Probability and Death Sentences

[benorin]

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

 

[StoneTemplePython]

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]

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:

1. 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.
2. 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):

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

@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]

@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]

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):

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

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]

... 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]

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.

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]

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]

"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.

1. You ask friend to show you a card. He shows you the Seven of Hearts. Your chances are still 1/52.
2. 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:

1. Revealed a different door than Monty Hall opened.
   
2. Pointed to a different prisoner than the guard pointed to.
3. 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]

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]

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:

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.​

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).

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.​

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]

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]

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:

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)]​

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:

Pr(H|PT) = Pr(PT|H) / [1 + Pr(PT|H)]
Pr(D|PT) = 1 / [1 + Pr(PT|H)]​

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:

1. Acceptance of Y and rejection of X.
2. Acceptance of Y but uncertainty why X isn't correct.
   
3. Insistence on X and rejection of Y.
4. Complete disillusionment over the field of Probability.
5. 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]

...
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]

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.

1. 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).
2. 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.
3. 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.

1. 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.
   
2. 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]

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]

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]

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]

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:

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]

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]

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]

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]

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]

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.

1. "He cannot do anything with the new information" is a probability analysis.
   
2. The same analysis, if you believe it, applies in the exact same way to the MHP.
3. 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.
   
4. 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:

1. Is the statistician right, and his chances change to 1/2?
2. Is the statistician wrong, and his chances are unaffected?
3. Is the statistician wrong, and his chances are affected but "change" to the same value as before?
   
4. 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]

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]

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]

-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]

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.