I meant strictly more, as in at least three.
If you know that there are more than two blue eyed people and you can only see two, you know that you have blue eyes.
If you can see 99 blue eyes, you know that all the other people see either 98 or 99 (if there's a total of 99 blue eyes) or 99 or 100 (if there's 100 blue eyes). However, if some people see 98 and know that there's at least 99 people with blue eyes, then they would all suicide. Since they don't, you know everyone sees either 99 or 100 blue eyes. Since you see 99, you must have blue eyes.
They wouldn't be counting to 100 days, because there's no point where everyone would start counting together (no one is allowed to reveal anything about eye color).
So you know there's no reason for all the blue eyed people to commit suicide together, so you can't reason anything out about the total number of blue eyed people.
Everyone commits suicide at noon in view of everyone else. So the lowest possible number of blue eyed people go up one per day.
Here's a longer explanation:
Since people who know their eye color suicides at noon in plain sight of everyone else, we can set up induction.
If there's only one blue eyed person, they would suicide the first day at noon. Since everyone sees > 1 blue eyed people, they know this wouldn't happen. Since this doesn't happen, it's common knowledge that there is more than 1 blue eyed person. So if you only see one other blue eyed person, you must have blue eyes (but you don't know this yet because you see two blue eyed people).
On the second day, if there are only two blue eyed people, they would both suicide at noon. You see two blue eyed people. No one commits suicide because they either see two blue eyed people (if they have blue eyes) or three blue eyed people (if they don't). Now, everyone knows that there are more than two blue eyed people.
Since no one committed suicide the day before, you reason that there must be three blue eyed people, with the 2 people you see plus yourself. Thus, you know your eye color.
The important point is that the outsider's comment "synchronizes" the islander's information.
On Day 2, say you are a blue eyed person. You can see two other blue eyed people. Since these two people don't commit suicide during that day, you know there must be three blue eyed people with one of them being you.
One player, the 'dictator', decides how to split a cash prize between themselves and the other player, the recipient. The recipient has no input into how the prize is split.
I don't think IRV is worse than first-past-the-post (FPTP) at all. IRV eliminates vote splitting, reduces strategic voting, and results in fewer wasted votes. Also, the fact that IRV ignores all rankings except the first until the first is eliminated means that you won't hurt your most preferred candidate by changing the order of your other preferences. If you're voting A>B>C, then if A wins, then you wouldn't really care whether you helped B win or not.
Approval voting has the problem of electing a candidate that's acceptable instead of one who the majority actually likes.
Also, IRV has a greater tendency to elect the Condorcet winner when compared to FPTP.
Every voting system has its problems but IRV is still a lot better than FPTP.
>I will personally send you a thief I think you'll like every day, entirely determined by your preferences. Each thief will come as a surprise, probably not at the same time every day.
If you can see 99 blue eyes, you know that all the other people see either 98 or 99 (if there's a total of 99 blue eyes) or 99 or 100 (if there's 100 blue eyes). However, if some people see 98 and know that there's at least 99 people with blue eyes, then they would all suicide. Since they don't, you know everyone sees either 99 or 100 blue eyes. Since you see 99, you must have blue eyes.