Eye colour problem...

Disclaimer: I cannot be held responsible for any blown fuses (*mwahahhaha*)...

200 people live on an Island. They are all perfect logicians. If they can work something out logically, they will do so instantly. 100 of these people have blue eyes. 100 have brown eyes. Nobody knows the colour of their own eyes, and there are no reflective surfaces or any other such means by which they can fathom the colour of their own eyes. Although, everyone on the island can see everyone else at all times (including eye colour), they are unable to just tell each other what colour eyes each has....

At noon one day, a stranger arrives on the island. She is a sage, and known to be completely honest, she has red eyes; and she proclaims "someone on this island has blue eyes", and then immediately leaves on the ferry. The ferry comes each day at noon; and takes anyone who has worked out what colour eyes they have, off the island.

How long before anyone (other than the red eyed sage) leaves the island, and who leaves?

I know :p

Paul wrote:I know :p

hehe!

Shall we keep it quiet and see if Sam's head explodes?

100 days, and all the blue-eyed boys and girls leave at the same time...

1) If they all see everyone's eyes and one guy doesn't see any blue then he knows he's the only one and goes on the next ferry

2) If two people each see one other guy with blue then they will wait a day - if the other guy goes then they know he was the only blue eyed boy and left, ergo they must also have blue eyes and both will leave on the following ferry

3) same process of elimination applies for 3 (day 3) all the way up to 100 (day 100) at which point all 100 of them will bugger off. Hope the boat is big enough...

DaveB wrote:100 days, and all the blue-eyed boys and girls leave at the same time...

1) If they all see everyone's eyes and one guy doesn't see any blue then he knows he's the only one and goes on the next ferry

2) If two people each see one other guy with blue then they will wait a day - if the other guy goes then they know he was the only blue eyed boy and left, ergo they must also have blue eyes and both will leave on the following ferry

3) same process of elimination applies for 3 (day 3) all the way up to 100 (day 100) at which point all 100 of them will bugger off. Hope the boat is big enough...

If the requirement for getting on the ferry is NOT seeing any blue eyes and they can see ALL eyes at the start how does the first person get on the ferry when there are (assuming they do have blue eyes) 100 ppl with brown and 99 people with blue eyes.

Mattel wrote:

DaveB wrote:100 days, and all the blue-eyed boys and girls leave at the same time...

1) If they all see everyone's eyes and one guy doesn't see any blue then he knows he's the only one and goes on the next ferry

2) If two people each see one other guy with blue then they will wait a day - if the other guy goes then they know he was the only blue eyed boy and left, ergo they must also have blue eyes and both will leave on the following ferry

3) same process of elimination applies for 3 (day 3) all the way up to 100 (day 100) at which point all 100 of them will bugger off. Hope the boat is big enough...

If the requirement for getting on the ferry is NOT seeing any blue eyes and they can see ALL eyes at the start how does the first person get on the ferry when there are (assuming they do have blue eyes) 100 ppl with brown and 99 people with blue eyes.

What Dave is trying to say is IF there was only one person with blue eyes they would leave immediately (as they would know that no one else had blue eyes) 
If there were 2 people with blue eyes,  they would both realise the next day when the blue eyed person they knew about didn't leave. 
Since there are 100 people with blue eyes it would take them 100 days to figure it out.

Mattel wrote:

DaveB wrote:100 days, and all the blue-eyed boys and girls leave at the same time...

1) If they all see everyone's eyes and one guy doesn't see any blue then he knows he's the only one and goes on the next ferry

2) If two people each see one other guy with blue then they will wait a day - if the other guy goes then they know he was the only blue eyed boy and left, ergo they must also have blue eyes and both will leave on the following ferry

3) same process of elimination applies for 3 (day 3) all the way up to 100 (day 100) at which point all 100 of them will bugger off. Hope the boat is big enough...

If the requirement for getting on the ferry is NOT seeing any blue eyes and they can see ALL eyes at the start how does the first person get on the ferry when there are (assuming they do have blue eyes) 100 ppl with brown and 99 people with blue eyes.

I didn't explain that very well.

1) if there was 1 person with blue eyes and 199 with brown and you were presented with absolute proof that there were folks with blue eyes on the island you'd know you were the one and would leave immediately (since you can't see anyone with blue, the only one you can't see is your own, ergo you must be blue).

2) If there were 2 people with blue eyes and 198 with brown then each would see one other person with blue and the rest are brown. That then leaves two possibilities - that either 1) that one person is the only one with blue eyes, if so they'll leave on the next ferry (as with 1) above) or b) I also have blue eyes. If the other guy doesn't leave then I must also have blue eyes. In which case both leave on day 2

...

100) If there are 100 people with blue eyes and 100 with brown then each of the blue-eyed folk will see 99 blue eyes, they will therefore wait 99 days per the above and if no-one leaves will work out that they must also have blue eyes and leave on the next ferry, on day 100.

The important thing is its not one person working this out, it's every blue-eyed person arriving at the same logical conclusion

Matt, the requirement for getting on the ferry, is knowing the colour of your own eyes.

Well done Dave, you got it! Was that the same answer you came up with Paul?

Indeed it was.

I think DaveB's explanation of the solution is really clear, but can you go from that to being able to answer this: What extra information did the sage give to the islanders?

At the start each individual had no way of knowing what colour their own eyes were. All she told them was that there was at least one blue-eyed person on the island and it seems that this is a fact which everyone there already knew as they could each see either 99 or 100 blue-eyed people.

She didn't need to have said that anyone had blue eyes, I think this is intended as a clue as to how to solve the puzzle.

Jamie wrote:Matt, the requirement for getting on the ferry, is knowing the colour of your own eyes.

Well done Dave, you got it! Was that the same answer you came up with Paul?

doh thanks, no wonder I couldn't solve it

Paul wrote:She didn't need to have said that anyone had blue eyes, I think this is intended as a clue as to how to solve the puzzle.

It's significant, otherwise, why wouldn't 100 brown eyed people leave, rather than 100 blue eyed people?

Tom, poses an excellent part 2 question to the problem...

If there was only 1 blue eyed person; when the sage says someone has blue eyes, the person who does have blue eyes, will see no-one else does, and so conclude that he must have blue eyes...

If 2 people (A and B) have blue eyes:
A sees B is blue eyed, and that B doesn't leave on day 1.
A will conclude that B saw someone else with blue eyes; and that because A only sees one person (B) with blue eyes, the person that B sees who has blue eyes, must be himself (A).

If 2 people (C and D) have green eyes (the sage doesn't seed these 2 people with an eye colour):
C sees D is green eyed, and that D doesn't leave on day 1. Why should they?
C will be none the wiser regarding the colour of his own eyes (they could be purple for all he knows).

Hmm. To be honest; I don't think I've given a satisfactory answer to Tom's question...

Jamie wrote:

Paul wrote:She didn't need to have said that anyone had blue eyes, I think this is intended as a clue as to how to solve the puzzle.

It's significant, otherwise, why wouldn't 100 brown eyed people leave, rather than 100 blue eyed people?

Tom, poses an excellent part 2 question to the problem...

No,  the information is totally superfluous. 
Ignoring the colours,  each person can see 99 people with one eye colour and 100 with the other eye colour. Using the extrapolated logic already discussed they know that if they share the eye colour of the 100,  the other 99 people will leave after 99 days. When they don't,  everyone knows there must be 100 of each colour and thus know which eye colour they have and leave on the 100th day.

The information the sage gave was how to leave by the sounds of it.

Personally, If I was on an island and there were 99 people with one eye colour and 100 people with a different eye colour, I'd leave on the first day and when they asked what eye colour I had, I'd just guess the same as the 99. Seems most likely and it's 50/50 anyway. I'm not waiting another 3 months to get outta there.

Kes wrote:The information the sage gave was how to leave by the sounds of it.

Personally, If I was on an island and there were 99 people with one eye colour and 100 people with a different eye colour, I'd leave on the first day and when they asked what eye colour I had, I'd just guess the same as the 99. Seems most likely and it's 50/50 anyway. I'm not waiting another 3 months to get outta there.

It's a logic puzzle dude, not a reflection on real life, and the rules state they can't tell each other

Yeah but my way is more fun

Amazing puzzle. I didn't get the answer but just tossing the answer around in your head until you grasp it intuitively is a mental work out. I'm still thinking about the puzzle as a whole (once you consider what the problem would be if there were 4 blue eyed people and 100 brown eyes people it becomes mind bending) but I want to say the sage's info is not superfluous, because nobody can follow the inductive reasoning required if they can't assume that if there was only one person with blue eyes he would leave if he didn't see anyone else with blue eyes (he wouldn't know that anyone at all had blue eyes without the sage's wisdom, so he would stay on the island). It seems the whole solution rests on that assumption.

Sam wrote:Amazing puzzle. I didn't get the answer but just tossing the answer around in your head until you grasp it intuitively is a mental work out. I'm still thinking about the puzzle as a whole (once you consider what the problem would be if there were 4 blue eyed people and 100 brown eyes people it becomes mind bending) but I want to say the sage's info is not superfluous, because nobody can follow the inductive reasoning required if they can't assume that if there was only one person with blue eyes he would leave if he didn't see anyone else with blue eyes (he wouldn't know that anyone at all had blue eyes without the sage's wisdom, so he would stay on the island).

Yes,  you're right, if there were only 1 person with blue eyes it would be vital information,  but there aren't and therefore everyone can see there are blue eyed people.

I think you're missing something Paul; and it's a something I'm pretty sure is there, I just don't know how to explain it. But let's consider this...

Why don't the any of the brown eyed people ever leave?

Also, if the sage didn't mention the colour blue; would any blue eyed person leave?

If not, why not?

Well, if there are two blue-eyed people they know there is another blue-eyed person so the Sage's advice sounds superfluous. So, can they both leave the island without the words of wisdom after two days? I think not, because neither of them can reason that if they have brown eyes the other person will leave. They are both stuck there looking into each others blue eyes, lovingly, forver.

Sam wrote:Well, if there are two blue-eyed people they know there is another blue-eyed person so the Sage's advice sounds superfluous. So, can they both leave the island without the words of wisdom after two days? I think not, because neither of them can reason that if they have brown eyes the other person will leave. They are both stuck there looking into each others blue eyes, lovingly, forver.

Yes, they will leave on the 2nd day when they realise that the only other blue eyed person  didn't leave. They are pure logicians Sam,  they have no time for you fuzzy minded emotions!

Jamie wrote:I think you're missing something Paul; and it's a something I'm pretty sure is there, I just don't know how to explain it. But let's consider this...

Why don't the any of the brown eyed people ever leave?

Also, if the sage didn't mention the colour blue; would any blue eyed person leave?

If not, why not?

They do,  they ALL leave. 

Colour is Irrelevant, imagine the same question but with brown being the stated colour,  same result,  they ALL leave.

Why would they assume the other blue eyed person would leave at all, if the sage didn't give the advice? It all comes back to that one blue eyed person situation. If nobody can assume that one blued eyed person who sees only brown eyed people will know that he has blue eyes (because the sage has told them he sees somebody with blue eyes) they can do bugger all.

Here's an xkcd page on the answer:

https://xkcd.com/solution.html

The questions he asks "that may force you to further explore the structure of the puzzle" are:

Randall Monroe wrote:What is the quantified piece of information that the Guru provides that each person did not already have?

Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?

Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?

I think I've got a grip on this now. Ready for the next one when you are, Jamie!