## eye color

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### Re: eye color

The key is the number of people with the same color eyes as you. If you have a color that's different from everyone else, alone with purple eyes, you're not getting off the island. Everyone else will leave on their "scheduled" day. But if you share colors with someone else, at least one, you'll see them not leave on their scheduled day and know you're with them.

Hence everyone but the guru leaves on the 100th day.

I can't understand why some think the brown eyed people can't get off the island. THey follow the SAME PATTERN as the blues. And they leave the same day.

What the guru does besides setting day zero I don't know.
WhoDatMath
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### Re: eye color

Eh, I guess I'll just say it.

Let's label the people with blue eyes $b_1, b_2, ..., b_{100}$. Before the guru speaks, $b_1$ sees 99 people with blue eyes. He therefore knows that $b_2$ can see at least 98 people with blue eyes. Along the same lines, since $b_2$ can see at least 98 people with blue eyes, $b_1$ knows that $b_2$ knows that $b_3$ can see at least 97 people with blue eyes. Carrying this on down the line, $b_1$ knows that $b_2$ knows that $b_3$ knows that.....that $b_{98}$ knows that $b_{99}$ can see someone with blue eyes. However, once the guru makes his statement and 99 days pass without anyone leaving,$b_1$ knows that $b_2$ knows that $b_3$ knows that.....that $b_{98}$ knows that $b_{99}$ knows that $b_{100}$ can see someone with blue eyes. (That was interesting to type...) The only person whose eyes $b_1$ can't see is himself; therefore, the person with blue eyes that $b_2$ knows that $b_3$ knows that....that $b_{99}$ knows that $b_{100}$ can see is him.

@WhoDatMath
Well, I'm not sure it's any easier to follow, but if you can work through the logic of this post, you'll see that the brown-eyed people don't have the same information as the blue-eyed people and that's why they can't leave.

DeathRowKitty
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### Re: eye color

That's REALLY abstract. I guess i get it. You're talking pure logic, while the alternative is an agreed-upon strategy.

But if they were perfectly logical, then the brown-eyed people would all agree that my idea would be the strategy they'd most likely agree upon, and follow through. Then they'd get off the island anyway.
WhoDatMath
Mathlete

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### Re: eye color

Well, the brown-eyed people could never be sure of their eye colors. They could see that everyone else had brown eyes and assume they did, but there would be no logical way to draw that conclusion.

DeathRowKitty
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### Re: eye color

DeathRowKitty wrote:Well, the brown-eyed people could never be sure of their eye colors. They could see that everyone else had brown eyes and assume they did, but there would be no logical way to draw that conclusion.

I was going on the example of the guess the number game with the clock. There you have an agreed-upon strategy. In this case, I took it as an agreed-upon strategy that they leave on day N when they see N-1 people of that eye color. Then, if you didn't have brown eyes, the 99 people would leave before day 100. (They all see 98 people)

I guess the solution doesn't allow that, it's more pure logic. But if they're thinking like I am, yes, they would know they had brown eyes when all the brown-eyed people remained on day 100.

EDIT: Here's the xkcd link that explains it. I understand my logic is wrong or at least informal. I was just explaining where my error was, in comparing it to a problem that allowed a pre-determined strategy (in which case the brown-eyed people COULD leave).
http://xkcd.com/solution.html
WhoDatMath
Mathlete

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### Re: eye color

DeathRowKitty wrote:Eh, I guess I'll just say it.

Let's label the people with blue eyes $b_1, b_2, ..., b_{100}$. Before the guru speaks, $b_1$ sees 99 people with blue eyes. He therefore knows that $b_2$ can see at least 98 people with blue eyes. Along the same lines, since $b_2$ can see at least 98 people with blue eyes, $b_1$ knows that $b_2$ knows that $b_3$ can see at least 97 people with blue eyes. Carrying this on down the line, $b_1$ knows that $b_2$ knows that $b_3$ knows that.....that $b_{98}$ knows that $b_{99}$ can see someone with blue eyes. However, once the guru makes his statement and 99 days pass without anyone leaving,$b_1$ knows that $b_2$ knows that $b_3$ knows that.....that $b_{98}$ knows that $b_{99}$ knows that $b_{100}$ can see someone with blue eyes. (That was interesting to type...) The only person whose eyes $b_1$ can't see is himself; therefore, the person with blue eyes that $b_2$ knows that $b_3$ knows that....that $b_{99}$ knows that $b_{100}$ can see is him.

@WhoDatMath
Well, I'm not sure it's any easier to follow, but if you can work through the logic of this post, you'll see that the brown-eyed people don't have the same information as the blue-eyed people and that's why they can't leave.

Isn't that basically the solution that I gave? :\

xander

xander
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### Re: eye color

Opening Post wrote:Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

That sums up pretty nicely why they can't come up with a plan (and if they could communicate, they could simply tell each other their eye colors).

@xander
Not sure if that's what you were attempting to get at, but the answer you gave was something they already knew. If you were to add 90something "that everyone knows" to what you gave, it would be equivalent to what I said.

DeathRowKitty
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### Re: eye color

I worked the problem out by simplifying it, and starting with the most simple situation: 2 people on the island, one blue, one brown. and then worked up. As people are arguing about whether or not the brown-eyes could leave lets do the same for them.
In the one blue one brown case, the blue eyed person would leave immediately. The brown eyed person would then know their eyes are not blue, but they would have no other colour to go on.
It is the same in a case where there are 2 brown eyes and say, 2 blues, they would leave on the second day, then those with brown eyes would then know that their eyes are not blue, but again, they could be any other colour, and that the other person has brown eyes, but they can't communicate this to them.
The only way for them to leave is if one of the things they knew to begin with was that there were only 2 possible eye colours, or for the Guru to say "I see a person with x coloured eyes" and start the process again until everyone but her has left and she could enjoy some peace and quiet.
TalksToRainbows
Kindergarten

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Joined: Fri Feb 11, 2011 5:54 am

### Re: eye color

sorry if i missed it, but there's proof by induction that the blue eyed guys will leave on the 100th day leaving the guru and the brown eyed guys to party forever (each brown eyed dude doesn't know he has brown eyes)
Math problems? Call 1-800-[(10x)(13i)2]-[sin(xy)/2.362x]

poochon
High School

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Location: Israel

### Re: eye color

A simple variation: the islanders all know that everyone's eye color is blue, brown, or green.
Spoiler! :
100 days after the Guru speaks, the blue-eyed people can all leave.
If at least one "Bluey" leaves then (it isn't stated that anyone wishes to leave), the brown eyes can leave 100 days later.
If at least one does, the Guru can leave the next day.
Gary
Mathlete

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Joined: Mon Nov 07, 2011 10:49 am

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