Well following along the discussion I’m similarly confused with coremelt. The reason I wouldn’t want to play your 3 player game is that I don’t see how the same logic applies to the 200 player game.
Of course if there are only 2 other people you can’t be sure they both see a blue eye. But with 200 people we know that everybody can see at least 1 pair of blue eyes, and we know that everybody is constantly aware of the eye colors because that was in the rules.
If every single person can see at least 99 pairs of blue eyes and I know they can all see them, per the rules, how does the guru change anything?
Yes exactly, thats why I didn’t play. The guru’s statement makes a difference if it’s possible for one player to only see one set of blue eyes. It makes no difference once the number of blue eyes seen is two or greater by each single player.
The mistake in the logic is thinking that each person has to inductively make assumptions about everyone else from what John knows that Fred knows about Bob. Which leads to the absurd assumption that someone on the island could think there was no blue eyes at all until the Guru spoke.
Actually as every single person was only missing one piece of information (their own eye color) it was never possible for anyone to believe that there was no blue eyes with N Blue = 100.
No, it leads to the correct conclusion (not assumption) that someone might think that someone might think that someone might think… that someone might think that someone might think there are no blue-eyed islanders.
Before the guru speaks, person one thinks he might have non-blue eyes. He also thinks person two might think person two has non-blue eyes. Drawing a valid conclusion from this, person one thinks that person two might think that there are only 98 blue eyed islanders. (It is not actually possible for person two to think this, but it is possible for person one to think person two might think this, because person one doesn’t know his own eye color.)
So, it’s established that P1 thinks P2 might think there are only blue-eyed islanders. Now, what might P1 think that P2 thinks about P3? Well, P1 thinks P2 might think there are only 98 blue eyed islanders, so P1 thinks P2 might think P3 sees only 97 blue eyed islanders. So, P1 thinks P2 might think P3 might think that there are only 97 blue eyed islanders.
Alright. Now, what does P1 think P2 might htink about what P3 thinks about P4’s beliefs? well, P1 thinkis P2 might think that P3 might think there are only 97 blue eyed islanders. So P1 thinks that P2 might think that P3 might think that P4 thinks there are only 96 blue eyed islanders.
I’m sure you see the pattern. Continuing on, we’ll get to:
P1 thinks that P2 might htink that P3 might think that P4 might think that P5 might think that P6 might think that P7 might think that P8 might think that P9 might htink that P10 might think that P11 … P12 … P13 … P14 … etc etc … that P98 might think that P99 might think that P100 might think that there are no blue eyed islanders.
Again, this is not to say that P1 thinks that P100 might think there are no blue-eyed islanders. P1 doesn’t think that. Instead he thinks the thought just given above, one thta takes an extremely long amount of time to express in English, that P2 might think that P3 might think that… that P100 might htink there are no blue eyed islanders.
And for reasons that have been explained, the guru’s saying “I see at least one blue-eyed islanders” makes P1 unable to believe this long thought anymore. He can’t think that anyone would think that anyone would think… that anyone would think that anyone would htink that there are no blue eyed islanders, because the guru has made it impossible for anyone to think there are no blue eyed islanders and P1 knows this and knows that everyone (being perfect logicians) knows this.
It seems to me that the critical jump is from 3 to 4 blue-eyed islanders (BEIs).
When there are 3 BEIs, each can see 2 others, and each wonders whether the total is actually 2, in which case the other 2 each would see just one. One BEI means the Guru’s announcement causes departure that day; failure to depart give important information: there’s more than one BEI.
But when there are 4 BEIs, all know that everyone can see at least 2 others - so there is no possibility there could be just one BEI, nor that anyone could think that. So all can conclude that the Guru’s announcement contains no information not already known to everyone. Importantly, all know there was no possibility of a departure on the first night, and thus no useful information can be derived from the fact that no one left.
I’m not fully convinced by the above argument, but I’m not sure how to get around it.
They should pose explaining this puzzle as an exercise for prospective logic teachers, because it is clearly very hard to explain! Monty Hall’s got nothing on this.
Xema, it looks like we simulposted. I have to go but late I’ll make a new version of the above post with a complete explication of all the reasoning involved in the 4-person case. That is, unless either someone else does it, or you read the above explication of the 100-person case and respond wiht s omething along the lines of “You know what? You’re right and when you’re right you’re right.”
It’s not a valid pattern. I don’t need to work out what someone might think might think etc.
Everyone is only missing one piece of information, and everyone knows that everyone else is only missing one piece of information.
You need to look at it from what the system knows, not what anyone one person can know, and if we can logically work out what the system as a whole knows (that the minimum number of blue eyes is 99) then so can your perfect logicians.
In my experience, there are plenty of students who will not be ready to understand a puzzle like this even after their first semester of logic. That’s not to say being able to explain this isn’t valuable for a logic teacher, but because it’s literally impossible to explain it to a lot of beginning students, it might not be a good test for quality of teaching, at least for teaching beginning students…
All know there is no possibility of a departure the first night, but don’t know that everyone else must see no significance in that. If there had been 3 BEIs instead of 4, those three would have been justified in concluding that a departure on the first night was possible, and thus the fact that it didn’t happen was significant.
It works, only if you assume a very specific type of stupidity by the islanders. They have to be smart enough to follow the inductive reasoning, but clueless enough not to have realised that it’s not possible for anyone else to believe that one person could possibly not see any blue eyes.
For this method to work, they would all have to communicate an agreement to play by these rules and not one of them can make the leap that there must be a minimum of 99 blue eyes and everyone can see that.
Since communication is banned, they can’t do that. No one leaves.
Maybe the best way to understand the logicians is this way:
You, and 199 other people know you are all going to be sent to the island, immediately after being anesthetized and undergoing eye-color reassignment surgery (which may reassign your eye color back to its original color).
But, you also know ahead of time that the Guru will show up one random day and make a similar statement, except you don’t know what eye color the Guru will specify. More importantly, before being sent to the island, you can meet all your future fellow islanders and discuss the puzzle and work out with them any kind of agreed system you want to get as many people off of the island as possible. As in the original puzzle, there is absolutely no communicating in any way once you’re on the island: no gestures, shaking heads, or anything. The only information from one person to another is eye color, and whether the person has left the island.
Now, what system would you agree to with your future fellow islanders? Can you come up with anything that in the situation in the puzzle gets people off the island sooner AND works for other situations?
If so, well congratulations, you’re right and have triumphed over the puzzle writer, who came up with the wrong solution. Please tell us what your system is. If not, then, the solution must be right, even if it’s hard to understand.
Thats a different problem. Without a prior system the problem as described here Is unsolvable:
Because the guru’s statement conveys no information that was not already known by the simple observation “the minimum number of blue eyes anyone can see is one less than what I can see”. This does not regress to zero, because everyone is missing only one piece of information. If you’re going to argue with me, at least acknowledge the point, that all islanders know everyone else is only missing one piece of information.
This suffices as a response to your last post to me, coremelt, so I’ll just second it.
It appears you follow the logic of the solution, but you think that the solution contains an illicit hidden premise that the logicians don’t take certain information into account. But the abovequoted post directly refutes your argument for that view.
Why don’t you guys try answering the question? I know Indistinguishable tried to start a game, but I did it at least a page before he did, so everyone owes me the $20. Now answer my question: How many do I think there could be?
I already told you how it’s possible for someone to conclude that the minimum someone else is thinking of is ‘98’. Do you not see how I arrived at that number?
The critical thing you’re missing is that when asked the question “What is this other person thinking?”, you have to include the options that you, yourself, know to be wrong. And if you ask the person “What do you think this other person thinks about that third person’s thoughts?”, you again have to include options that not only you know to be false, but also that the second person knows to be false. Why? Because the third person doesn’t know that.
Every time I try to explain this to someone, I ask them “What does A think about what B thinks about C?” and they always end up answering with what A thinks about C. Note that that wasn’t the question. The question was A-B-C, not A-C. It helps to understand this if you have the people record their thoughts. Then you ask the next guy “What do you think he wrote down?” instead of “What do you think he thinks?” That way, you can’t break the chain easily.
So let’s suppose I ask person E to write down what color the five blues on the island have, 1 for blue, 0 for brown.
E: “11110 or 11111”. Even though I know that the second one is right, he thinks it could be the former.
So now I go to D and ask him “What did E write down?” Notice that I didn’t ask what he sees. I asked what E wrote.
D: “Either ‘11110 or 11111’ if I have blue eyes, or ‘11100 or 11101’ if I don’t.” Notice that he is forced to give two options for a total of four patterns. He knows that both 11110 and 11100 are incorrect, but he correctly reasons that E could’ve written them down. He also knows that E didn’t write down both pairs of patterns. He knows E wrote down one or the other, each containing one error, but he doesn’t know which one.
Now suppose I go and ask C what D wrote down. And this is where most people go off the rails. They try to change the question to “Hey C, what did E write down?” But that’s not the question. The question is what D wrote about E. And he’ll answer:
C: Either " [11110, 11111] or [11100, 11101] " if I have blue eyes, or " [11010, 11011] or [11000, 11001] " if I have brown eyes.
Notice again that C knows that anything that isn’t xxx11 is wrong, and he knows that D knows that anything other than xxxx1 is wrong, but remember, we’re not talking about what D knows is fact. We’re talking about what D knows about what E wrote. He also knows that D wrote only a single set of four. And he knows that E only wrote a set of two. And he knows that I, the interviewer, knows that only one actual pattern is possible. He just doesn’t know which one D wrote about E’s writings.
Now it’s B’s turn. “What did C write down?”
B: Either [11110, 11111, 11100, 11101][11010, 11011, 11000, 11001] if I have blue eyes, or [10110, 10111, 10100, 10101][10010, 10011, 10000, 10001] if I have brown eyes.
Again, B knows that I know the single correct pattern. He knows that E wrote two things. He knows that D would also write two things if he were asked about the pattern itself, but that wasn’t the question. The question was about what E wrote, so he knows D wrote down four patterns. Similarly, he knows that C wrote eight patterns, even know B knows that C knows that D actually only wrote four patterns, not eight. C just can’t narrow down which set of four D actually wrote, so C was forced to write down eight patterns total. But B can’t narrow down what C put in the second slot, so B has to guess two distinct sets of eight for what C wrote down.
Got that so far? I know one pattern, E wrote two, D wrote four, C wrote eight, and B wrote 16. Now for A. “Focus, A, focus. I’m not asking about E. I’m not asking about the pattern. I’m not asking about what C wrote about D. I’m asking this- What did B write about what C wrote about what D wrote about E?”
A: “Either [11110, 11111, 11100, 11101][11010, 11011, 11000, 11001][10110, 10111, 10100, 10101][10010, 10011, 10000, 10001] or
[01110, 01111, 01100, 01101][01010, 01011, 01000, 01001][00110, 00111, 00100, 00101][00010, 00011, 00000, 00001],depending on whether or not I have blue eyes. But it’s either the first or second set of 16, not both sets.”
Do you see the pattern I bolded and underlined? That’s the thing right there. Their ignorance has stacked up so much that A can’t rule out that B wrote 00000, despite everyone knowing that it’s incorrect.
Now for the reveal. When the Guru speaks, A knows that the second set is no longer what could be written down by B. It kills the entire set of 16. He’s narrowed it down to 16. The next night, any set containing a single 1 with four 0s is also ruled out. And so on, until only 11111 is left. And they can all go on the boat.
Coremelt, tell me where you think this reasoning in the four-person case goes wrong. (Take careful note of indentation and quotation marks, both of which are intended to convey the same information about who is thinking who is thinking who is thinking what.)
If you don’t think the reasoning goes wrong, then explain why you think P1 is not licensed to believe it possible that P2 believes it possible that P3 believes it possible that P4 believes it possible there are no blue-eyed islanders. (If you grant the reasoning, it should be impossible for you to explain this, since the reasoning leads to exactly this as its conclusion.)
Befiore the Guru speaks, P1 thinks to himself the following indented thoughts:
"P2 might think it's possible there are only two blue-eyed islanders, because for all I know he only sees two pairs of blue eyes (P3's and P4's).
So, P2 might think to himself the following indented thoughts:
'P3 might think it's possible there is only one blue-eyed islander, since for all I know, P3 only sees one pair of blue eyes (P4's).
So, P4 might think to himself the following indented thought:
"It is possible there are no blue-eyed islanders, since I do not see any blue eyes and do not know my own eye color."'"
I therefore believe there to be either 99 or 100 blue eyed people.
However, when I consider what YOU must see and know, you MIGHT see me as brown-eyed. Thus you might consider that the island has either 98 or 99 blue-eyed people.
I know that if I have brown eyes, then you might think Indistinguishable would see the island as having either 97 or 98 blue-eyed people.
That, in a nutshell, is why you need the guru to speak to provide the logical bootstrap. Regardless of the fact I KNOW there are either 99 or 100 blue-eyed people, I cannot deduce what EVERYONE ELSE might know since they have different sets of limited information.