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The doomsday argument is controversial not because its conclusion is bleak but because it has some pretty hard to explain implications. Like the choice of reference class is arbitrary but affects the conclusion, it also gives some unreasonable predicting power and backward causations. Anyone trying to understand it would eventually have to reject the argument or find some way to reconcile with these implications. To me neither position are biased as long as it is sufficiently argued.
The post specifically explained why your properties cannot be used for predictions in the context of doomsday argument and sleeping beauty problem. I would like to know your thoughts on that.
I will just post the relationship between perspective reasoning and simulation argument here.
In 2003 Nick Bostrom published his paper “Are you living in a computer simulation?”. In that paper he suggested once a civilization reaches a highly developed state it would have enough computing power to run “ancestral simulations”. Such simulations would be indistinguishable from actual reality for its occupants. Furthermore, because the potential number and levels of such simulated realities is huge, almost all observers with experiences similar to ours would be living in such simulations. Therefore either civilizations such as ours would never run such ancestral simulations or we are almost certainly living in a simulation right now. Perhaps one of its most striking conclusions is once we develop a ancestral simulation, or believes we eventually would develop one, then we shall conclude we are simulated as well. This highly specific world creation theory, while seems very unlikely at first glance, shall be deemed as almost certain if we apply the probability reasoning described in the argument. I would argue that such probability reasoning is in fact mistaken.
The argument states if almost all of observers with experiences similar to ours are simulated, we shall conclude we are almost certainly simulated. The core of this reasoning is self-sampling assumption (SSA) which states an observer shall reason as if she is randomly selected from all observers. The top contender to SSA, which is used as a counter argument to one of its most (in)famous applications: doomsday argument, is self-indication assumption (SIA). SIA states an observer shall reason as if she is randomly selected from all potential observers. However if we apply SIA to this idea the result is even further confirmation that we are simulated. Whether or not we would be able to run an ancestral simulation is no longer relevant, the fact that we exist is evidence suggesting our reality is simulated.
However, if we apply the same perspective reasoning used in the sleeping beauty problem this argument falls apart. Perspective reasoning states due to the existence of perspective disagreement between agents, an observer shouldn't reason as an imaginary third party who randomly selected the observer from a certain reference class. Picture a third party (god) randomly chooses a person from all realities, it is obvious the selected person is most likely simulated if majority of the observers are. Without this logic however an observer could no longer make such conclusion. Therefore even after running an ancestral simulation our credence of being simulated would not instantly jump to near certain.
The immediate opposition to this would be: in the duplicating beauty problem upon learning the coin landed on T beauty’s credence of being the clone would rise from 1/4 to 1/2, then why our credence of being simulated does not rise accordingly once we run ancestral simulations? After all the former case confirms the existence of a clone while the latter case confirms the existence of many simulated realities. The distinction here is the clone and the original are in symmetrical positions, whereas our reality and the realities simulated by us are not. In case of duplicating beauty, although they can have different experience after waking up, the original and the clone have identical information about the same coin toss. Due to this epistemic equivalence beauty cannot tell if she is the clone or the original. Therefore upon learning the coin landed on T thus confirming the existence of a clone both beauties must reason she is equally likely to be the clone and the original. In another word, the rise of credence is due to the confirmed existence of a symmetrical counterpart not due to the mere existence of someone in an imaginary reference class to choose from. But running an ancestral simulation only confirms the latter. Putting it blatantly, we know for sure we are not in the simulations we run so no matter how many simulation we run our credence of being in an ancestral simulation should not rise.
In fact I would suggest following the logic of Bostrom’s argument we should reduce our credence of living in a simulated reality once we run an ancestral simulation. As stated in his paper, simulators might want to edit their simulations to conserve computational power. A simulated reality running its own subsequent levels of simulations is going to require exponential amount of additional computational power. It is in the simulator’s interest to edit their simulation so they never reach such an advanced state with high computational capabilities. This means a base level reality is more likely to produces ancestral simulations than the simulated ones. Therefore once we runs such ancestral simulations, or strongly believe we are going to do so, our credence of being simulated shall decease.
Yes, that's why I think to this day Elga's counter argument is still the best.
No problem, always good to have a discussion with someone serious about the subject matter.
First of all, you are right: statistic estimation and expected value in bayesian analysis are different. But that is not what I'm saying. What I'm saying is in a bayesian analysis with an uninformed prior (uniform) the case with highest probability should be the unbiased statistic estimation (it is not always so because round offs etc).
In the two urns example, I think what you meant is that using the sample of 4 balls a fair estimation would be 5 reds and 15 blues as in the case of B but bayesian analysis would give A as more likely? However this disagreement is due to the use of an informed prior, that you already know we are more likely to draw from A right from the beginning. Without knowing this bayesian would give B as the most likely case, same as statistic estimate.
Think of it like this: if Beauty opens 8 doors and they're all red, and then she goes to open a ninth door, how likely should she think it is to be red? 100%, or something smaller than 100%? For predictions, we use the average of a probability distribution, not just its highest point.
Definitely something smaller than 100%. Just because beauty thinks r=81 is the most likely case doesn't mean she think it is the only case. But that is not what the estimation is about. Maybe this question would be more relevant: If after opening 8 doors and they are all red and beauty have to guess R. what number should she guess (to be most likely correct)?
Thank you for the reply. I really appreciate it since it reminds me that I have made a mistake in my argument. I didn't say SSA means reasoning as if an observer is randomly selected from all actually existent observers ( past, present and /b/future/b/).
So how do you get Beauty's prediction? If at the end of the first day you ask for a prediction on the coin, but you don't ask on the second day, then now Beauty knows that the coin flip is, as you say, yet to happen, and so she goes back to predicting 50/50. She only deviates from 50/50 when she thinks there's some chance that the coin flip has already happened.
I think Elga's argument is beauty's credence should not be dependent on the exact time of coin toss. It seems reasonable to me since the experiment can be carried out exact the same way no matter if the coin is tosses on Sunday or Monday night. According to SSA beauty should update credence of H to 2/3 after learning it is Monday. If you think beauty shall give 1/2 if she finds out the coin is tossed on Monday night then her answer would be dependent on the time of coin toss. Which to me seems a rather weak position.
Regarding a betting odds argument. I have give a frequentist model in part I which uses betting odds as part of the argument. In essence, beauty's break even odd is at 1/2 while the selector's is at 1/3, which agrees with there credence.
Ok, I should have use my words more carefully. We meant the same thing. When I say beauty think the 8 rooms are unbiased sample I meant what I listed as C: It is an unbiased for the other 80 rooms. So yes to what you said, sorry for the confusion. it is obvious because it is a simple random sample chosen from the 80 rooms. So that part there is no disagreement. The disagreement between the two is about whether or not the 9 rooms are an unbiased sample. Beauty as a thirder should not think it is unbiased but bases her estimation on it anyway to answer the question from the selector's perspective. If she does not answer from selector's perspective she would use the 8 rooms to estimate the reds in the other 80 rooms and then add her own room in, as halfers does.
Regarding the selector chooses a room and finds out it is red. Again they agree on whether or not the 8 rooms are unbiased, however because the first room is always red for beauty but not so for the selector they see the 9 rooms differently. From beauty's perspective dividing the 9 rooms into 2 parts and she gets a unbiased sample (8 rooms) and a red room. It is not so for the selector. We can list the three points from the selector's perspective and it poses no problem at all.
A: the 9 room is an unbiased sample for 81 rooms
B: the first room is randomly selected from all rooms
C: the other 8 rooms is an unbiased sample for other 80 rooms.
alternatively we can divid the 9 rooms as follows:
A: the 9 rooms is an unbiased sample for 81 rooms
B: the first red room he saw (if he saw one) is always red
C: the other 8 rooms in the sample is biased towards blue
Either way there is no problem. In terms of the predicting power, think of it this way. Once the selector sees a red room he knows if he ignore it and only consider the other 8 rooms then the sample is biased towards blue, nothing supernatural. However, for beauty if she thinks the 9 rooms are unbiased then the 8 rooms she chooses must be biased even though they are selected at random. Hence the "supernatural". It is meant to point out for beauty the 9 and 8 rooms cannot be unbiased at same time. Since you already acknowledged the 9 rooms is biased (for her perspective at least), then yes she does not have supernatural predicting power of course.
I guess the bottomline is because they acquire their information differently, the selector and thirder beauty must disagree somewhere. Either on the numerical value of estimate, or on if a sample is biased.
About the perspectivism posts. The concept is actually quite simple: each beauty only counts what she experienced/remembered. But I feel maybe I'm not doing a good job explaining it. Anyway, thank you for promising to check it out.
Very clear argument and many good points. Appreciate the effort.
Regarding your position on thirders vs halfers, I think it is a completely reasonable position and I agree with the analysis about when halfers are correct and when thirders are correct. However to me it seems to treat Sleeping Beauty more as a decision making problem rather than a probability problem. Maybe one's credence without relating consequences is not defined. However that seems counter intuitive to me. Naturally one should have a belief about the situation and her decisions should depend on it as well as her objective (how much beauty cases about other copies) and the payoff structure (is the money reward depends only on her own answer, or all correct answers or accuracy rate etc). If that's the case, there should exist a unique correct answer to the problem.
About how should beauty estimate R and treat the samples, I would say that's the best position for a thirder to take. In fact that's the same position I would take too. If I may reword it slightly, see if you agrees with this version: The 8 rooms is a unbiased sample for beauty, that is too obvious to argue otherwise. Her own room is always red so the 9 rooms is obviously biased for her. However from (an imaginary) selector's perspective if he finds the same 9 rooms it is an unbiased sample. Thirders think she should answer from the selector's perspective, (I think the most likely reason being she is repeatedly memory wiped makes her perspective somewhat "compromised") therefore she would estimate R to be 27. Is this version something you would agree?
In this version I highlighted the disagreement between the selector and beauty, the disagreement is not some numerical value but they disagree on whether a sample is biased. In my 4 posts all I'm trying to do is arguing for the validity and importance of perspective disagreement. If we recognize the existence of this disagreement and let each agent answers from her own perspective we get another system of reasoning different from SIA or SSA. It provides an argument for double halving, give a framework where frequentist and bayesians agrees with each other, reject Doomsday Argument, disagree with Presumptuous Philosopher, and rejects the Simulation Argument. I genuinely think this is the explanation to sleeping beauty problem as well as many problems related to anthropic reasoning. Sadly only the part arguing against thirding gets some attention.
Anyways, I digressed. Bottomline is, though I do no think it is the best position, I feel your argument is reasonable and well thought. I can understand it if people want to take it as their position.
Yes, I have given a long run frequency argument for halving in part I. Sadly that part have not gotten any attention. My entire argument is about the importance of perspective disagreement in SBP. This counter argument is actually the less important part.
OK, I misunderstood. I interpreted the coin is biased 1/3 to 2/3 but we don't know which side it favours. If we start from uniform (1/2 to H and 1/2 to T), then the maximum likelihood is Tails.
Unless I misunderstood again, you mean there is a coin we want to guess its natural chance (forgive me if I'm misusing terms here). We do know its chance is bounded between 1/3 and 2/3. In this case yes, the statistical estimate is 0 while the maximum likelihood is 1/3. However it is obviously due to the use of a informed prior (that we know it is between 1/3 and 2/3). Hardly a surprise.
Also I want to point out in your previous example you said SIA+frequentist never had any strong defenders. That is not true. Until now in literatures thirding are generally considered to be a better fit for frequentist than halving. Because long run frequency of Tail awakening is twice as many as Head awakenings. Such arguments are used by published academics including Elga. Therefore I would consider my attack from the frequentist angle has some value.
Maximum likelihood is indeed 0 or Tails, assuming we start from a uniform prior. 1/3 is the expected value. Ask yourself this, after seeing a tail what should you guess for the next toss result to have maximum likelihood of being correct?
If halfers reasoning applies to both Bayesian and Frequentist while SIA is only good in Bayesian isn't it quite alarming to say the least?
Nothing shameful on that. Similar arguments, which Jacob Ross categorized as "hypothetical priors" by adding another waking in case of H, have not been a main focus of discussion in literatures for the recent years. I would imagine most people haven't read that.
In fact you should take it as a compliment. Some academic who probably spent a lot of time on it came up the same argument as you did.
Appreciate the effort. Especially about the calculation part. I am no expert on coding. But from my limited knowledge on python the calculation looks correct to me. I want to point out for the direct calculation formulation like this+choose+3)++((81-r)+choose+6)),+r%3D3+to+75)+%2F+(sum+(+((r)+choose+3)++((81-r)+choose+6)),+r%3D3+to+75)) gives the same answer. I would say it reflect SIA reasoning more and resemble your code better as well. Basically it shows under SIA beauty should treat her own room the same way as the other 8 rooms.
The part explaining the relationship between expected value and unbiased estimation (maximum likelihood) is obviously correct. Though I wouldn't say it is relevant to the argument.
You claim Bayesian's don't usually uses maximum likelihood or unbiased estimates. I would say that is a mistake. They are important in decision making. However "usually" is a subjective term and argument about how often is "usual" is pointless. The bottom line is they are valid questions to ask and bayesians should have an answer. And how should thirders answer it, that is the question.
For the priors,. I would consider Beauty's expectations from the problem definition before she takes a look at anything to be a prior, i.e. she expects 81 times higher probability of R=81 than R=1 right from the start.
In the original sleeping beauty problem, what is the prior for H according to a thirder? It must be 1/2. In fact saying she expects 2 times higher probability of T than H right from the start means she should conclude P(H)=1/3 before going to sleep on Sunday. That is used as a counter argument by halfers. Thirders are arguing after waking up in the experiment, beauty should update her probability as waking up is new information. T being 2 times more likely than H is a posterior.
If you think thirders should reject A based on your interpretation of SIA, then what is a fair estimation of R according to thirders? Should they use a biased sample of 9 rooms and estimate 27, or estimate 21 and disagree with the selector having the same information?
This argument is the same as Cian Dorr's version with a weaker amnesia drug. In that experiment a weaker amnesia drug is used on beauty if Heads which only delays the recollection of memory for a few minutes, just like in your case the memory is delayed until the message is checked.
This argument was published in 2002. It is available before majority of the literature on the topic is published. Suffice to say it is not convincing to halfers. Even supporter like Terry Horgan admit the argument is suggestive and could run a serious risk of slippery slope.
In both boy or girl puzzle and Monty hall problem the main point is "how" the new information is obtained. Is the mathematician randomly picking a child and mentioning its gender, or is he purposely checking for a boy among his children. Does the host know what's behind the door and always reveal a goat, or does he simple randomly opens a door and it turns out to be a goat. Or in statistic terms: how is the sample drawn. Once that is clear bayesian and statistics gives the same result. Of course if one start from a wrong assumption about the sampling process his conclusion would be wrong. No argument there.
But SIA itself is a statement regarding how the sample is drawn. Why we must only check its merit with bayesian but not stats? And if you are certain the statistic reasoning is wrong then instead of pointing to different probability puzzles why not point out the mistake?
With all these posts you haven't even mention whether you believe the thirder should estimate R=27 or not. While I have been explicitly clear about my positions and dissecting my arguments step by step I feel you are being very vague about yours. This put me into a harder and more labours position to counter argue. That's why I feel this discussion is no longer about sleeping beauty problem but more about who's right and who's better at arguing. That's not productive, and I am leaving it.
Both claims are very bold, both unsubstantiated.
First of all, SIA in bayesian is up to debate. That's the whole point of halfer/thirder disagreement. A "consistent" reasoning is not necessarily correct. Halfers are also consistent.
Second of all, the statistics involved is as basic as it gets. You are saying with a simple random sample of 9 rooms with 3 reds, it is wrong to estimate the population have 30% reds. Yet no argument is given.
Also please take no offence, but I am not going to continue this discussion we are having. All I have been doing is explaining the same points again and again. While the replies I got are short and effortless. I feel this is no longer productive.
Ok, let's slow down. First of all there are two type of analysis going on. One is bayesian analysis which you are focusing on. The other is simple statistics, which I am saying thirders and SIA are having troubles with.
If there are 100 rooms either red or blue. You randomly open 10 of them and saw 8 red and 2 blue. Here you can start a bayesian analysis (with an uniform prior obviously) and construct the pdf. I'm going to skip the calculation and just want to point out R=80 would have the highest probability. Now instead of going the bayesian way you can also just use simple statistics. You have a simple random sample of size 10 with 8 reds. So the unbiased estimation should be 8/10x100=80. You have applied two different ways of reasoning but got the same result, unsurprising since you used uniformed prior in bayesian analysis. So far I hope everything is clear.
Now let's consider SIA. It tells you how to interpret the fact your own room is red. It says you should treat your own room as randomly selected from all rooms and it happens to be red, which is new information. Now if you open another room, then both rooms are randomly selected from all rooms. Thirders bayesian reasoning is consistent with this idea as shown by the calculation in my last reply.
Now apply SIA to statistics. Because it treat both rooms as randomly selected it is a simple random sample, which is a unbiased sample. I am not supporting that, all I'm saying is that's what SIA suggests. The population to this sample, is all the rooms (including your own). Using statistics you can give an estimation of R the same way we gave the estimation of 80 before. Let's call it E1. If thirder think SIA is valid they should stand by this estimation.
But you know you randomly selected a room (from the other 2 rooms). Which is a simple random sample of the other 2 rooms. If it helps, the room(s) you randomly selected are "the beans in hand", all other rooms are "beans in bags". Surely you should expect their fraction of red to be about equal right? Well, as I have calculated in my last reply. If you stand by the above estimation E1, then you would always conclude the rest of the rooms have a higher fraction of red, unless all the room you randomly opened are red of course. Basically you are already concluding the sample is biased towards blue before the selection is made. Or if you prefer, before you grab you already know you are going to say it has lower fraction of red than the bag does.
In essence you cannot take a unbiased sample and divided it into two parts, claiming one part is biased towards red while the other part is unbiased. The other part must be biased towards opposite direction aka blue.
I hope you now see that the probability of 2/3 you calculated is not relevant. It is a probability calculated using bayesian analysis. Not a sample's property or the sample's "fraction" used in statistics. For what it is worth, yes I agree with your calculation. It is the correct number a thirder would give.
Thirder and the selector have the exact same prior and posteriors. Their bayesian analysis are exactly the same.
Think from the selector's perspective. He randomly opens 9 out of the 81 rooms and found 3 red. Say he decided to perform a bayesian analysis. As stated in the question he starts from an uniform prior and updates it with the 9 rooms as new information. I will skip the calculation but in the end he concluded R=27 has the highest probability. Now think from the thirder's perspective. As SIA states she is treating her own room as randomly selected from all the rooms. Therefore she is treating all of the 9 rooms as randomly selected from the 81 rooms. The new information she has is exactly the same as the selector. Starting from an uniform prior and updates on those new information she would get the same pdf as the selector with R=27 has the highest probability. The two of them must agree just as in the original sleeping beauty problem.
Now suppose instead of doing a bayesian analysis, the selector just want to perform statistical analysis. He wants to get a fair estimate of R. It is clear he has a simple random sample with sample size 9 and 3 of which are red. He can estimate the R of the population as 3/9x81=27. It is unsurprising he got the same number as his bayesian analysis since he started from an uniform prior. Until now, all good.
The problem, however, starts when the thirder wants to use his sample in simple statistics. If he uses SIA reasoning as he did in bayesian analysis he would treat the 9 rooms as a simple random sample the same way as the selector did and give R=27 as the unbiased estimation. By doing so he accepts the 9 rooms as an unbiased sample which leads to the problems I discussed in the main post.
Maybe it is my English. In this case, you wake up in a red room, and open another room and found it to be blue. As SIA states, you should treat both rooms as they are randomly selected from all rooms. So in the 2 randomly selected rooms 1 is red and 1 is blue. Hence 50%.
Sure. Although I want to point out the estimation would be very rough. That is just the nature of statistics with very small sample size.
The "beans in the hand" would be the random other room you open. The "beans in the bag" would be the two other rooms.
Let's say you open another room and found it red. If I'm correct as a thirder you would give R=3 a probability of 3/4, R=2 a probability of 1/4. This can be explained by SIA: this is randomly selecting 2 rooms out of the 3 and they are both red. 3 ways for it to happen if R=3 and 1 way for it to happen if R=1. This is the bayesian analysis.
A statistic estimation of R is also quite easy. As SIA states you have 2 randomly selected rooms out of 3. Both of the rooms are red. Simple statistics suggests unbiased estimate would be all 3 rooms are red, aka R=3. You can also only estimate the number of reds in the other 2 rooms. You have 1 random room chosen out of the 2 and it is red. Simple statistics dictate a fair estimate would be both rooms are red. In this case we have no problem.
Now suppose you open the other room and it's blue. Now as a thirder your probability of R=1 and R=2 are both 1/2. This is explained by SIA: two randomly selected room contains 1 red and 1 blue each. There are two possible ways for it to happen for R=1 and two ways for R=2 as well. This is the bayesian analysis.
Now if we apply a statistical analysis, as SIA dictates 50% of the room in the 2 randomly selected rooms are red. Therefore an unbiased estimation would be R=1.5 which means you are estimating there are 0.5 reds in the other 2 rooms. However, a simple random sample of size 1 is available for those 2 rooms. In that sample (which is just 1 room) there is no reds. So by committing to SIA one has to suggest the simple random sample is not representative of the population it's drawn from. Aka Red is under represented, biased towards blue. To make the matter worse unless the rooms you open are all reds (as in the first case) you would always conclude the room(s) you randomly chosen are biased toward blue. Hence comes the predicting power: you are more likely to choose a random sample that contains more blue than population average suggests. Even before you decide which rooms you want to open.
Very clear argument, thank you for the reply.
The question is if we do not use bayesian reasoning, just use statistics analysis can we still get an unbiased estimation? The answer is of course yes. Using fair sample to estimate population is as standard as it gets. The main argument is of course what is the fair sample. Depending on the answer we get estimation of r=21 or 27 respectively.
SIA states we should treat beauty's own room as a randomly selected from all rooms. By applying this idea in bayesian analysis is how we get thirdism. To oversimplify it: we shall reason as some selector randomly chose a day and find beauty awake, which in itself is a coincidence. However there is no reason for SIA to apply only to bayesian analysis but not statistical analysis. If we use SIA reasoning in statistical analysis, treating her own room as randomly selected from all 81 rooms, then the 9 rooms are all part of a simple random sample, which by definition is unbiased. There is no baye's rule or conditioning involved because here we are not treating it as a probability problem. Beauty's own red room is just a coincidence as in bayesian analysis, it suggest a larger number of reds the same way the other 2 red rooms does.
If one want to argue those 9 rooms are biased, why not use the same logic in a bayesian analysis? Borrowing cousin_it's example. If there are 3 rooms with the number of red rooms uniformly distributed between 1 and 3. If beauty wakes up and open another door and sees another red what should her credence of R=3 be? If I'm not mistaken thirders will say 3/4. Because by randomly selecting 2 room out of 3 and both being red there are 3 ways for R=3 and 1 way for R=2. Here thirders are treating her own room the same way as the second room. And the two rooms are thought to be randomly selected aka unbiased. If one argues the 2 rooms are biased towards red because her own room is red, then the calculation above is no longer valid.
Even if one takes the unlikely position that SIA is only applicable in bayesian but not statistical analysis there are still strange consequences. I might be mistaken but in problems of simple sampling, in general, not considering some round off errors, the statistical estimation would also be the case with highest probability in a bayesian analysis with an uniform prior. By using SIA in a bayesian analysis, we get R=27 as the most likely case. However statistics gives an estimate of R=21. This difference cannot be easily explained.
To answer the last part of your statement. If beauty randomly opens 8 doors and found them all red then she has a sample of pure red. By simple statistics she should give R=81 as the estimation. Halfer and thirders would both agree on that. If they do a bayesian analysis R=81 would also be the case with the highest probability. I'm not sure where 75 comes from I'm assuming by summing the multiples of probability and Rs in the bayesian analysis? But that value does not correspond to the estimation in statistics. Imagine you randomly draw 20 beans from a bag and they are all red, using statistics obviously you are not going to estimate the bag contains 90% red bean.
Imagine a bag of red and blue beans. You are about to take a random sample by blindly take a handful out of the bag. All else equal, one should expect the fraction of red beans in your hand is going to be the same as its fraction in the bag. Now someone comes along and says based on his calculation you are most likely to have a lower faction of red beans in your hand than in the bag. He is telling you this even before you deciding where in the bag you are going to grab. He is either (a) having supernatural predicting power or (b) wrong in his reasoning.
I think it is safe to say he is wrong in his reasoning.
What you said is correct. I'm arguing because the 8 rooms is obviously an unbiased sample, the 9 rooms cannot be. Which means beauty cannot treat her own room as a randomly selected room from all rooms as SIA suggests. The entire thought experiment is a reductio against thirders in the sleeping beauty problem. It also argues that beauty and the selector, even free to communicate and having identical information, would still disagree on the estimation of R.
First thing I want to say is that I do not have a mathematics or philosophy degree. I come from an engineering background. I consider myself as a hobbyist rationalist. English is not my first language, so pease forgive me when I make grammar mistakes.
The reason I've come to LW is because I believe I have something of value to contribute to the discussion of the Sleeping Beauty Problem. I tried to get some feedback by posting on reddit, however maybe due to the length of it I get few responses. I find LW through google and the discussion here is much more in depth and rigorous. So I'm hoping to get some critiques on my idea.
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 outside "selector" perspective which is different from beauty's answer. I concluded that the correct answer should be double-halving.
Because I'm new and cannot start a new discussion thread I'm posting the first part of my argument here see if anyone is interested. Also my complete argument can be found at www.sleepingbeautyproblem.com
Consider the following 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 another 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 he would 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.
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.
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 years, 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 would be a thought experiment countering SIA and Thirdism. Which I would post in a few days if anyone's interested.