I just came across this really interesting riddle that I thought I’d share here.
Exactly 100 of these persons have blue eyes, 100 have brown eyes, and 1 has green eyes. The inhibitants do not know what his/her color eyes is. Everyone is constantly aware of everyone elses eye color but no person knows that there are 100 blue eyed, 100 brown eyed, and 1 green eyed person on the island.
If a person finds out his/her own eye color she/he must leave the island at midnight of the day she/he finds out! There are no mirrors or reflections of any kind on the island. Also, nobody on the island ever speaks except the Guru, who is the person with the green eyes, (she does not know her eye color and if she found out she would have to leave the island at midnight). The Guru says one sentence every fifty years. One day the Guru arrives and tells everyone on the island the following: “I SEE SOMEONE WITH BLUE EYES.”
Who (if anyone) leaves the island and when?
This is not a trick question, and the answer is a very logical one. It just needs a bit of thinking. Give it a try.
I’ll post the answer in a couple of days time.
Marwen, did the Guru spoke to everybody at a time or to everyone separately ?
The Guru spoke in front of everyone at the same time.
Yalaa!
Ok ok, no need to push…
Consider the situation if there were only 2 people with blue eyes. Everybody can see everybody else, and everybody already knew that the Guru could see someone with blue eyes. However, Blue-1 didn’t know that Blue-2 could see someone with blue eyes. After the Guru has spoken, Blue-1 still doesn’t know whether Blue-2 can see someone with blue eyes, so it might appear that no new information has been imparted. However, when midnight comes round, the fact that Blue-2 is still on the island imparts new information to Blue-1 – it tells him that Blue-2 must be able to see someone with blue eyes. This new information is enough to allow Blue-1 to work out that it must be his own eyes that are blue. Blue-2 reasons in a similar way. Therefore the two blue-eyed people leave on day 2.
Now consider the situation if there were 3 people with blue eyes. Blue-3 can work out the logic in the previous scenario, and expects Blue-1 and Blue-2 to depart on day 2. The fact that they don’t leave on day 2 imparts new information to Blue-3 – it tells her that Blue-1 and Blue-2 can each see more than one person with blue eyes. This new information is enough to allow Blue-3 to work out that it must be her own eyes that are blue. Therefore the three blue-eyed people leave on day 3.
The same reasoning can be extended to the situation where there are 3, 4, 5, 6,… 100 people with blue eyes. Since they all argue perfectly logically, they can work out that N blue-eyed people will leave on day N.
I never did puzzle this out, but reading the solution, feel somewhat less stupid. 🙂
I neva’ did manage to find the sol, of teh riddle. I was having a kind of idea but not exactly wht had turned out!!!!! hehehehe….anyways…way to go MMM!!! GUD ONE!!
I don’t agree with this logic, it works with 2 people but not more.
Let’s take the example of 3 blue eyed people. You said :
“The fact that they (Blue-1 and Blue-2) don’t leave on day 2 imparts new information to Blue-3 – it tells her that Blue-1 and Blue-2 can each see more than one person with blue eyes.”
This is wrong. I’d rather say “…can each see at least one person with blue eyes.”
So if I was Blue-3, I’ll say “Blue-1 is still here at least because Blue-2 has blue eyes, so I can have blue eyes, too, but I’m not sure”
And so on with 100 people…
So it works if there were only two people with blue eyes. Well now let’s take that case with 3 blue-eyed people.
So this guy being totally logical knows that if there were only two guys with blue eyes, they’d have left on day 2. So if they haven’t left, what does it mean, it means that they’re still not sure about their eye colour because each one of them sees more than one person with blue eyes, and as he sees everyone else other than these two has brown eyes, he knows that it must be him who has blue eyes, and so they all leave on the third day.
And it goes on and on as you increase the number of people with blue eyes. Each logical blue eyes person making the same conclusion.
I seems to be OK 🙂 Nice riddle !
Ehh oui bravo .. le principe de la r
VERY TRUE BILEL!!! quite intelligent!!! ahahahaha!
There are a couple of little tricks in this puzzle, that lead you astray. That the guru says “I see someone with blue eyes” is a bit misleading, since the information is trivial: everyone sees someone with blue eyes, and could assign those words to the guru. “The guru sees someone with blue eyes, and someone with brown eyes” is universal knowledge. What the guru is really imparting that is vital is that they will sort for blue. If she says “I see someone with blue eyes, and someone with brown eyes” – equally true and self-evident – the puzzle fails. So, she isn’t really saying that she sees someone with blue eyes, but that blue is the group that is being selected. The other ‘trick’ is the number and eye-color of the non-blues. This is irrelevant. There could be 2,423 ‘others’ of every other eye-color imaginable, including stripes and polka dots. The size of the blue group is known by time, and you belong to it when there is one less in the group, than days elapsed. Presto, on day 99 every blue-eyed person sees one less blue than they ‘should’ and suddenly each one knows his eye color.
No one would leave because “If a person finds out his/her own eye color she/he must leave the island at midnight of the day she/he finds out!” because midnight marks the beginning of a new day, by the time the person finds out his or her own eye color..the midnight of the day they found out their eye color would have already passed
I was really happy with myself when I solved this in under 5 minutes. Then my friend solved it in about 30 seconds…
Wouldn’t brown-eyed people think the same thing?
“Oh they haven’t left yet, so there must be more people with blue eyes”
Since the logic works like that, it shouldn’t be considered the ‘correct’ answer. TRY AGAIN.
yes jester, so im thinking 200 people, 100 of blue 100 of brown leave the island on the 100th day. leaving the guru forever because the guru has no way of finding out his/her own eye colour
Jester, the brown-eyed are of course following the same logic(it’s a perfect logic island), but each brown-eyed person see 100, not 99 blue-eyed people making their tentative exit on day 101… so when all the blue-eyed people they see leave on day 100, they know they are not part of the group.
Except that all brown eyed people can see 100 blue eyed people and know that they will all leave on the 100th night if there are no more blue eyed people.
mmm– why didn’t you answer ttt’s challenge?
No…
the brown eyes would leave too. Everyone would leave. When two blue eyed persons were left and they reasoned that they were blue-eyed with the logic that the other blue eye was waiting on the other to leave, the brown eyes would think they were blue-eyed as well because they would logically conclude that the final blue-eye was waiting on them. Everyone would leave because everyone would think they were blue eyed. The riddle is meant to show that logic isn’t the most necessary component of life.
Maxwell you are wrong for the reason that the people with brown eyes that are left have no idea what color their eyes have. For all they know it could be purple or pink.
Maxwell– You are mistaken. The blue-eyed people all see 99 other blue-eyed people, so when day 99 passes and all of them are left, They each conclude that they must be the last blue-eyed person and leave on day 100. The brown-eyed people see 100 blue-eyed people and would be waiting for day 101, but the blue-eyes all leave on day 100, so the brown-eyes do not leave with them.
no one leaves