[tab]It isn’t the fact that all others are duplicated that allows to make that assumption. It is the fact that the problem states that it is solvable.
You said it yourself, you can’t know what your color is, the problem is unsolvable. But because it is specifically stated that the problem is solvable, you have to make the assumption that your color is duplicated at least once because if you can’t see your own color, you don’t have a constrained group of colors to work with.[/tab]
I sort of see what he means…[tab]It is fair to expect that when you are going to work on a problem, whoever gave you the problem must give you all the constraints of the problem. In this problem there is a constraint missing, but there is enough information for you to make an assumption on how to close the problem.
I suppose that you shouldn’t be making any assumptions in a logic problem. In this specific problem, the logic actually begins once you define the last constraint, but the logic itself is not the interesting part of it, it is how you arrive at the conclusion that allows you to define the problem.
Making assumptions or deductions or whatever = feminine thinking?
In any case it is a good skill to exercise, since there are very few problems for which all constraints are known.
I know the solution, I just don’t want to type it out. I already solved it so I want to move on 
… ok I’ll type it out in a little while, but for a quick and dirty version, read on.[/tab]
Quick and dirty answer:
[tab]color repeated once leaves on 1st bell, color repeated twice leaves on 2nd bell, so on so forth.[/tab]
Yup.
[tab]Once you constrain the problem to require at least two of each headband, it becomes the Blue Eye problem, except that the last color to leave leaves on the bell after the second to last color to leave, not matter how many of them there are.
I do find interesting the logic surrounding the premise that the problem is not impossible. It seems straightforward that the assumption to make is that there’s more than one of each color. But it might be that we’re also implying a least-necessary assumption. Are there other assumptions that would make the problem possible and allow everyone to accurately deduce their band color? I haven’t spent a lot of time on this question, but it’s one I’d like to explore.
This problem reminds me of the concept of meta-programming (with which I’m only passingly familiar), since you need to deduce some of the rules from other rules in order to solve it.[/tab]
Nahhh…[tab]Probably True [size=150]≠[/size] True
It’s an invalid logic problem unless you specify “Bayesian logic” or “Statistics”
“If we assume that God created the Earth for a purpose, then we can deduce…”
[/tab]
[tab]You’ll need to flesh that argument out, James. I never used the word “probably”, and proof by contradiction is not Bayesian.
I take it you don’t like this line of (non-probabilistic) argument:
If the headbands could be any possible color, the problem would be impossible to solve.
The problem is not impossible to solve.
Therefore, the headbands cannot be any possible color.
If the headbands could be any color other than those that each logician can see, the problem would be impossible to solve.
The problem is not impossible to solve.
Therefore, the headbands cannot be any color other than those that each logician can see.
Each logician can see all the possible colors of her own headband
Therefore, for all logicians of headband color X, there must be at least one other logician with headband color X
Therefore, for any color X, there must be either 0 or >1 logicians with that color headband.
So explain what’s wrong with it. Please don’t be sarcastic.[/tab]
lol, James… we can’t know everything, always, and yet, we have toys to deliver.
The problem is solvable because the Master says that it’s solvable. If he did not say that it was solvable, then it would not be solvable.
Makes sense to someone.
[tab]Obviously I am saying that the Master was lying.
For the problem to be solvable there must be one provable, unique pattern from which to deduce (or perhaps many that all yield the exact same resolve). To prove that the color pattern is totally unique (and also involves “many colors”) would be to prove that no other pattern is at all possible. That means that you have to disprove every conceivable notion, requiring the “mind of God”, the knowing of all knowable.
If you go back up to the triangle problem and consider that I had also stated, “And Carleas, you cannot solve this puzzle”, would that invalidate his answer?
Saying that the person can or cannot solve the problem does not make it so, although it infers something from which one can make a reasonable guess. But it is only a guess. Guessing is not logic.
Okay, I’ll buy that.
Prove that one.
How many colors do you know or can discern? How many do the members know or can discern? If every known color was represented but one, even though many were duplicated, can’t you assume that yours is the last remaining color not represented rather than the color that fits a pattern of multiple colors being represented?
Given an obvious pattern of colors, how can anyone know with 100% certainty that no other pattern is at all possible? How can the Master know that? Perhaps a member can spot that every other color seen in the circle is a middle color of the two at its sides. Perhaps if one divides Pi by the square root of 2, a number pattern will reflect the color spectrum that then infers a specific color is to be next in the circle.
How can anyone ever prove that it is totally impossible for any pattern to fit a picture other than a specific one? I am pretty certain that clever mathematicians can come up with an alternative pattern to ANY chosen pattern, given any picture.
Each member has to have faith that the Master knows each member’s abilities and color perceptions. All possible patterns must be known. That is technically an assumption, and thus not a logical deduction.
The Master’s statement is an invalid axiom. And most probably incorrect, although I cannot currently prove that it is impossible for a specific pattern to be the only possible reflection of a given picture.[/tab]
If the Master says that it’s unsolvable, then it is unsolvable. The headbands could be exactly the same for solvable and unsolvable. 
How does that make sense?
To prove/know something unsolvable, one merely has to provide a single alternative. But to prove/know something solvable, one has to prove the total lack of ALL possible alternatives. How can anyone know ALL possible alternatives in such a situation as this?
Having a master there to say it is solvable just saves you the trouble of going through “what if…”.
This is a little bit burden-of-proof-tennis-ish. One could just as well say that providing just one solution proves something solvable, and showing that it’s unsolvable requires showing that ALL possible solutions will fail.
Neither is particularly useful; they are rhetoric and framing, they lack rigor. And they don’t address the problem at hand: what is the flaw in my previous syllogism?
The problem that I have is that it seems that the problem is solvable if the Master says so but it’s unsolvable if the Master says so. Nothing physically changes, only the declaration of the Master changes and the Master does not seem to give any information.
Needless to say, these kinds of problems feel unsatisfying.
Emmm… no. Perhaps I didn’t word it perfectly, but we are talking about a puzzle that requires that members know the one and only possible color. That means eliminating ALL other possibilities of colors, proving it to be the only possibility.
It is NOT the same as having to come up with merely one possible color regardless of how many others might be there.
I disagree. I explained the problem and I asked for you to prove one particular assertion in your response. Why aren’t you doing so?
… and is this another … “Here we go again….”? 
Sorry, my fault, I somehow missed this post.
The Master has listed a set of premises, one of which is “this is not impossible.” This premise does convey information, in the same way that Bernard and Albert being able to figure out the problem convey necessary information in the Cheryl’s Birthday problem, and Peter and Sarah’s solving the problem allow us to solve the sum/product problem. Does your objection apply there? Albert/Bernard and Peter/Sarah tell us, effectively, “I can solve this problem”, and because they do, and only because they do, we too can solve the problem.
Same here: because the Master tells us that the Logicians can solve the problem, we can solve the problem (unless of course we decide to decry the Master as a liar; that is to create a different problem).
As I said earlier,
So, we can’t say what the ‘one unique pattern’ is, or how many logicians there are or what color their headbands are, but we can describe how the logicians would solve the problem – what the problem would look like to them and how it would be solved regardless of what the specific colors involved are, or how many of each color there is.
Some of your objections are interesting, but I don’t think discussing them is useful just yet, because this strikes me as a more fundamental problem. Let’s take the paradigm syllogism:
All men are mortal
Socrates is a man
Socrates is mortal.
No argument there, I hope.
What if instead we had the Master say:
“All men are mortal
Socrates is a man
Socrates is mortal.”
“But wait!”, says James. “The Master has never peered into the Socratic trou to verify the second premise! Clearly no logician could solve this unsolvable logic problem without irrational faith in the words of the Master!” This seems a poor objection, and a hypothetical logician hearing this statement by the hypothetical Master will have no problem certifying that the logic presented is sound.
Similarly here: the Master’s statement is a premise. It’s truth is a given. And because it is a given, we know that none of the edge cases that would make it false are solutions to the logic problem. No logician is blind, because if any were, the problem would be impossible. And it is a given that it is not impossible.
I asked someone this in relation to the Blue Eye problem and I don’t recall getting an answer, I wonder if you’d indulge me here: Do you understand the logic problem that this is attempting to present? Could you reword it to avoid the problems you have with it?
Your finite-set-of-colors possibility is an interesting counter. If we assume there are a finite number of colors, there is one logician with each color headband and they all know the entire set of possible colors, it does seem like each could conclude their own headband color, especially given that the Master said the problem was not impossible.
But is that a problem for the problem? Since our answer is general, and just describes the solution space, can’t we just say “>1 of each OR one each of all possible colors OR [whatever else we come up with]”? Is there a general defeater that will always introduce ambiguity for the Logicians?
Also, I mentioned before the idea of a “least-necessary assumption”. Is there a smaller assumption than that each logician’s headband is one of the colors she sees?
No. Those other puzzles are different in a very important, although subtle way. With each of those, the participants could actually be right when they contributed their bits of the puzzles. In this puzzle, I am saying that the Master cannot be right, not that he just happens to be wrong. Thus what the Master says is invalid as an premise to the puzzle. Imagine him saying “And btw, 2+2=3”.
Imagine if Sarah or Albert really didn’t have enough information to contribute their parts to those puzzles. What if the puzzles were a little different such that neither could really deduce anything for certain. But the person making the puzzle stated they they both announced that they had deduced something, even though we can see that they could not have. Even though we could deduce that “because they know, then we can know….”, the premise is inadvertently wrong because they could not actually know. The puzzles would be invalid.
What we have here is that very scenario. The Master CANNOT know that he is right, just as if Sarah or Albert COULD NOT know their part.
It is easy enough to prove me wrong simply by giving an example of a color pattern that is deducible with 100% certainty. You can provide patterns that certainly lead to a very probable answer for each member to assume. But I’m betting that you cannot provide one that yields a 100% certain answer for the members (a solution with a total lack of alternative possibilities).
I am not sure that such is true … yet. First lets settle on at least one possible pattern that could have been the case where the Master would have been provably right.
You are misunderstanding my objection. In the example that you gave here as well as the others, the person speaking COULD know his part. It is possible that the person speaking really did know, thus it is acceptable as a premise to the puzzle.
But in this puzzle, it is impossible for the Master to know what he said to be true, thus impossible for any members to accept his word as a certain premise for their own deductions. And if the Master cannot be absolutely certain, the members cannot be either, in which case the Master is certainty wrong.
Yes, like the Blue-eyed problem, I told you that I understand the intended answer, explained it to many other people on the thread, and was impressed with its ingenuity. But, like this one, after closer examination, I see that it wasn’t actually the correct resolve to the puzzle. And in this case, there isn’t one.
Well, except that you cannot assume an infinite number of logicians, nor that any one logician could discern every infinitesimal color variation.
What if each person saw this color chart with only one color missing?
Or This:
Of course there could be duplicates and triplicates of the seen colors as well. Do you assume a pattern from the replication, or from the obvious missing color?
On the second bell everyone leaves?
Yes. There is no data pattern which yields a projection that could not be yielded by a different pattern in that same data. Error correction codes (ECC) have that very issue. There can be no ECC that is 100% certain.
Yes. The color that is missing. Not that such is relevant because as soon as you make an assumption, you already lost.
Carleas, you keep doing the same thing:
You keep limiting the logicians to your own knowledge and skill base (biasing).
“Having one potential solution, there can be no other possible solution.”
[tab]the master is the only person who can see all headbands, therefore he is in position to provide a clue about the headband that is on your head. When he tells you the problem is solvable, he is indicating that the headband on our head can be known, that you are looking at it.
Also, james, while I understand your argument that using colors in a math problem is tricky because it involves human sensorial perception which is not exact, you can as an exercise think that the colors being used are primary colors and none of the logicians are colorblind. Once you arrive at an algorythm, you can increase your group to an infinite number of logicians and an infinite number of colors, and it would still work.[/tab]
To pin down our disagreement, let me ask you about a simpler setup: assume there are three logicians, and they are all wearing hats of the same color. Now the master lays out the problem just the same: leave when you know your hat color, and it’s not impossible. Do you still maintain that the logicians can’t deduce their own hat color?
It seems like whether the Master could know the premise to be true is exactly the question, so we can’t use it’s knowability or unknowability as a premise. Which is strange, because it presents the possibility that we have a system even weirder than what Phyllo described: if the Master could be right, then he is right, and if he couldn’t, then he isn’t. I have a vague recollection of a field of set theory where something similar happens. Does it ring any bells for anyone else?
Phoneutria, doesn’t using only primary colors increase James’ problem? Let’s say instead of colors, the logicians had cards on the heads with integers 0-9. If there are 10 logicians, and they each have a different number on their head, could they deduce that they have the last number? It seems to depend on whether they are told “each of you has an integer on her forehead” or “each of you has a integer 0-9 inclusive on her forehead”.
Again, I don’t think this breaks the problem, I think it expands the solution set. But James’ worry, if I understand it, is that there might always be another solution sufficient to introduce ambiguity, and that would break the problem.
It seems possible to have a configuration which is both solvable and unsolvable. If the Master says nothing or says that it’s unsolvable, then it becomes unsolvable. And he is neither lying nor incorrect in his statement.
Then the problem is unsolvable.
Here (assume that you don’t know that I only picked colors from the set you gave):
