f2f is on the right track.
My answer was, on the 100th day, all 100 blue eyed people will leave the island.
Think about if there was just one person with blue eyes on the island- he would know that they would leave the first night, becuase he knows he is the only person the Dr. was talking about. He would look around, see no one else, and know he should leave. So (THEOREM 1) if there is one blue eyed person, they would leave the first night.
Now consider there are two blue eyed people. On day 1, X thinks to himself, "Suppose that I don't have blue eyes. Since I can see one person (Y) with blue eyes, this must be the one-person case. Thus, Y, seeing nobody else with blue eyes, will leave tonight." (THEOREM 1) But Y doesn't leave that night, so on day 2, X thinks to himself, "Well, Y didn't leave, so Y must seen somebody else with blue eyes. That is, because they are all perfect logicians, Y must not have gone through the reasoning I went through. Thus, I must also have blue eyes." X does the same thing with Y, and they both leave the second night.
Thus, (THEOREM 2): If there are two blue-eyed people on the island, they will each leave the 2nd night.
This continues with the group leaving en mass on the corresponding night of the # of people in their group. When you reach the 99th night and no one has left, all 100 blue eyed people will leave together on the 100th night. Think of it as all the blue eyed people on the island playing a big game of chicken with each other, right up until the 100th night. Good thing they are all "perfect logicians!"
Yes, but the critical point is that he only speaks once. He doesn't reveal anything the people on the island don't already know, they can all look at each other and see someone with blue eyes. The Doctor speaking merely starts the cycle of blue eyed people leaving.
When I was talking it over last night, I was having fun asking, "What if there's a blind person on the island?"Also, when do they make a Wilson volleyball for company?
Bookmarks