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mvassilev
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posted November 07, 2013 07:35 PM 


Poll Question: The Sleeping Beauty Problem
In the spirit of the previously discussed Newcomb's Problem, I present the Sleeping Beauty Problem. Imagine you're knowingly undergoing the following experiment. On Sunday, you will be put to sleep. Then, you are awakened once or twice and interviewed. After each interview, you are put to sleep with a drug that makes you forget that you were interviewed. Then, regardless of whether you're woken up once or twice, you are awakened on Wednesday and not interviewed.
To determine whether you will be awakened once or twice, the experimenter tosses a fair coin. If it lands on heads, you will be awakened on Monday, and interviewed then. If it lands on tails, you will be awakened both on Monday and Tuesday, and interviewed on both days.
When you are interviewed, you are asked, "What is the probability that the coin landed on heads?" How do you respond? If this explanation sounds confusing, here is a hopefully simpler explanation of the sequence of events.
1. Sunday, you are put to sleep. > A coin is tossed. > If it lands on heads, go to 2, if it lands on tails, go to 3.
2. You are awakened and interviewed on Monday and asked "What is the probability that the coin landed on heads?" > You are administered a drug that erases your memory of the interview and puts you to sleep. > You are awakened on Wednesday and leave.
3. You are awakened and interviewed on Monday and asked "What is the probability that the coin landed on heads?" > You are administered a drug that erases your memory of the interview and puts you to sleep. > You are awakened and interviewed on Tuesday and asked "What is the probability that the coin landed on heads?" > You are administered a drug that erases your memory of the interview and puts you to sleep. > You are awakened on Wednesday and leave.
So, when you wake up, you don't know if it's Monday or Tuesday, but you know that it can only be Tuesday if the head landed on tails. How do you answer their question?
____________
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OhforfSake
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posted November 07, 2013 08:24 PM 


So in 50% of the times you're asked twice and in the other 50% you're asked once.
Therefore you expect to be asked 1.5 times, of which 2/3 of the times you can expect one outcome, and 1/3 of the times another.
But it's also a question of how the actual question is formulated. The chance of a fair coin landing heads/tails is always 50%, so if that's the question, the answer is 1/2.
If the question however is that what's the most probable value the coin took given the conditions presented afterwards, in 2/3 times you'd be right to guess one thing, and hence 1/3 for the other.
The coin isn't unfair, it's merely that if the experiment consisted of multiple people, not everyone would be asked the same amount of time. If everyone said one answer, and given the sample is large enough, either ~1/3 would be right or ~2/3 would be right, as long as the question is not if the coin is fair, but what's the most probable value the coin took given the conditions of the experiment.


mvassilev
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posted November 07, 2013 08:32 PM 


OhforfSake said: If the question however is that what's the most probable value the coin took given the conditions presented afterwards, in 2/3 times you'd be right to guess one thing, and hence 1/3 for the other.
But when you wake up, you don't know if it's Monday or Tuesday, so you don't have any new information (compared to Sunday), so you don't have any grounds on which to change the probability you assign to the coin having any particular result.
To modify the scenario slightly so it would be more illustrative, imagine that if the coin lands on tails, you're woken up a million times instead of twice (and all the other conditions of the problem remain the same). But because you have amnesia, you don't know if you're in the branch that has a million awakenings or just one, so you don't have any new information that would adjust your probability of how the coin landed. So, if on Sunday you think it had a 1/2 chance landing on tails, there's no reason for you to believe differently once you wake up.
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OhforfSake
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posted November 07, 2013 08:55 PM 


Probability is success / total.
Let's assume there's a lot of people participating, so many that the coin hits tails and heads 50% of the times respectively.
The total amount of guesses equals the total amount of times the total amount of people are woken up. That is 1.5 times the amount of people present. If everyone decides to guess on the same, obviously not 50% will be correct, but either 2/3 or 1/3. Hence there is 100% more success for one option than the other (or to say one option has only 50% of the success that the other has).
Again, it depends on how the question is stated, but if it's about guessing what coin flip group you belong to, it's not 50/50, because by not asking the same amount of times, and by having each guess be success / failure to guess, it's equivalent to have a fair coin and remove 50% of the times it hits one side. The toss is still 50/50, but the outcome is not.


JollyJoker
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posted November 07, 2013 09:49 PM 


The probability is 50%.
You can simply alter the setup: if heads > 1 wakeup; if tails > 1.000.000 wakeups.
IF you wake up, chances are FIFTY % that this will be the one wakeup of heads, and 50% that it will be one in a series if a million wakeups initiated by tails.


Corribus
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posted November 07, 2013 10:28 PM 


Do you want to know how I would respond or what the real probability is? If the former, it depends on how much information I am given. If the latter, it is easy to show mathematically and experimentally that the answer is 1/3. This is similar to a problem I presented in the old math thread.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg


OhforfSake
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posted November 07, 2013 11:09 PM 


Yay! I'm not alone


Corribus
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posted November 08, 2013 01:55 AM 

Edited by Corribus at 01:56, 08 Nov 2013.

Well it's assuming I'm reading the problem correctly. The description is convoluted.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg


friendofgunnar
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posted November 08, 2013 02:57 AM 


Interesting philisophical question.
But that's all it is, because this question assumes a world in which complete information loss can exist. For example are the experimenters going to shave my face, cut a millimeter off of all my body hair, cut my nails 1 micrometer, evacuate my bowels of my last dinner, replace the liquid in my bladder, inject all the dietary nutrients into my body that I would have gotten otherwise, and also rejigger my entire brain's neurons such that the activites of two days previous only felt like yesterday's? There's just no way that you can block out the passage of time on your body.
If I remember right Newcomb's paradox also depended on this possibility of complete information loss.


JoonasTo
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What if Elvin was female?

posted November 08, 2013 03:17 AM 


mvassilev said:
OhforfSake said: If the question however is that what's the most probable value the coin took given the conditions presented afterwards, in 2/3 times you'd be right to guess one thing, and hence 1/3 for the other.
But when you wake up, you don't know if it's Monday or Tuesday, so you don't have any new information (compared to Sunday), so you don't have any grounds on which to change the probability you assign to the coin having any particular result.
To modify the scenario slightly so it would be more illustrative, imagine that if the coin lands on tails, you're woken up a million times instead of twice (and all the other conditions of the problem remain the same). But because you have amnesia, you don't know if you're in the branch that has a million awakenings or just one, so you don't have any new information that would adjust your probability of how the coin landed. So, if on Sunday you think it had a 1/2 chance landing on tails, there's no reason for you to believe differently once you wake up.
Ah but that is wrong. You have new information, that is, you woke up and were asked a question.
Becuase you know that you cannot tell how much time has passed, it is more logical to assume the position of tails since this way you have more situations where you could be right. If you have any incentive to quess right that is. You can simply quess randomly since there's no effect from it.
Though if you're smart, you'll quess heads the first time and bite one of your nails or better yet, the inside of your mouth. Now you can always tell the next time, if you are past monday.
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mvassilev
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posted November 08, 2013 03:30 AM 


There are more situations, but each of them is less likely. There are three possible cases:
Heads and wake up on Monday  50%
Tails and wake up on Monday  25%
Tails and wake up on Tuesday  25%
So, while you're correct in saying there are more situations in which you wake up for Tails, each individual situation is less likely, so it adds up to being 50% heads, 50% tails.
Also, there's no getting out of the conditions of this problem. There is no way for you to tell whether you were woken on Monday when you're woken on Tuesday.
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JoonasTo
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What if Elvin was female?

posted November 08, 2013 05:14 AM 


To an outside observer, yes. But as you, the subject, have no way of knowing, it is 1/3 for you that the coin landed heads up.
Of course there is a way to get out of the conditions. There always is, that's just how the world works.
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Corribus
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posted November 08, 2013 05:52 AM 


Obviously, the probability of any given coin flip being heads is 50%. So if your question is "what was the probability that a single coin flip was heads?", then the answer is 50%. However the probability of any given interview happening coincident with a result of heads is 1/3. That is, only 33% of all interviews, after repeating the experiment many times, will happen after a result of heads. It's not really clear to me which case you're asking for here.
In the latter case, if you use a random number generator for the coin flip, you get an answer that approaches 1/3 for the probability of an interview occurring with the coin being a heads. Out of 31 tries, the number I got was 9. Not a large n, but sufficiently large enough to see the value approaches 1/3. As I said, you can show this mathematically as well, but it'd be a pain in the ass to type in the equations here. The probability reduces nicely to 1/3, however.


OhforfSake
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posted November 08, 2013 07:12 AM 


mvassilev said: There are more situations, but each of them is less likely. There are three possible cases:
Heads and wake up on Monday  50%
Tails and wake up on Monday  25%
Tails and wake up on Tuesday  25%
When the question was originally asked, I wondered what was the point of the weekdays, so I did consider the prob[given day], but I couldn't see its relevance to the question.
While the first line is correct, 50% of the total subjects will wake up on Monday (heads subjects), so will the other 50% (tails subjects), and 50% of the total subjects will wake up on Tuesday (tails subjects).
Anyway, I believe it is:
Prob[Heads] = .5
Prob[Tails] = .5
Prob[Wake up on Monday] = Prob[Heads or Tails] = 1
Prob[Wake up on Tuesday] = Prob[Tails] = .5
Therefore it follows the Prob[Monday] = Prob[Wake up on Monday] / (Prob[Wake up on Monday] + Prob[Wake up on Tuesday]) = 2 / 3
E.g. try to expand the problem like JJ did, and ask what's the probability the researcher will wake up a subject on a given day. First heads/tails is decided, and no matter what, the subject will be woken up on a Monday.
If heads we move on to the next subject, if tails we continue to wake up the subject a million imagined different weekdays.
From the researchers point of view, for any 2 subjects (you've equal many tails/heads) there'll in average be 2 wake ups on Mondays', and 1 on each of the other imagined million different weekdays.
Hence is Prob[Monday] = 1 / 10^6, and for all the other weekdays .5 / 10^6.
Going back to the initial set up, with only Monday and Tuesday, it reduces to Prob[Monday] = (1 / 3) / (1 / 3 + .5 / 3) = 2 / 3
For what I think is an obvious reason it's Tuesday 33% of the times, and not only 25% of the times, imagine how you would conduct this experiment.
If it was Tuesday 25% of the times, it follows that you have 3 Monday's for every Tuesday. That is possible if you conduct your experiment by:
Chose a subject (A), heads, wake up on Monday.
Chose a subject (B), tails, wake up on Monday. Don't move on to Tuesday yet!
Chose s subject (C), heads, wake up on Monday.
Back to subject (B), wake up on Tuesday.
Then the researcher would experience 3 Monday's for every Tuesday, hence prob[Today is Tuesday] becomes 25%, but there's the problem that by this procedure, you get twice as many heads subjects. But since the coin is fair, you've an equal amount of heads/tails subjects.


TsarIvor
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posted November 08, 2013 07:29 AM 


Corribus said: Obviously, the probability of any given coin flip being heads is 50%. So if your question is "what was the probability that a single coin flip was heads?", then the answer is 50%. However the probability of any given interview happening coincident with a result of heads is 1/3. That is, only 33% of all interviews, after repeating the experiment many times, will happen after a result of heads. It's not really clear to me which case you're asking for here.
mvassilev said:
When you are interviewed, you are asked, "What is the probability that the coin landed on heads?" How do you respond??
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OhforfSake
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posted November 08, 2013 07:35 AM 


@Tsar
It's words like the coin, and landed, which makes me assume it's not simply a question of what is the probability of a fair coin landing on one side.
Edit: That is "the coin" reads as "your coin", as in the coin used in your case, and "landed" reads as "given the conditions" as in, the coin has been flipped > conditions > question.
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JollyJoker
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posted November 08, 2013 09:26 AM 


You make this way too complicated  the problem is way simpler.
The question is for the probabili8ty of the COIN FLIP  not for the probability of the WEEKDAY you are interviewed, which would indeed make a difference, because it's more likely for you to wake up on a Monday, than on a Sunday.
You are also not giving any additional information on waking up, so there isn't any dynamic either.
That means, for the question there are ONLY TWO RELEVANT CASES HERE:
1) Heads 50%
2) Tails 50%
Everything resulting from that is to be seen as ONE event:
1a) Waking on Monday (AND Waking on Wednesday)
2a) Waking on Monday AND Waking on Tuesday (AND Waking on Wednesday)
To make things clear, ask this question: How do you rate the probability of waking up on a Monday? The WRONG answer is 2/3, although it might look good on first sight. Why?
BECAUSE THE CASES ARE NOT EQUAL IN PROBABILITY!
The decisive thing is THE COIN FLIP  it's dividing probabilities into two parts with 50% each, which means, you have 50% prob on Heads for waking up on a Monday; Tails shares its 50% into one half Monday and another half on Tuesday.
That means, there is a 3/4 probability for you to wake up on a Monday and only a 25% probability to wake up on a Tuesday (and not a 2 in 3 chance to wake up on a Monday).
You must not make the mistake to count the awakenings. Look at 100 persons doing the thing. 50 of them will have had Heads and 50 of them will have had Tails > 50% prob that when you awake it will be on Heads.
Is it important that there are 150 QUESTIONS ASKED and the right answer is 50 times Heads and 100 times Tails?
NOPE  that is inconsequential for the problem  it's basically a double answer: the probability had been decided with the FIRST awakening and the second is DETERMINED BY THEFIRST, NOT an independent event.
So what you ACTUALLY have is 50 times right answer Heads and 50 times the right DOUBLEANSWER Tails.


OhforfSake
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posted November 08, 2013 09:37 AM 


JollyJoker said: Is it important that there are 150 QUESTIONS ASKED and the right answer is 50 times Heads and 100 times Tails?
NOPE  that is inconsequential for the problem  it's basically a double answer: the probability had been decided with the FIRST awakening and the second is DETERMINED BY THEFIRST, NOT an independent event.
Sure you can do that, but let's say subject A was head and subject B was tails.
Let's say they both answer the same, either tails or heads.
Monday:
A: heads , B: heads
Tuesday:
B: heads
Correct guesses 1, 3 questions [1/3]
or:
Monday:
A: tails , B: tails
Tuesday:
B: tails
Correct guesses 2, 3 questions [2/3]
To do it like I understand you suggest, B's guesses are only weighted at half the importance of A's guesses, hence making it 50/50 ([1/3] > [1/2], [2/3] > [1/2]), but I do not think that's in any way clear of the content, which I understand as to try to guess what the coin landed at given the conditions, rather than anything about all guesses you make will be counted as 1 and weighted properly, which is analogous to asking if a fair coin is fair.
Also you get halfwrongs and halfrights in the case of a tails subject who changes his answer which is a bit... (unless you count 1 right and all other wrongs as a right/wrong, in which case it is not 50/50)


TsarIvor
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posted November 08, 2013 11:26 AM 


Why do you have two subjects?
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JollyJoker
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posted November 08, 2013 11:36 AM 


Your mistake is that you give B 2 chances, while you give only 1 to A.
You can assign (the same) probabilities only to EQUAL events/cases and those two ARE NOT EQUAL.
The Heads event is different than the Tails event.
That's all there is to it. It just LOOKS like it was the same event, but it ISN'T.
Take this setup:
A coin is flipped, and the situation is as follow:
1) You don't get to see the result;
2) If Heads came, you will be asked ONCE for a guess;
3) If Tails came you wille be asked TWICE for a guess (but you will forget your first guess, so it's like making a RANDOM guess twice).
We have the following cases here:
1a) you say Heads for right;
1b) You say Tails for wrong;
21a) You say Heads for Wrong
21b) You say Tails for right
22a) You say Heads for Wrong
22b) You say Tails for right
So if you count this out, strangely enough HEADS is right ONCE, and wrong TWICE, while Tails is wrong ONCE and right TWICE.
However, you mix an event with two cases and an even with 4 cases, and that doesn't allow to assign EQUAL probabilities.



