EDIT: It seems this is not getting much attention except for part II with the counter example to SIA. Let me try to intrigue you in to reading this. It provides a systematic argument for double halving in the sleeping beauty problem, if my reasoning is correct and perspective disagreement is indeed the key, we can rejects Doomsday Argument, and disagree with the Presumptuous Philosopher also reject the Simulation Argument.My argument is quite long but the concept is not hard to understand.
Different part of the argument can be found here: I, II, III, IV.
First thing I want to say is that I do not have a mathematics or philosophy degree. I come from an engineering background. So please forgive me when I inevitably messed up some concept. Another thing I want to mention is that English is not my first language. If you think there is any part that is poorly described please point them out. I would try my best to explain what I meant. That being said, I believe I found a good explanation for the SBP.
My main argument is that in case of the sleeping beauty problem, agents free to communicate thus having identical information can still rightfully have different credence to the same proposition. This disagreement is purely caused by the difference in their perspective. And due to this perspective disagreement, SIA and SSA are both wrong because they are answering the question from an outsider’s perspective which provides different answer from beauty’s. I concluded that the correct answer should be double-halving.
My argument involves three thought experiments. Here I am breaking it into several parts to facilitate easier discussion. The complete argument can also be found at www.sleepingbeautyproblem.com. However do note it is quite lengthy and not very well written due to my language skills.
First experiment:Duplicating Beauty (DB)
Beauty falls asleep as usual. The experimenter tosses a fair coin before she wakes up. If the coin landed on T then a perfect copy of beauty will be produced. The copy is precise enough that she cannot tell if herself is old or new. If the coin landed on H then no copy will be made . The beauty(ies) will then be randomly put into two identical rooms. At this point another person, let’s call him the Selector, randomly chooses one of the two rooms and enters. Suppose he saw a beauty in the chosen room. What should the credence for H be for the two of them?
For the Selector this is easy to calculate. Because he is twice more likely to see a beauty in the room if T, simple bayesian updating gives us his probability for H as 1⁄3.
For Beauty, her room has the same chance of being chosen (1/2) regardless if the coin landed on H or T. Therefore seeing the Selector gives her no new information about the coin toss. So her answer should be the same as in the original SBP. If she is a halfer 1⁄2, if she is a thirder 1⁄3.
This means the two of them would give different answers according to halfers and would give the same answer according to thirders. Notice here the Selector and Beauty can freely communicate however they want, they have the same information regarding the coin toss. So halving would give rise to a perspective disagreement even when both parties share the same information.
This perspective disagreement is something unusual (and against Aumann’s Agreement Theorem), so it could be used as an evidence against halving thus supporting Thirdrism and SIA. I would show the problems of SIA in the second thought experiment. For now I want to argue that this disagreement has a logical reason.
Let’s take a frequentist’s approach and see what happens if the experiment is repeated, say 1000 times. For the Selector, this simply means someone else go through the potential cloning 1000 times and each time let him chooses a random room. On average there would be 500 H and T. He would see a beauty for all 500 times after T and see a beauty 250 times after H. Meaning out of the 750 times 1⁄3 of which would be H. Therefore he is correct in giving 1⁄3 as his answer.
For beauty a repetition simply means she goes through the experiment and wake up in a random room awaiting the Selector’s choice again. So by her count, taking part in 1000 repetitions means she would recall 1000 coin tosses after waking up. In those 1000 coin tosses there should be about 500 of H and T each. She would see the Selector about 500 times with equal numbers after T or H. Therefore her answer of 1⁄2 is also correct from her perspective.
(Edit Aug 1. I think this is the easiest way to see the relative frequency from beauty’s perspective: Ignore the clones first, if someone experienced 1000 coin toss she would expect to see 500 Heads. Now consider how the cloning would affect her conclusion. If there is a T somewhere along the tosses thus a clone is created, both the original and the clone would have remembered exactly the same tosses happened earlier. They would also have the same expectation about coin tosses after they split. So the cloning does not changes her answer. Therefore after beauty wakes up remembering a large number of tosses she shall expect about half of those landed in Heads. She does not have to consider when she is physically cloned or if she is the original to reach this conclusion.
Alternatively, a more complicated way would be to think about all the coin tosses happened and her own position. For example a beauty is experiencing two coin tosses there are several ways this can happen from a third party’s perspective:
1. H-H: Both coin tossing landed on H. There is a 1⁄4 chance for this happening. Beauty experience 2 Hs.
2. H-T: H and then T. There is a 1⁄4 chance for this happening. Doesn’t matter if the beauty is any one of the two at the end she experienced 1H and 1T.
3. T-(H,H): First toss landed on T and both resulting beauties experience H at the second toss. There is 1⁄8 chance, doesn’t matter if beauty is any one of the two, she experienced 1 H and 1T.
4. T-(H,T): First coin landed on T and a H and T happened at the second round of tosses respectively. There is 1⁄4 chance. Half the time (1/8) beauty experienced 1H and 1T, while the other half (1/8) she experienced 2 Ts.
5. T-(T,T): All tosses landed on T. There is 1⁄8 chance, doesn’t matter which second toss is hers beauty experienced 2Ts.
The expected number of H from beauty’s perspective: (1/4)x2+(1/4)x1+(1/8)x1+(1/8)x1=1: half the number of tosses. Obviously this calculation can be generalized to a greater number of tosses.
Both the above methods correctly gives the relative frequency from beauty’s perspective. The most common mistake is for a beauty to think she is equally likely to be any one of the beauties at the end of multiple coin tosses. This seemingly intuitive conclusion is valid from a third party’s (the selector’s) perspective, aka if he randomly chooses one from all the ending beauties then all beauties are in symmetrical position. However from beauty’s first person perspective the ending beauties are not in symmetrical position especially since they have experienced different coin toss results. edit finish)
If we call the creation of a new beauty a “branch off”, here we see that from Selector’s perspective experiments from all branches are considered a repetition. Where as from Beauty’s perspective only experiment from her own branch is counted as a repetition. This difference leads to the disagreement.
This disagreement can also be demonstrated by betting odds. In case of T, choosing any of the two rooms leads to the same observation for the Selector: he always sees a beauty and enters another bet. However, for the two beauties the Selector’s choice leads to different observations: whether or not she can see him and enters another bet. So the Selector is twice more likely to enter a bet than any Beauty in case of T, giving them different betting odds respectively.
(EDIT Aug 1. I am aware there are lots of discussions about bets with beauty. Eg. When is the bet settled, base on whose record is it settled, when does beauty pays for the bets and when can she cancels the bet etc. If we consider the importance of perspective disagreement and both party must agree on the bet then these is only one simple way to setup the bet. The bet is settled after each coin toss, it is payed and received at the same time. Beauty’s money clones with her. This way cumulatively both selector’s and beauties’s money would be in accordance with their win/loss record. If this bet is entered whenever the selector sees a beauty in the chosen room then the different betting odds mentioned above would observed. edit ends)
The above reasoning can be easily applied to original SBP. Conceptually it is just an experiment where its duration is divided into two parts by a memory wipe in case of T. The exact duration of the experiment, whether it is two days or a week or five hours, is irrelevant. Therefore from beauty’s perspective to repeat the experiment means her subsequent awakenings need to be shorter to fit into her current awakening. For example, if in the first experiment the two possible awakenings happen on different days, then the in the next repetition the two possible awakening can happen on morning and afternoon of the current day. Further repetitions will keep dividing the available time. Theoretically it can be repeated indefinitely in the form of a supertask. By her count half of those repetitions would be H. Comparing this with an outsider who never experiences a memory wipe: all repetitions from those two days are equally valid repetitions. The disagreement pattern remains the same as in the DB case.
PS: Due to the length of it I’m breaking this thing into several parts. The next part can be found here. Part III here.
Sleeping Beauty Problem Can Be Explained by Perspective Disagreement (I)
EDIT: It seems this is not getting much attention except for part II with the counter example to SIA. Let me try to intrigue you in to reading this. It provides a systematic argument for double halving in the sleeping beauty problem, if my reasoning is correct and perspective disagreement is indeed the key, we can rejects Doomsday Argument, and disagree with the Presumptuous Philosopher also reject the Simulation Argument.My argument is quite long but the concept is not hard to understand.
Different part of the argument can be found here: I, II, III, IV.
First thing I want to say is that I do not have a mathematics or philosophy degree. I come from an engineering background. So please forgive me when I inevitably messed up some concept. Another thing I want to mention is that English is not my first language. If you think there is any part that is poorly described please point them out. I would try my best to explain what I meant. That being said, I believe I found a good explanation for the SBP.
My main argument is that in case of the sleeping beauty problem, agents free to communicate thus having identical information can still rightfully have different credence to the same proposition. This disagreement is purely caused by the difference in their perspective. And due to this perspective disagreement, SIA and SSA are both wrong because they are answering the question from an outsider’s perspective which provides different answer from beauty’s. I concluded that the correct answer should be double-halving.
My argument involves three thought experiments. Here I am breaking it into several parts to facilitate easier discussion. The complete argument can also be found at www.sleepingbeautyproblem.com. However do note it is quite lengthy and not very well written due to my language skills.
First experiment:Duplicating Beauty (DB)
Beauty falls asleep as usual. The experimenter tosses a fair coin before she wakes up. If the coin landed on T then a perfect copy of beauty will be produced. The copy is precise enough that she cannot tell if herself is old or new. If the coin landed on H then no copy will be made . The beauty(ies) will then be randomly put into two identical rooms. At this point another person, let’s call him the Selector, randomly chooses one of the two rooms and enters. Suppose he saw a beauty in the chosen room. What should the credence for H be for the two of them?
For the Selector this is easy to calculate. Because he is twice more likely to see a beauty in the room if T, simple bayesian updating gives us his probability for H as 1⁄3.
For Beauty, her room has the same chance of being chosen (1/2) regardless if the coin landed on H or T. Therefore seeing the Selector gives her no new information about the coin toss. So her answer should be the same as in the original SBP. If she is a halfer 1⁄2, if she is a thirder 1⁄3.
This means the two of them would give different answers according to halfers and would give the same answer according to thirders. Notice here the Selector and Beauty can freely communicate however they want, they have the same information regarding the coin toss. So halving would give rise to a perspective disagreement even when both parties share the same information.
This perspective disagreement is something unusual (and against Aumann’s Agreement Theorem), so it could be used as an evidence against halving thus supporting Thirdrism and SIA. I would show the problems of SIA in the second thought experiment. For now I want to argue that this disagreement has a logical reason.
Let’s take a frequentist’s approach and see what happens if the experiment is repeated, say 1000 times. For the Selector, this simply means someone else go through the potential cloning 1000 times and each time let him chooses a random room. On average there would be 500 H and T. He would see a beauty for all 500 times after T and see a beauty 250 times after H. Meaning out of the 750 times 1⁄3 of which would be H. Therefore he is correct in giving 1⁄3 as his answer.
For beauty a repetition simply means she goes through the experiment and wake up in a random room awaiting the Selector’s choice again. So by her count, taking part in 1000 repetitions means she would recall 1000 coin tosses after waking up. In those 1000 coin tosses there should be about 500 of H and T each. She would see the Selector about 500 times with equal numbers after T or H. Therefore her answer of 1⁄2 is also correct from her perspective.
(Edit Aug 1. I think this is the easiest way to see the relative frequency from beauty’s perspective: Ignore the clones first, if someone experienced 1000 coin toss she would expect to see 500 Heads. Now consider how the cloning would affect her conclusion. If there is a T somewhere along the tosses thus a clone is created, both the original and the clone would have remembered exactly the same tosses happened earlier. They would also have the same expectation about coin tosses after they split. So the cloning does not changes her answer. Therefore after beauty wakes up remembering a large number of tosses she shall expect about half of those landed in Heads. She does not have to consider when she is physically cloned or if she is the original to reach this conclusion.
Alternatively, a more complicated way would be to think about all the coin tosses happened and her own position. For example a beauty is experiencing two coin tosses there are several ways this can happen from a third party’s perspective:
1. H-H: Both coin tossing landed on H. There is a 1⁄4 chance for this happening. Beauty experience 2 Hs.
2. H-T: H and then T. There is a 1⁄4 chance for this happening. Doesn’t matter if the beauty is any one of the two at the end she experienced 1H and 1T.
3. T-(H,H): First toss landed on T and both resulting beauties experience H at the second toss. There is 1⁄8 chance, doesn’t matter if beauty is any one of the two, she experienced 1 H and 1T.
4. T-(H,T): First coin landed on T and a H and T happened at the second round of tosses respectively. There is 1⁄4 chance. Half the time (1/8) beauty experienced 1H and 1T, while the other half (1/8) she experienced 2 Ts.
5. T-(T,T): All tosses landed on T. There is 1⁄8 chance, doesn’t matter which second toss is hers beauty experienced 2Ts.
The expected number of H from beauty’s perspective: (1/4)x2+(1/4)x1+(1/8)x1+(1/8)x1=1: half the number of tosses. Obviously this calculation can be generalized to a greater number of tosses.
Both the above methods correctly gives the relative frequency from beauty’s perspective. The most common mistake is for a beauty to think she is equally likely to be any one of the beauties at the end of multiple coin tosses. This seemingly intuitive conclusion is valid from a third party’s (the selector’s) perspective, aka if he randomly chooses one from all the ending beauties then all beauties are in symmetrical position. However from beauty’s first person perspective the ending beauties are not in symmetrical position especially since they have experienced different coin toss results. edit finish)
If we call the creation of a new beauty a “branch off”, here we see that from Selector’s perspective experiments from all branches are considered a repetition. Where as from Beauty’s perspective only experiment from her own branch is counted as a repetition. This difference leads to the disagreement.
This disagreement can also be demonstrated by betting odds. In case of T, choosing any of the two rooms leads to the same observation for the Selector: he always sees a beauty and enters another bet. However, for the two beauties the Selector’s choice leads to different observations: whether or not she can see him and enters another bet. So the Selector is twice more likely to enter a bet than any Beauty in case of T, giving them different betting odds respectively.
(EDIT Aug 1. I am aware there are lots of discussions about bets with beauty. Eg. When is the bet settled, base on whose record is it settled, when does beauty pays for the bets and when can she cancels the bet etc. If we consider the importance of perspective disagreement and both party must agree on the bet then these is only one simple way to setup the bet. The bet is settled after each coin toss, it is payed and received at the same time. Beauty’s money clones with her. This way cumulatively both selector’s and beauties’s money would be in accordance with their win/loss record. If this bet is entered whenever the selector sees a beauty in the chosen room then the different betting odds mentioned above would observed. edit ends)
The above reasoning can be easily applied to original SBP. Conceptually it is just an experiment where its duration is divided into two parts by a memory wipe in case of T. The exact duration of the experiment, whether it is two days or a week or five hours, is irrelevant. Therefore from beauty’s perspective to repeat the experiment means her subsequent awakenings need to be shorter to fit into her current awakening. For example, if in the first experiment the two possible awakenings happen on different days, then the in the next repetition the two possible awakening can happen on morning and afternoon of the current day. Further repetitions will keep dividing the available time. Theoretically it can be repeated indefinitely in the form of a supertask. By her count half of those repetitions would be H. Comparing this with an outsider who never experiences a memory wipe: all repetitions from those two days are equally valid repetitions. The disagreement pattern remains the same as in the DB case.
PS: Due to the length of it I’m breaking this thing into several parts. The next part can be found here. Part III here.