If you want a free Hspin Pheonix The first person to get it right Wins the pheonix

A group of people with assorted eye colors live on an island. 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 night at midnight, a ferry stops at the island. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight. 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.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let’s say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

Every single blue eyed person leaves on the 100th night.

Edited to include explanation:

So, if there is only one blue eyed person on the island, he would look around, see that nobody else has blue eyes, know the guru is talking about him and leaves the island. However, if there are two blue eyed people, let’s call them person A and B, A will know that the guru could be talking about B or himself, and B will know that the guru could be talking about person A or himself. So if after the first night, A sees that B still hasn’t left, he knows that B must see one other blue eyed person (i.e. A) and B sees that A still hasn’t left for the same reason.

If there are three blue eyed people (let’s call them A, B and C) then person C will observe A and B going through the same process as they did previously. When they don’t leave on the second night, C knows that there must be one other blue eyed person on the island which can only be himself since he knows what colour everyone else’s eyes are. A and B are also following this same thought process as C. Therefore, they all leave together on the 3rd night.

This pattern extends right up to the 100th blue eyed person. Ergo, all 100 of them leave on the 100th night.

The answer is that on the 100th day, all 100 blue-eyed people will leave. It’s pretty convoluted logic and it took me a while to believe the solution, but here’s a rough guide to how to get there. Note – while the text of the puzzle is very carefully worded to be as clear and unambiguous as possible (thanks to countless discussions with confused readers), this solution is pretty thrown-together. It’s correct, but the explanation/wording might not be the best. If you’re really confused by something, let me know.

If you consider the case of just one blue-eyed person on the island, you can show that he obviously leaves the first night, because he knows he’s the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: If there is one blue-eyed person, he leaves the first night.

If there are two blue-eyed people, they will each look at the other. They will each realize that “if I don’t have blue eyes then that guy is the only blue-eyed person. And if he’s the only person, by he will leave tonight.” They each wait and see, and when neither of them leave the first night, each realizes “My HYPOTHESIS 1 was incorrect. I must have blue eyes.” And each leaves the second night.

So:If there are two blue-eyed people on the island, they will each leave the 2nd night.

If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities – “I have blue eyes” or “I don’t have blue eyes.” He will know that if he doesn’t have blue eyes, there are only two blue-eyed people on the island – the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes – THEOREM 2 says that if he didn’t, the other guys would have left. When he sees that they didn’t, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave.

This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they’ll each wait 99 days, see that the rest of the group hasn’t gone anywhere, and on the 100th night, they all leave.

skitrz
(⛷ Noisy Lurker ISO the Elusive Snow Weasel)
#5

No my grandpa is good at riddles and he told me what to type

skitrz
(⛷ Noisy Lurker ISO the Elusive Snow Weasel)
#7

This is the content on:

https://xkcd.com/solution.html
The answer is that on the 100th day, all 100 blue-eyed people will leave. It’s pretty convoluted logic and it took me a while to believe the solution, but here’s a rough guide to how to get there. Note – while the text of the puzzle is very carefully worded to be as clear and unambiguous as possible (thanks to countless discussions with confused readers), this solution is pretty thrown-together. It’s correct, but the explanation/wording might not be the best. If you’re really confused by something, let me know.

If you consider the case of just one blue-eyed person on the island, you can show that he obviously leaves the first night, because he knows he’s the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: [THEOREM 1] If there is one blue-eyed person, he leaves the first night.

If there are two blue-eyed people, they will each look at the other. They will each realize that “if I don’t have blue eyes [HYPOTHESIS 1], then that guy is the only blue-eyed person. And if he’s the only person, by THEOREM 1 he will leave tonight.” They each wait and see, and when neither of them leave the first night, each realizes “My HYPOTHESIS 1 was incorrect. I must have blue eyes.” And each leaves the second night.

So: [THEOREM 2]: If there are two blue-eyed people on the island, they will each leave the 2nd night.

If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities – “I have blue eyes” or “I don’t have blue eyes.” He will know that if he doesn’t have blue eyes, there are only two blue-eyed people on the island – the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes – THEOREM 2 says that if he didn’t, the other guys would have left. When he sees that they didn’t, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave.

This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they’ll each wait 99 days, see that the rest of the group hasn’t gone anywhere, and on the 100th night, they all leave.

With grammar and spelling like that, I am now certain you copied and pasted the answer you theoretically wrote. In the answer you theoretically wrote, you spelled the word “you” using the correct spelling; while in this quote, you spelled it “u.” Also, you misspelled “plagiarized”, while in your theoretical answer you spelled everything with good form.

The answer is that on the 100th day, all 100 blue-eyed people will leave. It’s pretty convoluted logic and it took me a while to believe the solution, but here’s a rough guide to how to get there. Note – while the text of the puzzle is very carefully worded to be as clear and unambiguous as possible (thanks to countless discussions with confused readers), this solution is pretty thrown-together. It’s correct, but the explanation/wording might not be the best. If you’re really confused by something, let me know.

skitrz
(⛷ Noisy Lurker ISO the Elusive Snow Weasel)
#19

After further scrutiny it would appear that AjColes initial post is also a copy/paste event. Case in point, both AjColes and Username1 seem to have utilized the exact wording from the same site.
Rather suspicious!

I doubt there is even a yoyo to be had.

I’m locking this thread until a few things get cleared up.

Remember, plagiarism is theft. Please reference sources when quoting or else it looks like the above. :-[