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Old 11-04-2011, 10:53 PM   #1
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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.
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Old 11-04-2011, 11:20 PM   #2
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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
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Old 11-04-2011, 11:27 PM   #3
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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.
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Old 11-04-2011, 11:39 PM   #4
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Re: Sleeping Beauty Problem

Philolsophers can **** anything up. SIA as presented in wiki is an obviously retarded misapplication of the indifference principle.
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Old 11-04-2011, 11:48 PM   #5
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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.
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Old 11-05-2011, 12:10 AM   #6
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Re: Sleeping Beauty Problem

The way I understand it I side with the P = 1/3 argument.
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Old 11-05-2011, 02:01 AM   #7
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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.
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Old 11-05-2011, 02:03 AM   #8
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Re: Sleeping Beauty Problem

50%

there's a 50% chance this is her first awakening, or a 50% chance it's her x awakening.
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Old 11-05-2011, 02:44 AM   #9
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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, and
    • 25% 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.
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Old 11-05-2011, 10:25 AM   #10
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Re: Sleeping Beauty Problem

Quote:
Originally Posted by atakdog View Post
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, and
  • 25% 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.
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Old 11-05-2011, 10:44 AM   #11
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Re: Sleeping Beauty Problem

Quote:
Originally Posted by CompleteDegen View Post
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.
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Old 11-05-2011, 12:08 PM   #12
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Re: Sleeping Beauty Problem

Quote:
Originally Posted by DarkMagus View Post
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.
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Old 11-05-2011, 12:29 PM   #13
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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.
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Old 11-05-2011, 12:32 PM   #14
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Re: Sleeping Beauty Problem

Quote:
Originally Posted by DarkMagus View Post
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?
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Old 11-05-2011, 12:40 PM   #15
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Re: Sleeping Beauty Problem

We had a huge thread on this back in 2007

Sleeping Beauty Paradox - 2+2 - 2007


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