No, much of the debate gets obscured by trying to ignore how SB’s information, while it includes all of the information that will be used it separate the experiment into two observations, also is limited to the “inside information” and she is in just one of those parts. That’s the purpose of the shopping spree I suggested. Correcting what you wrote, if the memory wipe is complete and SB has literally no way of knowin what information might apply to what is clearly a distinct part of the experiment, but she knows there is a different part than the current one, how can it not be “considered (by her) to be a different experiences.”
For the purposes of sampling, there are two parts to the experiment. What you call them is not relevant, only that SB knows that she is in one, and only one, of two parts. In the classic version, as Elga modifided it from the actual problem he proposed, one part must have a waking and one part has either a sleeping or a waking. Are you really claiming that the part where she is left asleep is not a part of the experiment? One that she knows is a possibility that is ruled out by her current state? And so fits the classic definition of “new evidence” that you deny? But if you really believe that, what about my version where she is taken shopping? It is even better as a classic example of evidence.
Or you could look at my answer. There, I explain how the problem you are solving actually is an “alternate formulation.” And, while there are still two parts, they are equivalent so we do not need to treat them separately. The answer is unequivocally 1⁄3.
Or try this: Instead of a coin, roll two six-sided dice. On Monday, ask her for the probability that the resulting sum is 7. On Tuesday, if the sum is odd, ask her the same question. But if it is even, ask her for the probability that the sum is 8.
If the room has a calendar, Beauty should certainly say Pr(S=7)=1/6 on Sunday and Monday. But on Tuesday, in answer to “what is the probability of 7,” she should say Pr(S=7)=1/3. But without a calendar, Pr(S=7)=1/6 can’t be right. Because it might be Tuesday, and Pr(S=7)=1/3 can’t be right. And it might be Monday, when Pr(S=7)=1/6 can’t be right. It has to be something in between, and that something is (2/3)*(1/6)+(1/3)*(1/3)=4/18=2/9. Yes, even though this is never the answer if she knows the day. That is how conditional probability works.
No, much of the debate gets obscured by trying to ignore how SB’s information, while it includes all of the information that will be used it separate the experiment into two observations, also is limited to the “inside information” and she is in just one of those parts. That’s the purpose of the shopping spree I suggested. Correcting what you wrote, if the memory wipe is complete and SB has literally no way of knowin what information might apply to what is clearly a distinct part of the experiment, but she knows there is a different part than the current one, how can it not be “considered (by her) to be a different experiences.”
For the purposes of sampling, there are two parts to the experiment. What you call them is not relevant, only that SB knows that she is in one, and only one, of two parts. In the classic version, as Elga modifided it from the actual problem he proposed, one part must have a waking and one part has either a sleeping or a waking. Are you really claiming that the part where she is left asleep is not a part of the experiment? One that she knows is a possibility that is ruled out by her current state? And so fits the classic definition of “new evidence” that you deny? But if you really believe that, what about my version where she is taken shopping? It is even better as a classic example of evidence.
Or you could look at my answer. There, I explain how the problem you are solving actually is an “alternate formulation.” And, while there are still two parts, they are equivalent so we do not need to treat them separately. The answer is unequivocally 1⁄3.
Or try this: Instead of a coin, roll two six-sided dice. On Monday, ask her for the probability that the resulting sum is 7. On Tuesday, if the sum is odd, ask her the same question. But if it is even, ask her for the probability that the sum is 8.
If the room has a calendar, Beauty should certainly say Pr(S=7)=1/6 on Sunday and Monday. But on Tuesday, in answer to “what is the probability of 7,” she should say Pr(S=7)=1/3. But without a calendar, Pr(S=7)=1/6 can’t be right. Because it might be Tuesday, and Pr(S=7)=1/3 can’t be right. And it might be Monday, when Pr(S=7)=1/6 can’t be right. It has to be something in between, and that something is (2/3)*(1/6)+(1/3)*(1/3)=4/18=2/9. Yes, even though this is never the answer if she knows the day. That is how conditional probability works.