B Probability and Death Sentences

[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.