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11-04-2011, 10:53 PM   #1
CompleteDegen
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Sleeping Beauty Problem

For those who haven't seen it, here is the wording:

Quote:
 Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details. On Sunday she is put to sleep. A fair coin is then tossed to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a dose of an amnesia-inducing drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday. The drug makes sure that she cannot remember any previous awakenings during the course of the experiment, but she will retain the ability to memories gained after the experiment is over. Any time Sleeping beauty is awakened and interviewed, she is asked, "What is your credence now for the proposition that the coin landed heads?"
There doesn't appear to be a mathematical consensus on the correct answer. Obviously, from the experimenter's perspective, the odds are exactly 50/50 for heads/tails. From Sleeping Beauty's perspective, however, she is awoken twice as many times, on average, for a tails toss than a heads toss. For example, from Wikipedia, if you toss the coin 1000 times, she will be awoken 500 times on Monday for a heads toss, 500 times on Monday for a tails toss, and 500 times on Tuesday for a tails toss, giving a total of 2/3 of the time she is awoken, tails was thrown.

However, as an outside observer, we know it had to be 50/50 in the actual toss, so where does the difference enter? I think I'm in the 1/3 camp, but can't fully justify it. What if she is awoken once after a heads toss, and 100,000 times after a tails toss, should she believe with near certainty that tails was thrown? Seems almost like she should be able to correctly guess 2/3 of the time that it is Monday, but the number of times she is awoken should have no effect on the perceived likelihood of what was tossed.

 11-04-2011, 11:20 PM #2 TomCowley Pooh-Bah     Join Date: Sep 2004 Posts: 5,626 Re: Sleeping Beauty Problem Your prior probability of the coin being heads is 50%. Your current new information, being awake with no memory, is uninformative because it happens with 100% certainty if it's heads and also with 100% certainty if it's tails. So by Bayes' theorem you don't change anything and it's still 50%. P(heads|awake) = P(awake|heads) * P(heads)/P(awake) = 1*.5/1 = .5
 11-04-2011, 11:27 PM #3 CompleteDegen veteran     Join Date: May 2008 Location: degenin' in AC Posts: 2,481 Re: Sleeping Beauty Problem That's one interpretation. The different answers seem to stem from the two different schools of anthropic probability, ie the Self-Indication Assumption and the Self-Sampling Assumption. I can see the argument from both sides. You are awake with no memory, but you know the details of the experiment, that you will be woken twice after a tails toss and once after heads. Or, let's say, you will be woken 100,000 times after a tails toss, and once after heads. Seems like any given time you're awoken, it was more likely to be after tails was tossed. I'm starting to think it's more an illusion from the sleeping person's perspective. Maybe solidly in the 50/50 camp now.
 11-04-2011, 11:39 PM #4 TomCowley Pooh-Bah     Join Date: Sep 2004 Posts: 5,626 Re: Sleeping Beauty Problem Philolsophers can **** anything up. SIA as presented in wiki is an obviously retarded misapplication of the indifference principle.
 11-04-2011, 11:48 PM #5 Aaron W. Carpal \'Tunnel   Join Date: Sep 2002 Location: Henderson, NV Posts: 28,969 Re: Sleeping Beauty Problem It doesn't seem that interesting to me. It's just a difference between whether you count each interview as one interview, or if you throw in a weighting factor based on the frequency with which such the question would be asked.
 11-05-2011, 12:10 AM #6 nullspace banned     Join Date: Jul 2010 Location: { } Posts: 1,859 Re: Sleeping Beauty Problem The way I understand it I side with the P = 1/3 argument.
 11-05-2011, 02:01 AM #7 TomCowley Pooh-Bah     Join Date: Sep 2004 Posts: 5,626 Re: Sleeping Beauty Problem Also, the doomsday argument seems demonstrably lolbad for the same reason. Assuming you're just dealing with total births, and not any function of people alive at a given time or anything, whatever your ratio of P(100b)/P(90b) is in your prior will be the same after 60b are observed born. Again by Bayes' theorem, P(100b|>=60b)=P(>=60b|100b)*P(100b)/P(>=60b) = 1*P(100b)/P(>=60b) = P(100b)/P(>=60b), and same for P(90b), so the ratio is now P(100b)/P(>=60b) / P(90b)/P(>=60b) = P(100b)/P(90b) still. It's obviously uninformative to the ratio. And you can't even assign a uniform prior on N to even come up with an estimate, so it has to be something totally pulled out of your ass. Last edited by TomCowley; 11-05-2011 at 02:22 AM.
 11-05-2011, 02:03 AM #8 housenuts Carpal \'Tunnel     Join Date: Jul 2004 Location: Come With Me If You Want To Lift Posts: 32,994 Re: Sleeping Beauty Problem 50% there's a 50% chance this is her first awakening, or a 50% chance it's her x awakening.
 11-05-2011, 02:44 AM #9 atakdog addicted     Join Date: Jan 2008 Location: vṛkṣāsana Posts: 49,998 Re: Sleeping Beauty Problem From her perspective when she wakes up: It's 50% likely that the flip was heads and it's Monday. It's 50% likely that the flip was tails, and25% likely that it's Monday, and 25% likely that it's Tuesday. That happens also to be what the experimenter thinks, but she knows the experiment and thus reasons the same (or, more likely, has it explained to her). In other words, the premise of the ⅔ – ⅓ argument, that there are three possible states — each equally likely — from her perspective when she wakes up, is false, and results derived therefrom also false. There are three states, but she knows darned well that they aren't equally likely. Last edited by atakdog; 11-05-2011 at 02:50 AM.
11-05-2011, 10:25 AM   #10
CompleteDegen
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by atakdog From her perspective when she wakes up: [LIST][*]It's 50% likely that the flip was heads and it's Monday.[*]It's 50% likely that the flip was tails, and25% likely that it's Monday, and 25% likely that it's Tuesday. .
This seems odd, because there is a 2/3 chance that it is Monday when she awakes, not 75%. Run the experiment 1000 times, she will awaken 500 times Monday from heads, 500 times Monday from tails and 500 times Tuesday from tails, giving a 1000/1500 awakenings, or 2/3, that it is Monday.

11-05-2011, 10:44 AM   #11
DarkMagus
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by CompleteDegen This seems odd, because there is a 2/3 chance that it is Monday when she awakes, not 75%. Run the experiment 1000 times, she will awaken 500 times Monday from heads, 500 times Monday from tails and 500 times Tuesday from tails, giving a 1000/1500 awakenings, or 2/3, that it is Monday.
I think this is the best way to look at it. If on awakening it is equally likely that the coin was heads or tails, then at each awakening Sleeping Beauty should be able to place an even money bet that it was heads, and break even. But clearly she won't break even.

11-05-2011, 12:08 PM   #12
TomCowley
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by DarkMagus I think this is the best way to look at it. If on awakening it is equally likely that the coin was heads or tails, then at each awakening Sleeping Beauty should be able to place an even money bet that it was heads, and break even. But clearly she won't break even.
She can, as long as she's only asked once. Imagine a slightly different game:

I declare that my intention is to always guess heads. You flip a coin and ask me. I clearly win half of these bets. Now, if and only if it's tails, you keep asking me to guess the result of *the same flip*, and I keep saying heads because that's how I roll. If there's money on the line each time, I obviously lose my ass, because the bets aren't independent. P(win first guess) = .5, but P(win 2nd guess | lost first guess)=0. Other than naive intuition, why would you expect the second bet to be breakeven? The mere act of being asked to make the bet means I'll lose it every time.

Last edited by TomCowley; 11-05-2011 at 12:19 PM.

 11-05-2011, 12:29 PM #13 DarkMagus Carpal \'Tunnel     Join Date: Jul 2007 Location: Canada Posts: 8,379 Re: Sleeping Beauty Problem In your situation, you have a 100% chance of making the first guess and a 50% chance of making the second guess. Also you have a 50% chance of winning the first guess and a 0% chance of winning the second guess. Your expected number of guesses is 1.5, and your expected number of wins is 0.5. Your expected number of wins per guess is therefore 1/3.
11-05-2011, 12:32 PM   #14
TomCowley
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by DarkMagus In your situation, you have a 100% chance of making the first guess and a 50% chance of making the second guess. Also you have a 50% chance of winning the first guess and a 0% chance of winning the second guess. Your expected number of guesses is 1.5, and your expected number of wins is 0.5. Your expected number of wins per guess is therefore 1/3.
No ****. Now, what does that have to do with my post?

 11-05-2011, 12:40 PM #15 PairTheBoard Carpal \'Tunnel   Join Date: Dec 2003 Posts: 8,545 Re: Sleeping Beauty Problem We had a huge thread on this back in 2007 Sleeping Beauty Paradox - 2+2 - 2007 PairTheBoard
11-05-2011, 12:41 PM   #16
Snaaak

Join Date: Apr 2007
Posts: 1,071
Re: Sleeping Beauty Problem

If she always answers tails she will be right a larger proportion of the time than if she always answers heads. As far as I understand probability that's means that P(heads)<P(tails) by definition...
Quote:
 Originally Posted by TomCowley Your prior probability of the coin being heads is 50%. Your current new information, being awake with no memory, is uninformative because it happens with 100% certainty if it's heads and also with 100% certainty if it's tails. So by Bayes' theorem you don't change anything and it's still 50%. P(heads|awake) = P(awake|heads) * P(heads)/P(awake) = 1*.5/1 = .5
She has the information that when she awakens there's a 1 in 3 chance it's a tuesday, and when it is tuesday there's a 100% chance the coinflip was tails

 11-05-2011, 12:42 PM #17 TomCowley Pooh-Bah     Join Date: Sep 2004 Posts: 5,626 Re: Sleeping Beauty Problem No.. that means P(being correct answering tails|asked the question)>P(being correct answering heads|asked the question). It's fallacious to conflate this with believing P(tails)>P(heads).
 11-05-2011, 12:45 PM #18 Snaaak adept   Join Date: Apr 2007 Posts: 1,071 Re: Sleeping Beauty Problem What is the difference between P(being correct answering tails|asked the question) and P(tails|asked the question)
11-05-2011, 12:46 PM   #19
DarkMagus
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by TomCowley No ****. Now, what does that have to do with my post?

11-05-2011, 01:54 PM   #20
TomCowley
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Join Date: Sep 2004
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by PairTheBoard We had a huge thread on this back in 2007 Sleeping Beauty Paradox - 2+2 - 2007 PairTheBoard
Looks like you nailed it then.

11-05-2011, 02:03 PM   #21
TomCowley
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Join Date: Sep 2004
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by Snaaak What is the difference between P(being correct answering tails|asked the question) and P(tails|asked the question)
That's not what I said. Read the archive link, it's all in there.

 11-05-2011, 02:09 PM #22 DarkMagus Carpal \'Tunnel     Join Date: Jul 2007 Location: Canada Posts: 8,379 Re: Sleeping Beauty Problem Ok I think I understand the problem a little better. I think the overriding fact is that being awakened doesn't give us any new information. We knew beforehand that we were going to be awakened. Everything we know about the current situation is exactly the same as what we knew before we were put to sleep. To us, the coinflip may as well not even have happened yet. So the correct answer is 50%. The gambling analogy is flawed because the number of times you place the bet is dependent on the outcome of the bet. In that kind of situation, EV equations don't work the same way.
11-05-2011, 03:29 PM   #23
PairTheBoard
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Re: Sleeping Beauty Problem

Quote:
 Originally Posted by TomCowley Looks like you nailed it then.
I think Jason1990's summary on the last page may be best. By the last page I had researced a little and come up with this paper by Nick Bostrom which I thought at the time was pretty creative and that he might be onto something. If he was he should by now have attracted some adherents to his view. But I suspect that's not the case.

Nick Bostrom on Sleeping Beauty

Bostrom differentiates between possible agent parts and actual agent parts. The idea being that in one trial of the experiment Sleeping Beauty will experience either exactly 1 or exactly 2 actual agent parts. Either 1 corresponding to heads or 2 corresponding to tails. While in a large number of trials She will experience aproximately twice as many actual agent parts corresponding to tails as corresponding to heads. He then claims Her credence for heads should be different in the two cases - 1 trial or many. I'm afraid it's a tough sell but who knows.

imo, jason1990 clarifies things nicely in his last post. For the 1/2's the problem comes with what Sleeping Beauty should think if she is awakened and informed the coin was tails. What should her credence be that the day is Monday? I believe the Indifference Principle is required for her to have credence 1/2. But there's no proof of that. Just an arbitrary appeal to a principle that may not apply. Jason1990 rejects that appeal to indifference. However his innovation (imo) is to not try to argue another value for what her credence should be in that case. Instead, he claims there is not enough information for Her to form any rational credence other than it must be either Monday or Tuesday. With that allowed, all the other credences conform to intuition without being contradictory.

So I think jason1990's summary is probably the best that can be done unless you want to intoduce Sleeping Beauty to Alice in Wonderland and take her down Nick Bostrom's probability rabbit hole of possible and actual agent parts.

PairTheBoard

 11-05-2011, 05:53 PM #25 masque de Z Carpal \'Tunnel     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 9,343 Re: Sleeping Beauty Problem Notice also that if you never told the beauty that there was a pill involved to make her forget , just that she would sleep and wake up and asked a question , her answer would be 1/2 always. However it would be also wrong. She would be imagining its always Monday when it can be any future day as well (depending on how many pills used). There is nothing wrong with that mistake she now makes. But we know better!

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