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Posted: Tue Feb 19, 2013 1:13 am
by EdLee
Hi Bill,
Bill Spight wrote:Do you mean for (2) If this stone turns out to be :white:.
In that case you are asking how often you pick the bowls with :white: :white:.
Thanks; yes -- I used the same wording as perceval's original.
Bill Spight wrote:Or do you want to ask, given that the chosen stone is :white:, how often is the other stone :white:? If so, you need to count how often the chosen stone is :white:.

(0) 3 bowls: ( :white: :white: ), ( :black: :black: ), and ( :white: :black: ).
(0a) Set X = 0.
(0b) Set Z = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) If this stone turns out to be :white:, increment Z.
(3a) IF second stone in the bowl is also :white:, increment X.
(4) Go back to (1); repeat this until Z = 1 million. What is the approx. value of X?
Also correct -- it's not me who's asking --
this is precisely why I found the original wording to be confusing:
This stone turns out to be :white:. What is the probability that the second stone in the bowl is also :white: ?
Is the original question asking for (X/N) or (X/Z) ? We know (X/N) = 1/3 as N -> Inf.
We also know (X/Z) = 1/2 as N -> inf. Which is why some people thought 50% to the original question.
Or, maybe the original question is asking for another ratio -- what is it?
And how are we supposed to determine which ratio the original question is asking for ?
Maybe the original wording is 100% unambiguous in the field of probability,
but it's been a few years since school, so I'm probably(*) very much out of it.
This is why I took the little green alien bit (which you replied to). Thanks. :)

(*) Stolen from Tom Stoppard.

Re:

Posted: Tue Feb 19, 2013 11:10 pm
by Bill Spight
EdLee wrote:Hi Bill,
Bill Spight wrote:Do you mean for (2) If this stone turns out to be :white:.
In that case you are asking how often you pick the bowls with :white: :white:.
Thanks; yes -- I used the same wording as perceval's original.
Bill Spight wrote:Or do you want to ask, given that the chosen stone is :white:, how often is the other stone :white:? If so, you need to count how often the chosen stone is :white:.

(0) 3 bowls: ( :white: :white: ), ( :black: :black: ), and ( :white: :black: ).
(0a) Set X = 0.
(0b) Set Z = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) If this stone turns out to be :white:, increment Z.
(3a) IF second stone in the bowl is also :white:, increment X.
(4) Go back to (1); repeat this until Z = 1 million. What is the approx. value of X?
Also correct -- it's not me who's asking --
this is precisely why I found the original wording to be confusing:
This stone turns out to be :white:. What is the probability that the second stone in the bowl is also :white: ?
Is the original question asking for (X/N) or (X/Z) ? We know (X/N) = 1/3 as N -> Inf.
We also know (X/Z) = 1/2 as N -> inf. Which is why some people thought 50% to the original question.
Do we know that? :)

Posted: Wed Feb 20, 2013 6:00 am
by EdLee
Bill, thanks, I'll have to think about (X/Z). But the main thing I was confused about
was whether the original question was asking for (X/Z) (versus X/N). Thanks! :)

Re: well known proba problem

Posted: Wed Feb 27, 2013 3:03 pm
by drmwc
The key issue with the sleeping beauty problem is that SB has no way to differentiate Monday and Tuesday. This leads to some conterinuitive features.

Consider the statement "Given that the coin landed head, what is the probability that today is Tuesday?" Intuitively, the answer is 1/2. However, this leads to problems.

Formally, we have the following statements we wish to be true:

(1) P(Mon or Tue) = 1
(2) P(heads and Tue) = 0
(3) P(heads | Mon or Tue) = 1/2
(4) P(heads | Mon) = 1/2
(5) P(Mon | tails) = 1/2

We know that 1 is true from the set up of the experiment.

Similarly. we know that 2 is also true.

We want 3 to be true because we are told that the coin is fair (and presumably SB verified this before the experiment.) Also, SB gains no information from waking.

We want 4 to be true, again because the coin is fair and SB gains no new information from waking on Monday.

We want 5 to be true for reasons such as the indfference principle.

However, the issue is that 1-5 above are inconsistent - they do not define a valid probability measure. (I've left the proof as an exercise to the reader.)

People who insist that the answer is 1/3 reject hypothesis 3 above, and insist that P(heads | Mon or Tue) = 1/3. This does give a consistent probability measure. However, changing the probabilty of heads when no new information is available is offensive to me, and feels hard to justify.

The neatest resolution is that 5) above has no defined probability. If, at the interview, SB is asked "What is the probability today is Monday?", her reply is: "I do not have sufficient information to answer".

The key issue with the question is the use of the word "today". This means the statement uses an indexical, and its truth value changes dependent on circumstances. Other indexicals include "I", "now", "this" etc.

Probability behaves badly when we attempt to assign probabilities to statements which include indexicals. Finding a replacement for the statement "Today is Monday" without fundamentally changing the problem is difficult - I have not seen it done.

For example, suppose an inspector arrvies either on Monday or Tuesday and attends the interview (if there is one). The day she arrives is chosen at random with probability 1/2, and is independent of the coin flip.

Now the symmetry of Monday and Tuesday collapses, and frequentists and Bayesians would both agree that probability of heads, given the inspector is present at the interview when asked, is 1/3. But this seems to violate the spirit of the oringinal experiment, where SB could not differentiate Monday and Tuesday.

Re: well known proba problem

Posted: Wed Feb 27, 2013 8:02 pm
by Bill Spight
drmwc wrote:The key issue with the sleeping beauty problem is that SB has no way to differentiate Monday and Tuesday. This leads to some conterinuitive features.

Consider the statement "Given that the coin landed head, what is the probability that today is Tuesday?" Intuitively, the answer is 1/2. However, this leads to problems.
Wait a second. Which version is this about? If it is the one in which Sleeping Beauty is awakened on Tuesday if and only if the coin came up tails, then intuitively, the answer is 0.
Formally, we have the following statements we wish to be true:
Starting with what we wish to be true does not always work. ;)

(1) P(Mon or Tue) = 1
(2) P(heads and Tue) = 0
(3) P(heads | Mon or Tue) = 1/2
(4) P(heads | Mon) = 1/2
(5) P(Mon | tails) = 1/2
I suppose you mean the conditions to be what Beauty knows.
We know that 1 is true from the set up of the experiment.

Similarly. we know that 2 is also true.

We want 3 to be true because we are told that the coin is fair (and presumably SB verified this before the experiment.) Also, SB gains no information from waking.
Of course Beauty gains information from waking. The probability that she wakes up on Monday is 1. The probability that she wakes up on Tuesday is 1/2. So when she wakes up, the odds are 2:1 that it is Monday. That's information.

Now, you may say that Beauty already had that figured out on Sunday. But on Sunday the probability that it was Monday was 0. It is easy to go astray with such statements as those about gaining or losing information when waking. The question is what are the conditions of the probabilities. On Sunday, the probability was 1/2 that the coin would come up heads. On Monday and Tuesday, the probability that it had come up heads does not have to be the same as it was on Sunday, because the conditions are different.
We want 4 to be true, again because the coin is fair and SB gains no new information from waking on Monday.
What information Beauty gains from waking on Monday is not the question. The point is that Beauty always wakes up on Monday, regardless of the result of the coin toss. But on awakening, Beauty is not sure that it is Monday. She has to be told in order to get the "Mon" condition for 4.
We want 5 to be true for reasons such as the indfference principle.
That is one reason that probability is empirical. :)
However, the issue is that 1-5 above are inconsistent - they do not define a valid probability measure. (I've left the proof as an exercise to the reader.)

People who insist that the answer is 1/3 reject hypothesis 3 above, and insist that P(heads | Mon or Tue) = 1/3. This does give a consistent probability measure. However, changing the probabilty of heads when no new information is available is offensive to me, and feels hard to justify.
The "no new information" phrase is a trap. The right way to think about these things is in terms of the conditions of the probabilities. True, in one interpretation of "no new information", Sleeping Beauty gains no new information. But the conditions of the probabilities are different, as indicated above. That is what is important.
The neatest resolution is that 5) above has no defined probability. If, at the interview, SB is asked "What is the probability today is Monday?", her reply is: "I do not have sufficient information to answer".

The key issue with the question is the use of the word "today". This means the statement uses an indexical, and its truth value changes dependent on circumstances. Other indexicals include "I", "now", "this" etc.

Probability behaves badly when we attempt to assign probabilities to statements which include indexicals. Finding a replacement for the statement "Today is Monday" without fundamentally changing the problem is difficult - I have not seen it done.

For example, suppose an inspector arrvies either on Monday or Tuesday and attends the interview (if there is one). The day she arrives is chosen at random with probability 1/2, and is independent of the coin flip.

Now the symmetry of Monday and Tuesday collapses, and frequentists and Bayesians would both agree that probability of heads, given the inspector is present at the interview when asked, is 1/3. But this seems to violate the spirit of the oringinal experiment, where SB could not differentiate Monday and Tuesday.
Since Beauty cannot tell the difference between Monday and Tuesday, we should regard them as simply labels for different cases. I don't think that there is a problem with "today", meaning "this case", any more than there is a problem, if we draw two cards from a well shuffled standard deck and leave them face down, of then pointing to one card and asking what is the probability that this card is a Jack.

But Monday and Tuesday are not symmetrical, anyway. Beauty just does not know which is which. :)

----

Here is a variant that restores the symmetry between Monday and Tuesday as far as Beauty's awakening goes, but retains the asymmetry of the original problem. I'll skip the setup. :)

On Monday after Beauty awakes she is asked what is the probability that the coin came up heads. She is given the amnesia drug to make her forget the events of Monday, but she is allowed to awaken on Tuesday.

On Tuesday after Beauty awakes, if the coin came up tails she is asked what is the probability that it came up heads. If it came up heads, she is not asked anything about how the coin came up.

Upon awakening, what probability should Beauty assign to the coin coming up heads? Plainly, 1/2. When asked, what probability should she assign? Plainly -- I trust --, 1/3. ;)

In this variation awakening does not affect her prior belief in the odds that the coin came up heads, but being asked (by the experimenters) does. She is more likely to be asked when the coin came up tails. When Beauty wakes up, Monday and Tuesday are symmetrical. It is the asking that breaks the symmetry. :)

If we eliminate the case in which Beauty wakes up but is not asked about how the coin came up, the symmetry is again destroyed.

Re: well known proba problem

Posted: Wed Feb 27, 2013 8:53 pm
by Bill Spight
drmwc wrote:Formally, we have the following statements we wish to be true:

(1) P(Mon or Tue) = 1
(2) P(heads and Tue) = 0
(3) P(heads | Mon or Tue) = 1/2
(4) P(heads | Mon) = 1/2
(5) P(Mon | tails) = 1/2
(2) - (4) are interesting. (2) implies

(2a) P(heads | Tue) = 0
(4) P(heads | Mon) = 1/2

Now the only way that

(3) P(heads | Mon or Tue) = 1/2

can be true is if it is Monday (and Beauty knows it). Given what Beauty knows, the cases that it is Monday or Tuesday are exhaustive and mutually exclusive. P(Mon) + P(Tue) = 1.

P(heads | Mon or Tue) = P(heads | Mon) * P(Mon) + P(heads | Tue) * P(Tue) , or

1/2 = 1/2 * P(Mon) + 0 * P(Tue) , so

1 = P(Mon)

Since this violates the givens of the problem, (3) is false.

Re: well known proba problem

Posted: Thu Feb 28, 2013 6:50 am
by drmwc
Bill Spight wrote:
Wait a second. Which version is this about? If it is the one in which Sleeping Beauty is awakened on Tuesday if and only if the coin came up tails, then intuitively, the answer is 0.
Whoops, I meant to write heads - it was a typo. As stated, you are clearly correct and the answer is 0.
Bill Spight wrote:
(2) - (4) are interesting. (2) implies

(2a) P(heads | Tue) = 0
(4) P(heads | Mon) = 1/2

Now the only way that

(3) P(heads | Mon or Tue) = 1/2

can be true is if it is Monday (and Beauty knows it). Given what Beauty knows, the cases that it is Monday or Tuesday are exhaustive and mutually exclusive. P(Mon) + P(Tue) = 1.

P(heads | Mon or Tue) = P(heads | Mon) * P(Mon) + P(heads | Tue) * P(Tue) , or

1/2 = 1/2 * P(Mon) + 0 * P(Tue) , so

1 = P(Mon)

Since this violates the givens of the problem, (3) is false.
Suppose we re-formulate the statements without indexicals. One was of doing is this is: "Mon is the event SB was interviewed on Monday; Tue is the event she was interviewed on Tuesday".

Under this approach, P(Mon)=1 and 1-4 from my post hold. There is no contradiction. We run into issues if we try to define Mon event along the lines "This interview is taking place on a Monday", since it's an indexical statement. I have not seen a satisifactory formulation without indexicals that gives 1/3 - it may be possible, but it's certainly not strightforward.

Re: well known proba problem

Posted: Thu Feb 28, 2013 8:48 am
by Bill Spight
drmwc wrote:
Bill Spight wrote:
Wait a second. Which version is this about? If it is the one in which Sleeping Beauty is awakened on Tuesday if and only if the coin came up tails, then intuitively, the answer is 0.
Whoops, I meant to write heads - it was a typo. As stated, you are clearly correct and the answer is 0.
Bill Spight wrote:
(2) - (4) are interesting. (2) implies

(2a) P(heads | Tue) = 0
(4) P(heads | Mon) = 1/2

Now the only way that

(3) P(heads | Mon or Tue) = 1/2

can be true is if it is Monday (and Beauty knows it). Given what Beauty knows, the cases that it is Monday or Tuesday are exhaustive and mutually exclusive. P(Mon) + P(Tue) = 1.

P(heads | Mon or Tue) = P(heads | Mon) * P(Mon) + P(heads | Tue) * P(Tue) , or

1/2 = 1/2 * P(Mon) + 0 * P(Tue) , so

1 = P(Mon)

Since this violates the givens of the problem, (3) is false.
Suppose we re-formulate the statements without indexicals. One was of doing is this is: "Mon is the event SB was interviewed on Monday; Tue is the event she was interviewed on Tuesday".

Under this approach, P(Mon)=1 and 1-4 from my post hold. There is no contradiction. We run into issues if we try to define Mon event along the lines "This interview is taking place on a Monday", since it's an indexical statement. I have not seen a satisifactory formulation without indexicals that gives 1/3 - it may be possible, but it's certainly not strightforward.
1-4 are consistent, but imply that P(Tue) = 0, which means that Beauty was not interviewed on Tuesday.

Re: well known proba problem

Posted: Thu Feb 28, 2013 2:09 pm
by shapenaji
Here's one:


A man is sentenced to death on Sunday, and told that he will be killed in the next 5 days, but as a condition of the sentence, he cannot be aware of the day of his execution.

As the man sits in jail and bemoans his fate, his lawyer comes rushing in with a huge grin on his face.

LAWYER: "I found a way around it, they can't put you to death"

The man raises his eyebrows.

LAWYER: "Suppose it's Thursday, and you haven't been put to death. You know that the last day to execute is Friday, invalidating the sentence. So you can't be executed on Friday. But if you can't be executed on Friday, then if it's wednesday night, and you haven't been executed, then you know you'll have to be executed on Thursday. So Thursday's out too!

If we continue this logic, you can't even be killed tomorrow!"


The man visibly relaxes and sighs deeply.

He is then killed on a Thursday and is very surprised.

Re: well known proba problem

Posted: Thu Feb 28, 2013 8:02 pm
by ez4u
shapenaji wrote:Here's one:


A man is sentenced to death on Sunday, and told that he will be killed in the next 5 days, but as a condition of the sentence, he cannot be aware of the day of his execution.

As the man sits in jail and bemoans his fate, his lawyer comes rushing in with a huge grin on his face.

LAWYER: "I found a way around it, they can't put you to death"

The man raises his eyebrows.

LAWYER: "Suppose it's Thursday, and you haven't been put to death. You know that the last day to execute is Friday, invalidating the sentence. So you can't be executed on Friday. But if you can't be executed on Friday, then if it's wednesday night, and you haven't been executed, then you know you'll have to be executed on Thursday. So Thursday's out too!

If we continue this logic, you can't even be killed tomorrow!"

The man visibly relaxes and sighs deeply.

He is then killed on a Thursday and is very surprised.
So did his lawyer kill him by convincing him that it could not happen? If the prisoner had accepted the logic and convinced himself each day, "They have to kill me today, since tomorrow the chain of logic running to Friday will be even stronger", would he have been safe under the terms of the sentence? (typical lawyer! :blackeye: )

Re: well known proba problem

Posted: Fri Mar 01, 2013 1:42 am
by TheBigH
Hmm, that one's an example of mathematical induction gone wrong.

The problem is that in this problem applying the inductive step invalidates the previous cases. That is, if you know that P(n) is true and use that to prove that P(n+1) is true, then P(n) becomes false.

Re: well known proba problem

Posted: Fri Mar 01, 2013 4:39 am
by drmwc
Bill Spight wrote:
1-4 are consistent, but imply that P(Tue) = 0, which means that Beauty was not interviewed on Tuesday.
I don't see why. Recall that I've re-defined Mon to be the event "SB was interviewed on Monday;" and similarly for Tue. So Tue is identical to the event "tails" from a probability point of view.

Specifically, I can take P(Tue and tails)=1/2 and end up with an indexical-free probability model:

(1) P(Mon or Tue) = 1
(2) P(heads and Tue) = 0
(3) P(heads | Mon or Tue) = 1/2
(4) P(heads | Mon) = 1/2
(5) P(Tue and tails) = 1/2

So P(Tue|tails) = P(Tue and tails)/P(tails) = (1/2)/(1/2) = 1.

And P(Tue) = P(Tue and tails) + P(Tue and heads) = 1/2 + 0 = 1/2.

I believe that this set up is mathematically consistent, and a vaild model for the experiment.

Note that 5) from my original model is no longer true: P(Mon|Tails) = 1.


I have not seen an indexical-free probability model under which P(heads|Mon)=1/3. It may be possible, but it will certainly be non-trivial to construct.

Re: well known proba problem

Posted: Fri Mar 01, 2013 6:31 am
by lightvector
TheBigH wrote:Hmm, that one's an example of mathematical induction gone wrong.

The problem is that in this problem applying the inductive step invalidates the previous cases. That is, if you know that P(n) is true and use that to prove that P(n+1) is true, then P(n) becomes false.
Actually, the heart of the problem is not really that the induction is wrong, it's that even the base case leads to problems. Consider the version where there's only 1 day instead of 5:

The judge tells the prisoner "The day of your execution will be one that you will not be able to know in advance, and it will be tomorrow."

What is the poor prisoner to conclude from this? If he concludes today that he will be executed tomorrow, then that means he knows it advance. But that contradicts the first part of the statement, so the day of his execution cannot be tomorrow. The only thing the prisoner can consistently logically conclude is the judge is lying and that at least one of the two parts of the judge's statement is false, but there is no a-priori reason to think that it should be one or the other - indeed, the judge could indeed be lying about it being tomorrow and execute the prisoner on some other day. So the prisoner cannot, with any logical certainty, conclude that he will be executed tomorrow.

But the judge nonetheless has a way to keep his word. He can have the prisoner executed tomorrow. Since the prisoner will be incapable of deducing this without logical contradiction, all the parts of the judge's statement are truthfully fulfilled. (If you prefer, you can replace "know" in the judge's statement above with "deduce without logical contradiction" to make this clearer).

The 5-day version of the puzzle is basically an elaborate form of the statement "X is true, but person A cannot logically deduce X from this statement without contradicting some part of this statement". And because of the way the statement refers to the state of A's knowledge, indeed person A cannot, even though other people can.

Re: well known proba problem

Posted: Fri Mar 01, 2013 6:58 am
by ez4u
lightvector wrote:...If you prefer, you can replace "know" in the judge's statement above with "deduce without logical contradiction" to make this clearer...
That's really the point, isn't it? Any dictionary will show you that "know" has many more definitions than "deduce without logical contradiction". Indeed the vast majority of what we 'know' in daily life does not meet that standard, right? So you can substitute one statement for the other, but the problem is different after you do.

Re: well known proba problem

Posted: Fri Mar 01, 2013 8:50 am
by Bill Spight
drmwc wrote:
Bill Spight wrote:
1-4 are consistent, but imply that P(Tue) = 0, which means that Beauty was not interviewed on Tuesday.
I don't see why.
Because for each possible interview event, P(Mon) + P(Tue) = 1. So if P(Mon) = 1 for each possible interview event, then P(Tue) = 0 for each possible interview event.
Recall that I've re-defined Mon to be the event "SB was interviewed on Monday;" and similarly for Tue. So Tue is identical to the event "tails" from a probability point of view.
OK, so you are switching to a frequentist model. You still need to examine cases.

Edit: Even if "tails" implies Tue, that does not mean that you can simply substitute tails for Tue in a probability. Or vice versa. Even if they were identical, you could not simply substitute one for the other.
----

BTW, with my variation where Beauty wakes up on both days but is interviewed on Tuesday only if the coin came up tails, do you think that Beauty should have the same belief on Tuesday -- OC, she does not know that it is Tuesday --, whether she is interviewed or not?