. There are multiple coin flips? — Michael
Actually that’s not right. Need to think about this. — Michael
Not sure what this is supposed to show? — Michael
It's worth noting that your provided sequence converges on 1/3. If the captive is not keeping track of the date, their credence should indeed be exactly 1/3. — Pierre-Normand
In this case you're in safehouse 1 if not tails yesterday and not tails today and you're in safehouse 2 if either tails yesterday or tails today. Obviously the latter is more likely. I think only talking about the preceding coin toss is a kind of deception. — Michael
Your credence in each possibility is based on the number of ways in which you could find yourself in your current situation given the possible outcomes of the specific coin toss. — Pierre-Normand
If you don't like to consider my extreme example because the numbers are too high then let's consider a less extreme version. Rather than a coin toss it's a dice roll. If 1 - 5 then woken once (or safehouse 1 with crocodiles for one day), if 6 then woken six times (or safehouse 2 with lions for six days). Any rational person would take the wooden plank, and 5 out of every 6 kidnapped victims would survive. — Michael
However, consider a different scenario where the hostage has a small, constant probability ε of discovering the means of escape each day (case-2). In this scenario, stumbling upon this means of escape would provide the hostage with actionable evidence that he could use to update his credence. Now, he would believe with a probability of 6/11 that he's in safehouse #2, thereby justifying his decision to pick up the torch. Consequently, given that 6 out of 11 kidnapped victims who find the means to escape are surrounded by lions, 6 out of 11 would survive. — Pierre-Normand
Also, as an aside, if you correctly reason that it's tails then you escape on the first day, and so you can rule out today being the second day (assuming you understand that you would also reason as if it's tails). — Michael
Then this is a different scenario entirely. If we consider the traditional problem, it would be that after the initial coin toss to determine which days she could be woken, a second coin toss determines if she will be woken (heads she will, tails she won't). — Michael
The only significant divergence lies in the frequency of opportunities: the hostage can't be provided with frequent chances to escape without invalidating the analogy, whereas Sleeping Beauty can be given the chance to guess (or place a bet) every single day she awakens without undermining the experiment.
However, we can further refine the analogy by allowing the hostage to escape unharmed in all instances, but with the caveat that he will be recaptured unknowingly and re-administered the amnesia-inducing drug. This would align the scenarios more closely. — Pierre-Normand
But even if the same participant were to repeat the experiment 2^100 times, they don't bet on 100 heads because they think it's more likely, they bet on 100 heads because they know that eventually they will win, and that the amount they will win is greater than the amount they will lose. — Michael
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