The “valid answers for different questions” position is a red herring, and I’ll show why. But first, here is an unorthodox solution. I don’t want to get into defending whether its logic is valid; I’m just laying some groundwork.
On Sunday (in Elga’s re-framing of the problem he posed—see my answer), the sample space for the experiment seems to be {T,H} with equal probability for each outcome. Both Monday and Tuesday will happen, and so cannot be used to specify outcomes. This is one of your “different questions.”
But when SB is awake, only one of those days is “current.” (Yes, I know some debate whether days can be used this way—it is a self-fulfilling argument. If it is assumed to be true, you can show that it is true.) Elga was trying to use the day as a valid descriptor of disjoint outcomes, so that in SB’s world the outcomes T&Mon and T&Tue become distinct. The prior probabilities of the three disjoint outcomes {T&Mon, T&Tue, H} are each 1⁄2, and the probability for H can be updated by the conventional definition:
Pr(H|T&Mon or T&Tue or H) = Pr(H)/[Pr(T&Mon)+Pr(T&Tue)+Pr(H)]
Pr(H|T&Mon or T&Tue or H) = (1/2)/[(1/2)+(1/2)+(1/2)] = 1⁄3
If you think about it, this is exactly what Elga argued, except that he didn’t take the extreme step of creating a probability that was greater than 1 in the denominator. He made two problems, that each eliminated an additional outcome. I suggest, but don’t want to get into defending, that this is valid since SB’s situation divides the prior outcome T into two that are disjoint. Each still has the prior probability of the parent outcome.
But this is predicated on that split. Which is what I want to show is valid. What if, instead of leaving SB asleep on Tuesday after Heads, wake her but don’t interview her? Instead, we do something quite different, like take her on a $5,000 shopping spree at Saks Fifth Avenue? (She does deserve the chance of compensation, after all). This gives her a clear way to distinguish H&Mon&Interview from H&Tue&Saks as disjoint outcomes. The use of the day as a descriptor is valid, since this version of the experiment allows for it to be observed.
The actual sample space, on Sunday, for what can happen in an “awake” world for SB is {T&Mon, H&Mon, T&Tue, H&Tue&Saks}. Each has a prior probability of 1⁄4, for being what will apply to SB on a waking day. When she is actually interviewed, she can eliminate one.
And now I suggest that it does not matter, in an interview, what happens differently on H&Tue. SB is interviewed in the context of an interview day, so the question is placed in that context. Claiming it has the context of Sunday Night ignores how the Saks option can only be addressed in a single-day context, and can affect her answer.
Much of the debate gets obscured in the question of outside (of SB’s perception) information. That’s really the purpose of the thought experiment—if the memory wipe is complete and SB has literally no way of knowing whether it’s monday or tuesday, why are they considered (by her) to be different experiences? For purposes of sampling, there is only one waking, the one she’s experiencing.
Alternate formulations (especially frequentist models where there is some “truth” to probability, as opposed to it being purely subjective) can be different, but only by introducing some distinction of experience to make them into two wakings, somehow.
My view is that to an outside observer, the probability is 1⁄2 before the flop, 1 or 0 after the flip. To Beauty, it remains 1⁄2 until she receives evidence, in the form of something she can observe that would be different with heads than with tails. She should WAGER1⁄3 in some formulations (for instance, if the wagers for Monday and Tuesday are summed, rather than being one evaluation).
I’m sure you’d agree that, if the room has a calendar, Beauty should assign 1⁄2 on Sunday and Monday, and 0 on Tuesday. Never 1⁄3, right?
Like a number of people (including Elga), you’re converting an almost-doable thought experiment into one that may be impossible in principle. If the experiment is done with some not-yet-invented but plausible memory-erasing drug, but is otherwise realistic, Beauty will not have the same experiences when woken Monday and when woken Tuesday. Various aspects of her sensed environment, as well as internal thoughts, will be different for the two awakenings. We just assume that none of these differences allow her to infer the day of the week.
Second, if she should wager as if the probability of Heads is 1⁄3, in what sense is the probability of Heads not actually 1/3? The only point of probabilities is to influence actions. (Also, it is not just “some” situations where she should wager on the basis that the probability of Heads is 1⁄3. That’s always what she should do.)
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.
The “valid answers for different questions” position is a red herring, and I’ll show why. But first, here is an unorthodox solution. I don’t want to get into defending whether its logic is valid; I’m just laying some groundwork.
On Sunday (in Elga’s re-framing of the problem he posed—see my answer), the sample space for the experiment seems to be {T,H} with equal probability for each outcome. Both Monday and Tuesday will happen, and so cannot be used to specify outcomes. This is one of your “different questions.”
But when SB is awake, only one of those days is “current.” (Yes, I know some debate whether days can be used this way—it is a self-fulfilling argument. If it is assumed to be true, you can show that it is true.) Elga was trying to use the day as a valid descriptor of disjoint outcomes, so that in SB’s world the outcomes T&Mon and T&Tue become distinct. The prior probabilities of the three disjoint outcomes {T&Mon, T&Tue, H} are each 1⁄2, and the probability for H can be updated by the conventional definition:
Pr(H|T&Mon or T&Tue or H) = Pr(H)/[Pr(T&Mon)+Pr(T&Tue)+Pr(H)]
Pr(H|T&Mon or T&Tue or H) = (1/2)/[(1/2)+(1/2)+(1/2)] = 1⁄3
If you think about it, this is exactly what Elga argued, except that he didn’t take the extreme step of creating a probability that was greater than 1 in the denominator. He made two problems, that each eliminated an additional outcome. I suggest, but don’t want to get into defending, that this is valid since SB’s situation divides the prior outcome T into two that are disjoint. Each still has the prior probability of the parent outcome.
But this is predicated on that split. Which is what I want to show is valid. What if, instead of leaving SB asleep on Tuesday after Heads, wake her but don’t interview her? Instead, we do something quite different, like take her on a $5,000 shopping spree at Saks Fifth Avenue? (She does deserve the chance of compensation, after all). This gives her a clear way to distinguish H&Mon&Interview from H&Tue&Saks as disjoint outcomes. The use of the day as a descriptor is valid, since this version of the experiment allows for it to be observed.
The actual sample space, on Sunday, for what can happen in an “awake” world for SB is {T&Mon, H&Mon, T&Tue, H&Tue&Saks}. Each has a prior probability of 1⁄4, for being what will apply to SB on a waking day. When she is actually interviewed, she can eliminate one.
And now I suggest that it does not matter, in an interview, what happens differently on H&Tue. SB is interviewed in the context of an interview day, so the question is placed in that context. Claiming it has the context of Sunday Night ignores how the Saks option can only be addressed in a single-day context, and can affect her answer.
Much of the debate gets obscured in the question of outside (of SB’s perception) information. That’s really the purpose of the thought experiment—if the memory wipe is complete and SB has literally no way of knowing whether it’s monday or tuesday, why are they considered (by her) to be different experiences? For purposes of sampling, there is only one waking, the one she’s experiencing.
Alternate formulations (especially frequentist models where there is some “truth” to probability, as opposed to it being purely subjective) can be different, but only by introducing some distinction of experience to make them into two wakings, somehow.
My view is that to an outside observer, the probability is 1⁄2 before the flop, 1 or 0 after the flip. To Beauty, it remains 1⁄2 until she receives evidence, in the form of something she can observe that would be different with heads than with tails. She should WAGER 1⁄3 in some formulations (for instance, if the wagers for Monday and Tuesday are summed, rather than being one evaluation).
I’m sure you’d agree that, if the room has a calendar, Beauty should assign 1⁄2 on Sunday and Monday, and 0 on Tuesday. Never 1⁄3, right?
Two points…
Like a number of people (including Elga), you’re converting an almost-doable thought experiment into one that may be impossible in principle. If the experiment is done with some not-yet-invented but plausible memory-erasing drug, but is otherwise realistic, Beauty will not have the same experiences when woken Monday and when woken Tuesday. Various aspects of her sensed environment, as well as internal thoughts, will be different for the two awakenings. We just assume that none of these differences allow her to infer the day of the week.
Second, if she should wager as if the probability of Heads is 1⁄3, in what sense is the probability of Heads not actually 1/3? The only point of probabilities is to influence actions. (Also, it is not just “some” situations where she should wager on the basis that the probability of Heads is 1⁄3. That’s always what she should do.)
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.