well known proba problem

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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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: )
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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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.
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Re: well known proba problem

Post 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?
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Re: well known proba problem

Post by hyperpape »

ez4u wrote:
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.
Good point, but not obviously a solution. The problem doesn't go away if you substitute "believe with justification".

In any case, it may be worth knowing that the paradox of the surprise exam has no known solution. I understand that debate on sleeping beauty is equally unsettled, but don't know enough to know that one side just isn't being silly.
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Re: well known proba problem

Post by ez4u »

hyperpape wrote:
ez4u wrote:
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.
Good point, but not obviously a solution. The problem doesn't go away if you substitute "believe with justification".

In any case, it may be worth knowing that the paradox of the surprise exam has no known solution. I understand that debate on sleeping beauty is equally unsettled, but don't know enough to know that one side just isn't being silly.

From Merrium-Webster online (we can skip definition 3. for now :blackeye: ):
"transitive verb
1
a (1) : to perceive directly : have direct cognition of (2) : to have understanding of <importance of knowing oneself> (3) : to recognize the nature of : discern
b (1) : to recognize as being the same as something previously known (2) : to be acquainted or familiar with (3) : to have experience of
2
a : to be aware of the truth or factuality of : be convinced or certain of
b : to have a practical understanding of <knows how to write>
3
archaic : to have sexual intercourse with"

As we can see 'justification' is also not part of the definition. Indeed, being able to answer the question, "How do you know that?", is not necessary to "know" something in common usage. You have to hijack the word with some special definition, turning it into a piece of logical or philosophical jargon for the problem to make sense.

All that said, the original problem was stated in terms of "be aware of", which is even easier as far as I can see.

"1
archaic : watchful, wary
2
: having or showing realization, perception, or knowledge"

Seemingly as long as the prisoner was the type of person who "lives in the moment" and is aware of the day, the judge would not have been able to execute him under that carelessly worded sentence. :mrgreen:

On second thought, maybe the easy way to carry out the sentence within the common meaning of awareness is to just kill him in his sleep? :batman:
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Re: well known proba problem

Post by hyperpape »

A dictionary exists to give you a rough idea of what a word means--in various cases, it may be arbitrarily precise or not. In this case, I would say the fact that the dictionary

Now, as you point out, justification is tricky. If we think of it as explicitly cognized, articulable reasons for belief, then a lot of things we know, we're not obviously justified in believing. For that reason, philosophers often talk about entitlement: we might be entitled to believe things based on the ordinary functioning of our cognitive capacities, even if we're not capable of articulating a justification for those beliefs. So you're right that justification is probably not the best concept to use. That said, in this case, since it's about what the prisoner can know via reasoning about the date of his execution, justification seems like it's probably ok.*

And nothing changes if we substitute "believe with some entitlement or justification". Arguable, it doesn't even "reasonably believe" in what I said. The last is the most tricky, because it's not at all clear what beliefs are reasonable (in ordinary speech, might we sometimes call an unjustified belief reasonable? Who knows?).

* We've been a little unclear about this, but the interesting versions of the problem are stated in terms of a student/prisoner's ability to know, not whether he actually knows. Not actually knowing is easy, like you said: maybe the prisoner just doesn't think about such things.
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Re: well known proba problem

Post by Bill Spight »

drmwc wrote: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.


This may be clearer. Given "Mon or Tue" and "heads or tails",

P(Mon, heads) + P(Tue, heads) + P(Mon, tails) + P(Tue, tails) = 1

(I have used a comma to indicate "and".)

We also have

P(Mon) = P(Mon, heads) + P(Mon, tails)
P(Tue) = P(Tue, heads) + P(Tue, tails)
P(heads) = P(Mon, heads) + P(Tue, heads)
P(tails) = P(Mon, tails) + P(Tue, tails)

P(Mon) + P(Tue) = 1
P(heads) + P(tails) = 1

According to your model,

(2) P(Tue, heads) = 0

That gives us

P(Tue) = P(Tue, tails)
P(heads) = P(Mon, heads)

According to your model,

(5) P(Tue, tails) = 1/2

That gives us

P(Tue) = 1/2
P(tails) = P(Mon, tails) + 1/2
P(Mon) = 1/2

According to your model,

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

Since "Mon or Tue" is given, that means

P(heads) = 1/2
P(tails) = 1/2
P(Mon, heads) = 1/2
P(Mon, tails) = 0

We can put that in the following table.

Code: Select all

          Mon  Tue  Σ
   heads  1/2   0  1/2
   tails   0   1/2 1/2
     Σ    1/2  1/2  1


But P(Mon, tails) <> 0, as we know. The coin could come up tails and Beauty be interviewed on Monday.

By the givens of the problem, P(Tue, heads) = 0. :) And, P(heads) = 1/2 is your contention, so we retain that. :) That means that (5) has to go. That leaves us with this table.

Code: Select all

          Mon  Tue  Σ
   heads  1/2   0  1/2
   tails   p    q  1/2
     Σ     r    q   1


By your model,

(4) P(heads | Mon) = 1/2

That means that P(Mon, heads) = P(Mon)/2.

And that gives us the following table.

Code: Select all

          Mon  Tue  Σ
   heads  1/2   0  1/2
   tails  1/2   0  1/2
     Σ     1    0   1


That's why I said that (1) - (4) imply that P(Tue) = 0. :)
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Re: well known proba problem

Post by ez4u »

hyperpape wrote:...

* We've been a little unclear about this, but the interesting versions of the problem are stated in terms of a student/prisoner's ability to know, not whether he actually knows. Not actually knowing is easy, like you said: maybe the prisoner just doesn't think about such things.

I suspect (but do not 'know') that we do not agree on what are the interesting versions of the problem. I will at this point leave the contemplation of the good bits to the philosophers. :salute:
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