The following is what no Halfer has ever responded to, except to say “Gee, that might be interesting to discuss. We’ll get back to it.” They never do.
The experiment:
Beauty will be put to sleep on Sunday Night, and a fair coin will be flipped. On each of the next two days (Monday and Tuesday), she will participate in one of two different procedures, named A and B. There are two requirements for them: #1: During either procedure, Beauty will be unaware of what may have, or will have, happened on the “other” day. #2: In procedure A she will be asked a probability (or credence, if you prefer) question about the coin flip. But no such question will be asked in procedure B.
On Monday, Beauty will participate in procedure A. On Tuesday, she will participate in procedure B if the coin landed on Heads, and in procedure A is it landed on Tails.
The question, that is only asked in procedure A, is “What do you believe is the probability that the coin landed on Heads?”
This is a simple conditional probability question. There are four situations during the experiment when Beauty could participate in a procedure. Before she is—or, if she won’t be—asked the question, Requirement #1 makes her current situation an independent sampling into that set of four possibilities. If she is asked the question, Beauty knows that she is participating in procedure A, and so it is either Monday after Heads, Monday after Tails, or Tuesday after Tails. The probability that the coin landed on Heads is 1⁄3. If she is not (yet) asked, well, does it really matter?
The popular version of the Sleeping Beauty Problem fits this outline. Halfers don’t want to acknowledge two facts about that implementation of my outline. First, that “new Information” does not mean that something happened that was not guaranteed to happen, or is all that Beauty can perceive. It means that something happened that eliminates a possibility from the sample space. Second, that the possibility of something happening on Tuesday makes everything that could happen on Tuesday a member of that sample space. EVEN IF SHE WOULD SLEEP THROUGH IT, Beauty knows that the events of Tuesday, after Heads, must be represented is the sample space.
There is no probability issue in the Sleeping Beauty Problem, just obfuscation about what belongs in a sample space.
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.
The following is what no Halfer has ever responded to, except to say “Gee, that might be interesting to discuss. We’ll get back to it.” They never do.
The experiment:
Beauty will be put to sleep on Sunday Night, and a fair coin will be flipped. On each of the next two days (Monday and Tuesday), she will participate in one of two different procedures, named A and B. There are two requirements for them: #1: During either procedure, Beauty will be unaware of what may have, or will have, happened on the “other” day. #2: In procedure A she will be asked a probability (or credence, if you prefer) question about the coin flip. But no such question will be asked in procedure B.
On Monday, Beauty will participate in procedure A. On Tuesday, she will participate in procedure B if the coin landed on Heads, and in procedure A is it landed on Tails.
The question, that is only asked in procedure A, is “What do you believe is the probability that the coin landed on Heads?”
This is a simple conditional probability question. There are four situations during the experiment when Beauty could participate in a procedure. Before she is—or, if she won’t be—asked the question, Requirement #1 makes her current situation an independent sampling into that set of four possibilities. If she is asked the question, Beauty knows that she is participating in procedure A, and so it is either Monday after Heads, Monday after Tails, or Tuesday after Tails. The probability that the coin landed on Heads is 1⁄3. If she is not (yet) asked, well, does it really matter?
The popular version of the Sleeping Beauty Problem fits this outline. Halfers don’t want to acknowledge two facts about that implementation of my outline. First, that “new Information” does not mean that something happened that was not guaranteed to happen, or is all that Beauty can perceive. It means that something happened that eliminates a possibility from the sample space. Second, that the possibility of something happening on Tuesday makes everything that could happen on Tuesday a member of that sample space. EVEN IF SHE WOULD SLEEP THROUGH IT, Beauty knows that the events of Tuesday, after Heads, must be represented is the sample space.
There is no probability issue in the Sleeping Beauty Problem, just obfuscation about what belongs in a sample space.
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.