This is a variation of the puzzle discussed in another thread http://www.sciencefo...rs-and-puzzles/.

 

Some of this is taken from http://www.xkcd.com/blue_eyes.html, edited for this variation.

 

 

A group of people with blue or brown eyes color live on an island. There are some unknown number N of blue. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every morning, a ferry stops at the island. Any islanders who had figured out the color of their own eyes the previous day must leave the island on that ferry, and the rest stay. 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.

 

An exception to the above, there is a guru with grey eyes who is allowed to speak once about eye color.

 

It is common knowledge that everyone else has either blue or brown eyes, but not common knowledge that there are at least one of either.

 

The guru says "I see someone with blue eyes."

 

 

It is common knowledge that the islanders are all unerringly truthful, and that all use identical reasoning.

 

 

Variation 1:

 

It is common knowledge that any islander who knows for certain (barring any unpredictable events including new outsider knowledge) what day they will leave, must announce it the following day, and tell everyone else what day they're leaving.

 

Variation 2:

After the guru's announcement, on that day and each day afterward, every islander must announce to all, their best guess of how long they think they'll stay on the island. This is also common knowledge.

 

What happens?

 

 

 

Hint:

This one is more like the unexpected hanging paradox.

 

Edited February 16, 201213 yr by md65536

I'll add two intermediate variations:

 

Variation 0:

As above, without the stuff from variation 1 or 2.

 

What happens?

 

I'll let it be taken for granted that (as I believe) the solution for the xkcd version also applies to this variation; All of the blue-eyed islanders leave on day N.

 

 

Variation 0.5:

As variation 0, with an additional rule: Any islander who knows what day she'll be leaving must immediately book a ticket off the island (ie. as far in advance as possible). The booking information is private; everyone may know that you're booking a ticket but no one else is able to find out what day you're booking for (unless by deduction).

 

What happens?

 

 

 

Alternate: Variation 1b:

 

It is common knowledge that any islander who knows for certain (barring any unpredictable events including new outsider knowledge) what day they will leave, must announce it immediately, and tell everyone else what day they're leaving.

 

What happens?

 

 

 

 

No need to solve all the variations yourself...