Probability and Death Sentences

[bahamagreen]

As long as the story is about suspects, a murder, a trial, and a sentence, one must assume that the verdict is not random, but based to some degree on the facts of real guilt or innocence of the suspects. If the self knowledge of guilt or innocence was not meant to be part of the puzzle, it would not be set as a legal proceeding, but rather a lottery.

This puzzle needs to have an added condition for the usual analyses to work; either the statistician suspect finds himself in this situation with amnesia, or the whole story is changed so not to include a crime and judgement scenario... something more like three strangers are abducted and subject to a homicidal lottery.

The statistician subject in the present story knows something very important that I have not seen anyone mention (unless I missed it)... he knows for a certainty whether he himself in fact did or did not commit the murder.

 

[sysprog]

As long as the story is about suspects, a murder, a trial, and a sentence, one must assume that the verdict is not random, but based to some degree on the facts of real guilt or innocence of the suspects. If the self knowledge of guilt or innocence was not meant to be part of the puzzle, it would not be set as a legal proceeding, but rather a lottery

This puzzle needs to have an added condition for the usual analyses to work; either the statistician suspect finds himself in this situation with amnesia, or the whole story is changed so not to include a crime and judgement scenario... something edlike three strangers are abducted and subject to a homicidal lottery.

The statistician subject in the present story knows something very important that I have not seen anyone mention (unless I missed it)... he knows for a certainty whether he himself in fact did or did not commit the murder.

.From the Wikepedia version of this problem:

Three prisoners, A, B and C, are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned.​

The problem as stated in this thread is silent on the actual guilt of any of the prisoners, but clearly states that 2 have been found guilty, and tacitly implies, by not saying otherwise, that 1 has not. Regardless of actual guilt, the prisoners do not know which of them has been found guilty and which not, so the starting conditions for each prisoner are 2/3 chance found guilty, and 1/3 not found guilty.

 

[bahamagreen]

I still think that the choice of the situation being a legal verdict suggests that self known guilt or innocence is intended to be an important part of the puzzle.

The Wiki version reflects a different situation because the governor's selection is random, unlike the deliberation of guilt and sentencing by a court. If the statistician suspect did not commit the murder, upon hearing that only two of the three suspects were found guilty, he would be extremely relieved, the best interpretation for his knowledge being good news - that his innocence was confirmed.

 

[sysprog]

I still think that the choice of the situation being a legal verdict suggests that self known guilt or innocence is intended to be an important part of the puzzle.

The Wiki version reflects a different situation because the governor's selection is random, unlike the deliberation of guilt and sentencing by a court. If the statistician suspect did not commit the murder, upon hearing that only two of the three suspects were found guilty, he would be extremely relieved, the best interpretation for his knowledge being good news - that his innocence was confirmed.

The problem as stated in this thread also contains the peculiar condition that of 3 persons, 2 are found guilty, but the findings are concealed from all 3. That's not what happens in a real court. I think the problem as stated in this thread is probably just a mis-recounting of the problem in the Wikipedia article or of a similar problem

 

[JeffJo]

I meant in the second of those three sentences to repeat the reference to the other prisoner in the problem as stated in this thread, ...

In whatever terms and/or references you can use, the statistician, Prisoner C, and the reader have the same knowledge of the information and choices in the problem. When I said that prisoner C was the personification of the reader, I meant that they have the same motivation - is the statistician right, or wrong?

In the problem as stated in this thread, there is no other prisoner who becomes privy to the new knowledge of the inquirer. You can't legitimately put the remaining non-statistician prisoner in the same knowledge position as that of prisoner C, or of us. He has not been told of what the guard has revealed, as prisoner C has, and as we have.

In the problem as stated in this thread, the reader becomes privy to the new knowledge of the statistician. I can put the statistician, the reader, and Prisoner C if mentioned in a superficially-different version, in the same knowledge position. Because they all, well, have the exact same knowledge. I never mentioned "the remaining non-statistician prisoner" - by whom I think you mean Prisoner B - in anything.

An expanded summery of what I have said - and you won't address - is that the problems represent the different real-world manifestations of the same underlying probability space. The minute differences in the presentation affects only how we might phrase an answer to address the explicit question, or the question we infer. The consequences of the outcomes in the real-world manifestations are irrelevant to how we address the problem.

 

[sysprog]

I never mentioned "the remaining non-statistician prisoner" - by whom I think you mean Prisoner B - in anything.

I don't mean prisoner B, who in the Wikipedia version of the problem is pointed out as guilty. I mean that prisoner who, in the problem as stated in this thread, is the one the two who is not the statistician, and who, of those two, is the one not pointed ot as guilty.

That prisoner corresponds to the Wikipedia problem's prisoner C, except that that the statistician prisoner in the problem as stated in this thread, does secretly not tell that prisoner the news, as A does to C.

I said that A telling C the news in the Wikipedia version was pivotal, and that unlike in the Wikipedia version, in the problem as stated in this thread, no prisoner other than the statistician is privy to the pointing out of one of the prisoners as guilty.

You said that doesn't matter because we know, and C is merely the personification of us.

But there is no such personification in the problem as stated in this thread, because the other prisoner (not the one pointed out as guilty and not the inquiring prisoner) does not have our knowledge, as C does in the Wikipedia version.

Consequently the problem as stated in this thread is not equivalent to the version of it in the Wikipedia article, and is not equivalent to the Monty Hall problem, as the Wikipedia version is.

 

[JeffJo]

I don't mean prisoner B, who in the Wikipedia version of the problem is pointed out as guilty. I mean that prisoner who, in the problem as stated in this thread, is the one the two who is not the statistician, and who, of those two, is the one not pointed out as guilty.

Yet what you said compared "the remaining non-statistician prisoner" and "prisoner C" in a way that precluded them being the same person ("can't put them in the same knowledge position"). That leaves only B. If you are going to nitpick the phrasing at such a pedantic level, you should really take more care to be accurate.

You said that doesn't matter because we know, and C is merely the personification of us.

No, you said that including C made the TTP different than the MHP because the statistician was unconcerned with whether his chances improved. For the equivalent of prisoner C in the MHP - the switch door - we do care. So prisoner C is the personification of the fact that we care that a door's probability improves, in the 100% equivalent Three Prisoner's Problem.

The problems represent the different real-world manifestations of the same underlying probability space. The minute differences in the presentation affects only how we might phrase an answer to address the explicit question (when there is one), or the question we infer (as is the case in the OP). The consequences of the outcomes in the real-world manifestations are irrelevant to how we address the problem.

 

[sysprog]

Yet what you said compared "the remaining non-statistician prisoner" and "prisoner C" in a way that precluded them being the same person ("can't put them in the same knowledge position"). That leaves only B. ...

Although the remaining non-statistician prisoner and prisoner C are not in the same knowledge position, they are otherwise in the same role and position: each of them is the not-inquiring not pointed out prisoner.

The fact that they are not in the same knowledge position is why the two versions of the 3 Prisoners problem are different from each other. The sole difference between them is that in the version of the problem as stated in this thread, the statistician does not tell the non-inquiring not pointed out prisoner the new knowledge, whereas in the Wikipedia version, prisoner A tells prisoner C, who is in that version the non-inquiring not pointed out prisoner, about the pointing out of B.

You said that doesn't matter because we know, and C is merely the personification of us.

No, you said that including C made the TTP different than the MHP because the statistician was unconcerned with whether his chances improved.

I didn't say that. I did say that the statistician can't swap verdicts, unlike the contestant, who can swap doors. So the problem as stated in this thread is not equivalent to the Monty Hall problem.

For the equivalent of prisoner C in the MHP - the switch door - we do care. So prisoner C is the personification of the fact that we care that a door's probability improves, in the 100% equivalent Three Prisoner's Problem.

Prisoner C, the non-inquiring not pointed out prisoner, being told by A, the inquiring prisoner, maps to the contestant having an option to switch doors in the Monty Hall problem. Those problems are equivalent. But because in the version of the three prisoners problem stated in this thread, the inquirer does not tell the non-inquiring not pointed out prisoner, the two versions of the 3 Prisoners problem are different: the version of the problem in this thread is not equivalent to the Monty Hall problem, and the version in the Wikipedia article is equivalent to the Monty Hall problem. And you can't legitimately drag our knowledge into the first problem to make it also equivalent to the Monty Hall problem.

 

[JeffJo]

The fact that they are not in the same knowledge position is why the two versions of the 3 Prisoners problem are different from each other.

And what I had said was that the reader is in the knowledge position that you attribute to Prisoner C. So the role of Prisoner C, as you describe him to be different than the non-pointed-to, non-statistician prisoner in the OP, is to have a person in the problem ("personify") with the reader's information state. Since such a person exists in the OP (the reader), and it is implied that that person (the reader) should examine the information the same way Prisoner C did, this is not a difference in the problems.

I didn't say that. I did say that the statistician can't swap verdicts, unlike the contestant, who can swap doors. So the problem as stated in this thread is not equivalent to the Monty Hall problem.

And I told you that the what the participants do with, or why they care about, the answers to the problem (which is to determine a valid probability space, which must be complete to be valid, regardless of what part they are most interested in) comes after the problem is solved. So it is superficial.

And you can't legitimately drag our knowledge into the first problem to make it also equivalent to the Monty Hall problem.

I can "drag" our knowledge of any problem into the solution of that problem; it's kinda the point.

The problems represent the different real-world manifestations of the same underlying probability space. The minute differences in the presentation affects only how we might phrase an answer to address the explicit question (when there is one), or the question we infer (as is the case in the OP). The consequences of the outcomes in the real-world manifestations are irrelevant to how we address the problem.​

 

[sysprog]

So the role of Prisoner C, as you describe him to be different than the non-pointed-to, non-statistician prisoner in the OP, is to have a person in the problem ("personify") with the reader's information state.

Keeping it within the problem as stated, it would be: with the "inquiring prisoner's" knowledge state; not the "reader's" knowledge sate.

Since such a person exists in the OP (the reader),

The reader is not in the problem as stated in the OP. problem. In the OP, no such person exists. That's why the problems are different.

and it is implied that that person (the reader) should examine the information the same way Prisoner C did, this is not a difference in the problems.

In the OP, you are relying on the knowledge of the reader to make it equivalent to the Monty Hall problem. In the Wikipedia version, you don't have to do that. That is how the two problems are different, and why it is pivotal, and not irrelevant, that in the Wikipedia version, A tells C the news.

And I told you that the what the participants do with, or why they care about, the answers to the problem

They don't care about our answers to the problem as we encounter it -- they don't know about us.

(which is to determine a valid probability space, which must be complete to be valid, regardless of what part they are most interested in) comes after the problem is solved. So it is superficial.

You're commingling the point of view of the subjects with our point of view. The problems as stated are different. In the problem as stated in this thread, the non-inquirer who is not pointed out as guilty is not told by the inquirer about the pointing out, so he doesn't have our knowledge, whereas in the Wikipedia version of the problem, prisoner C is told by prisoner A about the pointing out of B, and only because of that, his knowledge becomes the same as that of the reader. The two problems are consequently different in that regard.

I can "drag" our knowledge of any problem into the solution of that problem; it's kinda the point.

You can't legitimately put us in the first problem as a subject to hold the knowledge position of prisoner C in the second problem. As the recipients of the problem, we are ineligible to be considered as within it. The prisoners and the contestant are not faced with the problem, as we are, they are faced with their own predicaments. The problems as stated seek answers dependent only on the subjective knowledge within the problem. The difference between the one subject who knows what we know in the first problem, and the two subjects who know what we know in the second, is what makes the first problem not equivalent to the Monty Hall problem, and the second problem equivalent to the Monty Hall problem.

In your formulations, you named prisoner C of the second problem, as equivalent to the second door in the Monty Hall problem, and considering the second problem as having two subjects about whose updated knowledge and statements we are asked, that is a valid correspondence.

Just as the second subject in the first problem has the new knowledge, the contestant in the Monty Hall problem can by switching doors effectively consult the knowledge of the second door (considering "subject" functionally to mean "repositor(y) of knowledge" within the problem) to improve his own chances, which makes that door equivalent in that regard to prisoner C in the second problem. But in the first problem, there is only one subject whose post-revelation statement we are (implicitly or explicitly (in this case, implicitly, and in the second problem case, explicitly) asked to evaluate, and he can't switch verdicts, whereas in the Monty Hall problem, there is prima facie only one subject, but that subject can switch doors. So the two 3 Prisoners problems are not the same; the second is equivalent to the Monty Hall problem, and the first is not.

 

[sysprog]

... Just as a thought, there is an interesting angle to this in terms of generalising mathematical arguments.

The specific problem in front of us has only a few possibilities and we can crank through them all, as @JeffJo has done.

But, what if we generalise the problem: $n$ prisoners, of whom $m$ are to be executed.

Now, it's not so easy to dismiss a more general, logical argument in favour of a painstaking enumeration of all the options.

This succinct explanatory answer on stackexchange shows the results of a similar extension of the Monty Hall problem.

(That page includes this xkcd cartoon:)

"A few minutes later, the goat from behind door C drives away in the car."
(n.b. the goat would probably eat more shrubbery than grass)

 

[JeffJo]

The reader is not in the problem as stated in the OP. problem.

If a tree falls...

The reader is, presumably, asked to assess the correctness of the statistician's solution in the OP, in the same way prisoner C does in Wikipedias's rendition of the same problem. The Wikipedia version just inserts that role into the story. See "personification."

And regardless, it only affects the superficial point you brought up: that nobody in the OP, except the statistician, cares if the statistician was right. Yet you said the reader is supposed to care. See "personification."

In the OP, no such person exists. That's why the problems are different.

You said the reader is supposed to make the determination of whether the statistician is correct. You even claimed it was a different problem if the the question was "is he right?" or "is he wrong?" So such a person does exist, just not as a character in the story. See "superficial."

But this is becoming ridiculous. I have defined the conditions that make the problems equivalent, and you continue to ignore that definition in order to emphasize the window dressing devised to make the problem into a story problem.

The problems represent the different real-world manifestations of the same underlying probability space. The minute differences in the presentation affects only how we might phrase an answer to address the explicit question (when there is one), or the question we infer (as is the case in the OP). The consequences of the outcomes in the real-world manifestations are irrelevant to how we address the problem.​

What this means is that the same problem is described by a different story in each version.

+++++

But, what if we generalise the problem: $n$ prisoners, of whom $m$ are to be executed.

Now, it's not so easy to dismiss a more general, logical argument in favour of a painstaking enumeration of all the options.

My point at the beginning, before sysprog's absurd digression, was that what I think you call a "more general, logical argument" is an incorrect, non-mathematical approach that just happens to get the right answer.

In the following, I'm ignoring sysprog's kind of reasoning (where "two have guilty verdicts and one (at random) has a not-guilty verdict" and "three are convicted but on (at random) has a pardon", and asking an additional question, make it a different problem). Since the Wikipedia version is also the same problem, I'll use it to compare these solutions:

1. Prisoner A now knows that there are only two prisoners who could receive the pardon. Since each was equally likely to receive it in the first place, each now has a 1/2 probability.
2. Prisoner A didn't receive any information that could make a change, so his initial probability of 1/3 is unaffected.

Both of these make the same mistake, of not recognizing that "new information" affects not just one probability but the set as a whole. In the first it is one mistake - using an insufficient, but otherwise correct event partition {A pardon, B pardon, C pardon} with probabilities {1/3,1/3,1/3}. It then correctly set the probability of "B pardon" to zero and correctly normalizes what remains, {1/3,1/3,0} to {1/2,1/2,0}.

The sufficient partition is {A pardon point to B, A pardon point to C, B pardon point to C, C pardon point to B} with probabilities {1/6,1/6,1/3,1/3}. The same, correct procedure now removes the two events with "point to C", leaving {1/6,0,1/3,0} which normalizes to {1/3,0,2/3,0}.

The second not only makes the incorrect claim of no new information, it also claims that it cannot affect his probability. The reason many people don't understand problems like this (as opposed to confusing the problem with the story, as sysprog does), is because the information does affect prisoner A's chances. But in a way that returns it to the value it had before.

+++++

But I usually take your generalization one step further. All of these problems, and one more where many "experts" accept the statistician's logic, are variations of what I call the Generalized Bertrand's Box Problem. I'll point out that an odd transformation, that is perfectly valid but I'm fairly certain sysprog won't accept because it essentially uses different names for the cases, is necessary to apply it to Monty Hall and Three Prisoners. So the some of the numbers are reversed.

1. There are N boxes. Each box contains some coins. M<N/2 contain only gold coins, another M contain only silver coins, and the remaining N-2M have a mix of gold and silver coins. (In the MHP and TTP, and in Bertrand's actual story problem, N=3 and M=1).
   
2. A box is chosen at random. The probability that it has only one kind of coin is 2M/N.
3. I look in the box, and pull out a gold coin to show you. Now what are the chances the box has only gold coins?
   1. Some will say M/(N-M) since there are M only-gold boxes and N-M some-gold boxes.
   2. Bertrand argued that if M/(N-M) were correct, then it would also be correct if I had shown you a silver coin and asked the same question about silver coins. And if it is correct no matter what kind of coin I pull out, then I don't have to show you the coin I pulled out. You can claim the answer changes from 2M/N to M/(N-M) simply because I took one out. But that's a paradox.
      
   3. Since any answer other than 2M/N produces this paradox, it must still be 2M/N. This is actually the correct version of the more general, logical argument you mentioned.
   4. Note: many call the problem Bertrand's Box Paradox, but the paradox Bertrand referenced was the argument I just used.
4. The correct solution is that for M boxes, there was a 100% chance I'd pull out a gold coin. In N-2M, there was only a 50% chance. So the answer is [M]/[M+(N-2M)/2] = M/[M+N/2-M] = 2M/N.

Note: In the Three Prisoners, "Gold" means A is in a different state than C, and "silver" means A is in a different state than B. A is both Gold and Silver if he is to be pardoned. By pointing to B, the guard tells A that he is "gold," but he wants to be "silver" also. His probability, both before and after the pointing, is (N-2M)/N=1/3.

The other problem that this applies to is:

You know that a woman has two children, and that at least one of them is a boy. What are the chances that both are boys?​

Here, N=4 and M=1. Most books you see a problem like this in will say the answer is 1/3. Its author, Martin Gardner, said it was ambiguous because it doesn't tell you, or imply, why you know that one is a boy. My point is that any answer other than 1/2 leads to Bertrand's Box Paradox, and so is unacceptable. And that many versions, like "I have two children including at least one boy," do imply a 50/50 choice between my mentioning a boy instead of his sister, just like the guard pointing to Dick instead of Harry. So the answer is 1/2.

 

[PeroK]

 

[sysprog]

The reader is, presumably, asked to assess the correctness of the statistician's solution in the OP, in the same way prisoner C does in Wikipedias's rendition of the same problem.

Prisoner C doesn't present all that analysis. He presumably could, but he doesn't have to. He knows that he could have been pointed out, but wasn't.

The Wikipedia version ... inserts that role into the story.

and the version as stated in the thread does not, wherefore the problems are different.

And regardless, it only affects the superficial point you brought up: that nobody in the OP, except the statistician, cares if the statistician was right. Yet you said the reader is supposed to care. See "personification."

That's not what I said.

You said the reader is supposed to make the determination of whether the statistician is correct.

That's close enough to something I said.

You even claimed it was a different problem if the the question was "is he right?" or "is he wrong?"

No, I didn't.

So such a person does exist, just not as a character in the story.

No such person exists as a subject in the problem as stated in this thread, whereas in the problem as stated in Wikipedia, there is such a subject: Prisoner C.

In the problem as stated in this thread, the only legitimate candidate for a subject that corresponds to prisoner C in the Wikipedia version, is the non-inquiring not pointed out prisoner, and he does not, in terms of his knowledge, correspond to prisoner C, because unlike prisoner C, he has not been told by the inquirer of the pointing out.

You can't legitimately drag in the reader, who is not a subject in the problems, to take the place of a subject in the problem. Our knowing that a fact is known to two subjects in the second problem doesn't make the second problem equivalent to the first problem, in which we know there to be only one subject who knows that fact.

But this is becoming ridiculous.

It's ridiculous to take someone to task merely for pointing out a difference as a difference after you incorrectly pronounced two different problems to be the same.

Even after repeated precise identification of the difference and of its consequences, you persist in saying that two problems that are different are the same. You misstate and mischaracterize what your opposition says. You present as equivalent a problem that in most aspects, but not all, is equivalent, and then dismiss the inequivalency as superficial, as if it were at the level of superficiality of mere names, e.g. calling the inquirer in the first version the statistician, and calling the inquirer in the second version prisoner A.

You drag in a viewpoint (ours) that's not part of the problem, and pretend that the existence of that external viewpoint is a valid counterpart for a point of view that is internal to the problem. You try to excuse that by pointing out that in the second problem, prisoner C is in the same knowledge position, and that in the first problem, the statistician is in the same knowledge position. That's obviously not parallel.

I have defined the conditions that make the problems equivalent,

You are attempting to do that by decree. The problems already have their own pre-stated conditions. You can't legitimately arrogate to yourself the fiat to rule out a stated condition as unimportant in order to make yourself right and someone else wrong. The problems as originally stated are not the same.

and you continue to ignore that definition in order to emphasize the window dressing devised to make the problem into a story problem.

What I originally said was that the problem as stated in this thread was not equivalent to the Monty Hall problem, because the statistician doesn't have an option to swap verdicts, as the contestant has an option to switch doors.

You said that the 2/3 chance of the other non-inquiring not pointed out prisoner corresponded to the 2/3 chance of the not-opened door. I agreed, but only because in the second version of the 3 prisoners problem, the non-inquiring not pointed out prisoner -- prisoner C -- is told by the inquirer about the pointing out of the other non-inquiring prisoner. I said that the first version of the 3 prisoners problem was not the same as the Monty Hall problem, because unlike in the second problem, in which prisoner C being told mapped to the contestant being given an option to switch doors, there.was not only no such option; there was also no such informing to take its place.

You couldn't resort to the knowledge of prisoner A's first version counterpart, the statistician, because in that problem, the inquirer didn't tell the other non-inquiring not-pointed out prisoner the news. You then wound up resorting to the reader as the counterpart to prisoner C, because he too has the knowledge that prisoner A has. The problem with that, is that just like the statistician, we didn't tell the other non-inquiring not pointed out prisoner either.

The problems represent the different real-world manifestations of the same underlying probability space. The minute differences in the presentation affects only how we might phrase an answer to address the explicit question (when there is one), or the question we infer (as is the case in the OP). The consequences of the outcomes in the real-world manifestations are irrelevant to how we address the problem.​

What this means is that the same problem is described by a different story in each version.

It doesn't mean that, and even if it did, that wouldn't change the fact that that the two problems are different. Even if it be granted that from an external objective point of view, the probability spaces are the same, the problem in each case is to evaluate the statements of the subjects, and the result for the one subject in the first version does not match up with the results for the two subjects in the second version, there being a second subject in the second version, only because there is an added condition in the second version, that along with the requirement to evaluate the second subject's appraisal of the post-pointing out chances of the two subjects, makes the two problems different.

+++++

My point at the beginning, before sysprog's absurd digression,

No other person has ownership of your digression from your point. Putting that label on my disagreement with your assessment, doesn't make me wrong, any more than any of your other failed attempts does.

was that what I think you call a "more general, logical argument" is an incorrect, non-mathematical approach that just happens to get the right answer.

I think PeroK did as I did. He looked around for a prima facie reason to suppose that the statistician was right to think his chances had improved, and correctly seeing none, correctly concluded that the statistician still had the same chances as before.

In the following, I'm ignoring sysprog's kind of reasoning

You can ignore it, but you haven't refuted it.

(where "two have guilty verdicts and one (at random) has a not-guilty verdict" and "three are convicted but on (at random) has a pardon", and asking an additional question, make it a different problem).

That's yet another misstatement of what I said. You deliberately left out the distinction that in this thread's version of the 3 prisoners problem, the non-inquiring not pointed out prisoner is not told of the news about the pointing out of the other non-inquiring prisoner as guilty, while in the other version, prisoner A tells prisoner C the news. You say that's irrelevant, and I say it's pivotal. Pretending I said that something else was alone enough to make the two problems different, is not even a good faith effort at honest exposition.

Since the Wikipedia version is also the same problem,

It isn't the same problem, and saying "also" should means with something else. Presumably that should mean that you have just shown the 3 prisoners problem as stated in this thread to be equivalent to the Monty Hall problem, but you haven't, so I'll disregarding the bolded "also", and take this re-invocation of the Wikipedia version to mean that's the problem you'd like to discuss.

I'll use it to compare these solutions:

1. Prisoner A now knows that there are only two prisoners who could receive the pardon. Since each was equally likely to receive it in the first place, each now has a 1/2 probability.
2. Prisoner A didn't receive any information that could make a change, so his initial probability of 1/3 is unaffected.

1. 1. is incorrect, and 2. is correct.

   Both of these make the same mistake, of not recognizing that "new information" affects not just one probability but the set as a whole.

   1. "makes a mistake": the second sentence of 1. is a non sequitur. 2. does not evince any mistake.

   In the first it is one mistake - using an insufficient, but otherwise correct event partition {A pardon, B pardon, C pardon} with probabilities {1/3,1/3,1/3}. It then correctly set the probability of "B pardon" to zero and correctly normalizes what remains, {1/3,1/3,0} to {1/2,1/2,0}.

   If you "correctly normalize" from an "insufficient but otherwise correct" partition, because if it is insufficient, it is ipso facto incorrectly defined, wherefore a normalization predicated thereupon is of no value. There was no need in 2. to do a partitioning and normalization, because there was no prima facie indication that any of the new information could affect the chances for A.

The sufficient partition is {A pardon point to B, A pardon point to C, B pardon point to C, C pardon point to B} with probabilities {1/6,1/6,1/3,1/3}. The same, correct procedure now removes the two events with "point to C", leaving {1/6,0,1/3,0} which normalizes to {1/3,0,2/3,0}.

That is a not-incorrect showing of what is wrong with 1. It has no necessary bearing on 2.

The second not only makes the incorrect claim of no new information,

2. does not make the claim of no new information. In your statement of 2. you refer to "information that could make a change"; not to "new information" per se.

it also claims that it cannot affect his probability.

It correctly states that A does not receive "any new information that could make a change" (for A).

According to your statement of it, 2. says:

2. Prisoner A didn't receive any information that could make a change, so his initial probability of 1/3 is unaffected.
​

The new information would make a difference for A if and only if A could swap positions with C. Response 2. correctly observes that the new information cannot change anything for A, and correctly infers that his initial probability of 1/3 is unaffected.

The reason many people don't understand problems like this ... is because the information does affect prisoner A's chances. But in a way that returns it to the value it had before.

A single informational event is not a process that can change something and then change it back. Your subsequent analysis is a process, only inside of which A's chance changes and changes back. The process from input to output produces no external change for A's chance. The condition of A's chance before, during, and after the informational event remain exactly the same. The analysis you present is sufficient for recognizing that, but not necessary for recognizing it. It can be recognized without any such analysis.

If you add the condition that A tells C what the guard did, and ask also about C's updated chances and his new estimation thereof, as the Wikipedia version of the problem, which version you are at this juncture purporting to be referencing does, then and only then is some further analysis necessary, because the added condition that C is told the news, has changed the impact of the event, not on the chances themselves, but on C's ability to recognize them. He doesn't need to do an exhaustive anylysis, but he does need to recognize that he could have been pointed out but wasn't. Whether his estimation of his new chance as having improved to 2/3 while the chance of A remains the same 1/3 it was to begin with is correct, is part of what the Wikipedia version asks, that the problem as stated in this thread does not ask. Along with C being told the news, that part makes the two problems different.

(as opposed to confusing the problem with the story, as sysprog does)

I didn't confuse anything.
+++++

But I usually take your generalization one step further. All of these problems, and one more where many "experts" accept the statistician's logic, are variations of what I call the Generalized Bertrand's Box Problem. I'll point out that an odd transformation,

that is perfectly valid but I'm fairly certain sysprog won't accept because it essentially uses different names for the cases,

is necessary to apply it to Monty Hall and Three Prisoners.[/quote]You have no good reason to toss in this misleading jibe. I already expressly acknowledged the Wikipedia version of the 3 prisoners problem to be equivalent to the Monty Hall problem, given that after A has told C the news, C's new information, and the contestant's option to switch doors, make C mappable to the other unopened door.

 

[JeffJo]

That's not what I said.

I'm really getting tired of your insistence that any different way to ask about the results of the same analysis, makes the analysis different. While each one of the participants may be more interested in one part of the problem, in order to apply it to their part of the story, each should understand a complete analysis (correct or incorrect) in order to believe it is correct. Prisoner A/the statistician gets it wrong - but even if he we don't see that he thinks not-pointed-too/Prisoner C's probability is 1/2, he does. Not including that in the story does not make the problem different. That's why any mathematician you ask, except you, will say the problems themselves are the same. In his book about the MHP, Jeffrey Rosenhouse doesn't even mention which specifics he thinks are asked for in the TPP, he just says it is the predecessor of the MHP. Which is all I "originally brought up," and keep getting "taken to task" for.

You can't legitimately drag in the reader, who is not a subject in the problems,...

If you insist that which specifics are asked for makes the problems "different," then I most certainly can "drag in" anybody who is asked about specifics. Including the reader. And please recognize that in the OP, nothing was asked for. Get that? THERE WAS NO EXPLICIT QUESTION. So there is no "problem" to say is the same, or is different, unless you infer a question. And any of the question you say make the problems "different" can be inferred this way, not just the one you choose to say is the original problem.

Even after repeated precise identification of the difference and of its consequences, ...

You mean just like how you ignore how I have given repeated and precise descriptions of why the problem underlying the stories is the same? And similar repeated and precise descriptions for what the consequences are relevant only to the story, not the problem itself?

You drag in a viewpoint (ours) that's not part of the problem,...

So, you are saying that what we are asked for is not a part of "the problem?"

What I originally said was that the problem as stated in this thread was not equivalent to the Monty Hall problem, ...

And all I am saying, and have said over and over, is that any question that can be asked, about any probability in any version of either question, has an exact counterpart in all of them. This is true whether or not they are asked explicitly, implicitly, or seem unconnected to the fate of the character you choose to isolate from the others for some reason. In fact, their fates are relevant only to the story, not the problem itself.

[/quote]the statistician doesn't have an option to swap verdicts, as the contestant has an option to switch doors.[/quote]And how does that affect how I determine whether it is advantageous? Or whether the statistician's chances have changed? HOW DOES SWITCHING MAKE IT A DIFFERENT PROBLEM, instead of just a different consequence of the solution to the problem?

The problems represent the different real-world manifestations of the same underlying probability space. The minute differences in the presentation affects only how we might phrase an answer to address the explicit question (when there is one), or the question we infer (as is the case in the OP). The consequences of the outcomes in the real-world manifestations are irrelevant to how we address the problem.​

What you keep ignoring is that the story is just a vehicle for what I call the problem. The problem is what question is asked of us. The story can include questions asked of the characters and of us. But which questions are asked of the characters, and how they are affected in the story, is irrelevant to the problem asked of us. The reason you are wrong to call it a "different problem," when all that differs is parts of the story, is because part of what you say is part of the problem in the OP - what question we are supposed to answer - is not a part of the story at all.

 

[sysprog]

As an expedient, in this post, I'll call the three prisoners problem as stated in this thread, 3P1, and the non-equivalent three prisoners problem as stated in Wikipedia, 3P2, and similarly, I'll call the 3P1 counterparts of A, B, and C in the 3P2 problem, A1, B1, and C1, and I'll use MH to designate the Monty Hall problem, which problem is equivalent to 3P2 and not to 3P1, and MHC, to designate the contestant therein.

I'm really getting tired

Please feel free to take a nap.

of your insistence that any different way to ask about the results of the same analysis, makes the analysis different.

That's not what I'm insisting on. I said the problems were different. 3P1 does not require the same analysis as 3P2, because in 3P1, A1 does not tell C1 about B1 being pointed out, while in 3P2, A tells C about B being pointed out. I can't require you to agree with me about that, but saying I'm insisting on something else isn't a reasonable way to decry my insisting on what I am insisting on.

While each one of the participants may be more interested in one part of the problem, in order to apply it to their part of the story, each should understand a complete analysis (correct or incorrect) in order to believe it is correct.

That is another flawed attempt to make the problems equivalent, when they're not, by pretending that a correct description of the two problems does not have to include that there is an additional condition and related question in 3P2.

Regarding A1 and A we do not have to do anything more than recognize that the 1/3 chance that each has throughout his respective problem does not improve to 1/2 upon the guards in 3P1 and 3P2 pointing out B1 and B, respectively. 3P1 does not ask us about the change in the chance of C1, whereas 3P2 expressly asks about C's chance, as he determines it, having improved to 2/3.

That difference is part, but not all, of what makes the problems different. Again what is pivotal in that regard is that in 3P2, A tells C about B, while in 3P1, A1 does not tell C1 anything that he has learned about B1.

Analysis that yields the 2/3 chance for C is necessary for us to have foundation for correctly answering the question in 3P2, but in 3P1, we are not asked, either explicitly or implicity, about the changing chance of C1. It's not part of that problem, because C1 isn't told anything by A1 in that problem.

Prisoner A/the statistician gets it wrong - but even if he we don't see that he thinks not-pointed-too/Prisoner C's probability is 1/2, he does.

Each of A1 and A is said to have arrived at his (incorrect) assesment by first the (correct) observation that there are only 2 not pointed out prisoners left, and the incorrect inference that because he is now one of 2 instead of 1 of 3, his chance is 1 out of 2 instead of his original 1/3.

We need not diagnose the internals of the incorrect reasoning of A1 (or, in the 3P2 problem, of A) by recognizing that the chance fomerly held by B1 (or of B) has not distributed equally over A1 and C1 (or over A and C). Recognizing that the chance of A1 (and that of A) remains at 1/3, and does not change to 1/2 upon his seeing B1 (or B) pointed out, is all that is required of us for 3P1 (and all that's required for 3P2 regarding A).

Not including that in the story does not make the problem different.

A telling C about B and us being asked about whether C is right in assessing A's chance as remaining at 1/3 and his own chance to have improved to 2/3 makes 3P2 different from 3P1, in which A1 does not tell C1 about B1, and we are not asked about C1's chance having improved to 2/3, which we are not asked in 3P1, because in 3P1, C1 has not been told by A1 about B1, so he cannot know his chance to have improved to 2/3.

That's why any mathematician you ask, except you, will say the problems themselves are the same.^{[citation needed]}

That remark is obviously unfounded.

In his book about the MHP, Jeffrey Rosenhouse doesn't even mention which specifics he thinks are asked for in the TPP, he just says it is the predecessor of the MHP.

He was referring to 3P2, or an equivalent thereto; not to any equivalent of 3P1, which is not equivalent to MH.

Which is all I "originally brought up," and keep getting "taken to task" for.

This is not an example of a mathematician who holds that 3P1 is equivalent to MH or to 3P2; it is a reference to a mathematician who recognizes that 3P2 or some equivalent thereto, in which the inquiring prisoner tells the other not pointed out prisoner about the pointing out of the pointed out prisoner, is equivalent to MH.

You said that 3P1 was equivalent to MH, and after I disagreed, you cited 3P2 as equivalent to MH, which it is, and when I then said that 3P1 was not the same as 3P2, because of A telling C about B in 3P2, which corresponds to MHC being given an option to switch doors, you said that was irrelevant. I'm confident that you won't find that contention anywhere in Mr. Rosenhouse's work.

If you insist that which specifics are asked for makes the problems "different," then I most certainly can "drag in" anybody who is asked about specifics. Including the reader.

That's a non sequitur.What makes the problems different is not only that we are asked different questions, The additional condition that forms the basis for the additional question being asked is also part of what makes the problems different. Again, pivotally, 3P1's A1 does not tell C1 about B1, whereas 3P2's A tells C about B.

And please recognize that in the OP, nothing was asked for. Get that? THERE WAS NO EXPLICIT QUESTION. So there is no "problem" to say is the same, or is different, unless you infer a question. And any of the question you say make the problems "different" can be inferred this way, not just the one you choose to say is the original problem.

In 3P1, we are asked, albeit only implicitly, only whether A1 is right or wrong about his chance having changed from 1/3 to 1/2 after B1 is pointed out, whereas in 3P2, we are asked not only about whether A's chance has changed, but also about whether and how C's chance has changed, because unlike in 3P1, in which C1 has not been told about B1, in 3P2, C has been told about B.

Although in both problems we are asked to evaluate whether new information changes a probability, and although the answer is no in both problems regarding A and A1, only in 3P2 are we asked further about the prisoner whose chances from an objective perspective have improved to 2/3, because only C, and not C1, has been updated with the new information, wherefore only C's, and not C1's, subjective probability can have changed, and that again is the difference between the two problems.

It's an easily articulable difference, and your position that it is an irrelevant superficial difference does not make it not a functional difference, or in any way lend merit to your contention that the problems are the same. You cannot point to a corresponding element within 3P1 to take the place of A telling C about B, but in MH you can point to not only the unopened non-selected door, but also to MHC being offered the option to switch doors, which 2 elements together are necessary to make 3P2 equivalent to MH. 3P1 is missing the element of A1 telling C1 about B1, or of A1 being given an option to swap positions with C1, either of which which would make 3P1 elso equivalent to MH, but neither of which is present in CP1, wherefore only 3P2, and not 3P1, is equivalent to MH.

HOW DOES SWITCHING MAKE IT A DIFFERENT PROBLEM, instead of just a different consequence of the solution to the problem itself?

Again, the option to switch doors corresponds to A telling C about B, for which there is no counterpart in 3P1.

You drag in a viewpoint (ours) that's not part of the problem,...

So, you are saying that what we are asked for is not a part of "the problem?"

You are trying to drag in our external point of view to make up for C1 not gaining the knowledge that A1 has in 3P1, as C gains the knowledge that A has in 3P2. C1 is the 3P1 counterpart of C in 3P2. We are not the counterpart of anyone or anything in the problem, because we are not in the problem. You cannot legitimately use our having the same knowledge as A1 and A have, to rescue the two problems from being different, when C has the knowledge of A, while C1 does not have the knowledge of A1, and we are asked about C in 3P2, but we are not asked about C1 in 3P1.

And all I am saying, and have said over and over, is that any question that can be asked, about any probability in any version of either question, has an exact counterpart in all of them. This is true whether or not they are asked explicitly, implicitly, or seem unconnected to the fate of the character you choose to isolate from the others for some reason. In fact, their fates are relevant only to the story, not the problem itself.

Their knowledge is relevant to the problem. If C is asked about his chance he can say that it started as 1/3, and that after what A told him about B, it improved to 2/3, whereas if we ask C1 about his chance, he can say only that it is 1/3. He does know that there was a disclosure event regardin B1, and our knowing it doesn't make him know it, and isn't in any other way legitimately a counterpart in 3P1 to anything in 3P2

the statistician doesn't have an option to swap verdicts, as the contestant has an option to switch doors.

And how does that affect how I determine whether it is advantageous?

There is no counterpart of 3P1 to it being in the MH problem necessary to determine that. It's not part of 3P1. If it were, it would correspond to MHC being given an option to switch doors in MH. That would suffice to make 3P1 equivalent to MH. So would A1 telling C1 about B1, and us then being asked about what happens to the chance of C1, just as A telling C about B and us being asked about what happens to the chance of C in 3P2 makes that problem equivalent to MH.

Or whether the statistician's chances have changed? HOW DOES SWITCHING MAKE IT A DIFFERENT PROBLEM, instead of just a different consequence of the solution to the problem?

It's having the option to switch that makes it a different problem from 3P1. If that option were not there, and after the door was opened, the contestant said "thanks Monty, now my chance is 1/2", and we were asked whether or not he was right about that, then MH would be equivalent to 3P1, but not to 3P2, because without an option to switch, the contestant's chance would not change, and there would be no 2/3 chance necessary for us to discern for our answer.

The problems represent the different real-world manifestations of the same underlying probability space. The minute differences in the presentation affects only how we might phrase an answer to address the explicit question (when there is one), or the question we infer (as is the case in the OP). The consequences of the outcomes in the real-world manifestations are irrelevant to how we address the problem.​

Stated and restated; answered and reanswered.

What you keep ignoring is that the story is just a vehicle for what I call the problem.

Your apperception differs from mine.

The problem is what question is asked of us.

The knowledge conditions of the subjects regarding whom the questions are asked are also part of the problem. That's why 3P1 is concerned only with the 1/3 chance of A1, and whether or not it has improved to 1/2, as A1 supposes it has, whereas in 3P2, we are asked to assess, as C has, not only the unchanging 1/3 chance of A and his impression that it has improved to 1/2; because A has told C about B, we are also asked whether C's chance has improved to 2/3, as C is correct in recognizing that it has.

The story can include questions asked of the characters and of us. But which questions are asked of the characters, and how they are affected in the story, is irrelevant to the problem asked of us.

3P1 and 3P2 are two similar but different problems, asking two different question sets, based on two different subject knowledge sets.

The reason you are wrong to call it a "different problem," when all that differs is parts of the story, is because part of what you say is part of the problem in the OP - what question we are supposed to answer - is not a part of the story at all.

It's clear that in 3P1 we are to evaluate whether or not the chance of A1 has improved to 1/2, which A1 says he infers that it has, and which it hasn't. In 3P1, A1's chances are 1/3, before, during, and after B1 being pointed out, and that's all we need to establish. In 3P2, we have to establish not only that A's chance is 1/3 throughout, but also that C's chances are correctly discerned by C to be 2/3 after A tells him about B having been pointed out. That's why the two problems are different.

 

[JeffJo]

“The problem” (these points will be referenced later by the prefix “P” and the point’s label, like “P4.2”) is:

1. A situation has arisen where there are three equally likely possibilities.
2. One possibility has value X, and the other two have value Y.
   1. Call them X, Y1,and Y2.
   2. The point is that one is different, and we should be concerned with whether that one us the one that occurred. Not that it is “better” or “worse,” even if the problem statement distinguishes the possibilities that way.
3. An independent entity, who knows the actual result, eliminates one of possibilities with value Y.
   1. If it isn’t made explicit, the solver should assume that this entity does not discriminate between Y1 and Y2 if neither occurred. That’s the point of them having the same value.
   2. Since it doesn’t matter, let’s say that Y2 was revealed to not have happened.
4. The probabilities before the elimination are trivial; in fact, the are implied by the words “equally likely” in P1.
   1. The probabilities before the elimination are trivial; in fact, the are implied by the words “equally likely” in P1.
   2. Since Pr(X)+Pr(Y1)=1, stating either one implies that the solver knows what the other is. That is, “determining Pr(X) or Pr(Y1)” is also “determining Pr(X) and Pr(Y1)”.

In any version of the problem, there are two knowledge states (prefix “K”):

1. No knowledge of the elimination, whether because it has not yet been made, or a participant was not included.
2. Knowledge of which possibility is eliminated.

Although the exact details vary, there are generally three solutions (prefix “S”) for “the problem”:

1. There are two-equally possibilities left; X and Y1. Since Pr(X)+Pr(Y1)=1, each is 1/2.
   1. Note how this solution requires evaluating two probabilities, even if only one is stated as the answer.
2. The probability of X can’t change from what it was in state K1, because K2 does not include ""new information" about X. So Pr(X) is still 1/3.
   1. This is the only solution that doesn’t utilize Y1 in some way. That is why S2 is wrong even though it gets the right answer.
   2. Specifically: K2 does include "new information" about X, because the entity has to treat X and Y1 differently. Based on that difference, Pr(X) CAN CHANGE IN THEORY, but NOT UNDER P3.1. By not saying how this new information and P3.1 affect Pr(X) - in fact, most of the time P3.1 is ignored altogether - S2 is incorrect.
   3. The reason why it is important to point out that P1.3 is incorrect, is because it doesn't indicate why S1 is incorrect. A solver who accepts S1 will see S2 as a paradox, n
3. Since the entity does not discriminate between Y1 and Y2, the probability that Y2 would have been eliminated must be 1/2. This includes not only the 1/3 probability where Y2 did occur, but also the 1/6 probability where X occurred but Y1 would have been eliminated. So the chances for X, compared to Y1, are (1/6):(1/3), or 1:2. That makes Pr(X)=1/3 and Pr(Y1)=2/3.
   1. This shows how X and Y1 are different, which is why S1 is incorrect. No paradox.

The reason why variations of this problem remain controversial, is because both S1 and S2 seem to be intuitive and neither shows what is wrong with the other. Adherents to either deny its own flaws, while pointing out the flaws of the other.

My point in this thread, is that this “problem” can be expressed in many different forms. But all of them ask the reader for the exact same solution, regardless of whether they want the reader to use it to provide Pr(X), Pr(Y1), or both. Also regardless of whether they are contrasting S1 and S3 (or S2 if you can't understand why it is incorrect). That “the problem” is any story asking you to provide this solution. Asking the reader to evaluate whether a character in the story – whether with knowledge K1 or K2 - has made a correct solution, is the same as asking the reader for that solution and is, again, the same problem.

And finally, why the characters in the stories want a solution is completely irrelevant to how the reader finds a solution itself, and so not a part of the problem.

Sysprog, if you want to define what “the same problem” is in a different way, that ignores how all of the stories require the same solution (even if you can leave part of it out of the answer), you are free to do so. I just don’t think you can provide a definition of what makes it the same, and in fact it makes the expression “the same problem” meaningless. I have provided a definition of what I mean by "the same problem," and the Monty Hall Problem is the same as the Three Prisoner's Problem by that definition.

And finally, if you ask a mathematician what “The Three Prisoner’s Problem” is, he will not insist that it “is” any particular expression of it, like Wikipedia’s, the OP here, or Martin Gardner’s (which, incidentally, differs from what Wikipeda says it is in a way you have said is significant.) It is the problem from any story with the elements I described above, put in the context of three prisoners.

+++++

Other stories that present this problem include:

1. The original, the Bertrand Box Problem. Three boxes each contain two coins. Two of them contain only one kind (value Y), one only gold coins (Y1) and one only silver coins (Y2). One contains a gold coin and a silver coin (value X). I choose a box at random, open it, and take out a gold coin. What are the chances the coin still in the box is also gold?
   1. This is asking for Pr(Y1).
   2. Bertrand’s point was to show why S1 is incorrect. That you can’t apply the Principle of Indifference without determining that the set of cases are equivalent somehow.
   3. He pointed out that if Y1 had been eliminated, the answer has to be the same. So if S1 is correct, it applies regardless of which coin is taken out, making the revelation of the kind of coin irrelevant. Taking out a coin, but not looking at it, changes the probability of matching coins from 2/3 to 1/2. This paradox disproves S1.
   4. The paradox doesn’t, however, prove S2 or S3. So Bertrand gave S3.
2. Three pancakes. Absent-minded Cook has prepared three pancakes. Only one is perfect, another is burnt on one side, and the third is burnt on both sides. He puts them on different plates without looking to see if it is burnt, and you choose one plate at random. It has an unburnt side showing. What are the chances that it is the perfect pancake?
   1. Value Y is matching sides. Value X is half-burnt.
   2. You are asked for Pr(Y1).
   3. Is it “different” because the cook didn’t choose a side to reveal? That is, the “entity” was sheer chance?
   4. Does it matter how upset you’d be to eat a half-burnt pancake?
3. Three Card Swindle: A street hustler shows you three cards; one has a black dot on both sides, one has a red dot on both, and one has a red dot on one and a black dot on the other. He puts them in a hat, and let's you pick one which you place on a table without seeing the bottom side. The top side has a red dot. He wants to bet you even money that the bottom side is also a red dot, since it is obvious you have either the red-red or red-black card. Should you?
   
   1. Value Y is matching dots. Value X is different dots.
   2. In fact, this is pretty much the same as the Pancake problem, changing pancakes to cards and cooked-levels to dots. Does that make it different?
   3. But this asks you only if Pr(Y1) is 1/2,is less, or is more. Does that make it a different problem?
4. Monty Hall (the one that started it, not Marilyn vos Savant’s): On his game show, Monte [sic] Hall offers you the choice of three boxes labeled A, B, and C. One has the keys to a new car, the others are empty. You choose box B. Monte offers to buy that box back from you for increasing amounts of cash, but you decline. Finally, he says “I’ll do you a favor by eliminating an empty box." He opens box A, and it is empty. “Your chances are now 1/2” he claims, so he doubles his last cash offer for box B. But instead of saying “no” this time, you ask if you can trade it for box C.
   1. Value Y is "empty." Value X is "the keys."
   2. Monte suggested S1.
   3. Just like the OP here, no probability was asked for. But it is still implied that you should compare the solutions S1 and S3 (or S2).
   4. Incidentally, no game like vos Savant’s was ever played. Games like this were.
5. Two Child Problem (A little different, but the same fundamental issues. There are two case X’s. So each possibility starts at 1/4): “Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?”
   1. S1 says that Y2 is eliminated, so now Pr(Y1)=Pr(X1)=Pr(X2)=1/3. Even though this is just as wrong in this problem, as in any of the others here, it is by far the most common answer given in textbooks.
   2. Martin Gardner got burned by saying the answer to this exact question was 1/3. He retracted that, saying that because P3.1 was not implied, the statement was ambiguous. The answer could be either 1/3 or 1/2. He even referred to one of the textbooks I mentioned.
   3. To be fair, many such textbooks change the problem to say “chosen from all families of two including at least one boy.” This makes S1 correct, because P3.1 is no longer true.
   4. But then others change it in the other direction: “I have two children, and at least one is a boy.” Now P3.1 is implied the same way it is in vos Savant’s Game Show Problem.
   5. And there reason S1 is accepted, is because they ignore S2.2 and S2.3

These are all the same problem (one with a different number of cases), because they ask you to find the same solution. Usually in comparison to an incorrect one.

 

[sysprog]

In the Bertrand's Box problem, there is a 1/3 probability that the contents, i.e. the two coins it contains, of the randomly selected box are different (bimetallic -- one coin gold and one coin silver) and a 2/3 probability that the two coins contained in it are the same (monometallic -- both coins silver or both gold coins gold).

There is no change in that probability distribution after a coin is randomly picked from the randomly selected box and is then revealed.

What is established by the revealing is which of the two metals the box has a 2/3 chance of containing two coins of.

After the revealing, there is no longer a chance of the box having contained two coins of the non-revealed metal; there is only the original and continuing 1/3 chance of it having contained both metals, and the original 2/3 chance of it having contained exactly one of the two kinds of metals, and that kind has at that point been shown to be the revealed coin's kind of metal.

Whether the revealed coin is silver or gold, the chance of it having come from the bimetallic box is still 1/3, and the chance of it having come from a monometallic box is still 2/3, so if the revealed coin is silver, there is a 2/3 chance that the other coin in that box is silver, and if the revealed coin is gold, there is a 2/3 chance that the other coin in that box is gold.

 

[JeffJo]

In the Bertrand's Box problem, there is a 1/3 probability that the contents, i.e. the two coins it contains, of the randomly selected box are different (bimetallic -- one coin gold and one coin silver) and a 2/3 probability that the two coins contained in it are the same (monometallic -- both coins silver or both gold coins gold).

In the OP, there is a 1/3 chance that the inquisitive prisoner, A, is to be pardoned, and a 2/3 chance that he is not to be pardoned. That is, the distribution for {bimetallic,monometallic} is {1/3,2/3}, as is the distribution for A's fate {A pardoned, A not pardoned}.

But you are avoiding the fact that the event partition you can use to describe a probability space, and so establish a distribution, is not unique. The distribution for {bimetallic,gold,silver} is {1/3,1/3,1/3}, as is the distribution for {A pardoned, B pardoned, C pardoned}.

There is no change in that probability distribution after a coin is randomly picked from the randomly selected box and is then revealed.

There is no apparent change in the distribution for the partition you specifically choose to have that property. For the more "atomic" partition, it changes from {1/3,1/3,1/3} to {1/3,0,2/3}. The reason that the solution I called S2 is incorrect, is that you need a different method to prove that Pr(A not pardoned)=2/3 after the reveal. It is no longer a case of three equally-likely possibilities.

The reason S2 doesn't convince many adherents to S1 is that it doesn't explain the fact that there is a change. In fact, there is a quite blatant change: Pr(B pardoned) went from 1/3 to 0. There has to be some accompanying change to Pr(A pardoned) and Pr(C pardoned), and asserting "oh, well, only the other one changes" is not a solution.

So the above is a misleading argument that misrepresents the situation. I understand that it is a commonly-held opinion that it is correct. That doesn't make it so.

What is established by the revealing is which of the two metals the box has a 2/3 chance of containing two coins of.

What is established, is that there is "new information" that changes the probability space. What S2 says is that there is no change; you can claim that it only says this about one event, but that is the misleading claim I keep pointing out to you.

A distribution is a set, not a single value. The distribution {1/3,1/3,1/3} is affected by the information by first removing what that information says is impossible, making it {1/6,0,1/3}. If we use your distribution, it makes it {1/6,1/3}. Get that? Even if you ignore which prisoner is indicated, it still is a specific prisoner. Half of the probability in your distribution needs to be "eliminated," and S2 does not provide a way to do that.

So either way, the element that S2 says is a constant, is actually affected. Half of the time it is true a different prisoner will be indicated. The 1/3 that S2 says is constant is reduced to 1/6. The 2/3 that S2 says is constant is reduced to 1/3. The prior probabilities of the events that remain possible are {1/6,0,1/3} (or {1/6,1/3} if we use S2, but it can't say why). This makes the distribution improper - it no longer sums to 1 - so we normalize it by dividing by the affected sum. It becomes {1/3,0,2/3} (or {1/3,2/3}).

Whether the revealed coin is silver or gold, the chance of it having come from the bimetallic box is still 1/3, and the chance of it having come from a monometallic box is still 2/3,...

Yes, that is true. But simply asserting that it is so is not a correct solution, is it? It is affected by the information, but does not change in value as a result.

 

[sysprog]

In the OP, there is a 1/3 chance that the inquisitive prisoner, A, is to be pardoned, and a 2/3 chance that he is not to be pardoned. That is, the distribution for {bimetallic,monometallic} is {1/3,2/3}, as is the distribution for A's fate {A pardoned, A not pardoned}.

I agree with that much.

But you are avoiding the fact that the event partition you can use to describe a probability space, and so establish a distribution, is not unique.

The problem isn't presented as that of providing an event partition to describe a probability space. That's a problem-solving technique, and no such technique is necessary for answering the question about the second coin in the selected box.

Of the 3 boxes, one is bimetallic, and the other two are monometallic. The 1/3 probability of the bimetallic selection, and the 2/3 probability of a monometallic selection, don't change on the revealing; the revealing establishes only which metal is 2/3 likely to be what the other coin in the selected box is made of.

The distribution for {bimetallic,gold,silver} is {1/3,1/3,1/3}, as is the distribution for {A pardoned, B pardoned, C pardoned}.

Compared to the 1/3 chance of the bimetallic box, the second and third 1/3 chances start as 1/2 chance each of a 2/3 chance of one or the other monometallic box. After the revealing of one of the coins from the selected box, we know which one, but the chance that it is one or the other of the two monometallic boxes, and not the bimetallic box, remains 2/3.

There is no apparent change in the distribution for the partition you specifically choose to have that property.

The problem statement presents one bimetallic box and two monometallic boxes. That means a 1/3 + 2/3 distribution. That doesn't change.

For the more "atomic" partition, it changes from {1/3,1/3,1/3} to {1/3,0,2/3}.

The change is from [1/3 bimetallic and 2/3 [gold or silver]] to [1/3 bimetallic and 2/3 gold], or [1/3 bimetallic and 2/3 silver]. The 1/3 bimetallic and 2/3 monometallic probabilities don't change; the probability remains 1/3 bimetallic and 2/3 monometallic, and the revealing establishes only which one metal it is that the second coin in the selected box has a 2/3 chance of being made of.

The reason that the solution I called S2 is incorrect, is that you need a different method to prove that Pr(A not pardoned)=2/3 after the reveal. It is no longer a case of three equally-likely possibilities.

The 1/3 probability remains 1/3, without regard for whether the other 2/3 probability is collected or distributed.

The reason S2 doesn't convince many adherents to S1 is that it doesn't explain the fact that there is a change.

I didn't present your S2 as an answer, and I didn't present my answer as more convincing than any other correct answer. I observe that my answer is more parsimonious than the non-S2 answer you propose, but I don't think that necessarily makes it more perspicuous. Your answer is correct in saying that that the probability of the second coin being the same as the first is 2/3. Wherein your answer exceeds its charter, in order to pronounce to be incorrect my answer that more directly arrives at the same conclusion, your answer is incorrect, and mine is correct.

In fact, there is a quite blatant change: Pr(B pardoned) went from 1/3 to 0. There has to be some accompanying change to Pr(A pardoned) and Pr(C pardoned), and asserting "oh, well, only the other one changes" is not a solution.

It's not the mere offhanded dismissal you suggest it to be. The collection of the distribution of the 2/3 chance doesn't affect the 1/3 chance, and is adequately explained by reference to the fact that the possession of the 2/3 chance has gone from being distributed between 2 possessors to being collected into the possession of 1 possessor.

So the above is a misleading argument that misrepresents the situation.

That's entirely untrue. The chance of the other coin being the same as the revealed coin is 2/3. Recognizing that doesn't require provisionally acknowledging a basis for, and then rejecting on deeper analysis, an erroneous 50:50 assessment. That's required only for enabling those who do the deeper analysis in order to smugly gloat over the errancy of their less enlightened neighbors. As you recognize, getting the right answer doesn't require that.

I understand that it is a commonly-held opinion that it is correct.

It is correct.

That doesn't make it so.

Nothing is ever correct due merely to it being commonly held to be correct, but being commonly held doesn't disqualify any contention from being correct.

What is established, is that there is "new information" that changes the probability space.

What is established by the revealing of a coin, is only which metal -- it doesn't change the likelihood of the second coin being the same as the first. That remains 2/3, regardless of which metal is revealed. The likelihood of a second coin in any box being made of a different metal from that of which a first coin in the same box is made is 1/3, before, during, and after the revealing.

What S2 says is that there is no change; you can claim that it only says this about one event, but that is the misleading claim I keep pointing out to you.

According to you, S2 says:

The probability of X can’t change from what it was in state K1, because K2 does not include ""new information" about X. So Pr(X) is still 1/3.

The 1/3 original chance of selecting the bimetallic box doesn't change upon the showing of a coin. The chance that the coin is from a monometallic box remains 2/3. The revealing of the coin eliminates the possibility of it having come from a monometallic box containing coins of the non-revealed kind, but doesn't thereby change the 2/3 chance that the selected box has of being monometallic.

The selected box obviously cannot contain two coins of the kind that is not the kind of the revealed coin., but that does not affect the 2/3 probability that it is a monometallic box, or the 1/3 probability that it is a bimetallic box. The probability at the start for anyone coin being of either one of the two kinds is 1/2 + 1/2, because there are the same number of coins of each kind. We're not asked about the other boxes or the other coins; we're asked only about the second coin in the selected box. After the revealing, the probability of the second coin being of the same kind as that of the revealed coin is 2/3.

A distribution is a set, not a single value.

That's true -- the distribution of 1/3 bimetallic and 2/3 monometallic is a set of two values, not a single value.

The distribution {1/3,1/3,1/3} is affected by the information by first removing what that information says is impossible, making it {1/6,0,1/3}.^Incorrect.

That sums to 1/2. It can't be correct if it doesn't sum to 1, as 1/3 + 1/3 + 1/3 does, and as 1/3 + 0 + 2/3 does, and as 1/3 + 2/3 does.

If we use your distribution, it makes it {1/6,1/3}.^Incorrect.

The distribution [1/3 bimetallic, 2/3 monometallic], sums to 1, and doesn't change on the revealing -- which metal the second coin is 2/3 likely to be made of is established by the revealing.

Get that? Even if you ignore which prisoner is indicated, it still is a specific prisoner. Half of the probability in your distribution needs to be "eliminated,"^Incorrect. and S2 does not provide a way to do that.

None of the 1/3 + 2/3 probability needs to be eliminated. The half that is eliminated is the 1/2 chance that the revealed coin could have been of the other kind. The second coin continues to have the 1/3 chance of being from the bimetallic box, and the 2/3 chance of being from a monmetallic box. The metal type of the revealed coin determines only which monometallic box the second coin has a 2/3 chance of being from.

So either way, the element that S2 says is a constant, is actually affected.^Incorrect. Half of the time it is true a different prisoner will be indicated. The 1/3 that S2 says is constant is reduced to 1/6.^{Incorrect.[/sup}

The 1/3 original chance of selecting the bimetallic box doesn't change upon the showing of a coin.

The 2/3 that S2 says is constant is reduced to 1/3. The prior probabilities of the events that remain possible are {1/6,0,1/3} (or {1/6,1/3} if we use S2, but it can't say why).^Incorrect.

You immediately hereafter acknowledge that to be incorrect, but you haven't shown, as you apparently intended to, that my contention set is inclusive of or equivalent to S2, or that S2, or my contention set, requires or allows a change in the 1/3 + 2/3 probability distribution.

This makes the distribution improper - it no longer sums to 1 -^Correct. ...

That's correct, but we don't need the correction if we don't make the error in the first place.

so we normalize it by dividing by the affected sum. It becomes {1/3,0,2/3} (or {1/3,2/3}).

It doesn't become 1/3 + 2/3 -- it remains 1/3 + 2/3 -- the showing of a coin doesn't affect the 1/3 chance of selecting the bimetallic box, or the 2/3 chance of selecting a monometallic box. It establishes only which monometallic box continues to hold the 2/3 chance of a monometallic box having been the one box selected.

Whether the revealed coin is silver or gold, the chance of it having come from the bimetallic box is still 1/3, and the chance of it having come from a monometallic box is still 2/3,

Yes, that is true. But simply asserting that it is so is not a correct solution, is it?

After the more minimal analysis necessary, and without the more maximal and partly unnecessary analysis you present, the complete sentence was:

Whether the revealed coin is silver or gold, the chance of it having come from the bimetallic box is still 1/3, and the chance of it having come from a monometallic box is still 2/3, so if the revealed coin is silver, there is a 2/3 chance that the other coin in that box is silver, and if the revealed coin is gold, there is a 2/3 chance that the other coin in that box is gold.​

I think that's correct and satisfactory.

It is affected by the information, but does not change in value as a result.

The 1/3 + 2/3 distrubution is not affected by the revealing of a coin from the selected box. It affects only which monometallic box has the 2/3 chance.

 

[Stephen Tashi]

The guard's selection process doesn't have to be 50:50, as long as Harry has no knowledge of it.

That isn't an axiom of probability theory.

This problem is only problematic once the general population gets involved. To anyone who understands probability theory it's fairly trivial.

It's possible to get a specific answer to the Martin Gardner variation of the problem, but its not trivial to justify a particular answer to a version of the problem that omits some of Gardner's information.

You certainly don't have a monopoly on the "correct" solution.

As I read post #3 by @JeffJo, he is saying there is no unique solution to the problem unless its statement includes the information given by Gardner's variation. So I don't see that @JeffJo has proposed a (unique) correct solution to the non-Gardner variation.

If a problem in applying probability has a trivial solution then it should be easy to perform the analysis in the usual manner. The first step would be to define the "probability space" explicitly. Discussions of controversial applications of probability theory often go on-and-on while omitting this step. The probability space is assumed to be "obvious" and yet never explicitly described as a set of outcomes.

The most obvious way to formulate the probability space for the problem in the original post doesn't work. We can define a probability space S whose outcomes represent 2 out of 3 prisoners being executed. A point in set S is itself a set of 3 symbols:
S = { {AE,BE,C},{AE,B,CE}{A,BE,CE} }
The probability measure P0 assigns each outcome in S a probability of 1/3.

However, a probability space for the problem must be capable of representing the conditional probabilities that the problem asks about. The above probability space cannot represent events such as "Prisoner A asks the guard to designate another prisoner who will be executed and the guard designates C".

If someone asserts there is a unique answer to the problem, it should be possible for them to state explicitly what probability space they are using. The space S isn't sufficient.

 

[JeffJo]

The problem isn't presented as that of providing an event partition to describe a probability space.

And this is where we disagree in the fundamentals. The "problem" and the "presentation" are not the same thing. As Stephen Tashi (how do you get the link with the "@"?) points out, any formal solution to a probability problem requires making your probability space explicit. I don't think we need to be that formal here, but I have indicated mine in the "S3" solution. You are free to use whatever you want, but (again, as Stephen Tashi points out) to be a correct solution it needs to include the event where X occurs and Y1 is indicated, and remove it from the realm of the possible. Your entire argument is flawed because it does not do so; I could go through it point-by-point and say this each time, but I'm going to save time and space by not doing so. I'll just state the three solutions in a different way:

• Basis common to S1 and S2: The sample space is {X,Y1,Y2} (X,Y1, and Y2 are outcomes. A set of outcomes is an event. Probabilities are assigned to events. "Monometallic" is a legitimate event comprising two outcomes. The single-outcome events {Y1} and {Y2} are just as legitimate.) The knowledge in K2 affects only the event {Y2}. {. Pr({X})=Pr({Y1})=Pr({Y2})=1/3. We are given the information state K2, that {Y2} didn't occur.
  
• S1, the correct solution given that basis:
  • Formal version: K2 is the event {X,Y1}.
    • Pr({X}|{X,Y1}) = Pr({X}+{X,Y1})/Pr({X,Y1} = (1/3)/(2/3) = 1/2.
    • Pr({Y1}|{X,Y1}) = Pr({Y1}+{X,Y1})/Pr({X,Y1} = (1/3)/(2/3) = 1/2.
  • Informal version: X and Y1 are still equally likely, and their probabilities must change so that the sum to 1. So each is 1/2.
• S2, the incorrect solution given that basis:
  • No formal version exists.
  • Informal version: Gee, that means X is unaffected so its probability must still be 1/3. While the same argument could be applied to Y1, we make it change to 2/3 so the sum is still 1.
• Correct basis: We need outcomes that include all possibilities for state K2, not just what occurred. The sample space is {X1,X2,Y1,Y2}, where "X1" means that X occurred AND we are informed that Y1 did not.
  • The "X" in the incorrect basis is the union of X1 and X2. Pr({X1})=Pr({X2})=1/6 and Pr({Y1})=Pr({Y2})=1/3. Since X1 is affected in this union, the assertion in S2 that X is unaffected is incorrect.
• S3, the correct solution:
  • Formal version: K2 is the event {X2,Y1}
    • Pr({X1,X2}|{X2,Y1}) = Pr({X1,X2}+{X2,Y1})/Pr({X2,Y1}) = Pr({X2})/[Pr({X2})+Pr({Y1})] = (1/6)/(1/6+1/3) = 1/3
    • Pr({Y1}|{X2,Y1}) = Pr({Y1}+{X2,Y1})/Pr({X2,Y1}) = Pr({Y1})/[Pr({X2})+Pr({Y1})] = (1/3)/(1/6+1/3) = 2/3
  • Informal version: {X2} is half as likely as {Y1}, so the chances are 1/3 and 2/3.

As I read post #3 by @JeffJo, he is saying there is no unique solution to the problem unless its statement includes the information given by Gardner's variation. So I don't see that @JeffJo has proposed a (unique) correct solution to the non-Gardner variation

I didn't think I needed to be so formal. But I did say:

It doesn't really matter if these facts are included or not, since they are necessary assumptions anyway.

They are necessary because we don't know how the guard would choose between the other two, when they are equivalent.

 

[JeffJo]

If someone asserts there is a unique answer to the problem, it should be possible for them to state explicitly what probability space they are using. The space S isn't sufficient.

Note: I distinguish the answer "1/3 and/or 2/3 for the statistician and/or remaining prisoner" from the solution that produces that answer. The correct answer is unique, but there can be many solutions to get it. Including some that are wrong, or incomplete.

Sysprog distinguishes the presentations that ask only for one or the other part of the answer, or emphasize the importance of the answer to the characters over the actual values. I call them the same problem since every one requires the same kind of solution, and a correct one at least implies both parts of the answer.

The space that I described explicitly in post #3 had a typo pointed out by stonetemplepython, and is corrected here:

... there are four possibilities. Say the three prisoners are named Tom, Dick, and Harry (the statistician):

1. Tom is not guilty, the guard ignores the coin and points to Dick. Probability: 1/3
   
2. Dick is not guilty, the guard ignores the coin and points to Tom. Probability: 1/3
3. Harry is not guilty, the coin lands on heads and the guard points to Dick. Probability: 1/6
4. Harry is not guilty, the coin lands on tails and the guard points to Tom. Probability: 1/6

I never said that this is the only way to construct a correct probability space, just that any correct probability space needs to distinguish whatever passes for E3 and E4 in it. The conditional probability that Harry is not guilty, given that the guard pointed to Tom, is Pr(E4)/[Pr(E2)+Pr(E4)]=(1/6)/[(1/6)+(1/3)]=1/3. The conditional probability that Harry is guilty, given that the guard pointed to Tom, is Pr(E2)/[Pr(E2)+Pr(E4)]=(1/3)/[(1/6)+(1/3)]=2/3.

 

[Stephen Tashi]

I didn't think I needed to be so formal. But I did say:

JeffJo said: ↑
It doesn't really matter if these facts are included or not, since they are necessary assumptions anyway.

They are necessary because we don't know how the guard would choose between the other two, when they are equivalent.

There is an interesting inconsistency in the culture of mathematics when it comes to making assumptions. On the one hand, if we are in the context of teaching about indeterminate systems of equations, we would present examples like:
1) x + y + z = 4
2) x + 2y + 2z = 7
and emphasize to students: "You can't find a unique solution to this problem".

On the other hand when confronted with a mathematical "word problem", we are in cultural situation where the game proposed by the teacher (or the mathematical world at large) is to justify a unique solution. So it becomes culturally acceptable to make assumptions. Making assumptions is the required behavior.

For example, considering the above system of equations, it would be heresy if the teacher said "We don't know any distinction between y and z and they are arbitrary symbols, so we may assume y = z". By contrast, in the context of probability problem, it's common to hear arguments like "We don't know any distinction between events y and z and they are represented by arbitrary symbols, so we may assume Pr(y) = Pr(z)".

My behavior when considering controversial puzzles in probability theory may be an aberration, but I prefer to consider them in the same context as systems of simultaneous equations. From that point of view, what is needed is the explicit statement of the sample sample space and the constraints on how probabilities can be assigned to it.

Perhaps the previous posts about Tom,Dick, and Harry, or X1,Y1 etc. implicitly define sample spaces, but I can't keep track of them amidst the verbal argumentation. Controversy about what sample space is used is to be expected, but debate would be clearer if we distinguish whether a disagreement is about what sample space is used versus being about what probabilities are assigned to a commonly accepted sample space.

I suspect that if an adequate sample space for this problem is specified that there will be many different ways to assign probabilities to its outcomes. From the cultural viewpoint of simultaneous equations, this will show the problem is indeterminate. By contrast, from the cultural viewpoint of solving puzzles, most of the different solutions will be "unjustified" because the game in that context is to find solutions that satisfy assumptions with a simple verbal description.

 

[StoneTemplePython]

For example, considering the above system of equations, it would be heresy if the teacher said "We don't know any distinction between y and z and they are arbitrary symbols, so we may assume y = z". By contrast, in the context of probability problem, it's common to hear arguments like "We don't know any distinction between events y and z and they are represented by arbitrary symbols, so we may assume Pr(y) = Pr(z)".

In probability with a bounded (and in particular finite) distribution, unlike simultaneous equations, there are some interesting majorization (particularly entropy) reasons for choosing the uniform distribution. But... I quite liked the culture of puzzles vs the culture of simultaneous equations discussion. I'd estimate that $\gt \frac{3}{4}$ of the past posts were linguistic and not mathematical in nature.

Being explicit about sample spaces is a good way to go if people want to do the actual math. I like to re-frame every one of these problems in terms of betting as that clarifies the math. Reminds me of the sleeping beauty puzzle. If people can't agree on a betting formulation, then there's a problem.

 

[sysprog]

The problem isn't presented as that of providing an event partition to describe a probability space.

And this is where we disagree in the fundamentals. The "problem" and the "presentation" are not the same thing.

I didn't say they were. Correctly answering the Bertrand's box problem doesn't require the analysis you presented. It requires recognizing the fact that the chance of the second coin being the same as the first is 2/3 -- before, during, and after the revealing of the first coin. The revealing of the first coin doesn't change the 1/3 bimetallic and 2/3 monometallic probability distribution. It establishes only which metal the first coin is made of.

 

[JeffJo]

There is an interesting inconsistency in the culture of mathematics when it comes to making assumptions.

But we are discussing a puzzle, not a theorem. Unless the person who asked it is being juvenile, he thinks that enough information has been provided to result in an answer. If that information is to be inferred, an acceptable response is "If I assume X, and can justify (but not prove) assuming it, then the answer is Y".

This is even more common in probability puzzles, since by definition they must involve some situations that are not made explicit. And all too often, the only way to justify X is that you can't justify any other probability space.

There are even customary clues used to suggest such inferences. In the OP, "strangers to each other" is supposed to suggest that each prisoner has no information that would make his initial assessment of his chances different than either of the others, or make the others different. This establishes the initial probabilities at 1/3 each. The second part of that - the others can't be different - was indeed implied, but only before the prisoners came into communication with each other. But the first is not. For one thing, the statistician knows whether or not he is guilty, and should have been there when evidence was presented against him. Whatever his role was in the crime, or what evidence was, or was not , presented, he clearly can't think his initial chances were 1/3. But without knowing what he knows, we can't justify any probability distribution other than {1/3,1/3,1/3}.

My behavior when considering controversial puzzles in probability theory [is that] what is needed is the explicit statement of the sample space and the constraints on how probabilities can be assigned to it.

And my experience with such puzzles, is that we never have enough explicit information to debate what the sample space should be, when someone values the argument over a reasonable solution. But I will point out that I provided a sufficient description of a probability space (albeit with a typo that had an obvious correction) in my first post. Along with the reasons for the necessary assumptions.

I will also point out that this seemingly-endless debate has been over the meaning of "the problem"; or more specifically, "the same problem." Not how to address it (or them). And that my point in this seemingly-endless debate is that two problems are "the same problem" when the probability spaces, that you reminded us of, are equivalent. Sysprog wants problems to be "different" if the question (A) asks for a different element of your probability space that (B) he feels can be answered without your full probability space, or (C) the outcomes in the space have different implications to the characters.

 

[JeffJo]

Correctly answering the Bertrand's box problem doesn't require the analysis you presented. It requires recognizing the fact that the chance of the second coin being the same as the first is 2/3 -- before, during, and after the revealing of the first coin.

It "requires" an argument for why "the condition 'the second coin being the same' maintains a constant probability," is true, not just the assertion. And an argument for why "the unopened boxes maintain the condition of equiprobability" is false. I can only repeat that you have provided neither so many times.

The revealing of the first coin doesn't change the 1/3 bimetallic and 2/3 monometallic probability distribution.

Why not? Saying it is so does not demonstrate that it is. Repeating an assertion does not make the assertion stronger. It doesn't matter that your intuition tells you that the assertion is true - intuition can be wrong.

Repeating actual arguments can help - but only if you read them and are willing to accept them. Ignoring actual arguments does not disprove them, it just shows that you are unwilling to accept them. So I will repeat, again, in terms I hope are more agreeable to Stephan Tashi, the argument against your use of assertion:

New information affects the entire set that is the probability distribution in the probability space you choose to use, not just selected members of it. It affects the distribution by (except in problems like the Sleeping Beauty Problem) "zeroing out" the probabilities corresponding to some members of a proper sample space. As a result, the entire distribution needs to be re-normalized.

In Bertrand's Box Problem, the sample space (the set of all possible outcomes) requires at least four outcomes. Where the notation "XY-Z" means the outcome where a box has coins of metals X and Y, and a coin of metal Z is revealed, the sample space you need is {GG-G,SS-S,GS-G,GS-S}. The corresponding distribution is {1/3,1/3,1/6,1/6}.

The correct solution is that revealing a gold coin "zeros out" the probabilities corresponding to SS-S and GS-S. The affected distribution, {1/3,0,1/6,0} is re-normalized to {2/3,0,1/3,0}.

The event (a set of outcomes) you call "bimetallic" is the set {GS-G,GS-S}. The event you call "monometallic" is the set {GG-G,SS-S}. The "new information" affects both of these events, by revealing that one outcome in each did not happen. So whether or not your answer is correct, your assertion that the probabilities corresponding to these events are not affected must be wrong. The fact that they are affected proportionately can be shown, but only by recognizing all four outcomes in the sample space.

The incorrect solution to the Bertrand Box Problem works out to "zeroing out" only SS-S. So the affected distribution is {1/3,0,1/6,1/6}. This is the only mistake in the argument. The correct re-normalization of that distribution is {1/2,0,1/4,1/4}. The event "other coin gold," {GG-G}, ha probability 1/2. The event "other coin silver," {GS-G,GS-S}, has probability 1/4+1/4=1/2. People believe this answer because they don't see the one mistake they made, and you can't convince then that it is wrong without somehow using the full sample space.​

 

[sysprog]

Sysprog wants problems to be "different" if the question (A) asks for a different element of your probability space that (B) he feels can be answered without your full probability space, or (C) the outcomes in the space have different implications to the characters.

That is a mischaracterization. My recognition of a difference is not due to my wanting there to be one, and my answering a question with a more parsimonious sufficient analysis is not based on a feeling.

It "requires" an argument for why "the condition 'the second coin being the same' maintains a constant probability," is true, not just the assertion.

The problem statement establishes the 2/3 probability that the selected box is monometallic, and the 1/3 probability that it is bimetallic. Revealing one kind of coin drawn from the selected box obviously doesn't change the 1/3 chance of the box having bimetallic contents, just as revealing the other kind obviously wouldn't. And if the bimetallic probability remains 1/3, the monometallic probability has to remain 2/3, so that the sum of the probabilities will continue to be 1. The revealing has no effect on the 1/3 + 2/3 bimetallic plus monometallic probability distribution. It simply removes from further consideration the possibility of the selected box containing two coins of the other kind. The selected box therefore has 1/3 chance of being monometallic, and 2/3 chance of being bimetallic of the revealed kind.

And an argument for why "the unopened boxes maintain the condition of equiprobability" is false. I can only repeat that you have provided neither so many times.

There are 3 coins of each kind. The equiprobility is between the two possible kinds of coin that could have been first drawn. The probability that the selected box has two of the same kind as each other is 2/3. After the revealing eliminates one of the previously equiprobable two kinds of monometallic box, there remains only the 1/3 chance that the bimetallic box was selected to hold the possibility of the second coin being of the non-revealed kind, and the 2/3 chance that a monometallic box was selected, which at that point is still a 2/3 chance that the two coins are the same, so the revealing of the first coin establishes only which kind of metal of the original two possible kinds the second coin has a 2/3 chance of being made of.

The revealing of the first coin doesn't change the 1/3 bimetallic and 2/3 monometallic probability distribution.

Why not? Saying it is so does not demonstrate that it is. Repeating an assertion does not make the assertion stronger. It doesn't matter that your intuition tells you that the assertion is true - intuition can be wrong.

It establishes only which metal the first coin is made of.

The revealing of the coin eliminates the equiprobability of the two bimetallic boxes, but that elimination changes only the probability distribution within the bimetallic part of the overall distribution; it doesn't change the relation of that part to the whole. It makes the monometallic 2/3 chance no longer distributed between two equiprobable possibilities, but it obviously doesn't make the chance of the bimetallic box having been selected change from 1/3. The total probability has to add up to 1, and there are only two possibilities for which kind of metal the second coin is made of, so it has 1/3 chance of being of the other metal, and 2/3 chance of being of the same metal. Establishing which metal the first coin is made of affects only which kind of coin the second coin is 2/3 likely to be.

 

[JeffJo]

My recognition of a difference is not due to my wanting there to be one, ...

That is a misrepresentation of what I said. Not only because I was talking about the OP and not the Box problem, but also because it was about your claim of what makes problems "different" and not the analysis of the problem.

You think that that two stories, one that asks about the statistician's fate (even though it didn't, which you still ignore) and one that asks about prisoner C's fate, make the problems represented by those stories different. I think - AND HAVE DEFINED WHY I THINK - that having the same probability space makes them the same problem. Regardless of what part of that space is, or is not, asked for in the story.

You want others to accept your opinion as a statement of fact. I defined the conditions where mine is.

But your opinion is based on this fallacious argument about a different version of the same probability space:

...and my answering a question with a more parsimonious sufficient analysis ...

It is fallacious, because your analysis that addresses the probability of a bimetalic box requires an assessment of the probability of all three boxes. Even if your analysis explicitly mentions only one.

Specifically, the dismissed 1/3 probability that you have the SS box must be reconciled somehow. Your claimed "parsimonious sufficient analysis" is based entirely on how you think probabilities can change to accomplish that reconciliation. Being explicit only about what can't change is not the same thing as not being based on the fact that others consequently must. And your analysis is wrong, because it never defines why a set can, or can't, change. You just assert that one can't - probably so you can claim a "more parsimonious sufficient analysis" that doesn't mention half of what it depends on.

The equiprobility is between the two possible kinds of coin that could have been first drawn.

No, the equiprobability is between three different boxes of unknown content.

The probability that the selected box has two of the same kind as each other is 2/3.

And the probabilities that it is GG, or that it is SS, are still 1/3 each. Making a new category does not imbue that category with any special unchangableness.

Bertrand's entire point was that you can't base an analysis on categorization alone - you have to address how different members of the same category might be affected differently. In this case, if the choice between revealing G or S from a GS box is biased, then this circuitous argument is wrong:

After the revealing eliminates one of the previously equiprobable two kinds of monometallic box, there remains only the 1/3 chance that the bimetallic box was selected to hold the possibility of the second coin being of the non-revealed kind, and the 2/3 chance that a monometallic box was selected, which at that point is still a 2/3 chance that the two coins are the same, so the revealing of the first coin establishes only which kind of metal of the original two possible kinds the second coin has a 2/3 chance of being made of.

There are theorems that tell us how to determine conditional probabilities. The problem with ignoring them is that there is no mathematical justification for what we conclude. Why not:

After the revealing eliminates SS there remains only the 2/3 chance that GG or SS was selected. Since half of this "holds the possibility of the second coin being of the non-revealed kind," the revealing of the first coin establishes that there is a 1/2 probability that the second coin is of the non-revealed kind."​

But a mathematical analysis turns out to be "more parsimonious and sufficient"

• The probability of revealing G from GG is 1/3.
• The probability of revealing S from SS is 1/3.
  
• If the probability of revealing G from GS is, say, Q/3.
• Then probability of revealing S from GS is (1-Q)/3.
• Given that G is revealed, the probability of GS is (Q/3)/[(1/3)+(Q/3)]=Q/(1+Q)
• If Q=1/2, which is all we can assume, this is (1/6)/[(1/6)+(1/3)]=2/3

The point is that even if Q=1/2, your argument is still wrong because it is not based on proven mathematics. Only on the opinions of what can, and cannot, change. Opinions that are not correct in general, but turn out to match the actual results.

 

[sysprog]

The equiprobability is between the two possible kinds of coin that could have been first drawn.

No, the equiprobability is between three different boxes of unknown content.

Yes, there is also an original equiprobability between three different boxes of unknown content, but when you introduced the term "equiprobility" to the discussion, it was in reference to the false apperception of a 50:50 chance between the two possible kinds for the second coin in the selected box. The equiprobability between the two kinds coin that can be drawn, for the first coin exists before and after a box is selected and before the first coin is drawn, and does not exist for the second coin..

The probability that the selected box has two of the same kind as each other is 2/3

.
And the probabilities that it is GG, or that it is SS, are still 1/3 each.

Together those make 2/3 of the 3/3, with the bimetallic box making up the other 1/3.

Making a new category does not imbue that category with any special unchangableness.

I didn't make a new category, and the revealing of the kind of the first coin changes only which kind of coin the second coin is 2/3 likely to be; not how likely the second coin is to be of the same kind as the first. The 1/3 likely one of each kind category is established by the problem statement, and that establishes the 2/3 likely not one of each kind category, which can legitimately be called the two of the same kind category without that being an introduction of a new category.

When we reveal a coin, only its kind is new information, and what that tells us is obviously not that the box content must be bimetallic, so that chance is still 1/3, just as it would still be 1/3 if the other kind of coin had been revealed.

If the second coin is of a different kind, that means that the box selected was of the 1/3 likely one of each kind category. If it is of the same kind, that means it was not of the 1/3 likely one of each kind category, but was of the 2/3 likely not one of each kind category. Before the revealing of the second coin, and after the revealing of the first, the second coin is 2/3 likely to be the same as the first, no matter which kind is revealed for the first coin.

 

[Stephen Tashi]

But we are discussing a puzzle, not a theorem. Unless the person who asked it is being juvenile, he thinks that enough information has been provided to result in an answer.

Wouldn't it be juvenile to think that a probability puzzle that has caused lengthy controversy among experts has a unique answer? I agree that the "game" in considering such puzzles is pick a unique answer and provide some sort of justification for it.

If that information is to be inferred, an acceptable response is "If I assume X, and can justify (but not prove) assuming it, then the answer is Y".

Discussions would be reasonable if people were that modest. However, instead of saying "If I assume X..." we often hear the claim that "X is the only possible interpretation of the problem...".

There are even customary clues used to suggest such inferences.

Apparently there are disagreements about customs.

But I will point out that I provided a sufficient description of a probability space (albeit with a typo that had an obvious correction) in my first post. Along with the reasons for the necessary assumptions.

I'll critcize your approach. I take "1","2",3","4" as the notation for outcomes you intend to use (in post #3) then we have

$S = \{1,2,3,4\}$, Pr(1) = 1/3, Pr(2) = 1/3, Pr(3) = 1/6, Pr(4) = 1/6.

Interpreting information in the problem using that notation isn't straightforward because the verbal definitions of the elements don't explicitly state all the properties of a outcome that are mentioned in the problem - and you introduce a coin, which is not mentioned in the problem.

Assertions that can be made about a single outcome in the problem are.

1) Tom will be executed
2) Dick will be executed
3) Harry will be executed
4) The guard designates Tom
5) The guard designates Dick

An explicit way to define an outcome is as a vector of 5 truth values that specify whether properties 1) through 5) hold, or don't hold.

Constraints assign some outcomes a zero probability. For example, the constraint that only two prisoners are executed assigns the outcome (true, true,true, false, true) a probability of zero.

Abbreviate "true" and "false" by "T" and "F". Assume a truthful guard.

Listing only those outcomes that may have a non-zero probability, we get a sample space having 4 elements:

e[1] = (T,T,F,T,F)
e[2] = (T,T,F,F,T)
e[3] = (T,F,T,T,F)
e[4] = (F,T,T,F,T)

An event such as "Dick will be executed" is the set of the outcomes where the corresponding property is true - e.g. "Dick will be executed" = {e[1],e[2],e[4]}.

Use p[j] to denote the probability of outcome e[j].

The unambiguous constraints of the problem can be translated into algebra.

$0 \le p[j] \le 1 , j = 1,2,3,4$
$\sum_{j=1}^4 p[j] = 1$
Pr(Tom will not be executed) = p[4] = 1/3
Pr( Dick will not be executed) = p[3]= 1/3
Pr (Harry will not be executed) = p[1] + p[2] = 1/3

Which event and corresponding probability constitute the answer to the problem is controversial. However, once an event is precisely defined, its probability can be expressed in algebra.

For example:
Pr( Harry will not be executed given the guard designates Dick) =
P( Harry will not executed and the guard designates Dick)/ Pr (Guard designates Dick)
= p[2]/ ( p[2] + p[4]).

As I said before, I suspect the constraints explicitly given in the problem are insufficient to determine a unique value for that conditional probability. If that is really a source of controversy, we can try some examples.

On the other hand, if one interprets the problem as asking for:
Pr( Harry will not be executed given (the guard designates Tom or the Guard designates Dick))
= (p[1] + p[2]) / (p[1]+p[2]+p[3]+p[4]) = (1/3)/1 = 1/3

Further assumptions about the problem can add further constraints. Your (or Gardener's) fair coin approach advocates the assumption:

P(Guard designates Tom given Harry will not be executed) = P(Guard designates Dick given Harry will not be executed)
p[1]/(p[1]+p[2]) = p[2]/( p[1] + p[2])
which is sufficient to prove p[1] = p[2] = 1/6.

Aside from the possible mathematical interpretations of problem there is the psychological or philosophical question: Suppose Harry concludes there is not enough given information to compute P(Harry will not be executed given the Guard designates Dick). If the Guard designates Dick, does Harry assert the answer to the problem is the the unconditional probability P(Harry will not be executed)?

If we were to interpret such an assumption mathematically, it assumes:
p[2]/(p[2]+p[4]) = p[1] + p[2].

However, the assumption can be made in the non-mathematical sense of "I don't know how certain information affects the answer, so I'm going to ignore that information."

 

[JeffJo]

Wouldn't it be juvenile to think that a probability puzzle that has caused lengthy controversy among experts has a unique answer?

Well, there is only one correct answer. But who do you think said there is only one solution? All I said was that a correct solution has to recognize the choice made by whoever it is that provides the informaiton.

I'll critcize your approach.

So you insist on a well-defined sample space, despite the fact that I provided one from the start. And then criticize it?

[/quote]Interpreting information in the problem using that notation isn't straightforward because the verbal definitions of the elements don't explicitly state all the properties of a outcome that are mentioned in the problem - and you introduce a coin, which is not mentioned in the problem.[/quote]I also didn't mention whether the moon was waxing or waning, which you can include in a sample space if you want to. I did provide a partition of the event space that delineates the points of interest.

 

[Stephen Tashi]

Since the Bertrand Box Problem has come up in this thread, it would be useful to describe its probability space. Use the following version of the problem:

The original, the Bertrand Box Problem. Three boxes each contain two coins. Two of them contain only one kind (value Y), one only gold coins (Y1) and one only silver coins (Y2). One contains a gold coin and a silver coin (value X). I choose a box at random, open it, and take out a gold coin. What are the chances the coin still in the box is also gold?

I'll label the Boxes: Box 1, Box 2, Box 3.
The properties that are True or False about a particular outcome are

1) Box 1 is selected
2) Box 2 is selected
3) Box 3 is selected
4) The coin revealed is gold
5) The coin revealed is silver
6) The unrevealed coin is gold
7) The unrevealed coin is silver

Assume it is Box 2 that contains 2 gold coins, Box 3 that contains 2 silver coins and Box 1 that contains one silver coin and one gold coin.

An outcome is a vector of "T","F" values corresponding to the above list of properties , as in my previous analysis in post #67. The outcomes that may have non-zero probabilities can be defined as:

d[1] = (T,F,F,T,F,F,T)
d[2] = (T,F,F,F,T,T,F)
d[3] = (F,T,F,T,F,T,F)
d[4] = (F,F,T,F,T,F,T)

Use q[j] to denote the probability of outcome d[j]. By the usual interpretation of a box being selected "at random", we are given:

Pr(Box 1 is selected) = q[1] + q[2] = 1/3
Pr(Box 2 is selected) = q[3] = 1/3
Pr(Box 3 is selected) = q[4] = 1/3

By the usual interpretation of a coin being selected "at random" from a box, we are given:

Eq. 1) Pr( the revealed coin is gold | Box 1 is chosen) = q[1]/(q[1] + q[2]) = 1/2
Eq 2) Pr( the revealed coin is silver | Box 1 is chosen) = q[2]/(q[1] + q[2]) = 1/2

We can prove q[1] = q[2] = 1/6.

Is the Bertrand Box problem isomorphic to the prisoner problem analyzed in post #67 ?

The natural try for an isomorphism between the two problems is to map d[j] to e[j] for j = 1,2,3,4. ( e[j] being an event defined in post #67 )

The constraints on the q[j] map to the constraints on the p[j] except that the Eq. 1 and Eq. 2 above only have a corresponding constraint in the prisoner problem if we assume the problem informs us (in some way - e.g. the guard tossing a coin) that

p[1]/ (p[1] + p[2]) = p[2]/(p[1] + p[2])

Whether that constraint applies is a subjective matter in the literary interpretation of the prisoner problem.

For the two problems to be isomorphic, the isomorphism must apply to the questions being asked. In the version of the Bertrand Box problem given above, we are asked to find the value of Pr(The unrevealed coin is gold given the revealed coin is gold) = d[3]/(d[1] + d[3]).

The corresponding expression in the prisoner problem can be interpreted as Pr( Harry will be executed given the guard designates Tom). Knowing that value would allow us to find Pr(Harry will not be executed given the guard designates Tom) , which is one possible interpretation of what the prisoner problem asks for.

As mentioned in post #67 , what the prisoner problem wants as an answer is a matter of literary interpretation.
-----

The above analysis doesn't settle where the Bertram's Box problem is isomorphic to the prisoner problem, but it makes clear which mathematical relations are the focus of a literary debate.

 

[sysprog]

Since the Bertrand Box Problem has come up in this thread, it would be useful to describe its probability space. Use the following version of the problem:

The original, the Bertrand Box Problem. Three boxes each contain two coins. Two of them contain only one kind (value Y), one only gold coins (Y1) and one only silver coins (Y2). One contains a gold coin and a silver coin (value X). I choose a box at random, open it, and take out a gold coin. What are the chances the coin still in the box is also gold?

I'll label the Boxes: Box 1, Box 2, Box 3.
The properties that are True or False about a particular outcome are

1) Box 1 is selected
2) Box 2 is selected
3) Box 3 is selected
4) The coin revealed is gold
5) The coin revealed is silver
6) The unrevealed coin is gold
7) The unrevealed coin is silver

Assume it is Box 2 that contains 2 gold coins, Box 3 that contains 2 silver coins and Box 1 that contains one silver coin and one gold coin.

An outcome is a vector of "T","F" values corresponding to the above list of properties , as in my previous analysis in post #67. The outcomes that may have non-zero probabilities can be defined as:

d[1] = (T,F,F,T,F,F,T)
d[2] = (T,F,F,F,T,T,F)
d[3] = (F,T,F,T,F,T,F)
d[4] = (F,F,T,F,T,F,T)

Use q[j] to denote the probability of outcome d[j]. By the usual interpretation of a box being selected "at random", we are given:

Pr(Box 1 is selected) = q[1] + q[2] = 1/3
Pr(Box 2 is selected) = q[3] = 1/3
Pr(Box 3 is selected) = q[4] = 1/3

By the usual interpretation of a coin being selected "at random" from a box, we are given:

Eq. 1) Pr( the revealed coin is gold | Box 1 is chosen) = q[1]/(q[1] + q[2]) = 1/2
Eq 2) Pr( the revealed coin is silver | Box 1 is chosen) = q[2]/(q[1] + q[2]) = 1/2
We can prove q[1] = q[2] = 1/6.

...

That's elegant and correct, but in my view it's more analysis than is necessary to answer the question, and yet it doesn't answer the question, "What are the chances the coin still in the box is also gold?", the correct answer to which is 2/3.

My disagreement with JeffJo regarding the Bertrand's box problem was not regarding the correctness or completeness of his analysis, and he and I didn't disagree about 2/3 being the correct answer to the question of what the probability of the second coin being gold is if the first coin is gold.

JeffJo apparently takes the view that some analysis that is functionally equivalent to the one that he provided is necessary to arrive at a justification for a correct answer. He consequently distinguishes between an answer and a solution. I agree with him that in ordinary discourse, when something is presented as a problem, not only an answer, but some putative basis for why the answer is correct, is a more satisfactory response than the answer presented without explanation. Even so, although I view his analysis as correct and sufficient, I also regard it to be partly superfluous. I think a more parsimonious, less complete analysis suffices to justify a correct answer of 2/3 probability that if the first coin is gold the second coin is also gold.

In the terms your citing of the problem uses, the chance of one or the other of the two Y values being selected is 2/3, by simple recognition that they are two out of three, and when a coin is revealed, that establishes which of the two equally likely Y values, the Y1 value or the Y2 value, the selected box has a 2/3 chance of containing. It doesn't change the 2/3 chance that a Y value box, either Y1 or Y2, was selected. Whichever one box was selected, it had only a 1/3 chance of being selected, but there remains after the revealing, just as before the revealing, a 2/3 chance that a Y value box was selected. The revealing tell us nothing about whether the coin came from a Y value box or from the X value box, so that distribution remains 2/3 to 1/3.

Showing one of the coins obviously can't change the 1/3 chance of value X, because there's only one of the 3 boxes with value X, and its 1/3 likelihood is the same regardless of which kind of coin is revealed.

Labeling the box containing value X, Box 1, as you did for part of the analyses you presented: "Pr(Box 1 is selected) = 1/3", though clearly correct, doesn't require subdivision into the q[1] and [q2] individual probabilities for the revealed coin; doing that neither adds nor detracts from the fact that before and after a coin is revealed, the probability that whichever kind of coin it is, it came from Box 1, is 1/3.

If and only if the X value box is the selected box, the second coin can and must be different from the first. There being only a 1/3 chance that the X value box was selected, the second coin has only a 1/3 chance of being different, and therefore must have a 2/3 chance of being the same. Consequently, if the revealed coin is gold, the chance of the other coin in that box also being gold is 2/3.

JeffJo apparently prefers an approach which accounts for and corrects the specific error of thinking the second coin is 1/2 likely to be the same as the first. I regard that error to be a cognitive illusion that my approach avoids rather than addresses. I disagree with JeffJo in his contention that my approach only accidentally or by unsupported intuition arrives at the uniquely correct answer of 2/3; however, I do not as fully disagree with his apparent position that, if we wish to diagnose and correct the cognitive illusion by which some respondents arrive at the wrong answer of 1/2, we must do something equivalent to the analysis that he has proposed. I think the avoidance of error in my approach suffices for mere correction; I agree with JeffJo in his apparent idea that his approach offers better promise for convincing persons who apperceive the wrong answer of 1/2 to be correct.

I think the error of supposing the probability to be 1/2 for the second coin being gold if the revealed coin is gold, arises from first correctly recognizing that the elimination of the Y2 (both silver) value box, by the revealing as gold, of a coin drawn from the selected box, means that the selected box must be either the Y1 (both gold) value box, or the X value box, and then incorrectly redistributing the probability of the eliminated Y2 value box equally over the two still possible Y1 value and X value boxes.

If the revealing of one kind of coin increases to 1/2 the 1/3 likelihood of the X value box being the selected box, then so would the revealing of the other kind of coin, and that would mean that the X value box was already 1/2 likely to begin with, just as the revealed coin, before it is revealed, is 1/2 likely to be of one kind and 1/2 likely to be of the other kind, and the problem statement puts the starting likelihood of the X value box being selected at 1/3, wherefore adjusting it to 1/2 upon the revealing of a coin is unjustified.

That explanation doesn't go through the full analysis, inclusive of the 1/6 probabilities, as that provided by JeffJo, and by you, does, but it goes further in analysis than is necessary for, with adequate justification, arriving at the correct answer.