Puzzle Time

WIth the announcement of the overseer, we could effectively eliminate the red-eyed ppl, and only consider the blue's perspective. All of the answers are in the perspective of any single blue-eyed ppl

If we take a smaller population, let's say 1 red 1 blue 1 overseer, when the overseer announce, on the 1st day's 11am, the blue will suicide.

Now if we have a population of 2 blue 2 red 1 overseer, after the announcement, on the 1st day, no one will suicide, which implies 2 blues (since 2 are red and 1 overseer). Hence, 2 blue will suicide on the 2nd day.

Now with 3 blue 3 red 1 overseer, after the announcement, on the 1st day, no one will suicide, which implies that 2 or 3 of them are blue. Each blue-eyed ppl will see 2 other blue. So on the 2nd day, if the 2 "other" blue eye ppl did not suicide (according to the theory from the 2blue2red1overseer), you can deduce there are 3 blues, hence suiciding on the 3rd day.

goes on and on, Pn, where n = 100.

Using a reduction argument:

w/ 3 people, 1b 1r 1g (Overseer)

Overseer declares on Day 0: "I see at least one blue-eyed".

Blue-Eye knows he is blue-eyed (seeing 1g and 1r) and suicides the next day

Red-Eye knows ONLY that there is "at-least" one blue eyed; (seeing 1g and 1b), he cannot make any conclusion on his own eye color after blue suicides.

w/ 4 people (2r 1b 1g)

Blue eye instantly knows (2r 1g) he is blue-eyed and commits suicide on Day 1

Red eye knows that his color is not blue on day 1 after Blue suicides.

w/ 4 people (2b 1r 1g)

Blue knows: (1b 1g 1r) that if other blue eye sees 2 red, he would suicide tomorrow. Then he doesn't. He now knows that his eyes are blue, and suicides on day 2.

Red knows: (2b 1g) does not know his own color and would not suicide

w/ 5 people

Blue Eye: (1g 2r 1b) If not blue eyed, would expect other Blue to suicide day 1, yet he does not, seeing as nobody else has blue eyes he would know his is blue and would suicide day 2.

Red Eye: (1g 2b 1r) Does not know his color and thus would not suicide.

Since the argument can be extended to 100 blue eyes, one would expect blue eyes suicide on day 100, but knowing that they would deduce that by day 2 because nobody suicides on day 1...

Thus answer is ALL BLUE EYES SUICIDE ON DAY 2.

(Otherwise I believe that the Day 100 blue eye suicide argument is correct)

this is pretty much flawed, since if u extend it with 3 blue-eyes or more, from the anyone of the blue-eye perspective, they will see another 2, and expect them to die on 2nd day, which if they dont, he's the 3rd blue eye.. for your case, the any one of the blue-eyed will see one other, but if u have one blue-eyed who see 2 others? they wont suicide on the 2nd day..

so in short, you've not considered the "trend" to it.. it's like, if u have two points in the graph, if you join them straight, it comes a line.. but u cant deduce that if u are given only two points in an equation, it must be a line.. you can join it with a curved line. Points on any other parts of the line is different than if u assume a straight line.

Note why i said excluding the red-eye upon the announcement, since, they do not know how many colours are there, they may b purple, but u just assume that they know, which they dont.. (how to know ur eye colour if u are the only one with the unique colour? since overseer did not say other colour, so u can safely eliminate the reds)

another reason why the red eyes will not confuse themselves with the blue, is because, they see on additional blue-eyed ppl, as compared to the perspective of any of the blue-eyed ppl.. Hence, if they see 3 blue eyed ppl (while the blue eyed ppl see two amongsnt themselve), they can only deduce if they are blue or non-blue 1day more than the blue-eyed ppl.