Dead men make no observations. The equation you gave is fine for before the killing (for guessing what color you will be if you survive), not for after (when the set of observers is no longer the same).
Under a frequentist interpretation it is not possible for the equation to work pre-killing and yet not work post-killing: if one’s estimate of P(R|KS) = 0.01 is correct, that implies one has correctly estimated the relative frequency of having been red-doored given that one survives the killing. That estimate of the relative frequency cannot then change after the killing, because that is precisely the situation for which the relative frequency was declared correct!
The computer will interview all of the survivors. So in the 1-shot case, there is a 50% chance it asks the red door survivor, and a 50% chance it talks to the 99 blue door ones. They all get an interview because all survivors make observations and we want to make it an equivalent situation. So if you get interviewed, there is a 50% chance that you are the red door one, and a 50% chance you are one of the blue door ones.
I don’t agree, because in my judgment the greater number of people initially behind blue doors skews the probability in favor of ‘you’ being behind a blue door.
If you are having trouble following my explanations, maybe you’d prefer to see what Nick Bostrom has to say. This paper talks about the equivalent Sleeping Beauty problem. The main interesting part is near the end where he talks about his own take on it. He correctly deduces that the probability for the 1-shot case is 1⁄2, and for the many-shot case it approaches 1⁄3 (for the SB problem).
Reading Bostrom’s explanation of the SB problem, and interpreting ‘what should her credence be that the coin will fall heads?’ as a question asking the relative frequency of the coin coming up heads, it seems to me that the answer is 1⁄2 however many times Sleeping Beauty’s later woken up: the fair coin will always be tossed after she awakes on Monday, and a fair coin’s probability of coming up heads is 1⁄2.
In the 1-shot case, the whole concept of a frequentist interpretation makes no sense. Frequentist thinking invokes the many-shot case.
Reading Bostrom’s explanation of the SB problem, and interpreting ‘what should her credence be that the coin will fall heads?’ as a question asking the relative frequency of the coin coming up heads, it seems to me that the answer is 1⁄2 however many times Sleeping Beauty’s later woken up: the fair coin will always be tossed after she awakes on Monday, and a fair coin’s probability of coming up heads is 1⁄2.
I am surprised you think so because you seem stuck in many-shot thinking, which gives 1⁄3.
Maybe you are asking the wrong question. The question is, given that she wakes up on Monday or Tuesday and doesn’t know which, what is her creedence that the coin actually fell heads? Obviously in the many-shot case, she will be woken up twice as often during experiments where it fell tails, so in 2⁄3 or her wakeups the coin will be tails.
In the 1-shot case that is not true, either she wakes up once (heads) or twice (tails) with 50% chance of either.
Consider the 2-shot case. Then we have 4 possibilities:
coins , days , fraction of actual wakeups where it’s heads
It seems I was solving an equivalent problem. In the formulation you are using, the weighted average should reflect the number of wakeups.
What this results means is that SB should expect with probabilty 1⁄3, that if she were shown the results of the coin toss, she would observe that the result was heads.
No, it shouldn’t—that’s the point. Why would you think it should?
Note that I am already taking observer-counting into account—among observers that actually exist in each coin-outcome-scenario. Hence the fact that P(heads) approaches 1⁄3 in the many-shot case.
In the 1-shot case, the whole concept of a frequentist interpretation makes no sense. Frequentist thinking invokes the many-shot case.
Maybe I misunderstand what the frequentist interpretation involves, but I don’t think the 2nd sentence implies the 1st. If I remember rightly, a frequentist interpretation of probability as long-run frequency in the case of Bernoulli trials (e.g. coin flips) can be justified with the strong law of large numbers. So one can do that mathematically without actually flipping a coin arbitrarily many times, from a definition of a single Bernoulli trial.
Maybe you are asking the wrong question.
My initial interpretation of the question seems to differ from the intended one, if that’s what you mean.
The question is, given that she wakes up on Monday or Tuesday and doesn’t know which, what is her creedence that the coin actually fell heads?
This subtly differs from Bostrom’s description, which says ‘When she awakes on Monday’, rather than ‘Monday or Tuesday.’ I think your description probably better expresses what Bostrom is getting at, based on a quick skim of the rest of Bostrom’s paper, and also because your more complex description makes both of the answers Bostrom mentions (1/2 and 1⁄3) defensible: depending on how I interpret you, I can extract either answer from the one-shot case, because the interpretation affects how I set up the relative frequency.
If I count how many times on average the coin comes up heads per time it is flipped, I must get the answer 1⁄2, because the coin is fair.
If I count how many times on average the coin comes up heads per time SB awakes, the answer is 1⁄3. Each time I redo the ‘experiment,’ SB has a 50% chance of waking up twice with the coin tails, and a 50% chance of waking up once with the coin heads. So on average she wakes up 0.5×2 + 0.5×1 = 1.5 times, and 0.5×1 = 0.5 of those 1.5 times correspond to heads: hence 0.5/1.5 = 1⁄3.
I’m guessing that the Bayesian analog of these two possible thought processes would be something like
SB asking herself, ‘if I were the coin, what would I think my chance of coming up heads was whenever I’m awake?’
SB asking herself, ‘from my point of view, what is the coin about to be/was the coin yesterday whenever I wake up?’
but I may be wrong. At any rate, I haven’t thought of a rationale for your 2-shot calculation. Repeating the experiment twice shouldn’t change the relative frequencies—they’re relative! So the 2-shot case should still have 1⁄2 or 1⁄3 as the only justifiable credences.
This subtly differs from Bostrom’s description, which says ‘When she awakes on Monday’, rather than ‘Monday or Tuesday.’
He makes clear though that she doesn’t know which day it is, so his description is equivalent. He should have written it more clearly, since it can be misleading on the first pass through his paper, but if you read it carefully you should be OK.
So on average …
‘On average’ gives you the many-shot case, by definition.
In the 1-shot case, there is a 50% chance she wakes up once (heads), and a 50% chance she wakes up twice (tails). They don’t both happen.
In the 2-shot case, the four possibilities are as I listed. Now there is both uncertainty in what really happens objectively (the four possible coin results), and then given the real situation, relevant uncertainty about which of the real person-wakeups is the one she’s experiencing (upon which her coin result can depend).
I think I essentially agree with this comment, which feels strange because I suspect we would continue to disagree on a number of the points we discussed upthread!
Under a frequentist interpretation it is not possible for the equation to work pre-killing and yet not work post-killing: if one’s estimate of P(R|KS) = 0.01 is correct, that implies one has correctly estimated the relative frequency of having been red-doored given that one survives the killing. That estimate of the relative frequency cannot then change after the killing, because that is precisely the situation for which the relative frequency was declared correct!
I don’t agree, because in my judgment the greater number of people initially behind blue doors skews the probability in favor of ‘you’ being behind a blue door.
Reading Bostrom’s explanation of the SB problem, and interpreting ‘what should her credence be that the coin will fall heads?’ as a question asking the relative frequency of the coin coming up heads, it seems to me that the answer is 1⁄2 however many times Sleeping Beauty’s later woken up: the fair coin will always be tossed after she awakes on Monday, and a fair coin’s probability of coming up heads is 1⁄2.
In the 1-shot case, the whole concept of a frequentist interpretation makes no sense. Frequentist thinking invokes the many-shot case.
I am surprised you think so because you seem stuck in many-shot thinking, which gives 1⁄3.
Maybe you are asking the wrong question. The question is, given that she wakes up on Monday or Tuesday and doesn’t know which, what is her creedence that the coin actually fell heads? Obviously in the many-shot case, she will be woken up twice as often during experiments where it fell tails, so in 2⁄3 or her wakeups the coin will be tails.
In the 1-shot case that is not true, either she wakes up once (heads) or twice (tails) with 50% chance of either.
Consider the 2-shot case. Then we have 4 possibilities:
coins , days , fraction of actual wakeups where it’s heads
HH , M M , 1
HT , M M T , 1⁄3
TH , M T M , 1⁄3
TT , M T M T , 0
Now P(heads) = (1 + 1⁄3 + 1⁄3 + 0) / 4 = 5⁄12 = 0.417
Obviously as the number of trials increases, P(heads) will approach 1⁄3.
This is assuming that she is the only observer and that the experiments are her whole life, BTW.
This should be a weighted average, reflecting how many coin flips are observed in the four cases:
There are always 2 coin flips, and the results are not known to SB. I can’t guess what you mean, but I think you need to reread Bostrom’s paper.
It seems I was solving an equivalent problem. In the formulation you are using, the weighted average should reflect the number of wakeups.
What this results means is that SB should expect with probabilty 1⁄3, that if she were shown the results of the coin toss, she would observe that the result was heads.
No, it shouldn’t—that’s the point. Why would you think it should?
Note that I am already taking observer-counting into account—among observers that actually exist in each coin-outcome-scenario. Hence the fact that P(heads) approaches 1⁄3 in the many-shot case.
Maybe I misunderstand what the frequentist interpretation involves, but I don’t think the 2nd sentence implies the 1st. If I remember rightly, a frequentist interpretation of probability as long-run frequency in the case of Bernoulli trials (e.g. coin flips) can be justified with the strong law of large numbers. So one can do that mathematically without actually flipping a coin arbitrarily many times, from a definition of a single Bernoulli trial.
My initial interpretation of the question seems to differ from the intended one, if that’s what you mean.
This subtly differs from Bostrom’s description, which says ‘When she awakes on Monday’, rather than ‘Monday or Tuesday.’ I think your description probably better expresses what Bostrom is getting at, based on a quick skim of the rest of Bostrom’s paper, and also because your more complex description makes both of the answers Bostrom mentions (1/2 and 1⁄3) defensible: depending on how I interpret you, I can extract either answer from the one-shot case, because the interpretation affects how I set up the relative frequency.
If I count how many times on average the coin comes up heads per time it is flipped, I must get the answer 1⁄2, because the coin is fair.
If I count how many times on average the coin comes up heads per time SB awakes, the answer is 1⁄3. Each time I redo the ‘experiment,’ SB has a 50% chance of waking up twice with the coin tails, and a 50% chance of waking up once with the coin heads. So on average she wakes up 0.5×2 + 0.5×1 = 1.5 times, and 0.5×1 = 0.5 of those 1.5 times correspond to heads: hence 0.5/1.5 = 1⁄3.
I’m guessing that the Bayesian analog of these two possible thought processes would be something like
SB asking herself, ‘if I were the coin, what would I think my chance of coming up heads was whenever I’m awake?’
SB asking herself, ‘from my point of view, what is the coin about to be/was the coin yesterday whenever I wake up?’
but I may be wrong. At any rate, I haven’t thought of a rationale for your 2-shot calculation. Repeating the experiment twice shouldn’t change the relative frequencies—they’re relative! So the 2-shot case should still have 1⁄2 or 1⁄3 as the only justifiable credences.
(Edited to fix markup/multiplication signs.)
He makes clear though that she doesn’t know which day it is, so his description is equivalent. He should have written it more clearly, since it can be misleading on the first pass through his paper, but if you read it carefully you should be OK.
‘On average’ gives you the many-shot case, by definition.
In the 1-shot case, there is a 50% chance she wakes up once (heads), and a 50% chance she wakes up twice (tails). They don’t both happen.
In the 2-shot case, the four possibilities are as I listed. Now there is both uncertainty in what really happens objectively (the four possible coin results), and then given the real situation, relevant uncertainty about which of the real person-wakeups is the one she’s experiencing (upon which her coin result can depend).
I think I essentially agree with this comment, which feels strange because I suspect we would continue to disagree on a number of the points we discussed upthread!