You still seem to be holding on to the claim that there are as many observations after a head as after a tail; this is clearly false.
No. I never made that claim, so I cannot “hold on to it”. The number of observations after tails doesn’t matter here.
P(Monday ∩ H | W) = P(Monday ∩ T | W). Regardless of whether the coin came up heads or tails you will be woken on Monday precisely once.
Imagine we repeat the sleeping beauty experiment many times. On half of the experiments, she’d be on the heads path. On half of the experiments, she’d be on the tails path. If she is on the tails path, it could be either monday or tuesday. Thus, on an awakening Monday ∩ T is less likely than Monday ∩ H
The claim is implied by your logic; the fact that you don’t engage with it does not prevent it from being a consequence that you need to deal with. Furthermore it appears to be the intuition by which you are constructing your models of Sleeping Beauty.
Imagine we repeat the sleeping beauty experiment many times. On half of the experiments, she’d be on the heads path. On half of the experiments, she’d be on the tails path.
Granted; no contest
If she is on the tails path, it could be either monday or tuesday.
And assuredly she will be woken on both days in any given experimental run. She will be woken twice. Both events occur whenever tails comes up.P(You will be woken on Monday | Tails) = P(You will be woken on Tuesday | Tails) = 1
The arrangement that you are putting forward as a model is that Sleeping Beauty is to be woken once and only once regardless of the coin flip, and thus if she could wake on Tuesday given Tails occurred then that must reduce the change of her waking on Monday given that Tails occurred. However in the Sleeping Beauty problem the number of wakings is not constant. This is the fundamental problem in your approach.
I think we could make faster progress if you started with the assumption that I have read and understood the problem. Yes, I know that she is woken up twice when tails.
You agree that
On half of the experiments, she’d be on the heads path. On half of the experiments, she’d be on the tails path.
Given that she is awake right now, what should be her state of mind. Well, she knows if heads it’s Monday. She knows if tails it’s either Monday or Tuesday. The fact that she will (or has) been woken up on both days doesn’t matter to her right now. It’s either Monday or Tuesday. Given that she cannot distinguish between the two, it would make sense for her to think of those as equally likely at this awakening (under tails). Thus, P(Monday|T,W)=1/2, P(T|W)=1/2, P(Monday ∩ T | W)=1/4.
The problem with your 1⁄3 solution is you treat the data as if they are counts in a 3 by 1 contingency table (the 3 cells being Monday&H, Monday&T, Tuesday&T). If the counts were the result of independent draws from a multinomial distribution, you would get p(H|W)=1/3. You have dependent draws though. You have 1 degree of freedom instead of the usual 2 degrees of freedom. That’s why your ratio is not a probability. That’s why your solution results in nonsense like p(W)=3/2.
As I see it, initially (as a prior, before considering that I’ve been woken up), both Heads and Tails are equally likely, and it is equally likely to be either day. Since I’ve been woken up, I know that it’s not (Tuesday ∩ Heads), but I gain no further information.
Hence the 3 remaining probabilities are renormalised to 1⁄3.
Alternatively: I wake up; I know from the setup that I will be in this subjective state once under Heads and twice under Tails, and they are a priori equally likely. I have no data that can distinguish between the three states of identical subjective state, so my posterior is uniform over them.
If she knows it’s Tuesday then it’s Tails. If she knows it’s Monday then she learns nothing of the coin flip. If she knows the flip was Tails then she is indifferent to Monday and Tuesday. 1⁄3 drops out as the only consistent answer at that point.
As I see it, initially (as a prior, before considering that I’ve been woken up), both Heads and Tails are equally likely, and it is equally likely to be either day.
It’s not equally likely to be either day. If I am awake, it’s more likely that it’s Monday, since that always occurs under heads, and will occur on half of tails awakenings.
I know from the setup that I will be in this subjective state once under Heads and twice under Tails, and they are a priori equally likely. I have no data that can distinguish between the three states of identical subjective state, so my posterior is uniform over them.
Heads and tails are equally likely, a priori, yes. It is equally likely that you will be woken up twice as it is that you will be woken up. Yes. That’s true. But we are talking about your state of mind on an awakening. It can’t be both Monday and Tuesday. So, what should your subjective probability be? Well, I know it’s tails and (Monday or Tuesday) with probability 0.5. I know it’s heads and Monday with probability 0.5.
Before I am woken up, my prior belief is that I spend 24 hours on Monday and 24 on Tuesday regardless of the coin flip. Hence before I condition on waking, my probabilities are 1⁄4 in each cell.
When I wake, one cell is driven to 0, and the is no information to distinguish the remaining 3. This is the point that the sleeping twins problem was intended to illuminate.
Given awakenings that I know to be on Monday, there are two histories with the same measure. They are equally likely. If I run the experiment and count the number of events Monday ∩ H and Monday ∩ T, I will get the same numbers (mod. epsilon errors). Your assertion that it’s H/T with probability 0.5 is false given that you have woken. Hence sleeping twins.
No. I never made that claim, so I cannot “hold on to it”. The number of observations after tails doesn’t matter here.
Imagine we repeat the sleeping beauty experiment many times. On half of the experiments, she’d be on the heads path. On half of the experiments, she’d be on the tails path. If she is on the tails path, it could be either monday or tuesday. Thus, on an awakening Monday ∩ T is less likely than Monday ∩ H
The claim is implied by your logic; the fact that you don’t engage with it does not prevent it from being a consequence that you need to deal with. Furthermore it appears to be the intuition by which you are constructing your models of Sleeping Beauty.
Granted; no contest
And assuredly she will be woken on both days in any given experimental run. She will be woken twice. Both events occur whenever tails comes up.P(You will be woken on Monday | Tails) = P(You will be woken on Tuesday | Tails) = 1
The arrangement that you are putting forward as a model is that Sleeping Beauty is to be woken once and only once regardless of the coin flip, and thus if she could wake on Tuesday given Tails occurred then that must reduce the change of her waking on Monday given that Tails occurred. However in the Sleeping Beauty problem the number of wakings is not constant. This is the fundamental problem in your approach.
I think we could make faster progress if you started with the assumption that I have read and understood the problem. Yes, I know that she is woken up twice when tails.
You agree that
Given that she is awake right now, what should be her state of mind. Well, she knows if heads it’s Monday. She knows if tails it’s either Monday or Tuesday. The fact that she will (or has) been woken up on both days doesn’t matter to her right now. It’s either Monday or Tuesday. Given that she cannot distinguish between the two, it would make sense for her to think of those as equally likely at this awakening (under tails). Thus, P(Monday|T,W)=1/2, P(T|W)=1/2, P(Monday ∩ T | W)=1/4.
The problem with your 1⁄3 solution is you treat the data as if they are counts in a 3 by 1 contingency table (the 3 cells being Monday&H, Monday&T, Tuesday&T). If the counts were the result of independent draws from a multinomial distribution, you would get p(H|W)=1/3. You have dependent draws though. You have 1 degree of freedom instead of the usual 2 degrees of freedom. That’s why your ratio is not a probability. That’s why your solution results in nonsense like p(W)=3/2.
As I see it, initially (as a prior, before considering that I’ve been woken up), both Heads and Tails are equally likely, and it is equally likely to be either day. Since I’ve been woken up, I know that it’s not (Tuesday ∩ Heads), but I gain no further information.
Hence the 3 remaining probabilities are renormalised to 1⁄3.
Alternatively: I wake up; I know from the setup that I will be in this subjective state once under Heads and twice under Tails, and they are a priori equally likely. I have no data that can distinguish between the three states of identical subjective state, so my posterior is uniform over them.
If she knows it’s Tuesday then it’s Tails. If she knows it’s Monday then she learns nothing of the coin flip. If she knows the flip was Tails then she is indifferent to Monday and Tuesday. 1⁄3 drops out as the only consistent answer at that point.
It’s not equally likely to be either day. If I am awake, it’s more likely that it’s Monday, since that always occurs under heads, and will occur on half of tails awakenings.
Heads and tails are equally likely, a priori, yes. It is equally likely that you will be woken up twice as it is that you will be woken up. Yes. That’s true. But we are talking about your state of mind on an awakening. It can’t be both Monday and Tuesday. So, what should your subjective probability be? Well, I know it’s tails and (Monday or Tuesday) with probability 0.5. I know it’s heads and Monday with probability 0.5.
Before I am woken up, my prior belief is that I spend 24 hours on Monday and 24 on Tuesday regardless of the coin flip. Hence before I condition on waking, my probabilities are 1⁄4 in each cell.
When I wake, one cell is driven to 0, and the is no information to distinguish the remaining 3. This is the point that the sleeping twins problem was intended to illuminate.
Given awakenings that I know to be on Monday, there are two histories with the same measure. They are equally likely. If I run the experiment and count the number of events Monday ∩ H and Monday ∩ T, I will get the same numbers (mod. epsilon errors). Your assertion that it’s H/T with probability 0.5 is false given that you have woken. Hence sleeping twins.
That is Beauty’s probability of which day it is AFTER considering that she has been woken up.