B Probability and Death Sentences

benorin · Mar 15, 2018

Originally posted by Galileo in the thread I started called Bad Math Jokes on top of pg. 4:
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Not so much a joke as a brainteaser.

Three prisoners, strangers to each other, were suspects of a murder case. One day they came to hear that a sentence has been drawn. Two of them have been found guilty and will be executed, but they don't know which of the two . One guy, a statistician, figures his chances for survival are 1/3, so he goes to the bars of his cell and hails the guard: "Hey psst, do you know which of us has been sentenced?".
"Eh, yes.", says the guard, "But I'm not allowed to tell you.".
"Tell you what", says the guy, "I already know that 2 of us will executed, that means at least one of the other guys will be. I don't know them or anything, surely you can point to one which is guilty?". The guard sees no harm in that and points one of the prisoners, "He is guilty".
"Thanks!", proclaims the statistician, "my chances have just increased to 1/2".
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If you've got the time, throughout the next several pages of the original thread linked above starting on pg. 4 there are varying analyses of this 'brainteaser' (that do not all agree) and I felt after reading every relevant post on this that we hadn't really gotten to the bottom of it, or clearly I did not understand it if we did for I am left wanting a convincing analysis. Please explain, citing any formulae beyond the basics: I have taken an elementary probability and statistics course, and I know some analysis if need be.

Thanks for your time,
-Ben Orin

StoneTemplePython · Mar 15, 2018

This seems to be a thinly veiled version of the Monty Hall problem. Typical remedies:

1.) coding a simulation-- making the problem crystal clear from a computational procedure standpoint, and then run the simulation many times, averaging the result -- this is what convinced Erdos.

2.) After doing 1, re-writing the problem in a Bayesian Framework.
- - - -
I'd suggest you do number 1 first and share it. I can help in the Bayesian formulation after.

JeffJo · Mar 16, 2018

This is, in fact, the predecessor to the Monty Hall Problem. It was published by Martin Gardner in Scientific American in 1959, with two important facts that you left out:

If the statistician is to be set free, the guard should flip a fair coin to decide which of the other two prisoners he will point to.
The coin should be flipped even if the statistician is guilty, so the action doesn't give anything away.

It doesn't really matter if these facts are included or not, since they are necessary assumptions anyway. It is the recognition that this information is critical that makes it important. Gardner's inclusion of it is the reason his problem was not controversial and so is not as well remembered. Marilyn vos Savant did not include it in her version (which was not the original) of the Monty Hall Problem, and the controversy over it is a direct result of this omission.

The correct solution, seldom given for MvS's Monty Hall Problem even when the correct answer is arrived at, is that there are four possibilities. Say the three prisoners are named Tom, Dick, and Harry (the statistician):

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

Note that there is a 50% chance that the guard will point to either one of the other two prisoners. The point is that the information is not just that the guard can point to one - a guaranteed fact - but that the guard chose the one he pointed to. But that 50% is divided into two cases, and the case where Harry is guilty is twice as likely as the case where he is not guilty.

The usual solution - that Harry's chances can't change - is correct only if you make the assumption that a fair coin (or some other 50:50 mechanism) is used in cases 3 and 4. If a two-headed coin is used, Harry's chances are 1/2 if the guard points to Dick, and 0 if he points to Tom.

PeroK · Mar 16, 2018

@JeffJo although it's a neat approach, there is no need for the guard to toss a coin. He could simply go through the list of prisoners in alphabetical order. The important thing is that Harry doesn't know and can't guess the system.

Although, if the guard's system is not random, the problem doesn't stand up to repeated experiments, as in the Monty Hall game.

JeffJo · Mar 16, 2018

PeroK said:

@JeffJo although it's a neat approach, there is no need for the guard to toss a coin. He could simply go through the list of prisoners in alphabetical order. The important thing is that Harry doesn't know and can't guess the system.

And that is what I meant by "or some other 50:50 mechanism". The point is to separate cases #3 and #4, and establish the probabilities to be 1/6 probability each. The mistake is to ignore that they are separate cases. The significance of Gardner mentioning the coin is more in how it draws attention to separating them, than in establishing these probabilities.

Although, if the guard's system is not random, the problem doesn't stand up to repeated experiments, as in the Monty Hall game.

It is a common mistake to think "probability" means "under repeated experiments," and that "random" means "uniformly distributed." Mine is the correct solution, and solves the problem even if the coin is "random" but unfair. If the coin lands on heads with probability Q, Harry's chances are Q/(1+Q) if the guard points to Dick, and (1-Q)/(2-Q) if he points to Tom.

StoneTemplePython · Mar 16, 2018

JeffJo said:

The correct solution, seldom given for MvS's Monty Hall Problem even when the correct answer is arrived at, is that there are four possibilities. Say the three prisoners are named Tom, Dick, and Harry (the statistician):

Tom is not guilty, the guard ignores the coin and points to Dick. Probability: 1/3

Dick is not guilty, the guard ignores the coin and points to Tom. Probability: 1/3

Harry is not guilty, the coin lands on heads and the guard points to Dick. Probability: 1/6

Dick is not guilty, the coin lands on tails and the guard points to Tom. Probability: 1/6

Note that there is a 50% chance that the guard will point to either one of the other two prisoners. The point is that the information is not just that the guard can point to one - a guaranteed fact - but that the guard chose the one he pointed to. But that 50% is divided into two cases, and the case where Harry is guilty is twice as likely as the case where he is not guilty.

The usual solution - that Harry's chances can't change - is correct only if you make the assumption that a fair coin (or some other 50:50 mechanism) is used in cases 3 and 4. If a two-headed coin is used, Harry's chances are 1/2 if the guard points to Dick, and 0 if he points to Tom.

To be crystal clear, there are multiple ways of correctly solving this -- i.e. multiple approaches that get to the correct answer. The Bayesian approach gets you the same net result, and in that interpretation, the coin toss serves to clarify the likelihood function of the guard. I had suggested that OP first code up the problem as that process also forces one to think on the likelihood function.

Btw, I think your number 4 has a typo, and should say "4. ~~Dick~~ Harry is not guilty, the coin lands on tails and the guard points to Tom. Probability: 1/6"

PeroK · Mar 16, 2018

JeffJo said:

And that is what I meant by "or some other 50:50 mechanism". The point is to separate cases #3 and #4, and establish the probabilities to be 1/6 probability each. The mistake is to ignore that they are separate cases. The significance of Gardner mentioning the coin is more in how it draws attention to separating them, than in establishing these probabilities.

It is a common mistake to think "probability" means "under repeated experiments," and that "random" means "uniformly distributed." Mine is the correct solution.

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

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

You could have a furious TV debate about whether there are infinitely many primes, but that doesn't raise any mathematical issues, per se.

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

PeroK · Mar 16, 2018

... 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.

JeffJo · Mar 16, 2018

StoneTemplePython said:

To be crystal clear, there are multiple ways of correctly solving this -- i.e. multiple approaches that get to the correct answer.

Certainly. But they all must include a way to describe how the guard chooses between Tom and Dick, when Harry is not guilty, with a 50:50 probability distribution. Simply referring to "the Bayesian approach" does not do this, unless you describe how to do so.

The most commonly-claimed solution that gets the right answer is a variation of "Your initial probability can't change." Even though it gets the right answer, it is wrong unless it explains that there are two possible choices (Tom and Dick) that the guard could indicate, and they must be equally likely from Harry's point of view.

Btw, I think your number 4 has a typo, and should say "4. ~~Dick~~ Harry is not guilty, the coin lands on tails and the guard points to Tom. Probability: 1/6"

Oops. You are right. My fault for using cut-and-paste.

PeroK said:

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

If Harry recoginzes that there is a choice, but has no knowledge of it, then to him it is a 50:50 process. My point is that he has to recognize the choice, and realize that (to him) it is a 50:50 prospect.

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

There are many variations, but to be correct they must recognize the choice.

PeroK · Mar 16, 2018

JeffJo said:

There are many variations, but to be correct they must recognize the choice.

"Choice" is an interesting word. I've always thought in terms of "deliberate" and "accidental". As in the variation, where you draw a card from a normal deck and calculate the probability it is the Ace of Spades.

1) Your friend places the deck face down and draws the cards one at a time. As each card is revealed, the probability you have the Ace increases. "Accidentally" showing you cards that are not the Ace.

2) Your friend looks at the cards and shows you one by one cards that are not the Ace of Spades. You're still only 1/52, even after he's shown you 50 cards. "Deliberately" showing you cards that are not the Ace.

In both cases, your friend could be said to be "choosing" the cards, but it's whether he knows what they are or not that is the key difference.

In any case, my analysis of Monty Hall and its variants is that the key issue is to recognise which of these scenarios you are dealing with.

Often the wrong analysis of Monty Hall is simply a (valid) analysis of the wrong problem.

JeffJo · Mar 19, 2018

PeroK said:

"Choice" is an interesting word.

No, it is a word that has a very specific meaning, but is often used to mean something else.

The guard knows who will be set free. If it is not Harry, there is only one other prisoner he can point to. If it is, he must have a method to choose between Tom and Dick. You are right that, if Harry doesn't know this method (and that includes knowing that a coin flip is used; but not if the coin came up Heads or Tails, or which was assigned to Tom or Dick), then he can model the choice with a 50:50 distribution. But he is still modeling a choice.

1) Your friend places the deck face down and draws the cards one at a time. As each card is revealed, the probability you have the Ace increases. "Accidentally" showing you cards that are not the Ace.

Not quite.

There is a 1/52 chance that it is the first card. It is true that, if you draw a second card, there is a 1/51 conditional probability that it is the Ace of Spades. But there is only a 51/52 chance that you will draw it, making it a 1/52 chance of reaching that point.

2) Your friend looks at the cards and shows you one by one cards that are not the Ace of Spades. You're still only 1/52, even after he's shown you 50 cards. "Deliberately" showing you cards that are not the Ace.

Only if you assume he had free choice of what card to show you.

This is ambiguous. What is it that "you" have, that you are asking a probability for? So, say you draw a card, but don't look. Your chances are 1/52.

You ask friend to show you a card. He shows you the Seven of Hearts. Your chances are still 1/52.
You ask friend to show you the Seven of Hearts. He does. Your chances are 1/51 (because there is a 1/52 chance he can't).

The difference is that friend had a choice in the first variation, but not in the second. If you repeat this until N cards are shown, your probability stays 1/52 if he has a choice (see note below), but is either 1/(52-N), if he can show all the cards you name, or 0 if he can't.

In any case, my analysis of Monty Hall and its variants is that the key issue is to recognise which of these scenarios you are dealing with.

And mine is that the key issue you are recognizing is whether a choice is involved, that could have:

Revealed a different door than Monty Hall opened.
Pointed to a different prisoner than the guard pointed to.
Revealed a different card than was turned up.

+++++

Note: Only if the cards are revealed in an unbiased manner. If he reveals 2c, 2d, 2h, 2s, 2c, 3d, ... then at some point you need to assess his motivations.

sysprog · Apr 2, 2018

The statistician recognizes a group of 3 persons consisting of himself and 2 others, 2 of which group have been adjudged guilty. He requests of the guard a pointing out of one of the 2 guilty who is not the statistician. He says he already knows that (at least) one of the others is guilty, and again requests of the guard that he point out one of the other 2 who is guilty.

Assuming that the guard does know, and is being truthful, the statistician learns only that a specific one of the other 2 is guilty. He does not thereby know whether the guard pointed out, of the 2 prisoners who are not the statistician, the only guilty one, or one of 2 such. His chances after the identification are the same as before. Whether the guard chooses or selects among 2 qualifying possibilities, or merely points out the only qualifying possibility, is immaterial. The statistician can't do anything, predicated on the information thus given, that might affect the outcome.

The Monty Hall problem is different, in that the contestant makes a decsion to switch or not switch based on further information, whereas in the stated prisoner problem, no option for a prisoner is postulated. If the prisoner problem provided that the statistician was allowed to, after a guilty prisoner was pointed out, opt to swap his verdict, or not, with that of the non-pointed-out other prisoner, that would be similar to the Monty Hall problem.

Looking at the inverse/converse/dual: if there are 2 not guilty and only one guilty, the statistician with a similarly flawed analytic outlook would not ask the guard to point out a not guilty, because if the guard were to comply, the statistician's chances of being not guilty would reduce from 2/3 to 1/2.

If we revise the problem so that all 3 prisoners are statisticians, and so that each has, in isolation, unbeknownst at first to the other 2, the same conversation with the guard with similar ensuing actions of the guard, then we are left with the absurdity that each of the 3 prisoners, 2 of whom have been adjudged guilty, believes himself to have only 1/2 liklihood of having been adjudged guilty, while the guard silently considers how much easier to handle statisticians are compared to pessimists, until he decides to point out to each of them, each still in isolation, that he's done honestly as requested by each of them, while it remains that 2 of them are guilty, so no-one contemplating 50-50 chances can be right.

JeffJo · Apr 2, 2018

sysprog said:

Assuming that the guard does know, and is being truthful, the statistician learns only that a specific one of the other 2 is guilty.
...
His chances after the identification are the same as before.

He also learns that the guard chose that specific prisoner, which is still my point.

What if one of the other prisoners was accused of killing the guard's sister, and the guard sneered as he pointed to him and said "That one is guilty." Wouldn't that make you suspect he would always point to him, regardless of whether the statistician or the other prisoner is the one to be freed? In this case, the statistician's chances do change.

I'm not suggesting that this is the case, or that your result isn't the correct result. I'm saying that you need to recognize the 50:50 choice to understand why the answer is what it is. The more complete version of your logic is an analysis called Bertrand's Box Paradox. (And Bertrand used the word "paradox" to refer to this kind of an analysis, not to the problem he used to illustrate it).

What if the statistician's chances do change to 1/2 when the guard points to the prisoner named Tom? Then the same logic, and so the same answer, applies when he points to the prisoner named Dick. Since (as you point out) he can always point to one of them, and (part of what you left out) these are his only two options, the statistician can conclude that his chances RIGHT NOW are the average of his chances in those two cases. But that average is 1/2, so his chances RIGHT NOW must be 1/2. Since his chances are actually 1/3, there is something wrong in the logic used to deduce the change.

The subtle point here, is that this does not describe how his chances can still be 1/3 after the guard points out a prisoner, just that they can't change from what they were before.

Let T, D, and H be the events where Tom, Dick, and Harry are to be freed. And PT the event were the guard points to Tom after Harry's request. The start of the correct Bayesian analysis is:

Bayes Law says:

But the guard can't point to T if he is to be freed, so Pr(PT|T)=0:

Even if they don't go into this level of detail, people who don't recognize the difference between being able to point to Tom, and choosing to point to Tom, will "eliminate" the terms involving Pr(PT|T) in the right-hand side, and "keep" the terms involving Pr(PT|D) and Pr(PT|H):

Pr(H|PT) = Pr(H) / [Pr(D) + Pr(H)] = (1/3) / [(1/3) + (1/3)] = 1/2.
Pr(D|PT) = Pr(D) / [Pr(D) + Pr(H)] = (1/3) / [(1/3) + (1/3)] = 1/2.

My point is that this result is a conceptually valid Bayesian analysis, but with an error in it. My point is that these people don't see what that error is, just the valid analysis. Your result is based entirely on the assertion "His chances after the identification are the same as before." You provided no analysis or substantiation of that assertion.

If these people don't accept your assertion, all they see is that they performed a valid analysis where you did not, so their answer is preferable. If they do accept it, they see two differing answers from valid logic, so something must be wrong with the field of Probability. Either way, there is no reason to accept your result.

The error made in the Bayesian analysis, is that "eliminating" and "keeping" terms is incorrect - you need to use values for Pr(PT|D) and Pr(PT|H).

But Pr(PT|D)=1 and Pr(PT|H)=1/2, so

Pr(H|PT) = (1/2)*(1/3) / [(1)*(1/3) + (1/2)*(1/3)] = (1/6) / [(1/3) + (1/6)] = 1/3.
Pr(D|PT) = (1/2)*(1/3) / [(1)*(1/3) + (1/2)*(1/3)]= (1/3) / [(1/3) + (1/6)] = 2/3.

The Monty Hall problem is different, in that the contestant makes a decsion to switch or not switch based on further information, whereas in the stated prisoner problem, no option for a prisoner is postulated.

What the contestant, or the statistician, wishes to do with the updated set of probabilities has no impact whatsoever on how the probabilities are updated. They are the exact same problem.

sysprog · Apr 3, 2018

JeffJo said:

He also learns that the guard chose that specific prisoner, which is still my point.

What if one of the other prisoners was accused of killing the guard's sister, and the guard sneered as he pointed to him and said "That one is guilty." Wouldn't that make you suspect he would always point to him, regardless of whether the statistician or the other prisoner is the one to be freed? In this case, the statistician's chances do change.

That supposition is not consistent with the problem statement, which postulates that the guard always indicates a guilty one of the two other prisoners as guilty, without disclosing whether that prisoner is the only other prisoner who is guilty, or is one of two such. According to the problem statement, there is a 1/3 chance that anyone prisoner is not guilty. If the guard always chooses one particular one of the two other prisoners in the event that he has a choice, and the inquirer knows in advance which prisoner that is, if that prisoner is not indicated, that would mean that he was not guilty, and that the inquirer was definitely the other guilty prisoner.

I'm not suggesting that this is the case, or that your result isn't the correct result. I'm saying that you need to recognize the 50:50 choice to understand why the answer is what it is

I disagree with that contention. Although there is a 50:50 chance between the equally likely possibilities that both of the other prisoners are guilty, and that only one of them is, indication of one of the two other prisoners as guilty does nothing whatever toward establishing which of those possibilities is the case, and so does not in any way affect the 1/3 chance that the statistician has of not being guilty. Knowing whether the indicated prisoner is the only guilty one of the two others, or is, to the contrary, one of two such, would establish for certain whether the inquirer is guilty or not guilty, but without knowing that, knowing that one in particular of the two others is guilty does nothing to change the fact that together the two of them hold two of the 1/3 chances to be not guilty, while the inquiring statistician continues to hold the other 1/3 chance.

The more complete version of your logic is an analysis called Bertrand's Box Paradox. (And Bertrand used the word "paradox" to refer to this kind of an analysis, not to the problem he used to illustrate it).

What if the statistician's chances do change to 1/2 when the guard points to the prisoner named Tom? Then the same logic, and so the same answer, applies when he points to the prisoner named Dick. Since (as you point out) he can always point to one of them, and (part of what you left out) these are his only two options, the statistician can conclude that his chances RIGHT NOW are the average of his chances in those two cases. But that average is 1/2, so his chances RIGHT NOW must be 1/2. Since his chances are actually 1/3, there is something wrong in the logic used to deduce the change.

It is not true that the statistician may legitimately conclude that his present chances are the arithmetic mean average of "his chances in those two cases"; there are three cases to be tallied in calculating the average, and the total of the three 1/3 chances is 1, and 1 divided by 3 is 1/3, so the average is 1/3. It is from the outset not possible that niether of them is guilty, so the ostensive percept that one holder is now eliminated as a possible sole holder of the non-guilty 1/3 chance is specious, in that it disregards the initial bias in the selection process, in that the statistician was part of the initial probability distribution, and was not among the possible pointees when one of the guilty was pointed out.

The subtle point here, is that this does not describe how his chances can still be 1/3 after the guard points out a prisoner, just that they can't change from what they were before.

I regard that point as not subtle, but obvious, and the false analysis upon which its putative subtlety depends, as merely obfuscational.

Let T, D, and H be the events where Tom, Dick, and Harry are to be freed. And PT the event were the guard points to Tom after Harry's request. The start of the correct Bayesian analysis is:

You're calling the start after an error has already occurred. As you say later, "the guard can't point to T if he is to be freed", and that fact is not given due account prior to something subsequent which you pre-label as "correct".

Bayes Law says:

Pr(H|PT) = Pr(PT|H)*Pr(H) / [Pr(PT|T)*Pr(T) + Pr(PT|D)*Pr(D) + Pr(PT|H)*Pr(H)]
Pr(D|PT) = Pr(PT|D)*Pr(D) / [Pr(PT|T)*Pr(T) + Pr(PT|D)*Pr(D) + Pr(PT|H)*Pr(H)]
But the guard can't point to T if he is to be freed, so Pr(PT|T)=0:

Pr(H|PT) = Pr(PT|H)*Pr(H) / [Pr(PT|D)*Pr(D) + Pr(PT|H)*Pr(H)]
Pr(D|PT) = Pr(PT|D)*Pr(D) / [Pr(PT|D)*Pr(D) + Pr(PT|H)*Pr(H)]

Even if they don't go into this level of detail, people who don't recognize the difference between being able to point to Tom, and choosing to point to Tom, will "eliminate" the terms involving Pr(PT|T) in the right-hand side, and "keep" the terms involving Pr(PT|D) and Pr(PT|H):

Pr(H|PT) = Pr(H) / [Pr(D) + Pr(H)] = (1/3) / [(1/3) + (1/3)] = 1/2.
Pr(D|PT) = Pr(D) / [Pr(D) + Pr(H)] = (1/3) / [(1/3) + (1/3)] = 1/2.

A false premise entails anything.

My point is that this result is a conceptually valid Bayesian analysis, but with an error in it.

I reject that. It cannot be "conceptually valid" and also in error, unless the term "conceptually" is condescendingly being used to accommodate incorrect reasoning. If it's wrong, but a person unwilling or unable to apply adequate intellection regarding the matter might suppose otherwise, it's still wrong, and the prisoner is still just as 2/3 likely to be guilty, and just as 1/3 likely not to be, whether he is suscepted to a sophistry or not.

My point is that these people don't see what that error is, just the valid analysis.

I think I understand your intended anthropological point regarding "these people" not understanding, but I don't see a good reason for your descending into error with them by referring to their "valid analysis", and also distancing yourself from them by acknowledging that the analysis contains an error that renders it invalid. Proclaiming that you understand how others have erred in an analysis does not require any endorsement of a resultant invalid analysis as valid. The analysis presented is invalid from the outset, in that it presents the pointing out of one of the three prisoners as if it had no prior constraint, when in fact, the pointee can only have been one of T or D, and also cannot have been T if T were the one of the three prisoners who is not guilty.

Your result is based entirely on the assertion "His chances after the identification are the same as before." You provided no analysis or substantiation of that assertion.

It's based on the problem statement, and on simple showing of the invalidity of the statistician's incorrect analysis.

If these people don't accept your assertion,

I'm not running for office here.

all they see is that they performed a valid analysis

Again you refer to an analysis as a valid analysis despite your having also announced yourself to have recognized it to be invalid.

where you did not,

Disagree. I presented a correct critique of an invalid inferential construct. That is in fact, albeit minimally, a second order analysis.

so their answer is preferable.

There's no accounting for personal preference. What is true or untrue, valid or invalid, does not depend on preference, but you presumably already knew that; I view your remarks about preferences as anthropological opinions or observations.

If they do accept it, they see two differing answers from valid logic, so something must be wrong with the field of Probability.

A contradiction entails anything. In fact, they see one invalid analysis, and one valid critique thereof.

Either way,

You're apparently contending that there is a class of persons with some part of whose incorrect reasoning you sympathise but do not endorse, such that when they are confronted by my correct reasoning will either reject is as inconsistent with their own invalid reasoning that they incorrectly believe to be valid, or if they do accept it, they will do so only superficially, without accepting its consequence that the reasoning with which it conflicts must be incorrect, and will, despite accepting my reasoning as valid, also continue to accept as valid the conflicting reasoning they have already embraced, and so will perceive a contradiction, which they will attribute to the invalidity of the entirety of probability theory.

Apparently you discount the possibility that they might understand the explanation, recognize their previous reasoning as invalid, and accept the fact that the chance of the statistician of being the one not adjudged guilty, despite his specious reasoning by which he purports to establish otherwise, after the pointing out of one of the prisoners as guilty, continues to be 1/3, just as it was all along.

there is no reason to accept your result.

Disagree. The reason to accept the result (by the way, there cannot be a result of an analysis if there was no analysis) is that it is correct, and that if it is viewed as a result, it is a result of having showing a contrary result to have been predicated upon invalid analysis. I showed simply that the initial condition of 1/3 chance each of not being guilty was not altered by the interim pointing out to the inquirer of one of the two others as guilty, and that proposed departures from that were predicated upon invalid analysis, and I explained how so.

The error made in the Bayesian analysis, is that "eliminating" and "keeping" terms is incorrect - you need to use values for Pr(PT|D) and Pr(PT|H).

Pr(H|PT) = Pr(PT|H)*Pr(H) / [Pr(PT|D)*Pr(D) + Pr(PT|H)*Pr(H)]
Pr(D|PT) = Pr(PT|D)*Pr(D) / [Pr(PT|D)*Pr(D) + Pr(PT|H)*Pr(H)]
But Pr(PT|D)=1 and Pr(PT|H)=1/2, so

Pr(H|PT) = (1/2)*(1/3) / [(1)*(1/3) + (1/2)*(1/3)] = (1/6) / [(1/3) + (1/6)] = 1/3.
Pr(D|PT) = (1/2)*(1/3) / [(1)*(1/3) + (1/2)*(1/3)]= (1/3) / [(1/3) + (1/6)] = 2/3.

Regarding the Monty Hall problem, which you mentioned, in comparison to the prisoner problem, I said:

"The Monty Hall problem is different, in that the contestant makes a decsion to switch or not switch based on further information, whereas in the stated prisoner problem, no option for a prisoner is postulated."

and in your post you said:

What the contestant, or the statistician, wishes to do with the updated set of probabilities has no impact whatsoever on how the probabilities are updated.

In the prisoner problem, the statistician prisoner is not given an option, whereas in the Monty Hall problem, the contestant does not merely wish; the contestant decides whether or not to switch doors.

They are the exact same problem.

They are NOT the same problem. If the prisoner problem included the provision that the statistician, after one of the two other prisoners was pointed out as guilty, would be given the option to trade verdicts with the other prisoner not pointed out as guilty, then and only then, the two problems would be equivalent.

JeffJo · Apr 3, 2018

sysprog said:

That supposition is not consistent with the problem statement, ...

Didja see where I said "I'm not suggesting that this is the case" ? I was merely pointing out the role of the choice. You are using a strawman argument.

I disagree with that contention.

Didja see the Bayesian analysis? The one that provided the only actual calculation of a probability, instead of a handwaving rationalization for why it could be what it is? I'll repeat the result for you:

Most of these terms have fixed values at the point in time when Harry asks the guard for information:

Pr(H) = Pr(D) = 1/3 because there is no information about which prisoner is to be freed. These terms divide out.
Pr(PT|D) = 1 because there is no other prisoner that the guard can point to if Dick is to be freed.

These equations reduce to:

So I'll repeat my contention, with a bit more specificity. These equations - or something equivalent that includes the term Pr(PT|H) - represent the only mathematical solution for the conditional probabilities involved.

... indication of one of the two other prisoners as guilty does ... not in any way affect the 1/3 chance that the statistician has of not being guilty.

More accurately, we require such an indication that is equally likely to indicate either of the other two prisoners. All I'm saying is that you can't ignore what I added. Your version of the statement is an assertion, not a mathematical fact. To make it a mathematical fact, you need the implicit assumption you are avoiding.

Put another way, this is a matter of how to apply the Principle of Indifference: "The principle of indifference states that if the n possibilities are indistinguishable except for their names, then each possibility should be assigned a probability equal to 1/n." While I think that could be better worded, misapplying the part I emphasized is the root of the entire controversy over these two completely-equivalent problems. Those who think the answers are 1/2 are applying it to Dick and Harry (or the contestant's door and the switch door) without addressing whether they are "indistinguishable except for their names." You are applying it to the choice between pointing to Dick or to Tom without acknowledging why you do. You just assert the result that follows from it.

It is not true that the statistician may legitimately conclude that his present chances are the arithmetic mean average of "his chances in those two cases"; ...

I debated how much detail to put into that. I left one word out that would have required further exposition, but only if you contradiction yourself about necessary assumptions. I tend to be long-winded, and didn't expect you to contradict other things you have asserted. So I left it out.

The Law of Total Probability says that the probability of event A, when you consider whether event B may or may not occur, is Pr(A) = Pr(A|B)*Pr(B) + Pr(A|~B)*Pr(~B). Here, A is the probability RIGHT NOW that Harry (the statistician) is to be freed. Assuming the guard agrees to point to a prisoner, if B is "the guard will point to Tom", then ~B is "The guard will point to Dick". Since Pr(B)+Pr(~B)=1, Pr(A) is the weighted average of the answers Harry would arrive at in the two possibilities. Even if you contradict yourself and suppose Pr(B)<>Pr(~B), since you assert that we know Pr(A|B)=Pr(A|~B), it means that when the guard agrees to point to a prisoner, but before he does, Pr(A) changes to Pr(A|B).

I regard that point as not subtle...

And so you missed the subtle point entirely. All your analysis proves is that any set of numerical answers other than {1/3, 2/3} is incorrect. But it does not show that {1/3,2/3} is correct. The problem could be unsolvable.

You're calling the start after an error has already occurred. As you say later, "the guard can't point to T if he is to be freed", ...

With all due respect, you do understand how and when we use probabilities, don't you? If I flip a coin, but keep my hand over it after looking, to you the probability that it is Heads is 50%, but I know if it is, or isn't. The point is that Harry doesn't know if Tom is to be freed, so a correct analysis must use a probability to represent that event.

It cannot be "conceptually valid" and also in error, ...

It is conceptually valid to solve a conditional probability problem by enumerating cases, assigning probabilities to each, removing those that are proven to be impossible by the information you gain, and normalizing the probabilities of those that are still (completely) possible. The error is in not recognizing why some cases may not be completely possible. In our problem, this error is the result of not recognizing the case where the guard had a choice. Which was, and still, is, my original point.

And unless you can point out the error - which you don't - the result looks correct.

Apparently you discount the possibility that they might understand the explanation, recognize their previous reasoning as invalid, ...

I do no such thing. Among the many possibilities, after the initial impression that the answers are {1/2,1/2} (call this X) and hearing your analysis (call it Y), are:

Acceptance of Y and rejection of X.
Acceptance of Y but uncertainty why X isn't correct.
Insistence on X and rejection of Y.
Complete disillusionment over the field of Probability.
Others?

My point is that none of these result in a better understanding of probability, and many result in less understanding.

"The Monty Hall problem is different, in that the contestant makes a decsion to switch or not switch based on further information, whereas in the stated prisoner problem, no option for a prisoner is postulated."

And that decision depends on the probabilities we are asked to determine, based on a situation that is 100% equivalent to the Three Prisoners Problem. Anything that occurs after these probabilities are determined is irrelevant to how we determine them.

They are NOT the same problem.

They are the same problem. They suggest different courses of action after the problem is solved, but that has nothing to do with solving the problem.

sysprog · Apr 3, 2018

JeffJo said:

...
They are the same problem. They suggest different courses of action after the problem is solved, but that has nothing to do with solving the problem.

They are not the same problem. The problems are similar in many respects, including the similarity that in the prisoner problem, the chance of being not guilty of the one of the two other prisoners who was not pointed out as guilty goes to 2/3 after the pointing out, just as in the Monty Hall problem, the chance of the non-selected door having the good prize goes to 2/3 after one of the non-selected doors is opened to reveal a zonk prize, but the problems are in fact different, in that in the prisoner problem, the statistician is NOT offered the option to swap verdicts with the non-pointed-out-as-guilty other prisoner, whereas in the Monty Hall problem, the contestant IS offered the option to switch doors. The question in the prisoner problem is whether the statistician in stating his chances after the pointing out as having improved to 50% is right, to which the correct answer is NO, whereas the question in the Monty Hall problem is, after a zonk prize door is opened, and the contestant is then offered the option to switch doors, should the contestant switch, to which the correct answer is YES.

JeffJo · Apr 4, 2018

sysprog said:

They are not the same problem.

Please, identify one thing that is different, at the point in the story where you are supposed to determine a probability.

In each, there are three mutually exclusive cases; two of which are undesirable, and one that is very desirable.
1. In the Prisoners problem, Tom could be freed (undesirable), Dick could be freed (undesirable), or Harry could be freed (desirable).
2. In the MHP (assume you pick door #s), the car could be behind door #1 (undesirable), #2 (undesirable), or #3 (desirable).
In each, an outside agent, who knows which case it is, is required to reveal information that precludes one of the cases.
1. In the Prisoners problem, say the points to Tom.
2. In the MHP, say Monty Hall opens door #1.
In each, we are asked to update the probabilities for the two remaining cases.
1. In the Prisoners problem, there is still a 1/3 chance that Harry will be freed, but a 2/3 chance that Dick will be freed.
2. In the MHP, there is still a 1/3 chance that the car is behind door #3, but a 2/3 chance it is behind door #2.

THIS IS THE POINT IN THE NARRATIVE WHERE WE ARE ASKED TO PROVIDE AN ANSWER.

4. Based on that answer, the responses of the participants are indeed different.

If Harry understands probability, his well being is changed only if he cares about Tom's and/or Dick's future. If he doesn't, he gains false hope.
If the contestant understands probability, she will switch doors and double her chances. If she doesn't she may do either, but studies of human nature suggest she won't. Either way, her fortunes haven;t gotten any worse.

BUT WE AREN'T ASKED ABOUT HARRY'S FEELINGS, OR THE CONTESTANT'S FORTUNES. WE ARE ASKED TO DETERMINE THE PROBABILITIES.

The question in the prisoner problem is whether the statistician in stating his chances after the pointing out as having improved to 50% is right, to which the correct answer is NO, whereas the question in the Monty Hall problem is, after a zonk prize door is opened, and the contestant is then offered the option to switch doors, should the contestant switch, to which the correct answer is YES.

Really? The difference you insist exists is that you think one problem asks yes/no question, but the other asks the same question in "no/yes" form?

And so they would be the same problem if the Prisoners problem asked "Is the statistician wrong?" And please note, when you address this last bit, that neither question was asked in the OP.

PeroK · Apr 4, 2018

JeffJo said:

Please, identify one thing that is different, at the point in the story where you are supposed to determine a probability.

To the best of my knowledge no one has been executed on the Monty Hall show.

sysprog · Apr 4, 2018

JeffJo said:

Really? The difference you insist exists is that you think one problem asks yes/no question, but the other asks the same question in "no/yes" form?

No; as I said in in my first post in this thread (post #12):

The Monty Hall problem is different, in that the contestant makes a decsion to switch or not switch based on further information, whereas in the stated prisoner problem, no option for a prisoner is postulated. If the prisoner problem provided that the statistician was allowed to, after a guilty prisoner was pointed out, opt to swap his verdict, or not, with that of the non-pointed-out other prisoner, that would be similar to the Monty Hall problem.

JeffJo · Apr 4, 2018

sysprog said:

No; as I said in in my first post in this thread (post #12):

The Monty Hall problem is different, in that the contestant makes a decsion to switch or not switch based on further information, whereas in the stated prisoner problem, no option for a prisoner is postulated. If the prisoner problem provided that the statistician was allowed to, after a guilty prisoner was pointed out, opt to swap his verdict, or not, with that of the non-pointed-out other prisoner, that would be similar to the Monty Hall problem.

And as I said in my reply to that post (post #13):

What the contestant, or the statistician, wishes to do with the updated set of probabilities has no impact whatsoever on how the probabilities are updated. They are the exact same problem.

But what you pointed out in your last post, as the specific difference between the two problems was:

The question in the prisoner problem is whether the statistician in stating his chances after the pointing out as having improved to 50% is right, to which the correct answer is NO, whereas the question in the Monty Hall problem is, after a zonk prize door is opened, and the contestant is then offered the option to switch doors, should the contestant switch, to which the correct answer is YES.

I highlighted what you said were the questions that we are supposed to answer (and like you did, I'll ignore that one of them wasn't actually asked). Both are yes/no questions. Both are determined to be either "yes" or "no" based on a probability calculation that is identical in the two problems. The impact of those "yes" or "no" answers is not relevant to how you determine whether "yes" or "no" is the answer. The only difference in how that determination is made in the two problems is that the roles of "yes" and "no" are swapped.

So again I'll ask:

Really? The difference you insist exists is that you think one problem asks a yes/no question, but the other asks the same question in "no/yes" form?

sysprog · Apr 4, 2018

JeffLo said:

So again I'll ask:
Really? The difference you insist exists is that you think one problem asks a yes/no question, but the other asks the same question in "no/yes" form?

You know better. They're not the same question. Although the probabilities can be determined by the same reasoning in both cases, the question "is the statistician right in proclaiming his chances to be 50:50" is not the same question as the question "should the contestant switch doors". A correct answer to the first question is "no, his chance is still 1/3". A correct answer to the second question is "yes, because the chance of the originally selected door winning is still 1/3, but now that a non-winning door has been opened, the two 1/3 chances of winning for the other two doors consolidate behind the one of them remaining closed, so the chance for that door is now 2/3".

If you were to change the first question as "is the statistician wrong", or the second question to "should the contestant not switch, that would make both answers yes in the first case and both no in the second case, but the first and second question are still not exactly equivalent to each other; there is no option in the first problem scenario, as there is in the second.

You could modify one problem or the other to make them the same. You could change the 3 Prisoners problem to allow the statistician to swap verdicts after the revealing, or you could change the Monty Hall problem to not allow the contestant to swap doors after a non-winning door is opened. You could say that the modified problem was still no more or less the same problem as the other one, and that the modification was not such as to affect whether the problem was the same problem. I would disagree with that. Clearly, at least from the point of view of the statistician or of the contestant, the so-modified problem would not be the same.

I recognize that in saying that the two problems are the same problem, you are considering their isomorphism in terms of mathematical probability theory; however, in considering the complete problems, the heteromorphic element of option offered viv-a-vis no option offered is a legitimate basis for assertion that two problems are not the same.

The following is taken from a rosettacode.org MATLAB program for simulation of the Monty Hall problem. I have changed some of the commentary:

Code:

function montyHall(numDoors,numSimulations)

    assert(numDoors > 2);

    function num = randInt(n)
        num = floor( n*rand()+1 );
    end

    %The first column will tally wins; the second, non-wins
    switchedDoors = [0 0];
    stayed = [0 0];    for i = (1:numSimulations)

        availableDoors = (1:numDoors); %preallocate available doors
        winningDoor = randInt(numDoors); %define winning door
        playersOriginalChoice = randInt(numDoors); %player makes his initial selection

        availableDoors(playersOriginalChoice) = []; %remove selected door from available doors

        %select door to open from the available doors
        openDoor = availableDoors(randperm(numel(availableDoors))); %sort available doors randomly
        openDoor(openDoor == winningDoor) = []; %disallow opening of winning door
        openDoor = openDoor(randInt(numel(openDoor))); %randomly select door to open

        availableDoors(availableDoors==openDoor) = []; %remove opened door from available doors
        availableDoors(end+1) = playersOriginalChoice; %return originally selected door to available doors
        availableDoors = sort(availableDoors);

        %at the next line the program implements simulation of the exercise of the option to switch or not switch
        playersNewChoice = availableDoors(randInt(numel(availableDoors))); %select one of the available doors
                                                                           %(randomly switch or don't switch)
        %if the Monty Hall problem is modified to remove option to switch, change the previous line to:
                                                                           %playersNewChoice = playersOriginalChoice;

        if playersNewChoice == playersOriginalChoice %player does not switch doors
            switch playersNewChoice == winningDoor
                case true
                    stayed(1) = stayed(1) + 1; %if staying wins, add 1 to 'stayed and won' count
                case false
                    stayed(2) = stayed(2) + 1; %if staying does not win, add 1 to 'stayed and did not win' count
                otherwise
                    error 'ERROR'
            end
        else %player switches doors
            switch playersNewChoice == winningDoor
                case true
                    switchedDoors(1) = switchedDoors(1) + 1; %if switching wins, add 1 to 'switched and won' count
                case false
                    switchedDoors(2) = switchedDoors(2) + 1; %if switching does not win, add 1 to 'switched and did not win' count
                otherwise
                    error 'ERROR'
            end
        end
    end

    disp(sprintf('Switch win percentage: %f%%\nStay win percentage: %f%%\n', [switchedDoors(1)/sum(switchedDoors),stayed(1)/sum(stayed)] * 100));

end

montyStats(1e7)

The code as listed would display only the switch win and stay win percentages, but the page I took it from shows the following:

Output for 10 million samples:

no change change
win 33.3008% 66.6487%
win not 66.6992% 33.3513%

Also from that page, a Mathematica version, more succinct and perhaps less perspicuous, looks like this:

And one in php:

JeffJo · Apr 5, 2018

sysprog said:

You know better. They're not the same question.

I have explained to you why any difference is completely superficial. And you keep changing what you claim that superficial difference is.

the question "is the statistician right in proclaiming his chances to be 50:50" is not the same question as the question "should the contestant switch doors".

Both questions ask you evaluate two situations, and compare them. The fact that the primary subject of the comparison is different does not change that comparison.

But THERE WAS NO QUESTION IN THE ORIGINAL POST. You assumed a question was asked. No question was asked. If your argument is - as it seems to be - that a difference exists that is based on what question was asked, then you simply can't be right.

A correct answer to the first question is "no, his chance is still 1/3". A correct answer to the second question is "yes, because the chance of the originally selected door winning is still 1/3, but now that a non-winning door has been opened, the two 1/3 chances of winning for the other two doors consolidate behind the one of them remaining closed, so the chance for that door is now 2/3".

If you want to be that pedantic about it, the corresponding answer to the first is "no, because the chance that the originally-considered prisoner (the statistician) would be freed is still 1/3, but now that a guilty prisoner has been identified, the two 1/3 chances for the other two prisoners consolidate on the one who was not pointed to, so the chance for that prisoner is now 2/3."

And I'll point out that this phrasing for either problem, even though it gets the right answer, IS WRONG. Chances don't "consolidate." Once you remove probabilities based on information gained, you essentially normalize the probabilities that remain so that they add up to 1. The chances of {your door (#1) has a goat, Harry is guilty} and {Monty Hall opens a door (#3), the guard points to Tom} are 1/3. The chances of {your door (#1) has the car, Harry is not guilty} and {Monty Hall opens a door (#3), the guard points to Tom} are 1/6. You normalize these by dividing by 1/2, so the updates probabilities are 2/3 and 1/3, respectively.

If you were to change the first question as "is the statistician wrong" ...

If you were to change WHAT YOU ASSUMED WAS ASKED ...

... but the first and second question are still not exactly equivalent to each other; there is no option in the first problem scenario

There is no question in the first problem scenario, so basing this claim of a difference on what you assumed is absurd.

You could modify one problem or the other to make them the same. You could change the 3 Prisoners problem to allow the statistician to swap verdicts after the revealing, ...

Again: what the participants would do with the information gained from solving the problem does not affect the solution to the problem.

I recognize that in saying that the two problems are the same problem, you are considering their isomorphism in terms of mathematical probability theory; however, in considering the complete problems, the heteromorphic element of option offered viv-a-vis no option offered is a legitimate basis for assertion that two problems are not the same.

And now you are just rationalizing how the statement you want to be correct, should be.

sysprog · Apr 5, 2018

JeffJo said:

I have explained to you why any difference is completely superficial. And you keep changing what you claim that superficial difference is.

I disagree with the contention that the difference between the two problems is completely superficial. I have described the difference in more than one way, but I have not changed what difference I am describing: in the 3 Prisoners problem, there is no option to switch verdicts, whereas in the Monty Hall problem, there is an option for the contestant to switch doors.

Both questions ask you evaluate two situations, and compare them.

I disagree. In the 3 prisoner problem, there is no reason to do any before and after revelation probability analysis, because there's nothing that can change the outcome, whereas in the Monty Hall problem, the presentation of the option to switch is there to necessitate the probability analysis.

The fact that the primary subject of the comparison is different does not change that comparison.

To make up for there not being an option in the first problem, you're hinting at a secondary subject, viz, the prisoner whose chances go to 2/3, but the problem as presented does not at all suggest any reason for the statistician to understand anything beyond the fact that his own chances are still 1/3.

That's why the two problems are different. It's obvious that the two problems would be closer to the same if the prisoner had the option to switch verdicts, or the contestant had no option to switch doors, and that means that the two problems aren't really quite the same to begin with.

But THERE WAS NO QUESTION IN THE ORIGINAL POST. You assumed a question was asked. No question was asked.

You brought in the comparison to the Monty Hall problem, in which there is clearly a question. For the 3 Prisoner problem to be the same problem, as you have said it is, both problems would have to pose something seeking an answer. You distinguished between an answer and a solution, positing as a solution an exposition of a process contributory to derivation of a correct answer. If no answer is required, it's not really a problem. Any problem can be presented in the form of a question. The question in the 3 Prisoners problem as stated is implicit. That makes it susceptible to more than one formulation and interpretation.

I view the correction of the glaring error of the statistician, in proclaiming his post-revelation chances to be 1/2, to be what is sought in the 3 prisoner problem. The correction to 1/3 chance can be arrived at without recognition of the fact that the post-revelation chance of the remaining other prisoner is 2/3, so that recognition has no bearing on the outcome for the statistician, and so is not necessary for solution of the problem as presented, unless you arbitrarily decide that the problem is to determine the likelihood for each prisoner of each outcome.

If you take that approach, the problems would be equivalent; however, I think the correction of the glaring error is what the problem as stated seeks. To accomplish that correction, you need only note that the statistician cannot act on the new information, wherefore it is irrelevant for him. The expansion to the more complete analysis to include the 2/3 chance of the other remaining prisoner is necessary only to show why the statistician would be wrong to not swap verdicts if he were to be given such an option, which is not part of the problem as stated, whereas in the Monty Hall problem, the chances for the other door are explicitly sought, which is again why I consider the two problems to be different.

If your argument is - as it seems to be - that a difference exists that is based on what question was asked, then you simply can't be right.

If there were no question to be answered, there would be little point in articulating the before and after probabilities. Your model of the problem disregards anything in the narrative that occurs after the revelation event for either problem. I think that's an incomplete model, at least for the Monty Hall problem, and I think your insistence to the contrary is a consequence of you having pre-decided that the two problems are the same. I agree that in both problems the information present after the revelation is sufficient for any subsequent probability analysis, but in the first problem, the correction of the glaring error of the statistician does not require any such analysis, or anything else beyond recognition that his chances are still the same 1/3 that they were to begin with, because the new information doesn't change anything for him if he can't act on it.

... the corresponding answer to the first is "no, because the chance that the originally-considered prisoner (the statistician) would be freed is still 1/3, but now that a guilty prisoner has been identified, the two 1/3 chances for the other two prisoners consolidate on the one who was not pointed to, so the chance for that prisoner is now 2/3."

That's true, but the originally-considered prisoner can't switch verdicts with the other remaining prisoner, so it doesn't matter what that prisoner's chances are. In the Monty Hall problem, the contestant can switch doors, so it does matter what the chances for the other unopened door are. That's the difference between the problems again, and you've again shown that you in fact recognize it. Your opting to not acknowledge it as a significant difference, is in my view an expression of your preference; not a vindication of your position that the two problems are, for all legitimate interpretations, precisely the same.

And I'll point out that this phrasing for either problem, even though it gets the right answer, IS WRONG.

So you concede that there is a right answer for the 3 Prisoner problem, which entails that there is a question.

Chances don't "consolidate."

If I buy 10 lottery tickets, and then distribute them at random to 10 other persons, and they then play liar's poker using the serial numbers on the tickets, with the winner of that game being given the other 9 tickets, it's reasonable to say that the lottery chances represented by the tickets have, after the initial distribution and subsequent game, been consolidated into the possession of the 1 person now holding all 10 tickets.

Once you remove probabilities based on information gained, you essentially normalize the probabilities that remain so that they add up to 1. The chances of {your door (#1) has a goat, Harry is guilty} and {Monty Hall opens a door (#3), the guard points to Tom} are 1/3. The chances of {your door (#1) has the car, Harry is not guilty} and {Monty Hall opens a door (#3), the guard points to Tom} are 1/6. You normalize these by dividing by 1/2, so the updates probabilities are 2/3 and 1/3, respectively.

The descriptional language employed is a matter of preference.

If you were to change WHAT YOU ASSUMED WAS ASKED ...

There is no question in the first problem scenario, so basing this claim of a difference on what you assumed is absurd.

If there's no question in the first problem, and there is in the second, then they're not the same problem. You are presuming that what is sought in each problem is a post-revelation probability analysis, which is necessary in the Monty Hall problem, because the contestant is given a choice, but is not necessary in the 3 Prisoner problem, because the statistician is not given a choice.

Again: what the participants would do with the information gained from solving the problem does not affect the solution to the problem.

You appear to be trying to obscure a difference that you've already conceded exists, and that you've also ably elucidated. The inclusions and exclusions you impose are arbitrarily selected to make the problems the same, when by a plain reading of them, they're clearly similar but not the same.

And now you are just rationalizing how the statement you want to be correct, should be.

Your dismissal of the rest of my post with that characterization doesn't legitimately dispose if it. In my view, the MATLAB code snippet, including the comments, definitively shows the difference between the two problems. Eliminate the option to switch doors, and the Monty Hall problem becomes equivalent to the 3 prisoner problem. Retain that option, and it remains not equivalent.

JeffJo · Apr 5, 2018

sysprog said:

I disagree with the contention that the difference between the two problems is completely superficial.

Then point out a difference in the solution to the problems, not where one asks a "yes/no" question and the other, well, doesn't ask one at all but you assume it asked the exact same question in "no/yes" form.

In fact, you could start by acknowledging the difference you insist is there is based on something that IS NOT THERE.

I have described the difference in more than one way, ...

And I described why each is superficial.

in the 3 Prisoners problem, there is no option to switch verdicts, ...

And the problem solution comes before the impact of the correct solution. That is different in the two problems, but has nothing to do with how to answer the problems. See "superficial."

In the 3 prisoner problem, there is no reason to do any before and after revelation probability analysis, because there's nothing that can change the outcome.

But the point of that problem is not whether the statistician will be freed, will not be freed, or can change either outcome. It is, and I quote from post #16 but add punctuation, "Whether the statistician, in stating his chances after the pointing out as having improved to 50%, is right." Again, see "superficial."

whereas in the Monty Hall problem, the presentation of the option to switch is there to necessitate the probability analysis.

Do you mean the analysis that is identical to the analysis that shows whether the statistician is right? The analysis that let's us answer either of questions you said we were asked?

You brought in the comparison to the Monty Hall problem, ...

I quoted a comparison that is well known in the literature that addresses these problems. Wikipedia: "The Three Prisoners problem appeared in Martin Gardner's "Mathematical Games" column in Scientific American in 1959.[1][2] It is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively, and also equivalent to, and presumably based on, Bertrand's box paradox." Quora: "This is the famed 'Monty Hall problem' in different clothes." Jerffrey Rosenhouse's book The Monty Hall Problem, which is probably the most complete study of it: "The Monty Hall problem in its modern form goes back to 1975, ... For sixteen years prior to that it was traveling incognito as the Three Prisoners Problem."

For the 3 Prisoner problem to be the same problem, as you have said it is, both problems would have to pose something seeking an answer.

And there is an implied question - it just isn't the "yes/no" one you said it was in post #16, when you said the difference was that the MHP asked a "no/yes" one. It asks "right or wrong?", and that determination is the same one that is used to answer "switch or stay?"

The question in the 3 Prisoners problem as stated is implicit. That makes it susceptible to more than one formulation and interpretation.

I agree entirely. The implicit question is "what solution is the correct one for the identical probability problems?" That can be interpreted many different ways - yes you should switch, no the statistician is wrong, etc. - but they are all the same answer. And every other difference you have pointed out is superficial, because they all interpret the aftermath of the correct solution.

I view the correction of the glaring error of the statistician, in proclaiming his post-revelation chances to be 1/2, to be what is sought in the 3 prisoner problem.

And the correction of the glaring error, that a contestant would consider not switching, is the same thing.

To accomplish that correction, you need only note that the statistician cannot act on the new information, ...

Maybe he is smarter than you, then. The correct solution to either does "act on" the new information, by changing the set of probabilities from {1/3,1/3,1/3} to {1/3,2/3,0}. The statistician just acted on it incorrectly.

That's true, but the originally-considered prisoner can't switch verdicts with the other remaining prisoner, ...

Which is still a superficial issue,since the implied question in both is "what is the correct set of probabilities?" Not the superficial "what will happen afterwards?"

Your dismissal of the rest of my post ...

is because it doesn't address the issues in the problems, it addresses the superficial points you use to obfuscate it.

sysprog · Apr 5, 2018

JeffJo said:

But the point of that problem is not whether the statistician will be freed, will not be freed, or can change either outcome. It is, and I quote from post #16 but add punctuation, "Whether the statistician, in stating his chances after the pointing out as having improved to 50%, is right."

You don't need to do any probability analysis to know that the correct answer is that he is not right. All you need do is recognize that he cannot do anything with the new information that would change his chances.

Do you mean the analysis that is identical to the analysis that shows whether the statistician is right? The analysis that let's us answer either of questions you said we were asked?

That analysis is sufficient, but not necessary, for answering the question as you just quoted it. One can arrive at the correct answer with or without it.

I quoted a comparison that is well known in the literature that addresses these problems. Wikipedia: "The Three Prisoners problem appeared in Martin Gardner's "Mathematical Games" column in Scientific American in 1959.[1][2] It is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively, and also equivalent to, and presumably based on, Bertrand's box paradox."

The problem as described in the Wikipedia article is not the same as the one in this thread. In the Wikipedia article there is an additional condition, and a different question:

Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2, as it is now between him and C. Prisoner A secretly tells C the news, who is also pleased, because he reasons that A still has a chance of 1/3 to be the pardoned one, but his chance has gone up to 2/3. What is the correct answer?

[emphasis added -- it is pivotal that A secretly tells C the news]

The correct answer would be "A is wrong and C is right". Answering "A is wrong" by itself would lead to the followup question "what about C?", to which the correct answer would be "C is right", by which the fact that A is wrong, which he already was before he talked to C, is entailed. Unlike the problem as stated in this thread, the problem in the Wikipedia version includes a question that, in any reasonable interpretation, is asking about both A and C regarding who is right, whereas the problem as stated in this thread asks only about the statistician.

Once you include the condition that the inquiring prisoner discloses the information to the other prisoner, who recognizes that he could have been selected but was not, the problem changes in a way that makes it not significantly different from the Monty Hall problem.

The distinction between the inquiring prisoner knowing he could not have been chosen, versus the secretly-informed prisoner knowing that he could have been chosen and was not, maps inversely to the accidental versus deliberate distinction illustrated in PerotK's post #10 in this thread.

And there is an implied question - it just isn't the "yes/no" one you said it was in post #16, when you said the difference was that the MHP asked a "no/yes" one.

You're misstating what I said.

It asks "right or wrong?", and that determination is the same one that is used to answer "switch or stay?"

If it asks "right or wrong" about both A and C, yes it's the same determination. If it asks "right or wrong" only about the statistician, it is not the same.

The implicit question is "what solution is the correct one for the identical probability problems?"

That purposely begs the question whether the problems are identical.

That can be interpreted many different ways - yes you should switch, no the statistician is wrong, etc. - but they are all the same answer. And every other difference you have pointed out is superficial, because they all interpret the aftermath of the correct solution.

"No, A is wrong and yes, C is right", is not correspondentially the same as "no, the statistician is wrong". The Wikipedia version differs from the one in this thread, just as the one in this thread differs from the Monty Hall problem.

... the implied question in both is "what is the correct set of probabilities?" ...

That's true if and only if you substitute the Wikipedia version (or some other version) of the 3 Prisoners problem for the version in this thread.

JeffJo · Apr 6, 2018

sysprog said:

You don't need to do any probability analysis to know that the correct answer is that he is not right. All you need do is recognize that he cannot do anything with the new information that would change his chances.
That analysis is sufficient, but not necessary, for answering the question as you just quoted it. One can arrive at the correct answer with or without it.

"He cannot do anything with the new information" is a probability analysis.
The same analysis, if you believe it, applies in the exact same way to the MHP.
It isn't right. The information does change something. It tells you that one probability changes from 1/3 to 0, so all of the other probabilities ARE AFFECTED somehow. You are ignoring the fact that you think it is trivial to deduce that the probability is still 1/3. That is also a probability analysis.
You need further analysis to (correctly) prove what the affect is on the statistician's chances.
1. They are reduced, relative to the third prisoner's, because if the statistician is to be freed the guard could have pointed to either of the others.
2. The re-normalization step affects it again, and can make it any value between 0 and 1/2.
3. Only a correct probability analysis shows that it becomes 1/3 again.

The problem as described in the Wikipedia article is not the same as the one in this thread. In the Wikipedia article there is an additional condition, and a different question:

Since there is no explicit question in the OP, this is an assertion that has no basis in fact.

There is an implicit question tho. You can choose whatever form you want for it, but it is functionally equivalent to the one in Wikipedia.

It is pivotal that A secretly tells C the news

It is irrelevant. **WE** are told this same news. Wikipedia's "C" just personifies us.

If it asks "right or wrong" about both A and C, yes it's the same determination. If it asks "right or wrong" only about the statistician, it is not the same.

And if it asks neither, you can't dismiss any of the possible questions:

Is the statistician right, and his chances change to 1/2?
Is the statistician wrong, and his chances are unaffected?
Is the statistician wrong, and his chances are affected but "change" to the same value as before?
Is the statistician wrong, and his chances change to something that is neither 1/2 nor 1/3?

(Note: the correct answer is "3".)

Now, which the actual case for the OP? Does it ask about both A and C, does it ask "right or wrong" only about the statistician, or does it ask neither?

You're misstating what I said.

You: "The question in the prisoner problem is whether the statistician in stating his chances after the pointing out as having improved to 50% is right, to which the correct answer is NO, whereas the question in the Monty Hall problem is, after a zonk prize door is opened, and the contestant is then offered the option to switch doors, should the contestant switch, to which the correct answer is YES."

Me: "[the question] isn't the "yes/no" one you said it was in post #16, when you said the difference was that the MHP asked a "no/yes" one.

The statistician is right ("YES" to your first question) if a probability analysis says his chances change to 50%, and wrong ("NO" to your first question") if a probability analysis - even a trivial one that omits important details, like the one you say isn't a probability analysis - says anything else. There is no reason to switch ("NO" to your second question) if a probability analysis says your door's chances change to 50%, but your chances will improve ("YES" to your second question) if they change to anything less than 50%. So the only difference is if you consider the possibility that your door's chances change to something greater that 50%. But that isn't under consideration, is it?

sysprog · Apr 6, 2018

JeffJo said:

sysprog said:

It is pivotal that A secretly tells C the news

It is irrelevant. **WE** are told this same news. Wikipedia's "C" just personifies us.

The prisoners don't all know what we know -- subjective conditional probabilities and updates thereto that are predicated upon new knowledge events are based on what each subject knows; not on what we know.

For example: the players in a televised Poker game aren't allowed to see the other players' pocket cards on TV, as the TV viewers are. The odds calculations that are sometimes displayed for the viewer are based on knowing all the players' cards. None of the players can calculate those same odds subjectively. The odds calculations done for the viewer are presumably accurate, and we could make side bets based on them, but the players can act only on what they see.

In the Wikipedia version of the problem, the additional condition that A secretly tells C the news, maps to the Monty Hall problem condition that the host offers the contestant a choice to switch doors or not. In both cases, the condition brings in the relevancy of the updated chance -- in the prisoner problem, for the other not-pointed-out-as-guilty prisoner, and in the Monty Hall problem, for the other unopened door. In the problem as stated in this thread, there is no such mapping. You apparently view that difference as superficial or irrelevant, but it is clearly part of why the Wikipedia article asserts the two problems to be mathematically equivalent.

The Monty Hall problem would not be the same problem if the host did not offer an option to switch doors, because what the contestant does or does not do, or can or cannot do, or should or should not do with the information, though irrelevant to the objective probability consequences of a non-winning non-selected door being opened, is of the essence in the problem as stated, which is to determine, based on the updated subjective probability, whether the contestant should switch doors.

Cathr · Apr 6, 2018

benorin said:

Originally posted by Galileo in the thread I started called Bad Math Jokes on top of pg. 4:
_____________________________________________________________________________________________________________________
Not so much a joke as a brainteaser.

Three prisoners, strangers to each other, were suspects of a murder case. One day they came to hear that a sentence has been drawn. Two of them have been found guilty and will be executed, but they don't know which of the two . One guy, a statistician, figures his chances for survival are 1/3, so he goes to the bars of his cell and hails the guard: "Hey psst, do you know which of us has been sentenced?".
"Eh, yes.", says the guard, "But I'm not allowed to tell you.".
"Tell you what", says the guy, "I already know that 2 of us will executed, that means at least one of the other guys will be. I don't know them or anything, surely you can point to one which is guilty?". The guard sees no harm in that and points one of the prisoners, "He is guilty".
"Thanks!", proclaims the statistician, "my chances have just increased to 1/2".
_____________________________________________________________________________________________________________________

If you've got the time, throughout the next several pages of the original thread linked above starting on pg. 4 there are varying analyses of this 'brainteaser' (that do not all agree) and I felt after reading every relevant post on this that we hadn't really gotten to the bottom of it, or clearly I did not understand it if we did for I am left wanting a convincing analysis. Please explain, citing any formulae beyond the basics: I have taken an elementary probability and statistics course, and I know some analysis if need be.

Thanks for your time,
-Ben Orin

I am a bit late in this conversation and I haven't read all of the comments, but I've got an easy way to explain this - after this the situation will seem pretty clear, and the brainteaser not paradoxal at all.

Actually the stat guy's survival probability was 1/2 right from the start. The 1/3 chance is given from the perspective of another person who actually cares of the combinations of people, let's say the people's initials are A, B, S (S for our statistician), so the combinations are AB, AS and BS. But S doesn't care at all whether he will die with A or B - he doesn't know them anyways. So he can view the other two people as a pack, and the cases AS and BS are reduced to a single one, because he only cares if he will die or not.

Here's my view - and even if, I guess, the original target was to explain this with formulae (that I don't have) that's a pretty good way to visualize how a change of the point of view can change the estimated probability.

Cathr · Apr 6, 2018

benorin said:

-Ben Orin

Oh and by the way, let's consider that the total number of suspects increases - let's say to 15. If one person is eliminated from the equation by the guardian, the probability "becomes" 1/14. But wasn't it 1/14 right from the beginning since the stats guy knew right from the beginning that 2 people are going to die? What changes when he finds out who actually dies?

sysprog · Apr 6, 2018

Cathr said:

I am a bit late in this conversation and I haven't read all of the comments, but I've got an easy way to explain this - after this the situation will seem pretty clear, and the brainteaser not paradoxal at all.

Actually the stat guy's survival probability was 1/2 right from the start. The 1/3 chance is given from the perspective of another person who actually cares of the combinations of people, let's say the people's initials are A, B, S (S for our statistician), so the combinations are AB, AS and BS. But S doesn't care at all whether he will die with A or B - he doesn't know them anyways. So he can view the other two people as a pack, and the cases AS and BS are reduced to a single one, because he only cares if he will die or not.

Here's my view - and even if, I guess, the original target was to explain this with formulae (that I don't have) that's a pretty good way to visualize how a change of the point of view can change the estimated probability.

How do you account for the fact that from the problem statement, at the outset, 2 of the prisoners are guilty, and only 1 is not, while each of the 3 has the same chances? That means each prisoner has a 1/3 chance of being guilty, and a 2/3 chance of not being guilty. If S is guilty -- we can use Sg to denote that, the chance is 50:50 between SgAg and SgBg, but S not caring about which doesn't make the chance of Sg or not Sg change from 2/3 to 50:50.

sysprog · Apr 6, 2018

Cathr said:

Oh and by the way, let's consider that the total number of suspects increases - let's say to 15. If one person is eliminated from the equation by the guardian, the probability "becomes" 1/14. But wasn't it 1/14 right from the beginning since the stats guy knew right from the beginning that 2 people are going to die? What changes when he finds out who actually dies?

It looks like you're changing the number of prisoners without being clear about how many of the new number are guilty, and what any of the prisoners knows about that. If there are still only 2 guilty, and everyone knows that the two could be any of them, the chance of being guilty, for each prisoner at the outset, would be 2/15. If one of the 2 guilty is pointed out by the guard at the request of a prisoner, and the guard could not have pointed out the inquiring prisoner, the chances of the inquiring prisoner does not change. But if that prisoner tells another prisoner about it, that prisoner then knows that he could have been pointed out but wasn't, so his chances of guilty change from 2/15 to 1/15, while those of the inquirer, given that he knows he could not have been pointed out as guilty, now knows that the chance of each of the remaining other 13 prisoners is 1/15, and that his own chance remains 2/15.

JeffJo · Apr 7, 2018

sysprog said:

The prisoners don't all know what we know ...

Prisoner C does. Your objection to the comparison of the MHP to the TPP was that there was no equivalent to Prisoner C in the MHP. But there is - us. Prisoner C is the personification of the outside observer who knows all of the details, but not the specific outcomes.

If you are going to continue to deliberately ignore all of the evidence that contradicts you, there is no point in continuing. The probability spaces underlying the two problems are identical. The facets of that probability space are presented differently - but not very differently - in the two. The consequences of the outcomes are irrelevant to how we address the problem, a fact you continue to ignore.

The problems are the same because the underlying probability spaces are the same. The minute differences in the presentation affects only how you phrase the answer.

sysprog · Apr 7, 2018

JeffJo said:

sysprog said:

The prisoners don't all know what we know ...

Prisoner C does. Your objection to the comparison of the MHP to the TPP was that there was no equivalent to Prisoner C in the MHP. But there is - us. Prisoner C is the personification of the outside observer who knows all of the details, but not the specific outcomes.

I said:

The prisoners don't all know what we know -- subjective conditional probabilities and updates thereto that are predicated upon new knowledge events are based on what each subject knows; not on what we know.

and

In the Wikipedia version of the problem, the additional condition that A secretly tells C the news, maps to the Monty Hall problem condition that the host offers the contestant a choice to switch doors or not. In both cases, the condition brings in the relevancy of the updated chance -- in the prisoner problem, for the other not-pointed-out-as-guilty prisoner, and in the Monty Hall problem, for the other unopened door. In the problem as stated in this thread, there is no such mapping. You apparently view that difference as superficial or irrelevant, but it is clearly part of why the Wikipedia article asserts the two problems to be mathematically equivalent.

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 statistician in the same knowledge position as that of prisoner C, or of us. He has been not been told of what the guard has revealed, as prisoner C has, and as we have.

If you are going to continue to deliberately ignore all of the evidence that contradicts you, there is no point in continuing.

I deny that there is any such, but obviously whether you continue is not up to me.

Emphasis added to highlight a point on which we disagree:

The probability spaces underlying the two problems are identical. The facets of that probability space are presented differently - but not very differently - in the two. The consequences of the outcomes are irrelevant to how we address the problem, a fact you continue to ignore.

The problems are the same because the underlying probability spaces are the same. The minute differences in the presentation affects only how you phrase the answer.

From my prior post (emphasis added):

The Monty Hall problem would not be the same problem if the host did not offer an option to switch doors, because what the contestant does or does not do, or can or cannot do, or should or should not do with the information, though irrelevant to the objective probability consequences of a non-winning non-selected door being opened, is of the essence in the problem as stated, which is to determine, based on the updated subjective probability, whether the contestant should switch doors.

The 3 Prisoners problem as stated in this thread, is not equivalent to the Wikipedia version of it, because objective probability is not the same as subjective probability, and there is in that version, no prisoner C being informed by prisoner A to make the sum of the subjective knowledge the same as the objective knowledge, and we can't legitimately impute our own objective knowledge to the other prisoner to make up for that.

Unlike the version in this thread, the Wikipedia version is the same as the Monty Hall problem, because the informing of prisoner C by prisoner A in that version, maps to the contestant being given an option to switch doors. If the statistician were to to be given an option to switch verdicts, that would equally well make the problems equivalent, but he isn't given such an option, so the problems are not equivalent.

JeffJo · Apr 7, 2018

I wasn't going to respond again, but I just can't abide such blatant misrepresentation. Multiple times.

sysprog said:

You can't legitimately put the statistician in the same knowledge position as that of prisoner C, or of us.

I didn't. I put Prisoner C in the same knowledge position as us. But now that you mention it, the statistician has the same knowledge as well. He just applied it incorrectly.

The problems are the same because the underlying probability spaces are the same. The minute differences in the presentation affects only how you phrase the answer. The consequences of the outcomes are irrelevant to how we address the problem, a fact you continue to ignore.

sysprog · Apr 7, 2018

JeffJo said:

sysprog said:

You can't legitimately put the statistician in the same knowledge position as that of prisoner C, or of us.

I didn't. I put Prisoner C in the same knowledge position as us.

But now that you mention it, the statistician has the same knowledge as well. He just applied it incorrectly.

That de-contextualizing of what I said makes it misleading, whereas in the context of the immediately preceding and succeeding sentences, it is not:

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 statistician in the same knowledge position as that of prisoner C, or of us. He has been not been [sic] told of what the guard has revealed, as prisoner C has, and as we have.

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, as should be clear from the preceding sentence, and as would also be clear from the next sentence, had I not accidentally typed "been not been" instead of "not been" in that sentence. So to correct:

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.

bahamagreen · Apr 7, 2018

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 · Apr 7, 2018

bahamagreen said:

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 · Apr 7, 2018

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 · Apr 7, 2018

bahamagreen said:

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 · Apr 8, 2018

sysprog said:

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 · Apr 8, 2018

JeffJo said:

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 · Apr 8, 2018

sysprog said:

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 · Apr 8, 2018

JeffJo said:

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.

JeffJo said:

sysprog said:

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 · Apr 9, 2018

sysprog said:

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 · Apr 9, 2018

JeffJo said:

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 · Apr 10, 2018

PeroK said:

... 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 · Apr 10, 2018

sysprog said:

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.

+++++

PeroK said:

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:

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.
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.

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).
A box is chosen at random. The probability that it has only one kind of coin is 2M/N.
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.
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 · Apr 10, 2018

sysprog · Apr 11, 2018

JeffJo said:

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:

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.

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

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 · Apr 12, 2018

sysprog said:

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.

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