B Probability and Death Sentences

[sysprog]

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

I'm really getting tired

Please feel free to take a nap.

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

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

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

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

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

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

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

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

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

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

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

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

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

That remark is obviously unfounded.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Stated and restated; answered and reanswered.

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

Your apperception differs from mine.

The problem is what question is asked of us.

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

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

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

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

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

 

[JeffJo]

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

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

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

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

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

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

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

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

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

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

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

+++++

Other stories that present this problem include:

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

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

 

[sysprog]

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

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

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

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

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

 

[JeffJo]

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[sysprog]

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

I agree with that much.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

It is correct.

That doesn't make it so.

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

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

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

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

According to you, S2 says:

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

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

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

A distribution is a set, not a single value.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I think that's correct and satisfactory.

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

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

 

[Stephen Tashi]

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

That isn't an axiom of probability theory.

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

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

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

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

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

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

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

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

 

[JeffJo]

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

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

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

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

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

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

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

 

[JeffJo]

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

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

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

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

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

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

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

 

[Stephen Tashi]

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

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

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

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

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

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

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

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

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

 

[StoneTemplePython]

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

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

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

 

[sysprog]

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

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

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

 

[JeffJo]

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

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

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

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

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

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

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

 

[JeffJo]

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

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

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

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

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

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

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

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

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

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

 

[sysprog]

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

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

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

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

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

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

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

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

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

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

 

[JeffJo]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[sysprog]

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

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

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

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

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

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

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

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

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

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

 

[Stephen Tashi]

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

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

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

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

There are even customary clues used to suggest such inferences.

Apparently there are disagreements about customs.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[JeffJo]

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

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

I'll critcize your approach.

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

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

 

[Stephen Tashi]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[sysprog]

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

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

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

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

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

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

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

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

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

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

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

...

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

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

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

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

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

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

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

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

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

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

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

 

[Stephen Tashi]

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

When discussing a problem (given in words) where each outcome may have certain properties, I think the sample space should be detailed enough to exhibit whether each outcome does or does not have each property. If this is not done then the issues of mathematics and the issues of literary interpretation get jumbled together. People make verbal assertions about phenomena. These assertions refer to properties of outcomes. Unless the sample space is detailed enough to represent whether each property does or does-not apply to each outcome, there is no way to translate the verbal assertions into mathematical relations.

I agree with the conclusion that Pr(Box 1 is selected given the revealed coin is gold) = Pr(Box 1 is selected), however I don't understand what principle you would use to justify that. The general idea that Pr(A) = Pr(A|B) is clearly wrong, so one would need an argument, mathematical or literary, to justify why that pattern applies to selection of Box 1 in the Bertram Box problem.

One can make a literary argument that sounds like physics - e.g. "Once the box is selected, nothing that is done to examine it can change which box it is". I like that argument! However, as a generality, it often fails to account for the numerical behavior of probabilities.

There being only a 1/3 chance that the X value box was selected, the second coin has only a 1/3 chance of being different, and therefore must have a 2/3 chance of being the same.

In that fragment of your argument, you are reasoning about outcomes that involve properties of the "second" coin. So it seems to me that if you were to explicitly describe the sample space to which your argument applies, you'd need a description of outcomes that was detailed enough to describe the properties that may or may-not apply to the first and second coin in each outcome. That sample space might not be like my sample space, but it would be more detailed than simply having 3 outcomes, each defined only by which box is selected.

As I see it, you are formulating an argument that explains why a simple 3-element sample space can answer the question by implicitly discussing another sample space that has more detail.

 

[sysprog]

When discussing a problem (given in words) where each outcome may have certain properties, I think the sample space should be detailed enough to exhibit whether each outcome does or does not have each property. If this is not done then the issues of mathematics and the issues of literary interpretation get jumbled together. People make verbal assertions about phenomena. These assertions refer to properties of outcomes. Unless the sample space is detailed enough to represent whether each property does or does-not apply to each outcome, there is no way to translate the verbal assertions into mathematical relations.

I agree with the conclusion that Pr(Box 1 is selected given the revealed coin is gold) = Pr(Box 1 is selected), however I don't understand what principle you would use to justify that. The general idea that Pr(A) = Pr(A|B) is clearly wrong, so one would need an argument, mathematical or literary, to justify why that pattern applies to selection of Box 1 in the Bertram Box problem.

One can make a literary argument that sounds like physics - e.g. "Once the box is selected, nothing that is done to examine it can change which box it is". I like that argument! However, as a generality, it often fails to account for the numerical behavior of probabilities.

From my post:

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

And from your post:

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

In that fragment of your argument, you are reasoning about outcomes that involve properties of the "second" coin. So it seems to me that if you were to explicitly describe the sample space to which your argument applies, you'd need a description of outcomes that was detailed enough to describe the properties that may or may-not apply to the first and second coin in each outcome. That sample space might not be like my sample space, but it would be more detailed than simply having 3 outcomes, each defined only by which box is selected.

The first of the three sentences in the paragraph the remaining two sentences of which you excerpted and presented as a fragment of my argument, "If and only if the X value box is the selected box, the second coin can and must be different from the first.", I think makes it clear that the only property of the second coin I am reasoning about is whether or not it is the same as the first, which is equivalent to whether it is from a Y value box or from the X value box.

As I see it, you are formulating an argument that explains why a simple 3-element sample space can answer the question by implicitly discussing another sample space that has more detail.

As I see it:

The problem statement distinguishes two classes of box by their internal constituencies, viz, one class of box that has two constituents that are of the same kind as each other, and the other class that has two constituents that are of different kinds from each other, and two types of box constituent, viz, gold or silver, and also distinguishes the two members of the first class from each other as containing either only the first constituent type or only the second constituent type, viz both coins gold or both coins silver.

[It may be imperspicuous that the classes are not named in alphabetical order, and that the types are named before the second class is named, but that is the order in which the problem statement names them.]

Given 3 boxes, one of which is selected at random, the chance that the selected box is of the first class, class Y, which class comprises two of the three boxes, is 2/3, and the chance that the selected box is of the second class, class X, which class comprises one the three boxes, is 1/3.

Although there can be counted four outcomes for the 2 coins, viz, GG, SS, GS, and SG, the first 2 outcomes are pre-comprised in the first class, Class, Y, which has two of the three members (boxes), by 2/3 probability, and the latter two outcomes are of the second class, Class X, which class comprises only one of the three members (boxes), by 1/3 probability, so that although the probability space could be stated as

Y = 2/3(GG ⊗ SS) and X = 1/3(GS ⊗ SG),

it does not, for purposes of this problem, have to be so stated, otherwise equivalently stated, because the Class Y set has two members, and the Class X set has only one member; the two distributional expressions within X are not distinct set members, as the two members within Y are; they are different poset members, and as such are merely differently ordered expressions of the same set member, and the distinction between Class Y being a two-member set, and Class X being a one-member set, albeit one analyzable as a two-member poset, is sufficient to justify writing the probability space, for purposes of the problem, either as

Y = 2/3(GG ⊗ SS) and X = 1/3 ($\neg$(GG ⊗ SS)),

or as

Y = 2/3(Y1 ⊗ Y2) and X = 1/3 ($\neg$Y),

or in some other way that does not place the two members of Class Y in the same standing as that of the two orderings of the one member of Class X.

That avoids obscuring the fact that the two members of Class Y have together as Class Y twice the probability of being the selected box as the one member of Class X has; obviously 2/3(GG ⊗ SS) is equivalent to 2/3( $\neg$GG ⊗ $\neg$SS) but not to 1/3($\neg$(GG ⊗ SS)).

We already know that Y and X are mutually exclusive, and that Y1 and Y2 are also mutually exclusive, so we know that

Y ⇒ ($\neg$X ∧ ((Y1 ⇔ $\neg$Y2) ∧ ($\neg$Y1 ⇔ Y2))),

and when a coin is revealed to be of type G (gold), we know that

$\neg$Y2 ∧ (X ⇔ $\neg$Y1),

but that only affects the type possibilities within Class Y and the type order possibilities within Class X; it doesn't change the external probability of either class.

We knew from the start that Class Y had 2/3 of the probability of containing the selected box, and that only one of the two members of that class could be the selected box, and knowing which member it can't be has no effect on whether it is one of them.

Although the revealing of a coin type eliminates one of GG or SS within Class Y, it does not change the 2/3 probability of Class Y compared to the 1/3 Class X; it does not determine which of (GG ⊗ SS) ⊗ $\neg$(GG ⊗ SS) is true, and therefore renders a change in neither the probability of Y nor that of X.

Again, if a gold coin is revealed, that changes the internal probability share within Class Y of Y2 from 1/2 to 0 and of Y1 from 1/2 to 1, and removes the internal possibility of order SG within Class X, but it does not change the external 2/3 probability of Class Y, or the external 1/3 probability of Class X, containing the selected member, so after the revealing of a gold coin from the selected box, the probability that the other coin in the same box is also gold, is 2/3.

 

[Stephen Tashi]

As I see it:

The problem statement distinguishes two classes of box by their internal constituencies, viz, one class of box that has two constituents that are of the same kind as each other, and the other class that has two constituents that are of different kinds from each other, and two types of box constituent, viz, gold or silver, and also distinguishes the two members of the first class from each other as containing either only the first constituent type or only the second constituent type, viz both coins gold or both coins silver.

...etc.

As I see it, your treatment does not clearly define a probability space whose outcomes have all the properties you mention and you use terminology such as "the internal probability share within Class Y" that has no standard definition.

I agree that it is possible to discuss situations in real life (or imagined real life) without explicitly defining a probability space. However to apply mathematical probability theory coherently, a probability space is necessary. Part of this probability space is the set of outcomes. To define a set, one must specify the elements of the set. Your style of exposition does not make the set of outcomes clear. ( It may be that you don't intend for your arguments to be based on mathematical probability theory - in which case I don't know what standard to use in judging them.)

 

[sysprog]

@Stephen Tashi In my post (#72) I mistakenly used ⊗ (tensor product) where I meant ⊕ (XOR) -- I temporarily forgot that the encircled plus sign was in reference to modulo 2 binary addition, and didn't recall that the encircled X referred to matrix or tensor multiplication, and not to eXclusive, as the X in XOR does -- the edit time window is closed now, so I can't edit the correction into the post.