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Sleeping Beauty Problem Sleeping Beauty Problem

12-03-2011 , 07:13 PM
This all shows the disconnect between the thought processes of people who think of probabilities only in terms of frequencies, or think probabilities are defined by frequencies or are only meaningful when formulated as frequencies, and the thought processes of those who think of probability as information/state of knowledge. The frequency people have to define a process to model "an awakening" when one isn't specified, and then make a reverse appeal to the law of large numbers (that the frequencies from your process come from an iid sequence of whatever you're trying to assign probabilities to (awakenings), and I pointed out 200 posts ago that this step is fallacious here because they don't), and then call the frequency a probability.

The information/state of knowledge people don't have to do anything of the sort. It's as simple as post 2. Or post 374 if you want it written in formal mathematical language. 1/3 isn't "an alternative answer that can also be the probability". It's just an answer to a different question- "what percentage of total awakenings are the result of heads?"- that could only be an answer to the OP if experiments were made of iid awakenings, but they're not.
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12-03-2011 , 07:18 PM
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Originally Posted by PairTheBoard
If I understand you, you would be describing a different experiment where waking on Heads-Tuesday Beauty is still in the New-Experiment but not in the Old part of the New Experiment. I don't think this changes anything anymore than if you added another 365 days outside the Old experiment for heads and 573 days outside the Old experiment for tails. When she awakens within the Old part of whatever New experiment you cook up, she reasons as before. After all, in the original Old Experiment, when she awoke she also knew she was not waking outside the Old Experiment.

PairTheBoard
The fact that it doesn't change anything underscores the 1/3 side's argument on this, and it's something the 1/2 side will have to deal with, you say ~E doesn't happen, but it happens in a different version of the experiment that -according to you- doesn't change anything. The argument is exactly that it doesn't change anything, and that this is therefore the way she should be reasoning inside of the original problem aswel.

Note that this is basically still the same argument as in post #107. To which your rebuttal was 'that is one way to use the indifference principle, but I find another more convincing'.

Restricting your view to 'the old part' is basically cheating, you are now saying, well maybe this new experiment is different, but let's choose to ignore the part that this argument is about.

When she awakes in the old part, she doesn't learn that she's not-not in the old part because your premise (eg that she's in the old part) doesn't allow it.

It's like saying that she can see the coin when it flipped tails, then we wake her up in a tails awakaning, we determine she has seen the coin, but now we can't use this information because we asserted that this was in fact a tails-awakening to begin with, therefore ~E -that is, not seeing the coin having flipped tails- could not have occured. So it's still 50/50.

fwiw, I'm starting to see the 1/2-side's arguments but I don't think this is sufficiently explained/refuted to be convinced towards the other side.
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12-03-2011 , 08:46 PM
Quote:
Originally Posted by TomCowley
This all shows the disconnect between the thought processes of people who think of probabilities only in terms of frequencies, or think probabilities are defined by frequencies or are only meaningful when formulated as frequencies, and the thought processes of those who think of probability as information/state of knowledge. The frequency people have to define a process to model "an awakening" when one isn't specified, and then make a reverse appeal to the law of large numbers (that the frequencies from your process come from an iid sequence of whatever you're trying to assign probabilities to (awakenings), and I pointed out 200 posts ago that this step is fallacious here because they don't), and then call the frequency a probability.

The information/state of knowledge people don't have to do anything of the sort. It's as simple as post 2. Or post 374 if you want it written in formal mathematical language. 1/3 isn't "an alternative answer that can also be the probability". It's just an answer to a different question- "what percentage of total awakenings are the result of heads?"- that could only be an answer to the OP if experiments were made of iid awakenings, but they're not.
Tom, how do you argue with this?

When she awakens:

P(heads) = P(heads|Mon)*P(Mon) + P(heads|Tues)*P(Tues)

= (1/2)*(2/3) + 0*(1/3)

= 1/3.

It would seem you would want to reject P(Mon) = 2/3 for the same reasons you have given. But then what do you replace it with, 1? That doesn't make sense. If you leave it undefined, then the answer is also undefined.
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12-03-2011 , 08:52 PM
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Originally Posted by BruceZ
Tom, how do you argue with this?

When she awakens:

P(heads) = P(heads|Mon)*P(Mon) + P(heads|Tues)*P(Tues)

= (1/2)*(2/3) + 0*(1/3)

= 1/3.
This is the answer. There is no argument.

If you look at all of the attempts to create a 1/2 solution, they have one thing in common. They try to define the "event" as the experiment, not the awakening. If you define the event to be the experiment, then you lose the ability to deal with the two awakenings on tails. You essentially reduce the problem to one where she is awakened once on heads and once on tails, although the tails awakening could occur on Monday or Tuesday. This is obviously flawed, because SB's pov is that of awakenings, not experiments.

There is a test that I have shown before, that demonstrates that her credence in heads must be less than 1/2. When she awakens she knows that if it is Monday, then her credence in heads is 1/2. If it is Tuesday, her credence in heads is 0. Since she does not know the day, there is a finite chance of either Monday or Tuesday. Thus, from her point of view her credence in heads in that case must be less than zero, unless the possibility of Tuesday is zero. But it is not zero.
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12-03-2011 , 09:12 PM
Quote:
Originally Posted by BruceZ
Tom, how do you argue with this?

When she awakens:

P(heads) = P(heads|Mon)*P(Mon) + P(heads|Tues)*P(Tues)

= (1/2)*(2/3) + 0*(1/3)

= 1/3.
1) Mon isn't a proposition.
2) Even pretending it is, you don't get P(mon)=2/3 to begin with unless you make the same kind of reverse appeal to the law of large numbers (2/3 of total awakenings are mon, therefore P(mon)=2/3) that I mentioned in my last post.
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12-03-2011 , 09:27 PM
Quote:
Originally Posted by TomCowley
1) Mon isn't a proposition.
2) Even pretending it is, you don't get P(mon)=2/3 to begin with unless you make the same kind of reverse appeal to the law of large numbers (2/3 of total awakenings are mon, therefore P(mon)=2/3) that I mentioned in my last post.
How about this argument?

Quote:
There is a test that I have shown before, that demonstrates that her credence in heads must be less than 1/2. When she awakens she knows that if it is Monday, then her credence in heads is 1/2. If it is Tuesday, her credence in heads is 0. Since she does not know the day, there is a finite chance of either Monday or Tuesday. Thus, from her point of view her credence in heads in that case must be less than 1/2, unless the possibility of Tuesday is zero. But it is not zero.
Doesn't this show that her credence must be less than 1/2? Where is the error?
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12-03-2011 , 09:28 PM
Quote:
Originally Posted by TomCowley
1) Mon isn't a proposition.
2) Even pretending it is, you don't get P(mon)=2/3 to begin with unless you make the same kind of reverse appeal to the law of large numbers (2/3 of total awakenings are mon, therefore P(mon)=2/3) that I mentioned in my last post.
I agree that you have to reject 2/3 for your reasons. But when she wakes up, what does she think is the probability that it is Monday? Undefined? Then how can P(heads) = 1/2 or anything else if it depends on an undefined quantity? The answer would have to also be undefined. Are you saying that equation is not valid?
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12-03-2011 , 09:48 PM
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Originally Posted by BruceZ
I agree that you have to reject 2/3 for your reasons. But when she wakes up, what does she think is the probability that it is Monday? Undefined? Then how can P(heads) = 1/2 or anything else if it depends on an undefined quantity? The answer would have to also be undefined. Are you saying that equation is not valid?
OK, I think I know what you will say. "Today is Monday" isn't a proposition because Today isn't a random variable. You can't take a probability of it being equal to something. Therefore that equation

P(heads) = P(heads|Mon)*P(Mon) + P(heads|Tues)*P(Tues)

isn't valid.
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12-03-2011 , 09:57 PM
Quote:
Originally Posted by BruceZ
OK, I think I know what you will say. "Today is Monday" isn't a proposition because Today isn't a random variable. You can't take a probability of it being equal to something. Therefore that equation

P(heads) = P(heads|Mon)*P(Mon) + P(heads|Tues)*P(Tues)

isn't valid.
Nah, I'm not buying it. Why isn't the day of the week a random variable from her point of view? "Today is Monday" is certainly a proposition as much as "The next card to be turned over is a spade". Therefore you can't justify an answer of 1/2 unless you set P(Mon) = 1 which is ridiculous.
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12-03-2011 , 10:17 PM
It's kind of funny. That is basically what I'd say, but it pisses me off to leave it as undefined as well, because it seems like something you should be able to say something about. I've been exploring some ideas in that area. I'm pretty convinced that the laws of probability don't apply to "indexical propositions", but I'm not yet convinced that there's nothing we can say about them.
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12-04-2011 , 01:12 AM
Quote:
Originally Posted by TomCowley
It's kind of funny. That is basically what I'd say, but it pisses me off to leave it as undefined as well, because it seems like something you should be able to say something about. I've been exploring some ideas in that area. I'm pretty convinced that the laws of probability don't apply to "indexical propositions", but I'm not yet convinced that there's nothing we can say about them.
Why is "Today is Monday" from SB's point of view any less of a proposition than than "The next card to be dealt off the top of a deck is a spade"? Both things are only random events because they are unknown, but they both have an actual truth value. I see no difference. So if P(Mon) exists but is unspecified, then the answer must be unspecified too, and it can't be 1/2. It can only be 1/2 if we specify P(Mon) = 1. Jason supports an answer of 1/2, and I don't see how.

Last edited by BruceZ; 12-04-2011 at 01:33 AM.
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12-04-2011 , 01:20 AM
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Originally Posted by ZeeJustin
Each interview is accompanied by a wager. Halfers agree that SB will lose money if she bets 1:1 on heads.
I haven't read this thread in depth, but isn't this the only practical question to be asked? It's quite obvious that her EV is 0 if she bets 2:1 that the coin landed tails.

If you want to say that it the coin has a 2/3 chance of landing tails to her based on that, or a 1/2 based on some other logic, isn't it just some mostly meaningless metaphysical question?
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12-04-2011 , 01:46 AM
Also, P(Mon) must be allowed since

P(Mon or Tues) = 1

according to Jason, and certainly

P(Mon or Tues) = P(Mon) + P(Tues)

He also says that we could have P(Mon | tails). He chooses to leave it unspecified, but the fact that it's allowable means that it is the probability of a valid proposition.
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12-04-2011 , 02:09 AM
If they're allowable, then there's a paradox. No argument. I think jason has updated his views on this since 2007 (he's on record later denying indexical propositions), so I wouldn't categorize him as a believer quite yet.
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12-04-2011 , 02:46 AM
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Originally Posted by PairTheBoard
In the Original the question is, "While IN the experiment, does SB learn something from awaking such that it's possible for her, While IN the experiment, to learn the negation of that thing?"

No. She learns she has awoken within the experiment and,While IN the experiment, she cannot learn that she has not awoken with the experiment.
Lets say SB is awakened once on tails and 0 times on heads. Upon awakening in the experiment, what is her credence that the coin landed tails. If you answer anything other than 1/2, please tell me the information E that she learned that it was possible to "learn the negation of while in the experiment"
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12-04-2011 , 03:31 AM
Quote:
When she awakens:

P(heads) = P(heads|Mon)*P(Mon) + P(heads|Tues)*P(Tues)

= (1/2)*(2/3) + 0*(1/3)

= 1/3.

It would seem you would want to reject P(Mon) = 2/3 for the same reasons you have given. But then what do you replace it with, 1?
Noone mentioned replacing it with 3/4? (and P(heads|Mon) w 2/3)
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12-04-2011 , 04:06 AM
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Originally Posted by Banzai-
Lets say SB is awakened once on tails and 0 times on heads. Upon awakening in the experiment, what is her credence that the coin landed tails. If you answer anything other than 1/2, please tell me the information E that she learned that it was possible to "learn the negation of while in the experiment"
In that case, using jason1990's notation the information of "an awakening if and only if tails" would be part of the experiment's settup information J. So the probability space going in would include P(t|J and an awakening) = 1.

PairTheBoard
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12-04-2011 , 04:16 AM
Quote:
Originally Posted by PairTheBoard
In that case, using jason1990's notation the information of "an awakening if and only if tails" would be part of the experiment's settup information J. So the probability space going in would include P(t|J and an awakening) = 1.

PairTheBoard
But she would still grant the coin a priori a 1/2 chance of being tails, why do you allow for these to differ in this scenario but not the original one?
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12-04-2011 , 04:20 AM
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Originally Posted by Double Ice
Noone mentioned replacing it with 3/4? (and P(heads|Mon) w 2/3)
That's been well discussed. The problem is that the coin could just as well be flipped Monday night as Sunday night. It makes no difference since she is always awakened on Monday anyway.

For SB to form a credence for the Day she must make an assumption, usually with an appeal to the indifference principle. But appeal to the indifference principle is not dictated by the information J of the experiment. So the information J does not compel a credence for the day. In other words, she does not have enough information in J alone to from a credence for the day. So she forms no credence P(M) or P(T) or P(M|t). Of course she does have P(M|h)=1 because that information is available in J.

PairTheBoard
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12-04-2011 , 05:23 AM
Quote:
Originally Posted by ActionJeff
Bostroms paper had 2 possible rebuttals for the betting argument based on long run averages. The first rebuttal is that the betting proposition isn't "fair" or an indication of credence, and then provides a compelling example o this effect. The second rebuttal is based on his solution for the problem that "seemingly violates Bayesian conditionalization" and treats the first experiment and long run averages of the experiment differently because of self-locating information. I can post the transcript later
"Now, we already know from other examples that when the number of bets depends on whether the proposition betted on is true, then the fair betting odds can diverge from the correct credence assignment. For instance, suppose you assign credence 9/10 to the proposition that the trillionth digit in the decimal expansion of π is some number other than 7. A man from the city wants to bet against you: he says he has a gut feeling that the digit is number 7, and he offers you even odds – a dollar for a dollar. Seems fine, but there is a catch: if the digit is number 7, then you will have to repeat exactly the same bet with him one hundred times; otherwise there will just be one bet. If this proviso is specified in the contract, the real bet that is being offered you is one where you get $1 if the digit is not 7 and you lose $100 if it is 7. That you should reject this bet is quite unproblematic and does not in any way undermine your original assessment that the probability of the trillionth digit being 7 is 1/10. A similar situation can arise in a more subtle way. We can construct a scenario where, even though no “catch” is explicitly part of the contract, you nevertheless know that you will be put in a position where you will end up betting a hundred times if you are wrong but only one time if you are right. This could happen e.g. if there is a machine that will determine the correct answer and then, on the basis of what this answer is, will decide whether to repeatedly administer an amnesia drug to you that makes you forget whether you have already betted. The machine could do this in such a way that you end up making a larger number of bets if you are wrong. If you believe that you are facing a situation of this kind, you should take corrective action to limit the distortive effects of the memory erasure on your decision-making. In particular, you may decide to reject bets that seem fair to you and that may have been perfectly acceptable in the absence of the forced irrationality constraint." p19 http://www.anthropic-principle.com/p.../synthesis.pdf
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12-04-2011 , 05:26 AM
Quote:
Originally Posted by PairTheBoard
That's been well discussed. The problem is that the coin could just as well be flipped Monday night as Sunday night. It makes no difference since she is always awakened on Monday anyway.

For SB to form a credence for the Day she must make an assumption, usually with an appeal to the indifference principle. But appeal to the indifference principle is not dictated by the information J of the experiment. So the information J does not compel a credence for the day. In other words, she does not have enough information in J alone to from a credence for the day. So she forms no credence P(M) or P(T) or P(M|t). Of course she does have P(M|h)=1 because that information is available in J.

PairTheBoard
To further expand, in the paper I link above, Bostrom provides a mathematical equation demonstrating that SB's credence of Monday is undefined:
Quote:
Originally Posted by Nick Bostrom p12
"One may still wonder what conditional credence Beauty should assign, before being informed about it being Monday, to HEADS given that she is currently an agent- part that knows that it is Monday:
P(HEADS | H1M ∨ T1M) = ? However, there is no need to assign a value to this expression. Note that:
P(HEADS | H1M ∨ T1M) = P(HEADS & [H1M ∨ T1M]) / P(H1M ∨ T1M)
Since P(H1M ∨ T1M) = 0, this expression is undefined. And so it should be."

Last edited by ActionJeff; 12-04-2011 at 05:32 AM.
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12-04-2011 , 05:36 AM
Quote:
Originally Posted by PairTheBoard
That's been well discussed. The problem is that the coin could just as well be flipped Monday night as Sunday night. It makes no difference since she is always awakened on Monday anyway.

For SB to form a credence for the Day she must make an assumption, usually with an appeal to the indifference principle. But appeal to the indifference principle is not dictated by the information J of the experiment. So the information J does not compel a credence for the day. In other words, she does not have enough information in J alone to from a credence for the day. So she forms no credence P(M) or P(T) or P(M|t). Of course she does have P(M|h)=1 because that information is available in J.

PairTheBoard
The second part is more complicated and I probably agree that she cant form a credence P(whatever). Havent thought about it that much.

But the first part I disagree with, I dont think "the coin could be flipped later" to be a sound rebuttal. I agree that there is no difference to the experiment if the coin is flipped later, but presumably she can tell no difference between monday and tuesday and knows this. This makes it so that when it is monday, the coin is objectively 50/50 but from her viewpoint she can formulate a different probability.

Lets do the following experiment that clarifies my viewpoint: We flip a coin but noone looks at it. We then interview her once in a red room. Then we drug her, look at the coin and interview her in a blue room if the coin is tails. When she wakes up in a red room, I say she should think heads = 2/3, and the reason is that she could have woken up in a blue room and that affects the odds.

Actually now that I reflect on my own views above, I am not confident in them and I am very willing to accept what you mentioned instead... that she cant formulate a P(mon|tails), to be the better answer. Gotta think about it a bit more.
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12-04-2011 , 07:13 AM
Quote:
Originally Posted by ActionJeff
To further expand, in the paper I link above, Bostrom provides a mathematical equation demonstrating that SB's credence of Monday is undefined:
P(Mon|Awake)+P(Tue|Awake)=1

P(Mon|Awake) is undefined.

Is it comparable with the following example?

I have a red or blue car. What is your credence of me driving a red car?
P(Red)+P(Blue)=1
However P(Red) is undefined.
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12-04-2011 , 07:54 AM
Quote:
Originally Posted by PairTheBoard
In that case, using jason1990's notation the information of "an awakening if and only if tails" would be part of the experiment's settup information J. So the probability space going in would include P(t|J and an awakening) = 1.

PairTheBoard
Then it must include P(t|J and (not an awakening)) < 1/2 (in fact its clearly 0), but this contradicts your post I quoted, in that it is not possible to learn "not an awakening" while in the experiment.
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12-04-2011 , 08:03 AM
Quote:
Originally Posted by ActionJeff
"Now, we already know from other examples that when the number of bets depends on whether the proposition betted on is true, then the fair betting odds can diverge from the correct credence assignment. For instance, suppose you assign credence 9/10 to the proposition that the trillionth digit in the decimal expansion of π is some number other than 7. A man from the city wants to bet against you: he says he has a gut feeling that the digit is number 7, and he offers you even odds – a dollar for a dollar. Seems fine, but there is a catch: if the digit is number 7, then you will have to repeat exactly the same bet with him one hundred times; otherwise there will just be one bet. If this proviso is specified in the contract, the real bet that is being offered you is one where you get $1 if the digit is not 7 and you lose $100 if it is 7. That you should reject this bet is quite unproblematic and does not in any way undermine your original assessment that the probability of the trillionth digit being 7 is 1/10. A similar situation can arise in a more subtle way. We can construct a scenario where, even though no “catch” is explicitly part of the contract, you nevertheless know that you will be put in a position where you will end up betting a hundred times if you are wrong but only one time if you are right. This could happen e.g. if there is a machine that will determine the correct answer and then, on the basis of what this answer is, will decide whether to repeatedly administer an amnesia drug to you that makes you forget whether you have already betted. The machine could do this in such a way that you end up making a larger number of bets if you are wrong. If you believe that you are facing a situation of this kind, you should take corrective action to limit the distortive effects of the memory erasure on your decision-making. In particular, you may decide to reject bets that seem fair to you and that may have been perfectly acceptable in the absence of the forced irrationality constraint." p19 http://www.anthropic-principle.com/p.../synthesis.pdf
I dont buy this logic. If she is offered the option to bet at evens heads vs tails every interview, and she uses a randomisation technique to determine whether to accept the bet or not, you can show that the result of the randomisation technique affects her EV by exactly as much as if the coin was weighted 2 vs 1, which cannot possibly be due to the bet being 'the same as being lose 2 or win 1" since the result of the randomisation technique effects only THIS bet.
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