This is the second part of my argument. It mainly involves a counter example to SIA and Thirdism.
Different part of the argument can be found here: I, II, III, IV.
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 (since she used it in statistical estimation), 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.
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.
(EDIT Aug 1. I think the best answer thirders shall give is accept C since it is obviously a simple random sample. Adding another red room to this will make the 9 rooms biased from his perspective. However they can argue if a selector saw the same 9 rooms through random selection then it is unbiased from the selector's perspective. Thirders could argue she must answer from the selector's perspective instead of her own. Main reason being she is undergoing potential memory wipes so her perspective is somewhat "compromised". However, with this explanation thirder must confirm the perspective disagreement between beauty and the selector about whether or not the 9 rooms are biased. It also utilize perspective reasoning followed. In another word perspective disagreement is not unique to halfers and shall not be treated as a weakness.)
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.
(EDIT Aug 1. We can show this another way. Let the selector, halfer beauty and thirder beauty do a large number repeated estimation on the same set of rooms. The selector and halfer's estimations would be concentrated around the true value of R, where as thirders answer would be concentrated on some value larger.)
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.
Part III can be found at here.
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.
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:... (read more)