A Solution To The Sleeping Beauty Problem

Edit on 2017-12-03: I suggest new visitors read this post first. It is a newer (and much shorter) entry which better summarize my main argument.

ABSTRACT:I argue perspective disagreements in the setup of Sleeping Beauty Problem(SBP) is logically sound. This suggests double-halving is the correct answer. Counter examples against traditional Halfers and Thirders are also provided. As a result I propose the rejection of both Self-Sampling Assumption(SSA) and Self-Indication Assumption(SIA).  Because an observer reasoning as if she is randomly selected from a reference class switched to an outsider’s perspective which alters the answer. This could be important in anthropic reasoning. 

THE SLEEPING BEAUTY PROBLEM

The problem can be briefly described as follows:

Sleeping Beauty Problem(SBP):

Beauty is undergoing an experiment in which she will be waken up once or twice based on a fair coin toss. If the coin landed on tails (T) she will wake up twice with a memory wipe in between so that the 2 awakenings are indistinguishable. If the coin landed on heads (H) she will not be subjected to any memory wipe so only her first awakening needs to be considered. What should her credence of H be when she wakes up not having memories of a previous awakening.

There are two main camps to the problem. Halfers think the probability should be 1/2 and thirders argue the probability is 1/3. As of right now majority of the literature seems to lean towards the thirders’ view. There are some disagreements among halfers as what would the probability of H be after beauty learns that she is indeed in the first awakening. Here I will simply call those who changes their answer to 2/3 as halfers and call whose answers remains at 1/2 as double-halfers.

It is worth pointing out that halfers agrees with the Self-Sampling Assumption(SSA): that all else equal, an observer should reason as if she is randomly selected from the set of all actually existent observers. Thirders on the other hand agrees with the Self-Indication Assumption(SIA): that all else equal, an observer should reason as if she is randomly selected from the set of all possible observers. With these in mind, let us take a look at the perspective disagreement in SBP.

THE DISAGREEMENT

To see where the disagreement lies I purpose to consider the following problem:

Duplicating Beauty:

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 put into two identical rooms respectively. What should their probability of H be once awake fully knowing the setup.

I imagine many would agree the Duplicating Beauty and the original SBP are equivalent. Both camps have literatures using cloning examples to advance their argument. I will focus on Duplicating Beauty here since it facilitate easier description and discussion on the disagreement. However my argument does not depends on this equivalency. So I will later dedicate a brief section on the procedures leading to the same disagreement in original SBP for anybody suspicious of the equivalency.

Now suppose another person is observing the experiment. He will not be shown the coin toss result nor will he ever be cloned. However after beauty(ies) are put into the two rooms he need to randomly choose one room and enter. Let us call him the selector. Now suppose after entering the random room he sees a beauty inside. Since the chosen room is not empty by very simply bayesian updating his probability of H shall be 1/3.

On the other hand, per halfers, the beauty in the same room should have p(H)=1/2 after waking up. Since her room has equal chance of being selected regardless of H or T the probability remains unchanged when she sees the selector. Now the beauty and the selector has a disagreement about p(H). The disagreement remains when they are free to communicate. Here we have two people giving different probabilities to the same proposition while having the same information, in violation of Aumann’s agreement theorem.

One might use this against halfers since thirders’ interpretation will not lead to such disagreement. I will address the problems with thirders’ reasoning and the pitfall of such agreement later with a counter example. For now I would argue both the selector and beauty are correct in their reasoning and this perspective disagreement is valid.

We can see this perspective disagreement with a frequentist approach. What if the experiment is repeated by a large number, say 1000 times. From the selector’s point of view, as long as someone undergoes the experiment and let him choose one out of two rooms it is counted as a repetition. After 1000 repetitions by his count he should expect to find about 750 non-empty rooms, 250 times following H and 500 following T. Consistent with his p(H) of 1/3. From beauty’s point of view each repetition means undergoing the experiment and wakes up again waiting for the selector’s choice.  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, about 500 times she sees the selector half following H the other half following T. This is in accordance with her p(H)=1/2. Note the beauty waking up remembering 1000 coin tosses might be created by any one of those coin tosses just as the beauty disagreeing with the selector after the first toss might be a copy. Not being physically the same person should not affect the answer. To see this we can easily modify the Duplicating Beauty problem so the original beauty is always destroyed after each toss with either one or two copies created depending on H or T. In the modified version all resulting beauties are newly created yet it is hard to argue their answer would be different from the unmodified version. Another point worth noting is if all beauties in the experiment, old and new, each recall 1000 repetitions then by the selector’s count the number of repetition would be way greater. This is unsurprising since multiple beauties would be created in the process and each of them require additional tosses to complete their 1000 repetitions.

The perspective disagreement can also be shown by betting odds. To see this however beauties’ money must be duplicated when she is duplicated. In another word each person’s money follow each one’s perspective. This way beauty will not gain any information simply by checking her wallet meaning her money mimics her epistemic state which determines her answer. For the selector, the fair payout of a 1-dollar bet on H should be 3 dollars. Remember in the 1000 repetitions , he will see a beauty and enter the bet about 750 times. In which 250 times following H. If he pays 1 dollar to enter the bet each time, a 3 dollar payout on H would make his expected return 0. On the other hand the fair payout of a 1-dollar bet on H should be 2 dollars for beauty. In her 1000 repetitions she will be seeing the selector and entering the bet about 500 times, 250 times following H andT each –  a 2 dollar payout will make her expected return 0. The apparent disagreement is because the selector is doubly likely to enter the bet than beauty after T. Since from his perspective seeing any one of the two beauties is the same observation. In both cases he enters the bet. Whereas from beauties’ perspective which exact beauty the selector sees leads to different observations.  Although it gives her no information about the coin toss it decides whether or not she will enter the bet.

The above reasoning can be easily applied to original SBP. Although the execution would be rather cumbersome. In SBP the selector must lose track of the time himself in order to randomly select a day out of two. This can be done by using a pill that makes him sleep through beauty’s first awakening. He must choose between the sleeping pill and a undistinguishable placebo and go to sleep together with beauty when she first enters sleep. This way when he wakes up he can check if beauty is awake as well. Now he is in the same position as if he has just entered a random room in Duplicating Beauty. 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. It would cause each awakening to be increasingly short, making large number of repetitions impractical. However the methodology is still valid, since the length of the awakenings has no effect on beauty’s answer. To correctly see the effect of betting odds mentioned above her money should follow her epistemic state as well. Therefore any change to her wallet should be reversed when the relating memory is erased. In summary beauty and the selector would also be having disagreements in SBP with valid reasons.

ARGUMENT AGAINST SELF-INDICATION ASSUMPTION

The above arguments are of little importance if beauty should assigns p(H)=1/3 to begin with. In that case although she does not change her belief upon seeing the selector, unlike her counterpart does, both of them would still agree on p(H)=1/3. This can be explained by SIA, where she shall reason finding herself exist is the same as randomly selecting a potential observer(a room in this case) and then find the observer actually exist(the room is not empty). This put beauty in the same position as the selector, causing them to agree. To point out the weakness in this reasoning, consider the following example:

The 81-Day Experiment(81D):

There is a building with a circular corridor connected to 81 rooms with identical doors. At the beginning all rooms have blue walls. Then a painter randomly selects an unknown number of 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 red room is treated differently. As SIA states, finding herself awake is  as if she chose a random room from all 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.

I believe the above calculations are straightforward. The same numbers would be obtained as the most likely cases in a bayesian analysis with uniform priors. Notice if an outside selector randomly checks 9 rooms and happens to open the exact same 9 rooms beauty knows (her own room and the 8 rooms she checked), he would estimate R=27. Because beauty and the selector has exact same information about the rooms, he would be in disagreement with beauty according to halfers and in agreement with beauty according to thirders even if the two of them are free to communicate. The disagreement/agreement pattern is the same as in SBP.

However there are contradictions in thirders’ answer. First of all, before opening any doors if beauty is told that R=21 she should expect to see 2 reds if she opens 8 doors. According to thirders after opening 8 doors and actually seeing the expected number of 2 red rooms beauty must estimate R=27 instead of 21. This change of mind cannot be explained. Secondly, estimating R=27 means if beauty opens another 8 random doors she should expect to see 24/72×8=2.67 red rooms. This means after beauty saw 2 reds in the first 8 random rooms she would expect to see about 3 reds if she choose another 8 rooms. I fail to see how can this be justified. Last but not least, because beauty believes the 9 rooms beauty knows is a fair sample of all 81 rooms, it means red rooms (and blue rooms) are not systematically over- or under-represented it. 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 average. This supernatural predicting power is a conclusive evidence against SIA and thirders’ argument.

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 halfers 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.

To make this point clearer, it might be beneficial to understand how people reach to an agreement in an ordinary problem. Consider the following case:

Balls In Urns(BIU):

Suppose there is a urn filled with either 2 blue balls and 1 red ball(BBR) or 2 red balls and a blue ball(BRR) with equal chances. Andy randomly picked 2 balls from the urn and finds one ball of each colour. He correctly conclude the probability of BBR is 1/2, same as  the probability of BRR. Afterwards Bob asked for a red ball and was given one, he then randomly picked 1 ball from the 2 remaining balls in the urn and saw a blue one. He correctly concluded the probability of BBR is 2/3. It turns out however Andy and Bob actually saw the exact same 2 balls. The two of them are free to communicate and argue. Suppose both of them are rational can they reach to an agreement? Who should change his answer?

Again we can use a frequentist approach to solve this problem. Suppose the experiment is repeated many times with equal number of BBR and BRR. We can count the total number of occurrences when they both see the same 1 red and 1 blue balls. The relative frequency of BBR and BRR among these occurrences would indicate the correct probability. Both Andy and Bob should have no problem agreeing with this method. Here it becomes apparent that the exact procedure of the experiment determines whose initial probability is correct and who need to adjust his answer. More specifically how is the red ball given to Bob determined. Scenario 1: Bob is always given the red ball that has been picked by Andy. In this case Andy is correct, Bob should adjust his answer of BBR from 2/3 to 1/2. This is because for both BBR and BRR Andy has the same chance to pick a red and a blue ball. Given the same red ball Andy had picked Bob would have equal chance to pick the same blue ball again regardless the colour of the last ball left. The relative frequency of BBR and BRR given the occurrences would be about the same. Another case would be Scenario 2: Any red ball in the urn can be given to Bob. In this case Bob’s initial judgement is correct and Andy would have to change his probability for BBR to 2/3. Because all else equal, Bob is twice more likely to have the same red ball as Andy if there is only 1 red ball in the urn. The relative frequency of BBR to BRR with the occurrences would be 2:1. Since only one out of the two Scenarios can be true Andy and Bob must agree with each other as long as there is no ambiguity about he experiment procedure.

However, if we duplicate Bob (either by cloning or memory wiping) in the case of BRR and give each Bob a different red ball suddenly both Scenarios become true. To Andy the red ball he picked will always be given to Bob. To Bob the red ball given to him can be any one from the urn. Neither person would have reason to adjust his own probability fully knowing the experiment procedure. In this case the the two person having exact same information will remain in disagreement even if they are free to communicate. I believe the parallel between BIU, 81D and SBP is obvious enough to show why the selector, just as Andy, shall not adjust his probability.

ARGUMENTS AGAINST SELF-SAMPLING ASSUMPTION

I think the most convincing argument against SSA was presented by Elga(2000). He purpose the coin toss could happen after the first awakening. Beauty’s answer ought to remain the same regardless the timing of the toss. As SSA states, an observer should reason as if she is randomly selected from the set of all actual observers. If the selector randomly choose a day among all waking day(s) he is guaranteed to pick Monday if the coin landed on H but only has half the chance if T. From the selector’s perspective clearly a bayesian updating should be performed upon learning it is Monday. A simple calculation tells us his credence of H must be 1/3. As SSA dictates this is also beauty’s answer. Now beauty is predicting a fair coin toss yet to happen would most likely land on T. This supernatural predicting power is a conclusive evidence against SSA.

However, if we recognize the importance of perspective disagreement then beauty is not bound to give the same answer as the Selector. In fact I would argue she should not perform a bayesian update base on the new information. This can be explained in two ways.

One way is to put the new information into the frequentist approach mentioned above. In Duplicating Beauties, when a beauty wakes up and remembering 1000 repetitions she shall reason there are about 500 of H and T each among those 1000 tosses. The same conclusion would be reached by all beauties without knowing if she is physically the original or created somewhere along the way. Now suppose a beauty learns she is indeed the original. She would simply reason as  the original beauty who wakes up remembering 1000 tosses. These 1000 tosses would still contain about 500 of H and T each. Meaning her answer shall remain at 1/2.

Another way to see why beauty should not perform a bayesian update is to see the agreement/disagreement pattern between her and the selector. It is worth noting that beauty and the selector will be in agreement once knowing she is the original. As stated earlier, one way to understand the disagreement is after T, seeing either beauty is the same observation for the selector while it is different observations for beauties. This in turn causes the selector to enter twice more bets than beauty. However once we distinguish the two beauties by stating which one is the original the selector’s observation would also be different depending on which beauty he sees. To put it in a different way, if a bet is only set between the selector and the original beauty then the selector would no longer be twice more likely to enter a bet in case of T. He and the original Beauty would enter the bet with equal chances. Meaning their betting odds ought to be the same, they must be in agreement regarding the credence of H.

To be specific, the disagreements/agreements pattern can be summarized as follows. If the selector randomly chooses one of the two rooms as described by SIA. Upon seeing a beauty in the room the selector’s probability for H will change from 1/2 to 1/3. Beauty’s probability remains at 1/2 as described above. The two of them would be in disagreement. Once they learns the beauty is the original, the selector’s probability of H increases back to 1/2 from 1/3 by simple bayesian updating while beauty’s probability still remains at 1/2. This way the two would be in agreement. The selector can also randomly chooses one beauty from all existing beauty(ies) as described by SSA (here the total number of beauties should be shielded from the selector to not reveal the coin toss result). In this case seeing a beauty gives the selector no new information so his probability for H would remain unchanged at 1/2. On the other hand, from beauty’s perspective she is twice more likely to be chosen if there exist only one beauty instead of two. Therefore upon seeing the selector her credence of H would increase to 2/3. The two of them would also be in disagreement. Once they learn the beauty is the original the selector’s credence for H would increase to 2/3 by bayesian updating. Again beauty would not update her probability and it remains at 2/3. This way the two would agree with each other again.

As shown above, for the two to reach an agreement beauty must not perform a bayesian update upon the new information. This holds true in both cases regardless how the selection is structured.

Beauty’s antibayesianism, just like the perspective disagreement, is quite unusual. I think this is due to the fact that there is no random event determining which beauty is original/clone. While the coin toss may create a new copy of beauty, nothing could ever turn herself into the copy. The original beauty would eventually be the original beauty. It is simply tautology. There is no random soul jumping between the two bodies. Beauty’s uncertainty is because of the structure of the experiment which is purely due to lack of information. Compare this to the selector’s situation. The event of him choosing a room is random. Therefore learning the beauty in the chosen room is original gives new information about the random event. From his perspective a bayesian update should be performed.Where as from beauty’s perspective, learning she is the original does not give new information about a random event, for the simply fact there is no random event to begin with. It only gives information about her own perspective. So she should not performing a bayesian update as the selector did.

DISCUSSIONS

With the above arguments in mind we can clearly see the importance of perspectivism in SBP. It does not matter whether the selector follows SSA or SIA, his answer could not always correctly reflect beauty’s. Once beauty switch to selector’s perspective her answers would change. Therefore it is important for us to consciously track our reasoning process to make sure it contains no change of perspective. As shown above, beauty’s credence of H should be 1/2 when she wakes up and remains at 1/2 once learned it is Monday, aka double-halfers are correct. In another word, learning she exists does not confirm scenarios with more observers, learning she is the first does not confirm scenarios with less observers. Applying the same logic means we should reject Doomsday Argument while disagree with the Presumptuous Philosopher.

There are more interesting implication for perspective reasonings. One thing worth noting is that disagreements can also arise among beauties themselves. Consider DB with 2 repetitions for all beauties. Beauty wakes up in a room remembering taking part of 2 coin tosses. Now the experimenter tells her that including herself there are currently 3 beauties exist in total. Very soon the beauty figured out to get three beauties after 2 rounds means the first coin toss must be T. One of the resulting beauties would then experience another T while the other resulting beauty would experience another H. For ease of discussion lets randomly label the two beauties after the first toss T1 and T2 since they both experienced a T. Suppose T1 experience a H and T2 experience another T on the second round. We can label the the beauty who experienced a H asT1H1, and randomly label the two beauties experienced another T as T2T1 and T2T2. By indifference principle beauty shall reason she is equally likely to be T1 or T2 at her first awakening. Because T1H1 is a direct product of T1, her probability of being T1H1 at second awakening is the same as being T1 at the first awakening which is 1/2.  Because T2T1 and T2T2 are random labels for the direct products of T2, she shall reason the combine probability of her being either T2T1 or T2T2 is the same as being T2 at the first awakening. So her probability of being T2T1 and T2T2 must be 1/4 each. All three beauties would reach to the same conclusion since they have the same information. Notice here they would be in disagreement regarding their probability of being T1H1. Each of them would think herself is twice more likely to be T1H1 (1/2) than the other two beauties(1/4 each).This disagreement might seems alarming. However it is also valid. As discussed above, in problems involving duplications people with the same information can have different probabilities. The reason this disagreement seems more suspicious is because the resulting beauties appears to be in symmetrical positions thus should not disagree with each other. However this symmetry is only valid from an selector’s perspective. For the selector, any resulting beauty has an equal chance of being picked. In another word T1H1, T2T1 and T2T2 are only different in names during a random selection process. Therefore it is correct a randomly chosen beauty has equal chance of being T1H1, T2T1 or T2T2(1/3). From beauty’s perspective however such symmetry does not exist. The three labels, T1H1, T2T1 and T2T2, are not just different in name. If the coin landed H then she would be labeled H1 for certain. If the coin landed on T she would be randomly labeled T1 or T2. Therefore from beauty’s perspective being labeled as H1 is twice likely to happen than being label T1.

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14 Responses to A Solution To The Sleeping Beauty Problem

  1. Ioannis Mariolis says:

    Dear Darren,
    You have presenting a very interesting and clear answer to the SB problem. I completely agree with you that there is no reason for SB to update her belief on Heads when she is informed that it is Monday. Due to her uncertainty on the day of the week she can devise a hypothetical random experiment where in case of Tails her awakening occurs either on Monday or on Tuesday with equal probability, assigning P(Mon)=3/4 and P(Tue)=1/4 (since in case of Heads her -hypothetical randomly selected- awakening occurs always on Monday). However, learning that it is Monday cannot constitute evidence that a Monday awakening outcome of the hypothetical random experiment has occurred, therefore she should not update P(H). In case you are interested, a more detailed analysis of the above argument can be found in the following paper I have written: https://arxiv.org/abs/1409.3803 . Elga’s two main arguments against halfers are a) the higher frequency of Tuesday awakenings vs Monday awakenings, and b) the bayesian updating upon learning that it is Monday, with a) being easy to rebut. However, your analysis (as well as the one I present in the aforementioned paper) clearly indicate that there is no justification for updating P(H) upon learning that it is Monday.

    On a side note, I would like to comment on one of your conclusions. You are arguing that: “Here we have two people giving different probabilities to the same proposition while having the same information, in violation of Aumann’s agreement theorem. “. Although I am far from an expert on this field (I am an AI and machine learning post-doctoral researcher with sound knowledge of modern probability theory and a lot of experience in its application, but I have not systematically studied philosophy of probability), I think that since the Aumann’s agreement theorem requires equal priors (for the two agents) of the world being in a certain state, there is no violation in the example you describe. In your example, for the observer the prior probability of finding Room1 occupied is 3/4, which is equal to the one for Room2. SB priors on the other hand are different, the probability for her room (let’s say Room1) to be occupied is 1, whereas the probability for Room2 to be occupied is 1/2 (room labeling is arbitrary and both the observer and SB know the label of each room). Thus, when the observer sees SB in Room1 updates P(Room1)=1 and P(Room2)=2/3, whereas SB does not update.

    Best,
    Yannis

    • Darren Gao says:

      Thank you for the kind words. Actually I have read your paper ambient quite some time ago. I also think beauty should not update her credence after learning it is Monday.

      It is an interesting angle about Aumann’s agreement theorem. You are right about them having different priors about the occupancy of the two rooms. However, if we treat the proposition as “the coin landed on heads” then they do have a common prior (1/2). After their meeting they would “agree to disagree” about the probability of coin toss: meaning while beauty’s probability of H is 1/2 she would say the selector’s is right to say the probability is 1/3 from his perspective, vice versa. This to me seems to be a contradiction about Aumann’s theorem.

      Also Katja Grace argued against self-sampling assumption by pointing out it would contradict Aumann’s agreement theorem. (it can be found in here: https://meteuphoric.wordpress.com/2011/03/09/agreement-on-anthropics/). I think her argument about Aumann’s theorem works for halfers in general. Although I am a (double) halfer and reject SSA for completely different reasons.

      I just think the disagreement is valid in this case. Because beauty and the selector interpret their meeting differently. Selector interpret it as a meeting between him and A beauty (any beauty) while beauty interpret it as a meeting between the selector and THE beauty (herself). Whether or not it is technically in violation to Aumann’s theorem is not important to my argument per say. I personally think it is against the theorem.

      • Ioannis Mariolis says:

        Dear Darren,
        Thank you for your quick reply. I will not insist on the violation (or not) of the Aumann’s agreement theorem, since I agree that it is not important to your argument. As I said I am no expert on philosophy of probability, thus I may interpret wrong the definition of the theorem that says “equal priors (for the two agents) of the world being in a certain state”. To me this means all priors of their current world state should be equal for the two agents including their credence on P(Room1 is occupied), and P(Room2 is occupied). But I may miss something since I have not any hands on experience on the application of this theorem.

        More importantly, I would like to discuss our views on why SB should not perform Bayesian updating once she is informed that it is Monday. In your analysis on this matter you are arguing (to be fair, among other things) that:

        “Beauty’s uncertainty is because of the structure of the experiment which is purely due to lack of information. Compare this to the selector’s situation. The event of him choosing a room is random. Therefore learning the beauty in the chosen room is original gives new information about the random event. From his perspective a bayesian update should be performed.Where as from beauty’s perspective, learning she is the original does not give new information about a random event, for the simply fact there is no random event to begin with. It only gives information about her own perspective. So she should not performing a bayesian update as the selector did.”

        I think it can be misleading if we use the above wording. More precisely, if we focus on the structure of the experiment and the lack of information in the case of SB, one can use the same argument for the Selector as well. For example in case the Selector uses a coin toss to select a room, it is his lack of information and the experiment setup that do not allow him to predict the outcome (since there is no true randomness in the setup). Notice that the same applies for the original coin toss that defines the number of copies.

        As I argue in my paper (addressing the original formulation of the SB problem) the problem is that the random experiment accounting for SB’s uncertainty on the current (awakening) day is hypothetical. By “hypothetical”, I am not meaning that the procedure modeled as a random experiment (i.e. the selection of the awakening day) is not truly random and only accounts on lack of information, but that it is not actually performed (since in case of Tails two awakenings occur, thus no selection has been performed). Although SB can use the hypothetical random experiment to model her uncertainty and estimate the probability of the current day being Monday, she cannot use as evidence of Monday selection the information that it is Monday, since no selection has been performed.

        In the copy version of SB problem you analyze (which I think its a fair analogy to the original), informing SB that she is the original is analogous to informing her that it is Monday, whereas informing her that she is the copy is analogous to informing her that it is Tuesday. Similarly, SB can define a hypothetical random experiment where her “originality” is selected and assign P(orig)=3/4 [since P(orig)=P(orig|Heads)*P(H) + P(orig|Tails)*P(Tails)=1*1/2 + 1/2*1/2]. However, once learning that she is the original no evidence of the “orig” event is provided, since no selection has been performed. Thus, she should not use Bayesian updating.

        Adding a Selector into the problem is analogous to having him choose randomly Monday or Tuesday, use the same drug SB takes in case of Tails and wake him up in the room on the selected day (of course he does not remember and he is not informed which day it is, although he remembers the setup of the experiment). Before opening his eyes he estimates there is 3/4 probability to see SB. After opening his eyes if SB is in the room he should update to P(H|SB)= 1/3 like in your copy analogy. Should SB update upon seeing (or not seeing) the Selector? Notice that this is a different situation than the updating related to what (awakening) day it is, yet it also includes a hypothetical random experiment. The hypothetical part is again that in case of Tails she should randomly choose one of the two awakenings to be the current one and assign P(Sel|Tails)=1/2, even though in reality the Selector (in case of Tails) will be in the room with her for certain, either on Monday or on Tuesday (meaning that no such selection of current awakening ever takes place). Thus, even though SB should estimate P(Sel)=1/2, she cannot use as evidence for a “Sel” event the fact that she sees the Selector in her room.

        Best,
        Yannis

        • Darren Gao says:

          Dear Yannis,

          I often find myself struggling to effectively describe ideas. Allow me to explain what I meant by beauty’s uncertainty is purely due to lack of information. There is no random event causing the original beauty to be the original. Compare this to the selector’s case where there is an actual random event (random choosing a room among the two) determining which beauty he sees. That is why upon learning the beauty is the original the selector learnt something about the random experiment which merit a bayesian update. Beauty on the other had learnt nothing about any random experiment since herself being the original is not random, aka there is no random event in the first place. Her initial uncertainty is entirely due to the structure of the experiment which causes the original and the copy to have the same information, i.e. she lacks any info to confirm she is the original.

          If I understand correctly this is in line with the spirit of your argument as well. You are arguing after waking up beauty could use a hypothetical experiment to calculate the probability of her being the original. However because this hypothetical experiment is never actually performed, learning she is the original should not cause her to perform a bayesian update on the probability of the coin toss. In my argument this hypothetical experiment can be performed by a third party (the selector), and the selector should perform a bayesian update upon learning she is the original. However, due to the existence of perspective disagreement this does not necessarily means beauty should also do a bayesian update. In fact once we consider the agreement/disagreement pattern it suggests the contrary, which further supports beauty shouldn’t update her credence.

          • Ioannis Mariolis says:

            Dear Darren,
            It is a very interesting approach you propose on SB and the agreement/disagreement patterns, as well as the implications on SSA and SIA frameworks. I think it can work complementary to the “hypothetical” random experiment approach I propose, which is based on rigorously defining the random experiments involved, their sample spaces, the corresponding random events, as well as their mapping to real world events, and concludes that SB should not update P(H) upon learning that it is Monday. In your analysis you also include a brief explanation of the reasoning behind not updating based on random events and lack of information. Let me further clarify my objections on that explanation. In my approach a random event is a mathematical entity defined within a random experiment (another mathematical/abstract entity) and which may (or may not) be mapped to real world events. According to this approach, it is not rigorous to label real world events as random or not. Thus, even though for convenience some one may say that a Heads coin toss is a random event, if we want to be rigorous, a random experiment should be defined with Heads and Tails outcomes and assigned probabilities, and a real world heads toss should be mapped to the Heads random event. This is important, since even though the real event is part of the common reality of the involved agents, the random event must be associated to a corresponding random experiment and the lack of information that this experiment is addressing. Thus, different agents based on the information they have may use different random experiments with different random events to model their common reality with these random experiment’s events. This is why I believe it might be confusing if you try to explain the difference between SB and the Selector with arguments like: ” beauty’s uncertainty is purely due to lack of information. There is no random event causing the original beauty to be the original.” According to the approach I have presented it makes no sense to consider a “random event” as cause to other events (random or not). Moreover, “lack of information” is a valid (and when not accounting for quantum phenomena, the main) reason for defining a random experiment in the first place. In SB case I argue that the problem is not the “lack of information” or “randomness” , but the tricky mapping between the random events of SB’s random experiment that accounts for her lack of information and the real world events, since in the real world two awakenings occur in case of Tails, whereas in “her” random experiment (i.e. the random experiment she defines to model her lack of information) only one awakening occurs even in case of Tails. I understand that I am referring to very subtle differences, but I hope you can still see the importance behind these differences. I think that using the definitions in my approach can help demonstrating more clearly when Bayesian updating should not be performed and why.
            In your last post you are saying about the hypothetical random experiment:
            “In my argument this hypothetical experiment can be performed by a third party (the selector), and the selector should perform a Bayesian update upon learning she is the original.”
            Again if we want to be rigorous, this is how we can describe the situation (I am using the original SB problem, but similar description can be applied for the “copies” version): The Selector defines a random experiment (that accounts for his lack of information). The random experiment he defines is different to the “hypothetical” random experiment of SB. The Selector’s awakenings are independent from the coin toss outcome. Thus, even on Tails the Selector can wake up on Tuesday in an empty room with P(Tue)=P(Tue|Tails)=1/2. In SB’s random experiment(that accounts for her lack of information) P(Tue)=P(Tue|Tails)=1/4, since she cannot have a Tuesday awakening in case of Heads.

            I would like to conclude by thoroughly describing the correspondences between the two SB problem versions. I think it can help showing the equivalence between them if we add the Selector to the original SB experiment and analyze, but also provide more insights on the different perspectives. Thus, the added Selector is put to sleep on Sunday, and independently from the original coin toss, he is awaken only once either on Monday or on Tuesday with equal probability (perhaps using another independent coin toss). Moreover, SB learning that it is Monday is analogous to learning that she is the original Beauty, whereas if the Selector learns that it is Monday is analogous that he learns he has selected the room that has the original Beauty. Moreover, let “SB seeing the Selector” be denoted as a Sel event, and the “Selector seeing SB” be denoted as a SB event. With that in mind we can calculate (with Pc() denoting P() in the “copies” version) for SB upon awakening:
            P(H)=Pc(H)=P(T)=Pc(T)=1/2
            P(Mon)=Pc(Orig)=3/4
            P(Sel)=Pc(Sel)=1/2
            P(Mon,Sel)=Pc(Orig,Sel)= 3/8 =P(Mon)*P(Sel)=Pc(Orig)*Pc(Sel)
            P(H|Mon)=Pc(H|Orig)= 2/3 (according to the random experiment, but learning that it is Monday is NOT evidence of a Mon event, so no Bayesian updating can be performed and P(H|learning it is Monday)=P(H)=Pc(H)=Pc(H|learning she is the original)=1/2)
            P(H|Sel)=Pc(H|Sel)=P(H)=Pc(H)=1/2 (H and Sel are two independent events, thus, no updating due to independency)
            P(H|Mon,Sel)=Pc(H|Orig,Sel)=P(H|Mon)=Pc(H|Orig)=2/3 (No Bayesian updating is justified as explained for P(H|Mon) )
            For the random experiment modeling the Selector’s perspective:
            P(H)=Pc(H)=P(T)=Pc(T)=1/2
            P(Mon)=Pc(Orig)=1/2
            P(SB)=Pc(SB)=3/4
            P(Mon,SB)=Pc(Orig,SB)=P(Mon)=Pc(Orig)=1/2
            P(H|Mon)=P(H)=Pc(H)=Pc(H|Orig)=1/2 (H and Mon, independent events)
            P(H|SB)=Pc(H|SB)= 1/3 Bayesian Updating upon seeing SB!
            P(H|Mon,SB)=P(H|Mon)=Pc(H|Orig)=Pc(H|Orig,SB)=1/2

            The above analysis also makes your arguments of difference in perspective more clear, especially the point where you identified the difference between the Selector seeing A Beauty (in the copies version) and not THE Beauty. It is clear from the above analysis that there is no symmetry between “SB seeing the Selector” (it is denoted as a Sel event) and the “Selector seeing SB” (it is denoted as a SB event). For SB a “Heads toss” and “seeing the Selector” are two independent events, whereas for the Selector a “Heads toss” and “seeing SB” are dependent events. Notice that for SB P(H|Sel)=P(H)=1/2, whereas for the Selector P(H|SB)=1/3. Notice however, that SB not updating upon seeing the Selector is a result of the independence between the two events. If you want to rebut Elga’s second argument you should analyze also why SB should not update using P(H|Mon)=Pc(H|Orig)=2/3 when she learns that she is the original. For the Selector Pc(H|Orig)=1/2, with H and Orig being independent events (whereas for SB H and Sel where independent events). However, because SB shouldn’t update Pc(H) to 2/3 upon learning that she is the original (she should still assign Pc(H)=1/2) there is no perspective disagreement between her and the Selector in that case. Thus, it looks like the perspective disagreement you have presented is an argument against Elga’s view that upon awakening SB should update P(H) to 1/3 because there are twice as much Tails awakenings than Heads awakenings. By a third person’s perspective like the Selector’s its twice more probable to be in a Tails awakening than to a Heads awakening when he shes SB, but this reasoning cannot apply to SB herself. For the second Elga argument on the Bayesian updating when learning that it is Monday it is the lack of evidence that explains why SB should not update.

            To sum up, in the copies version, when SB sees the Selector, she doesn’t update p(H) because the Sel event is independent from the H event, whereas when she learns that she is the original she should not update p(H) to 2/3 because she does not have Bayesian evidence for an Orig random event (since the real world event of her being the original is not a result of the real world process that maps onto the random experiment that accounts for her her lack of information about her originality) . Regarding agreement/disagreement, in the first case she is in perspective disagreement with the Selector, whereas in the second case she is in agreement.

            Best,
            Yannis

            • Darren Gao says:

              Dear Yannis,
              Thank you very much for the comment, especially formulating everything in equations clearly and concisely. Although I am no expert in probability/philosophy by any means (I’ve only got a master degree in engineering), I tend to agree with this:

              In SB case I argue that the problem is not the “lack of information” or “randomness” , but the tricky mapping between the random events of SB’s random experiment that accounts for her lack of information and the real world events

              and this:

              when she learns that she is the original she should not update p(H) to 2/3 because she does not have Bayesian evidence for an Orig random event (since the real world event of her being the original is not a result of the real world process that maps onto the random experiment that accounts for her her lack of information about her originality) .

              In my opinion this trouble in mapping her own random event to the real world event is because there is no real world event at all. Aka, nothing actually causes the original to continue being the original, or the day following Sunday being Monday. It’s just tautology. Knowing she is the original or it is Monday she hasn’t learned anything about “what happened” but only learned about “who I am”.

              I think your argument is more structured and vigorous and quite convincing to me. However I lack the expertise to judge if it should be “the” argument for double halving. Therefore I want to continue using my original argument in this blog for the time being. For the reason because I understand it better personally.

              I think perspective disagreement is the key because not only it help explains beauty’s non-bayesianism it also explains the source of the dilemma. Basically halfer are answering the question from a first person perspective while thirders are answering it from a third person perspective. Their disagreement is just like the disagreement between beauty and the selector. It also explains why most people upon first hearing about the problem would be puzzled why the answer could be anything other than 1/2, but after “taking a step back” and giving more thought many switched to thirders’ camp.

  2. JeffJo says:

    And I didn’t just assert that the 81D was different, I gave a reason. I may not have explained it thoroughly, but it should have been good enough. The point is that the calculation you are trying to describe is just an estimation of F, the fraction of rooms that are painted. Not a calculation of the total number N, or of F itself. I don’t want to get into hairy calculations, so imagine the simple case where 80 rooms get painted (so F=80/81=98.765%), and SB opens 79 “other” rooms,leaving just one unopened.

    If SB opens the unpainted room (very likely at 79/80), SSA estimates F by F’=78/79=98.734%, and SIA by F’=79/80=98.75%. If she opens 80, these numbers are both 100%. So SSA is not better in either case.

    But the important number is the expectation over the 80 ways she could choose 79 doors. For SSA, that’s (78*79)*(79/80)+(1)*(1/80)=98.750%. For SIA, it’s (79/80)*(79/80)+(1)*(1/80) = 98.766%. Which did a better job of estimating? This difference should be repeatable for any set of numbers, but I’m not going to bother.

    • Darren Gao says:

      Hi, Jeff

      Since the discussion we are having digresses from this blog I emailed you my response to your latest post couple days ago. In case it is not your main email I’m also posting it here. Given your interest and effort of presenting you idea I still suggest you to post on reddit /r philosophy or lesswrong discussion broad. That way it would get more feedback. I would also join the discussion if you choose to do so.

      Thank you for carefully explaining the thought experiment. Here is the part I think halfer would disagree:

      “C) So (as in my first comment) write those four names on index cards, and pick one at random. Flip the coin on Sunday, and wake SB either once, or twice, on the following two days as indicated by that card and the procedures I just outlined. She doesn’t even need to know which card is picked. Her confidence that the coin landed on the result indicated by the first letter on the card has to be the same as the original SB’s confidence in Heads.”

      If I understand correctly in your experiment there is no memory wiping. In the original problem the memory wipe causes SB’s uncertainty about her location. In your experiment this uncertainty is replaced by a random card draw. It is a critical claim that beauty’s answers must be the same between the two problems. For problems involving memory wiping or cloning lead to disagreements between halfer and thirders, while no such disagreement exists for a simply random selection problem. Halfers (and myself) would think this equivalency is wrong. This is another reason why I think Michael Titelbaum’s technicolor beauty is a better argument than yours. It does not require a complete different experiment, retains the memory wipe and still argues for the symmetry for the 3 outcomes. I strongly recommend you to check it up.

      Of course I am more interested in your rebuttal to the 81D problem.

      First of all let me point out the mistake in your calculation. For SSA the calculation of F was wrong (again I want to stress I am against both SIA and SSA, but SSA gives correct answer in this problem). To calculate the total fraction F, it requires sample for two parts. One part is the room she wakes up in, which she has the perfect knowledge it is red. The other part is the other 80 rooms which she has to estimate base on her sample. so if the unpainted room is in the 79 rooms she opens then F=(78/79)*(80/81)+1*(1/81)=0.987498. If we also consider the case when the unpainted room is not in the 79 rooms she opens then F= 0.987498*(79/80)+1*(1/80)=0.98765432 or the exact value of 80/81. Your calculation for the SIA is correct which is larger than the exact value. So your argument actually shows SIA gives estimation biased towards red while SSA gives unbiased estimation.

      The disagreement lies only in how was her own room treated. As she opens more doors the weight of her own room decreases. That’s why the difference in above numbers are small because in your example beauty opens large amount of doors other than her own.

      “The point is that the calculation you are trying to describe is just an estimation of F, the fraction of rooms that are painted. Not a calculation of the total number N, or of F itself.”

      This is just a different wording of the same problem. In the SBP you can also ask beauty to guess the fraction of waking days, being either 50% or 100%, it is the same as asking beauty to guess H or T. May be it helps to consider the following question:

      1. If the painter toss a fair coin and decides to paint 21 red rooms if H and 27 red rooms if T. Again beauty saw 2 red among 8 other rooms, does H or T has a higher probability?

      2. Now the painter randomly generate a number between 1 and 81 with equal chances and paint the corresponding number of rooms. After beauty saw 2 red among 8 other rooms, does N=21 or N=27 has a higher probability?

      3. Same setup as in 2. How should beauty estimate the total number of red rooms?

      If you can give clear answers to the above three problem maybe the shortcomings of SIA would be clear.

  3. JeffJo says:

    I’m sorry, my bad. I use two versions of the same solution, and I mixed them up in the two comments. Since it seems I’m not being clear, I’ll go slower:

    A) It’s essentially the same problem, with the same answer, if SB is always wakened on Tuesday, and wakened on Monday only if Tails is flipped. Call the original HTue, and this equivalent variation HMon. These names describe the situation where SB will stay asleep one day. She will be wakened on the other regardless of the coin.
    B) There are two more variations, called TTue and TMon. In them, SB is wakened only once after Tails is flipped, sleeping through the day indicated in the name. Only one modification is needed to make this essentially the same problem: ask her about her confidence in Tails.
    C) So (as in my first comment) write those four names on index cards, and pick one at random. Flip the coin on Sunday, and wake SB either once, or twice, on the following two days as indicated by that card and the procedures I just outlined. She doesn’t even need to know which card is picked. Her confidence that the coin landed on the result indicated by the first letter on the card has to be the same as the original SB’s confidence in Heads.
    D) But this problem is trivial to solve. There were originally four cards, and one is ruled out by the fact that she is awake. Since each was equally likely to have been picked, the updates probability for each is 1/3.
    E) If you doubt this solution (for “how could SB reject an unknown card,” see note below), use four volunteers (as in my second comment) instead of one. Distribute the four cards randomly among them, without showing them. On Monday, wake all but the one with the XMon card, where X is the result of the coin flip. On Tuesday, wake all but the one with the XTue card. Each is wakened at least once, and maybe twice. But on either day, exactly three are wakened, and exactly one of those three has a card that starts with X. Since none have any information about which of the three it, each one’s confidence that she is the one can only be 1/3.

    Note: An unknown is a random variable. Let Y be a random variable representing the unknown card. Let M be the event we are looking for, that the first letter matches the coin. Surely Pr(M|Y=HMon)=Pr(M|Y=HTue)=Pr(M|Y=TMon)=Pr(M|Y=TTue); call this value Q. And jsut as surely, Pr(Y=HMon)+Pr(Y=HTue)+Pr(Y=TMon)+Pr(Y=TTue)=1. So the Law of Total Probability says Pr(M)=Q, without even needing to know any other probabilities.

  4. JeffJo says:

    “In the original SBP beauty knows for sure she would wake up. It is not so in the modified version.” If you think this is true, then you obviously did not understand my version.

    One of the Beauties, the one dealt the card (H,Tue), is in an experiment that is identical to the original SBP in every way except one: she knows that there are three others. Those others are in an experiment that differs in only one other way: it is symmetric with the original, since it substitutes a different day, a different coin result, or both, for the one day she stays asleep.

    Specifically, each Beauty knows for certain that she will be wakened once, and that whether she will be awakened a second time is determined by the coin. The answer in my version cannot be different from the answer in the original SBP, and that answer is 1/3.

    The 81D experiment is an entirely different kind of problem, since you are estimating a value by random sampling and not constructing an entire sample space. It’s not my area, so I won’t address how perspective should be applied. But the Beauty in the original SBP is uncertain how to, or if even she can, construct that entire sample space because of the issue of perspective. My variation makes it trivial to do so, and perspective has to address why 1/3 is correct.

    • Darren Gao says:

      After carefully reading your post several times I’m still not 100% sure what’s the experiment procedure. I initially assumed you mean to have 4 beauties each given one card and wake 3 of them, except the one given (H, Tue). It seems that is not what you have in mind. Then I assume you mean to have only 1 beauty and a random card is drawn by the experimenter, and beauty sleep through on the combination shown on the card. But then you said there are more than one beauties in the reply. So I’m guessing you mean to have 4 beauties, each assigned 1 card, without them knowing which card they have. The experimenter then toss a coin and make sure the beauties with sleep through on Monday or Tuesday according to her card if the card and the coin toss result are the same. I’m assuming this is the experiment setup?

      You are suggesting beauty shall reject one unknown card since she won’t be awake in that case and among the rest 3 cards only 1 is the same as the coin toss result therefore the probability is 1/3. I am puzzled by how could beauty reject an unknown card. Assuming we tell beauty the coin toss result is H which card can beauty reject? (H, Mon) or (H, Tue)? If the answer is “one of those two” then all this experiment does is to give tangible ways (3 cards) to represent the 3 possible positions beauty can be in. That is 1 unknown card represents the same result while 2 unkown cards represents different results. In that case a much cleaner example arguing for thirdism was already presented by Michael Titelbaum’s technicolor beauty. Suppose in the original SBP the room is randomly painted red or blue on Monday and the other color on Tuesday. When beauty wakes up say she sees red, there are three possibilities leads to this, coin landed on H and the room is painted red on Monday, the coin landed on T and the room is painted red on Monday, and the coin landed on T and the room is painted red on Tuesday (painted blue on Monday). Therefore she should conclude the probability of H is 1/3. The conclusion is the same if she sees blue due to symmetry. I won’t go into its rebuttal but suffice to say this argument does not end the debate. Your experiment to me just unnecessarily complicate the situation.

      In fact halfer can propose a similar experiment to yours to argue for their point. Instead of drawing 4 cards all at once draw 2 cards saying (H) or (T) first. The experimenter then toss the coin. If the toss result and the card are the same the experiment then would roll a die. If the die end up even beauty would sleep through Monday otherwise she would sleep through Tuesday. Now ask beauties credence if the card and the coin resulted the same. Here the die roll and which day beauty sleep through is independent of the coin and card result, so by symmetry the answer should be 1/2, supporting their answer.

      Saying 81D is completely different from SBP seems to be a common argument used by thirders. But that is misguided. Beside that SBP has 2 possible outcomes and 81D has 81, the structure of the question is exactly the same. The difference is that 81D also enable beauty to attain samples beside her own location. Thirders argue beauty’s own location is randomly selected from all possible locations, compare that with the true random selections (the simple random sample of 8 doors out of 80) it is clear that they are different and cannot be mixed. Whether the question is only asking for the most likely case or the complete pdf is irrelevant. We can also just ask beauty which coin toss result is most likely in the original SBP. Thirders will say T, halfers will say they are the same because no information. In 81D with a uniform prior thirder will say R=27 is most likely and halfers will say R=21 is most likely.

      I get the feeling you are very confident about your solution for the SBP. Base on your replies’ focus you seems to be more interested in presenting them rather than critiquing the ideas of this blog. That is quite fine by me, and I am happy to have such discussions. May I suggest you post your solution on reddit r/philosophy or lesswrong board? That way you would get more feedback. I am happy to discuss the problem with you there. For the purpose of this site I wish to keep the replies relevant to the blog. Thank you.

  5. JeffJo says:

    Modify the experiment in on inconsequential way: instead of fixing the circumstances for non-waking as (H, Tue), write the four possibilities (H, Mon), (H, Tue), (T, Mon), and (T, Tue) on different cards, and pick one at random on Sunday night. Without telling Beauty. Then, when she is awake, ask her for her confidence that the coin result and the card have the same face-value, H or T.

    Let X be the confidence Beauty should have in the original experiment. If she could assume that the chosen card had (H, Tue) written on it, then she could give that answer to this new one. But since the four possibilities are symmetric, it is also the answer for any of the cards. So the Law of Total Probability says the answer here is also X.

    But we can address the new one without the need for the perspectives discussed above. Instead of asking whether the random coin result shows the value written on the card, ask whether the random card has the value of the coin.

    There are four equally likely cards. One is eliminated, since Beauty would not be awake if it was chosen. Of the three remaining, only one matches the coin. X=1/3.

    • Darren Gao says:

      Reply to JeffJo

      The modified experiment illustrate SIA reasoning which unsurprisingly concludes 1/3 as the correct answer. The 4 cards represents the 4 potential positions beauty in the original experiment, and waking up eliminates 1 position. Therefore the remaining 3 card are equally likely, as if beauty is randomly selected from the corresponding 3 potential observers, same as SIA states.

      Halfers would simply disagree that the modification is inconsequential at all. In the original SBP beauty knows for sure she would wake up. It is not so in the modified version. While beauty clearly gained new information in your experiment it is not so in the original SBP. This discrepancy is very important since the major argument between halfer and thirders is whether or not beauty gained new information upon waking up. In this sense the purposed experiment would not settle the debate at all.

      To argue my point of perspective importance consider this. The 4 cards are symmetrical to beauty in your experiment because there is a random selection process where each card has an equal chance of getting picked. However in the SBP from beauty’s perspective there is no such random process determining whether it is her first or second awakening. I am arguing imagining a selection process from the perspective of an outsider, either by SSA or SIA, are mistaken.

      Another point I want to point out is that the modified experiment does not solve the question posed by 81D. It would still conclude the number of red rooms as 27 and treat a simple random sample (8 out of 80) as biased.

    • Darren Gao says:

      Hi, Jeff

      Since the discussion we are having digresses from this blog I emailed you my response to your latest post couple days ago. In case it is not your main email I’m also posting it here. Given your interest and effort of presenting you idea I still suggest you to post on reddit /r philosophy or lesswrong discussion broad. That way it would get more feedback. I would also join the discussion if you choose to do so.

      Thank you for carefully explaining the thought experiment. Here is the part I think halfer would disagree:

      “C) So (as in my first comment) write those four names on index cards, and pick one at random. Flip the coin on Sunday, and wake SB either once, or twice, on the following two days as indicated by that card and the procedures I just outlined. She doesn’t even need to know which card is picked. Her confidence that the coin landed on the result indicated by the first letter on the card has to be the same as the original SB’s confidence in Heads.”

      If I understand correctly in your experiment there is no memory wiping. In the original problem the memory wipe causes SB’s uncertainty about her location. In your experiment this uncertainty is replaced by a random card draw. It is a critical claim that beauty’s answers must be the same between the two problems. For problems involving memory wiping or cloning lead to disagreements between halfer and thirders, while no such disagreement exists for a simply random selection problem. Halfers (and myself) would think this equivalency is wrong. This is another reason why I think Michael Titelbaum’s technicolor beauty is a better argument than yours. It does not require a complete different experiment, retains the memory wipe and still argues for the symmetry for the 3 outcomes. I strongly recommend you to check it up.

      Of course I am more interested in your rebuttal to the 81D problem.

      First of all let me point out the mistake in your calculation. For SSA the calculation of F was wrong (again I want to stress I am against both SIA and SSA, but SSA gives correct answer in this problem). To calculate the total fraction F, it requires sample for two parts. One part is the room she wakes up in, which she has the perfect knowledge it is red. The other part is the other 80 rooms which she has to estimate base on her sample. so if the unpainted room is in the 79 rooms she opens then F=(78/79)*(80/81)+1*(1/81)=0.987498. If we also consider the case when the unpainted room is not in the 79 rooms she opens then F= 0.987498*(79/80)+1*(1/80)=0.98765432 or the exact value of 80/81. Your calculation for the SIA is correct which is larger than the exact value. So your argument actually shows SIA gives estimation biased towards red while SSA gives unbiased estimation.

      The disagreement lies only in how was her own room treated. As she opens more doors the weight of her own room decreases. That’s why the difference in above numbers are small because in your example beauty opens large amount of doors other than her own.

      “The point is that the calculation you are trying to describe is just an estimation of F, the fraction of rooms that are painted. Not a calculation of the total number N, or of F itself.”

      This is just a different wording of the same problem. In the SBP you can also ask beauty to guess the fraction of waking days, being either 50% or 100%, it is the same as asking beauty to guess H or T. May be it helps to consider the following question:

      1. If the painter toss a fair coin and decides to paint 21 red rooms if H and 27 red rooms if T. Again beauty saw 2 red among 8 other rooms, does H or T has a higher probability?

      2. Now the painter randomly generate a number between 1 and 81 with equal chances and paint the corresponding number of rooms. After beauty saw 2 red among 8 other rooms, does N=21 or N=27 has a higher probability?

      3. Same setup as in 2. How should beauty estimate the total number of red rooms?

      If you can give clear answers to the above three problem maybe the shortcomings of SIA would be clear.

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