Another twist on the blue-eyed puzzle

In Variants to blue eyes puzzle we explored what would have happened if the oracle of the Blue Eyes puzzle had said something different with different numbers of islanders. I've got another one for you to ponder:

On this island there are 100 blue-eyed people, 50 brown-eyed people, 50 green-eyed people, and one hazel-eyed oracle. When the oracle speaks, she tells them

"I see more people with blue eyes than people with brown eyes."

Other than the difference in what the oracle says and the eye color distribution of the islanders, all the rules are the same as in the original puzzle.

Under these conditions, when will each group leave (if at all) and why?

Note: even though the oracle's statement is very similar to the one in variant 2, the solution is not the same.

• They leave if they know the color of their eyes whatever the color of their eyes? Apr 27 '15 at 20:22
• @kaine Yes, as with the original puzzle if they know for certain what their eye color is they will leave. Apr 27 '15 at 20:23
• I"ve read a few different versions over the years...double checking... thank you. Apr 27 '15 at 20:24
• I think these kind of problems fall down unless each individual is forced to make a choice in turn. eg. say there are 3 blue and 1 brown eyed person. on day one the boat turns up and no one makes a move for the jetty.. a silent moment passes, you can see two blue eyes. why wait for tommorow?
– Ewan
Apr 27 '15 at 21:09
• @Ewan I don't think that part is any less believable than that all of the islanders are perfect logicians. If you really want a in-story explanation, assume they have to get a ticket before the boat arrives, and people do so discreetly so you can't assume that you will know in advance if someone has gotten a ticket. Apr 27 '15 at 21:40

If there were $51$ blue eyed people on the island, any blue eyed person would know from the statement that he had blue eyes. They would see $50$ blue eyed and brown eyed people and saying $51>50$ is the only way to rectify this. All blue eyed people would, therefore, leave on day $1$. If there are more than $51$ then on day $2$ they know there are more than $51$ blue eyed people on the island. If they only see $51$ other blue eyed people, they will leave on day $2$.

This means that blue eyed people will leave on day $L-R$ where $L$ and $R$ are the numbers of bLue and bRown eyed people respectively.

When all blue eyed people don't leave on day $49$ the brown eyed people who see $100$ blue eyed people know that $L-R>49$. They believe that could still have blue eyes or green eyes, however, as $101-49>49$ and $100-49>49$.

On day $50$ blue eyed people look around and see only $99$ blue eyed people. They realise, therefore, that they must have blue eyes. They all leave at the end of that day. The brown eyed people still think there could be $101$ blue eyed people so don't yet know their eye color.

When the brown eyed people wake up the next day they realise $L-R=50$ which means $R=50$. As they only see $49$ other brown eyed people, they realise their eye color and leave at the end of that day.

The green eyed people never leave...they could have hazel, purple, or amber eyes.

So:

Blue eyed people leave on day 50. Brown eyed people leave on day 51. Green eyed people don't leave until another oracle visits.