I’m sleepy, and I just had a drinky. I rather wish I hadn’t openened this thread. I’m so glad I know my eye color. Huh? Perhaps it will make more sense in the morning. Carry on.
To clarify for those viewing at home:
Statement: “There are between 97 and 101 blue-eyed Islanders.”
Reality that we know: There are exactly 100 blue-eyed Islanders.
If the natives assume the stranger/guru doesn’t lie then:
If there were 97 people on the island with blue eyes, they’d all leave the next day. But there are not. So no one leaves.
If there were 98 people with blue eyes, they’d all leave in two days. They’d rationally assume that were there only 97, they’d be gone in one way or another on the first day. Because that didn’t happen, they know that there were 98. But there are not just 98 blue eyes. So no one leaves.
If there were 99 blue eyed folks, they’d disappear on the third day because they recognize the above. But there are not just 99. So everyone stays.
But on the fourth day, these blue eyes (who now know there can only be 100 because they can dismiss higher numbers) have realized their true eye color. It’s off to the altar or the exit boats.
Now, it seems to me the conditional statement “If there is a blue-eyed native, then there is a brown-eyed native” sets up the brown-eyes first for eye-color self-realization. But otherwise, it’s not any different then saying, “There is a brown eyed native.” At least I don’t think so…
Finally, saying that there are both brown and blue-eyed natives would decrease the number of days to eye group death/departure by one because the plurality removes the necessity of a first day eliminating the possiblity of a single native with the certain singular eye color realizing his/her ocular hue. Depending on which group is the smallest (blue or brown eyes), that smallest group commits suicide/leaves one day earlier than if a single member of the eye-color group is first mentioned. Or, to put it another way, the smallest group leaves on a day number 1 less than the number of members of that group.
Now if the stranger’s eye color is not green and must be taken into account…ah forget it!
Ya it is a pain in the ass because the only reason they don’t self destruct in the resting state is that they don’t know that there are only two eye colors they can be. Once they can’t assume they are a third eye color they go boom.
Oh and why it generalizes, and this is non-obvious it just occured to me, is that everyone knows the logic and everyone is playing by the rules. Instead of doing my paper I’m going to lay it out.
In the case of a perfect logician whose a bit wacked out about eye color living on an island with a bunch of clueless happy natives who agree to his eye color ban out of respect if someone comes and makes the assanine statmenet nothing happens. This is because the logician knows that the natives don’t care to bother with even the n=2 case and so they won’t leave.
In the case of a perfect logician who is delusional about his neighbors in the situation he will go poof on the appropriate day not realizing that everyone else wasn’t actually giving a shit.
In the opposite scenarios where in there is one weirdo among the perfect logicians who is unknown the information means nothing because that individual breaks the progression. They can’t be exchanged for any other individual in the rational and as a result will show up when ever.
If the defector is unknown and of the color in question everyone but him will off themselves on the critical day.
Breaking the homoganity of logic destroys the puzzle and it is only because everyone essentially agrees to the n=2 and n=3 case, and thus to the N case, that the mass deaths occur.
In summary thinking may be bad for your health and the real thing to do is to shout back to the guru, “So do I you green-eyed lunatic.”
The Tim: You just made me laugh up my morning coffee into my nose. (Hmm. Suppose there were an island where some natives have coffee up their nose, while others do not…)
I like the many senarios/problems this puzzle presents. Here’s some more:
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Suppose the stranger/guru had said: “There are more brown-eyed natives than blue-eyed natives.”
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Suppose a blue-eyed native is blind in the original puzzle.
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If a blue-eyed native dies of natural causes during the original puzzle, is it all off?
I’m with shijinn. Why does the guru’s (or stranger’s) statement lead to a mass exodus (suicide) when everyone on the island already knew for certain that there was “at least one person with blue eyes”? If these people are are perfect logicians, and this statement prompts all the hullabaloo, why wouldn’t the same progression have have occurred without the statement? If the logic leading to the mass suicide is accurate, then these knuckleheads ain’t perfect logicians, not if they needed a stranger/guru to remind them of something they all knew perfectly well (knowledge they likewise knew every other person on the island possessed).
See what I mean? Good, 'cause I’m not sure I do at the moment. Anyways, somebody 'splain this seeming disconnect, if you please.:dubious:
I’m with Bob Cos. The guru isn’t telling them anything they didn’t already know.
no, but he is starting the counter. Without that statement, the Islanders don’t know where in the logical thought process everyone else is, and so cannot say “It’s day 99, everyone’s still here, I guess I’m blue eyed”
Don’t know why that would be at all. Can you explain? The day before the guru spoke, no one knew for sure what eye color he or she had (and he or she still would not have known 100 days hence, assuming the guru had kept his mouth shut). That’s accurate, right? If it is, why would a guru making a statement that added zero new information to anyone on the island get everyone started on an inexorable path to suicide?
Someone answer quick. My township is considering installing a similar “eye color” rule and I’m beginning to think it’s a bad idea.
Ok basically everyone, by being pefect logicians and knowing that everyone else is, agrees that this scenario works because it can be shown to work for the n=1 and n=2 case, and more generally the n case and n+1 case. Being perfect logicians they know that if you can prove it in the n case and n+1 case it must work for any number and so they have doomed themselves once the information is given.
Maybe a better way to explain why everyone offs themself on the same day is to rephrase the scenario in the following way. 100 inmates in a futuristic prison have cortical bombs implanted in their brain. These bombs give the individual perfect logic and makes them believe they have always had such perfect logic and do not have a cortical bomb. If anyone discovers they have a cortical bomb it goes off and their head explodes. The properties of cortical bombs are known to all. Each prisoner knows that the others have a cortical bomb because of it announces itself with a unique identifier to the others psychically evey second. Each prisoner knows the others are assuming that they are the ones without it and are also assuming that others assume that and so on. This produces the logic chain that there exists some hypothetical prisoner whose bomb should go off when the existence of a prisoner with a bomb is made known. Any prisoner can be that prisoner. The only way for their bomb not to go off when the decleration is made is if they sense a cortical bomb. Likewise there exists a prisoner in the chain who must only sense one individual with a cortical bomb and after the revelation of the existence of prisoners with cortical bombs and the subsequent non-head popping of the first they realize they must have one as well. Again any prisoner can be this prisoner. So it goes up the assumption chain, and the only way for a particular actual prisoner to be right is on the 99th second after revelation for everyone else’s head to explode.
So one day the warden realizes that hyperlogical prisoners aren’t the best plan and says to them over the loud speaker “There is a prisoner that has a cortical bomb.”
Now things are set into motion and prisoner X watches with great interest, knowing all of them are doomed save he. In the first second no one explodes, because the last prisoner in prisoner X’s chain prisoner (X+99)%100 can sense someone else with a cortical bomb. In the second second no one explodes, because the second to last prisoner in prisoner X’s chain prisoner (X+98)%100 can sense another who isn’t prisoner (X+99)%100. So it goes through the seconds with each individual on the chain’s assumption being violated yet not having their head explode and therefor violating the higher assumption. At 99 seconds prisoner X notices that his clothing is disturbingly clean for all the heads that should be exploded and at 100 seconds prisoner X’s head explodes along with all the other prisoner’s heads. It is important to note that any number from 0 to 99 can be used for X and it works and that is why the chain of assumptions collapses with fatal results.
After this however the warden’s assistant says “Well that shouldn’t have happened. They already knew that there was a prisoner with a cortical bomb. Those things can’t work very well.”
The warden tells him the following “Each prisoner knew of the general case for these things and knew the others did as well. Each one knew all the others were making a faulty assumption but did not know that all the others knew that he himself was making a faulty assumption because if he knew that he would know he had a cortical bomb and his head would explode. Thus each followed the assumption chain in each possible order, knowing that it would create a situation like that bizarre island riddle. As a result each knew that a situation was created in which the logic followed but only because each knew that all the others knew it.”
That’s the best I can do for explaining it.
Forgive me, dude, but your explanation just gave me a cortical-bomb-like headache.
Fair enough but I really can’t think of any more ways to explain it so the explanations are simply going to start becomming rehashes of previous posts.
As I understand the objection people are caught on the fact that even though it generalizes from n to n+1 there are so many people someone can see with blue-eyes that it can’t possibly generalize to the riddle’s case.
The thrust is, if the specific logic isn’t working to grasp the answer, that all the islanders know about the riddle and knowing about it forces them to accept that it applies in their case even though there are high numbers. The new information imparted by the statement is that a particular case is not true, and that every relevant individual now knows the case isn’t true. This is what causes the chain reaction of logicing themselves to death.
Bob cos,
Here’s as simple as I can make it:
**If the islanders know that there are exactly 100 blue-eyed natives on the island, then they can determine if they themselves have blue eyes. If an Islander were to only see 99 blues and he knew there were supposed to be 100 blues, then he would know that his own eyes must also be blue. Right?
The islanders do not know there are only 100 blues, but they do know this:
Because all blues can see 99 others with blue eyes, they (the blues) know that there can only be two possibilities: That there are only 99 blues or that there are 100 blues–and they are the 100th. No other possibilities.
Browns can see 100 sets of blue eyes. They know that there are 100 blue eyes only…or that there are 101 sets of blue eyes, and they are the 101st.
Now enter the stranger. The stranger or guru says, “I see someone with blue eyes.”
This starts the entire group at the same place. And they can figure out exactly how many natives have blue eyes by watching the actions of the others.
If there were only 1 native on the island with blue eyes, that person would instantly know the stranger was talking about him/her. That person would be gone the next day.
Because no one leaves on this next day, there must be more than one person with blue eyes. (Forget for a minute that everyone already knows this.)
Now suppose there were only 2 natives on the island with blue eyes. If that were so, each of these natives would only be able to see the one other set of blue eyes. Because we have determined from the lack of anyone’s exit on the first day that there is not simply one native with blue eyes, each of these two would realize on the second day that their own eyes are blue. They could not know this on the first day. They had to see if the other blue eye reacted. But on the second day they now know their eye color, and both blues would exit.
But no one exits on the second day. There must be more than 2 blues.
Similarly, if there were only 3 blues they’d figure this out on the third day because there was no reaction on the second day.
If there were four blues, they figure it out on the fourth day. Five blues figure it out on the fifth day. And so on.
When the 99th day occurs without incident, our actual blues (who formally only knew that there were either 99 or 100 blue eye natives can be certain that there are not just 99 blues on the island. Had there been, these folks would have offed themselves on the 99th day. They of course know, by looking around, that there cannot be more than 100 blues. Therefore there must be exactly 100 natives with blue eyes. Each blue-blue eyed native realizes that he/she is the 100th. Their knowledge, therefore, leads to their doom on the next day.
Go through it slowly and you will see that this is so.
thing is, we understand the logic behind the n=1 and n=2 case. but it requires a gradual introduction of the natives over a span of days as if they have never met before for it to be logical.
the guru had told them nothing new in front of everyone else so there can be no trigger made.
so i can only conclude that
a) the natives are robots badly programmed by a mad hat mathematician,
b) the one who came up with the riddle need a better scenario
I agree with shijinn. The guru told them nothing new. Everyone knew there were blue-eyed people on the island. So the guru told them that there was a blue-eyed person on the island. If there is more to it, I am very confused.
shijinn and BluMoon, consider it again but with only two blue eyed people.
Before the guru speaks, they both already know that there is at least one blue eyed person, but they don’t know that the other blue eyed person already knows this. The guru saying that there is at least one blue eyed person, indirectly lets them know, through their actions (i.e. not leaving the island), what the other knows.
So the new information that the guru gives (indirectly) is knowledge of what the other blue eyed person knows.
If you think about it, the same logic can be applied to 100, or any number of blue eyed people.
I get it when there are two people. I am still confused about the 100 blue eyed people though. How does the first person find out that they are blue eyed, and since the number of brown eyed people is not given, how does the pattern continue?
They all find out at the same time, so there is no ‘first person’.
You understand if there are two, so let’s say that there are three blue eyed people.
If there were two blue eyed people, they would have shipped out on day two, so the fact that nobody shipped out means there are more than two blue eyed people.
Since each blue eyed person can only see two other people with blue eyes, and they now know there must be more than two, each one must assume that there are actually three blue eyed people, and that they’re the third, and must leave the island.
(again, this will work with any number, not just three)
Listen, I strongly suspect that everything makes sense and I’m just being dense. And I do appreciate all the attempts at explanation. But I’m still hung up on a couple of things and no one has addressed them specifically and together (I don’t think).
Before the guru spoke, is it true that all the perfect logicians on the island had no way of knowing what their own eye color was?
Didn’t the guru provide information that was already known to everyone on the island (information that everyone also knew everyone else possessed, unlike the “2 people with blue eyes” scenario, for example)?
Can anyone reconcile these two statements (or point out how they’re wrong)?
I tried to understand this. I really tried. Never before have the words “brain” and “hurts” been more appropiate. Wow! 
A big stumbling block seems to be ‘But he told them something they already know.’
Baraqiyal already said this, but to reiterate (because I’m proud of figuring it out)
Consider the case with 3 blue eyes:
Blue-1 already knows that someone has blue eyes.
But doesn’t know that blue-2 knows someone has blue eyes until the guru tells him.
(And the same with any blues in place of blue-1 and blue-2)