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Comment author: cousin_it 20 July 2017 03:48:13PM *  0 points [-]

In the example I outlined with three rooms, can you give numerical values for "the fraction in the hand" and "the fraction in the bag"?

Comment author: Xianda_GAO 21 July 2017 02:16:42AM 0 points [-]

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.

Comment author: Manfred 20 July 2017 05:19:04PM *  0 points [-]

Are the rooms biased towards blue? Let's do this with balls from an urn.

Suppose there are four balls in an urn. 1,2,3,or 4 of them can be red, I've chosen the number (R) by a die roll. The rest are blue. We draw the balls from the urn in order.

Conditioned on the first ball being red, do we see a similar effect between the first two and last two balls?

Conditioning on the first ball being red means that it's more likely that my die roll was high - the distribution becomes proportional to R. Then if I see the second ball is blue, the probability distribution over R is proportional to R*(4-R), which is symmetrical about 2. So I expect that about one of the other balls will be red. Which means that the "other observed room" (ball 2) is systematically less red than an unknown room.

So this effect also occurs with drawing balls out of an urn! But what does it mean in this case? Does it mean we made a mistake with our math? No - you can do the experiment yourself easily. Hmm. Let's go back to your three possible desiderata:

A: Are the first two balls an unbiased sample of the balls? No - we conditioned that the first one was red. Even if R=1, we cannot get two blues in the first two balls. That's bias.

Note that when conditioning on the fact that the first ball was red, we used Bayes' rule for conditioning. We didn't assume that someone looked into the urn and pulled out a red ball, instead there was in some sense a fair process that "just happened" to give us a red ball first. But this does not mean that the first two balls are an unbiased sample. The fairness of the hypothetical process that gave us a red ball first does not carry over in any important way once we condition on it giving us a specific result. I think this may be a tricky point.

B: The first ball is guaranteed to be red. Yup.

C: The second ball is an unbiased sample of the other three balls. Yes. Even though it's less red than unknown ball in this example. But on the other hand, if it was red, it would be more red than an unknown ball.

Wait. I think you're not doing the math properly, which means that last statement doesn't hold in your post.

Back to the rooms and Sleeping Beauty. I think you're claiming that no matter how many red rooms S.B. sees, she expects even more, and therefore expects her surroundings to always be magically blue-biased. But if you do things correctly, then if Sleeping Beauty opens up 8 more rooms and they're all red, she doesn't jump straight to a belief that R=81. Instead, she thinks that R is on average somewhere around 75, and also she got a red-biased batch of rooms. But this only works if you keep track of these possibilities for R.

Comment author: Xianda_GAO 20 July 2017 11:53:09PM 0 points [-]

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.

Comment author: cousin_it 20 July 2017 02:33:07PM *  0 points [-]

Can you explain exactly what the supernatural powers are? For simplicity let's assume three rooms, with the number of red rooms uniformly distributed between 1 and 3. After waking up in a red room, and just before opening a random other room, a thirder expects it to be red with probability 2/3. You seem to say that it can't be right, but I can't tell if you consider it too low or too high, and why?

Comment author: Xianda_GAO 20 July 2017 03:14:47PM 1 point [-]

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.

Comment author: cousin_it 20 July 2017 12:03:02PM *  0 points [-]

(I'm not sure this is right anymore, deleted.)

Comment author: Xianda_GAO 20 July 2017 02:05:15PM 0 points [-]

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 probability of R.

Sleeping Beauty Problem Can Be Explained by Perspectivism (II)

2 Xianda_GAO 19 July 2017 11:00PM

This is the second part of my argument. It mainly involves a counter example to SIA and Thirdism.

First part of my argument can be found here.

 

The 81-Day Experiment(81D):

There is a circular corridor connected to 81 rooms with identical doors. At the beginning all rooms have blue walls. A random number R is generated between 1 and 81. Then a painter randomly selects R rooms and paint them red. Beauty would be put into a drug induced sleep lasting 81 day, spending one day in each room. An experimenter would wake her up if the room she currently sleeps in is red and let her sleep through the day if the room is blue. Her memory of each awakening would be wiped at the end of the day. Each time after beauty wakes up she is allowed to exit her room and open some other doors in the corridor to check the colour of those rooms. Now suppose one day after opening 8 random doors she sees 2 red rooms and 6 blue rooms. How should beauty estimate the total number of red rooms(R).

For halfers, waking up in a red room does not give beauty any more information except that R>0. Randomly opening 8 doors means she took a simple random sample of size 8 from a population of 80. In the sample 2 rooms (1/4) are red. Therefore the total number of red rooms(R) can be easily estimated as 1/4 of the 80 rooms plus her own room, 21 in total.

For thirders, beauty's own room is treated differently.As SIA states, finding herself awake is  as if she chose a random room from the 81 rooms and find out it is red. Therefore her room and the other 8 rooms she checked are all in the same sample. This means she has a simple random sample of size 9 from a population of 81. 3 out of 9 rooms in the sample (1/3) are red. The total number of red rooms can be easily estimated as a third of the 81 rooms, 27 in total.

If a bayesian analysis is performed R=21 and R=27 would also be the case with highest credence according to halfers and thirders respectively. It is worth mentioning if an outside Selector randomly chooses 9 rooms and check them, and it just so happens those 9 are the same 9 rooms beauty saw (her own room plus the 8 randomly chosen rooms), the Selector would estimate R=27 and has the highest credence for R=27. Because he and the beauty has the exact same information about the rooms their answer would not change even if they are allowed to communicate. So again, there will be a perspective disagreement according to halfers but not according to thirders. Same as mentioned in part I. 

However, thirder's estimation is very problematic. Because beauty believes the 9 rooms she knows is a fair sample of all 81 rooms, it means red rooms (and blue rooms) are not systematically over- or under-represented. Since beauty is always going to wake up in a red room, she has to conclude the other 8 rooms is not a fair sample. Red rooms have to be systematically underrepresent in those 8 rooms. This means even before beauty decides which doors she wants to open we can already predict with certain confidence that those 8 rooms is going to contains less reds than the average of the 80 suggests. This supernatural predicting power is a strong evidence against SIA and thirding. 

Another way to see the problem is ask beauty how many red rooms would she expect to see if we let her open another 8 rooms. According to SIA she should expect to see 24/72x8=2.67 reds. Meaning even after seeing 2 reds in the first 8 random rooms she would expect to see almost 3 in another set of randomly chosen rooms. Which is counterintuitive to say the least. 

The argument can also be structured this way. Consider the following three statements:

A: The 9 rooms is an unbiased sample of the 81 rooms.

B: Beauty is guaranteed to wake up in a red room

C: The 8 rooms beauty choose is an unbiased sample of the other 80 rooms.

These statements cannot be all true at the same time. Thirders accept A and B meaning they must reject C. In fact they must conclude the 8 rooms she choose would be biased towards blue. This contradicts the fact that the 8 rooms are randomly chosen. 

It is also easy to see why beauty should not estimate R the same way as the selector does. There are about 260 billion distinct combinations to pick 9 rooms out of 81. The selector has a equal chance to see any one of those 260 billion combinations. Beauty on the other hand could only possibility see a subset of the combinations. If a combination does not contains a red room, beauty would never see it. Furthermore, the more red rooms a combination contains the more awakening it has leading to a greater chance for a beauty to select the said combination. Therefore while the same 9 rooms is a unbiased sample for the selector it is a sample biased towards red for beauty.

One might want to argue after the selector learns a beauty has the knowledge of the same 9 rooms he should lower his estimation of R to the same as beauty’s. After all beauty could only know combinations in a subset biased towards red. The selector should also reason his sample is biased towards red. This argument is especially tempting for SSA supporters since if true it means their answer also yields no disagreements. Sadly this notion is wrong, the selector ought to remain his initial estimation. To the selector a beauty knowing the same 9 rooms simply means after waking up in one of the red rooms in his sample, beauty made a particular set of random choices coinciding said sample. It offers him no new information about the other rooms. This point can be made clearer if we look at how people reach to an agreement in an ordinary problem. Which would be shown by another thought experiment in the next part.  

Sleeping Beauty Problem Can Be Explained by Perspectivism (I)

0 Xianda_GAO 19 July 2017 10:11PM

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 is different from beauty's answer. 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. 

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 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.

Comment author: Xianda_GAO 16 July 2017 11:42:24PM 2 points [-]

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.