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[craze3]

The Blue Eyes Puzzle (Known as toughest puzzle)

[GuyP]

And one more thing: You aren't allowed to just guess your eye colour, get told it's not, then guess the other eye colour the next night. That's just not cricket!

Anyway, I had a long and involved debate about this with someone the other day. Yes, I understand how it's meant to work and everything, but nonetheless, it still seems to me that no real new information has been imparted. Is this like the Monty Hall problem (it seems like it shouldn't work, but it does) or does it just not really work?

[daveb]

I'm not convinced by the logic in that explanation

It works for 2 people with blue eyes, and for 3 people, but the logic doesn't scale for 4 in the same way.

Involved explanation follows later, but there is a simple flaw in the puzzle described:
Given the guru announces this amazing revelation that they can see someone with blue eyes - however, everyone has always known this information - because everyone can see at least 3 people with those colour eyes - so this doesn't change the situation (3 means that they know everyone else can see at least one other person with blue eyes). Meaning if it were possible to know with that information, they should have left the island already!

Involved explanation:
If there are 4 people with blue eyes, and 4 people with brown eyes.
Let's ignore the brown eye people - they aren't important.
So the 4 blue eyed people are called A, B, C and D
A knows there are either 3 or 4 people with blue eyes (3 if they have don't have blue eyes, and 4 if they do)
A knows that B, C and D all think there are either 2 people with blue eyes (if A doesn't, and B/C/D don't think they do), 3 with blue eyes (if A does, and B/C/D don't think they do OR A doesn't and B/C/D think they do), or 4 with blue eyes (if A does, and B/C/D think they do)

Now this is the case for A, B, C and D - there is not enough information to work out what colour eyes you have (everyone is in the same situation, and none of them are going to be leaving the island because they are all in the same position as you)

The logic in the given solution works for 2 blue and 2 brown eyed people. The nice bit of logic for 3 of each is clever (using the information that no one left allows you to deduce more)

But if someone can see a flaw in my logic - let me know

Hehe - that's my lunchhour well spent

[Tonamel]

The way I scaled it to four is that each person sees three blue-eyed people. Using their Perfect Logic, they know that it's going to take them three nights to leave, but when they're all still there on the fourth night, they all realize they're the fourth, and they all leave.

For five, they would all know to wait to see if anyone leaves on the fourth night, and so on.

[kittie.]

ok this was totally not the hardest logic puzzle ever.