I believe what you’ve described is just the original problem again with an extra possible waking of no question asked on Tuesday, but that doesn’t impact the credences of her waking and observing the question. The answer is still 1⁄2.
<Sigh>. It is very difficult to get halfers to understand that “Monday, after Tails” and “Tuesday, after Tails,” ARE DIFFERENT OUTCOMES. So are “Monday, after Heads” and “Tuesday, after Heads.” All four are outcomes, even the one Beauty sleeps through. Yes, each pair occurs in the same timeline, but Beauty’s amnesia makes the probability experiment be, to her, a sampling of the four possible situations. Monday, after Tails and Tuesday, after Tails, are independent outcomes. I can make that clearer, I just can’t make halfers accept it.
This is a repeat of something I posted previously. But, as I predicted, that was ignored. Suppose there are three gourmet restaurants in town, randomly (and unknown to Beauty) designated “A”, “B”, and “C”. Call Denny’s “D”. {A,B,C} are associated, in that order, with {T1,T2,H1}. D is associated with H2. Beauty is wakened both days (1 and 2, if that wasn’t clear), and will be taken to the assigned restaurant.
Before being taken to a restaurant, Beauty can relate the sample space {A,B,C,D} with the probability distribution {1/4,1/4,1/4,1/4}. She can also use the equivalent sample space {T1, T2, H1, H2}, because she doesn’t know which restaurant is which (except Denny’s).
Steff’s answer above uses two events, not four. The sample space is {A&B,C&D} or {T1&T2,H1&H2}. The probability distribution is {1/2,1/2}.
When Beauty arrives at a gourmet restaurant (remember, she doesn’t know if it is A, B, or C), she can update the distribution for the four-event sample space to {1/3,1/3,1/3,0}. If I understand Steff’s answer (“an extra possible waking … doesn’t impact the credences of her waking and observing the question”), the two-event distribution is still {1/2,1/2}. But that can’t be—Beauty knows she could be at C, but can’t be at D. That it could be H1, but not H2.
The Tails path works differently. It could be A, or B, but not A&B. It could be T1, or T2, but not T1&T2. Most Halfers want us to believe that the events T1, T2, and T1&T2 are all the same event, with the same probability of 1⁄2.
I’m sorry, that is wrong. The probabilities for {A,B,C,D}={T1,T2,H1,H2} update to {1/3,1/3,1/3,0}. Steff’s statement that the extra possible waking can’t impact anything is wrong.
And my point from before is the nothing in this calculation changes if Beauty is left asleep on H2, but sees that she has arrived at a gourmet restaurant. Or is simply wakened and questioned. “New information” is an informal term in probability that, while not defined (because it is informal), does not mean “what happened was not guaranteed to happen.” It means that we learned something which depends on certain elements of the probability experiment happening, and rules out others.
In the canonical version of the problem, Beauty knows that she will not be awake during certain portions of the experiment. That doesn’t remove that outcome from the experiment, as Halfers imply, it removes its probability from the distribution requiring an update.
I believe what you’ve described is just the original problem again with an extra possible waking of no question asked on Tuesday, but that doesn’t impact the credences of her waking and observing the question. The answer is still 1⁄2.
<Sigh>. It is very difficult to get halfers to understand that “Monday, after Tails” and “Tuesday, after Tails,” ARE DIFFERENT OUTCOMES. So are “Monday, after Heads” and “Tuesday, after Heads.” All four are outcomes, even the one Beauty sleeps through. Yes, each pair occurs in the same timeline, but Beauty’s amnesia makes the probability experiment be, to her, a sampling of the four possible situations. Monday, after Tails and Tuesday, after Tails, are independent outcomes. I can make that clearer, I just can’t make halfers accept it.
This is a repeat of something I posted previously. But, as I predicted, that was ignored. Suppose there are three gourmet restaurants in town, randomly (and unknown to Beauty) designated “A”, “B”, and “C”. Call Denny’s “D”. {A,B,C} are associated, in that order, with {T1,T2,H1}. D is associated with H2. Beauty is wakened both days (1 and 2, if that wasn’t clear), and will be taken to the assigned restaurant.
Before being taken to a restaurant, Beauty can relate the sample space {A,B,C,D} with the probability distribution {1/4,1/4,1/4,1/4}. She can also use the equivalent sample space {T1, T2, H1, H2}, because she doesn’t know which restaurant is which (except Denny’s).
Steff’s answer above uses two events, not four. The sample space is {A&B,C&D} or {T1&T2,H1&H2}. The probability distribution is {1/2,1/2}.
When Beauty arrives at a gourmet restaurant (remember, she doesn’t know if it is A, B, or C), she can update the distribution for the four-event sample space to {1/3,1/3,1/3,0}. If I understand Steff’s answer (“an extra possible waking … doesn’t impact the credences of her waking and observing the question”), the two-event distribution is still {1/2,1/2}. But that can’t be—Beauty knows she could be at C, but can’t be at D. That it could be H1, but not H2.
The Tails path works differently. It could be A, or B, but not A&B. It could be T1, or T2, but not T1&T2. Most Halfers want us to believe that the events T1, T2, and T1&T2 are all the same event, with the same probability of 1⁄2.
I’m sorry, that is wrong. The probabilities for {A,B,C,D}={T1,T2,H1,H2} update to {1/3,1/3,1/3,0}. Steff’s statement that the extra possible waking can’t impact anything is wrong.
And my point from before is the nothing in this calculation changes if Beauty is left asleep on H2, but sees that she has arrived at a gourmet restaurant. Or is simply wakened and questioned. “New information” is an informal term in probability that, while not defined (because it is informal), does not mean “what happened was not guaranteed to happen.” It means that we learned something which depends on certain elements of the probability experiment happening, and rules out others.
In the canonical version of the problem, Beauty knows that she will not be awake during certain portions of the experiment. That doesn’t remove that outcome from the experiment, as Halfers imply, it removes its probability from the distribution requiring an update.