Superrationality has already been tried on Rethinking Blue Eyes - A Logic Puzzle

I actually read that first. It seems like he misunderstood the concept of superrationality. You do not need to plan ahead. Superrationality depends on like minds given the same information coming to the same conclusions, at least that is how I understand it. Collusion is not necessary for superrationality.

It requires 1) everyone is rational 2) everyone is given the same information 3) everyone knows everyone else is rational and has the same information. All of these happen to be true for Blue Eyes.

No idea if the above underlined statement is even correct.The problem is that you are trying to solve the puzzle from the point of view of an outside observer

Actually, my whole point was trying to show all the islanders can logically deduce a minimum without additional knowledge. Going back over my logic, I made several mistakes in explaining. Trying again below in an abstract way with some math.

So the "minimum safe assumption" only works if you already know that there's either 99 or 100 blue eyed people on the island. If you are already on the island and have blue eyes, you see 99 blue eyed people, and your superrationality doesn't help, because you don't know if all the blue eyed people you see are seeing 98 other blue eyed people and have some "minimum safe assumption" of 97.

As an islander, you would know. If I can see 99 blues, then I know that everyone on the island will see at least 98 blues. The important part is that

I know that they can deduce I will see at least 97 blues.I will try to explain better below with some math for the generic case then the island case.

Which at first causes some "minimum safe assumption" of 96, but then it leads down the rabbit hole and you go up to a "minimum safe assumption" of 0, which changes when the guru speaks.

I tried to explain this in my first post. I referred to it as a cascade. I just did a terrible job of explaining why there is no cascade necessary due to superrationality. Refer to the abstract mathematical way in which I try to explain it. Starting with an arbitrary amount of colors and individuals, should provide for a more rigorous proof.

1) Let's say I am a random person on the island. There are two possible states for any color. I am some random unknown color and I see x blues. I might be blue so that means the total will either match what I see or be one higher (x + 0||1).

2) By the rules, I know that everyone else can see everyone else on the island except their own, same as me. This means that they can conclude #1.

3) This means that compared to everyone else on the island, I will know 1 less than them (my color) and 1 more than them (their color).

A) If they are color AA, I can assume that they will see one less color AA and one additional unknown (mine). This works for any color that I can see.

B) Let's say I see x of color AA, so they will see at least x-1.

I) Let's say we have the same color. Then they will see the same number of that color as me, x. x is at least x-1.

II) Let's say we have different colors. Then they will see one less than I see, x-1. x-1 is at least x-1.

III)

Lower bound: they will see at least as much as I do minus one, x-1.4)

I know everyone on the island is super logical just like me, so they can figure out #1, #2, and #3. This can all be figured out as a person on the island. No need for outside observations.To take it a few steps further, let's consider what I can conclude about what others see and what I can conclude about other people's conclusions. This will establish the superrational conclusions necessary.

5) What I think others see: Let's say I see x of color AA, y of color BB and some arbitrary z unknowns. Based on #3, I can conclude the following about people around me.

A) Focusing on just AA and BB, I can assume the following

I) AA will see x-1 AA, y BB, z+1 unknowns.

II) BB will see x AA, y-1 BB, z+1 unknowns.

III)

I can assume everyone I see will see at least x-1 of color AA and y-1 of color BB. C) If I

only see two color groups, then I know there are

at most three color groups on the island. This means

everyone on the island sees at most three colors. This leads to #6.

6) What others think I see, based on my perspective: We must consider what they can conclude about me. To make things less abstract (and closer to the actual problem), I now see x blues and y browns. Based on #5, I know that

A) Any one of the x blues sees at least x-1 blues and y browns and 1 unknown.

I) If I am blue, that blue knows I will see at least x-2 blues and y browns.

II) If I am brown, that blue knows I will see at least x-1 blues and y-1 browns.

III) Not blue and not brown, that blue knows I will see at least x-1 blues and y browns.

IV)

Any of the x blues I see knows regardless of my color, I will see at least x-2 blues and y-1 browns. V) Any of the x blues can make this assumption about every other person they see since color does not matter.

B) Any one of the y browns sees x blues and y-1 browns and 1 unknown.

I) If I am blue, that brown knows I will see at least x-1 blues and y-1 browns.

II) If I am brown, that brown knows I will see at least x blues and y-2 browns.

III) Not blue and not brown, that brown knows I will see at least x blues and y-1 browns.

IV)

Any of the y browns I see knows regardless of my color, I will see at least x-1 blues and y-2 browns. V) Any of the x brown can make this assumption about every other person they see since color does not matter.

D) Combined, it becomes a minimum of x-2 blues and y-2 browns.

C)

This means that everyone I see can make the following assumption about people they see (denoting this group as we because it includes me). We will see a minimum of[/u] x-2 blues and y-2 browns based on my seeing x blues and y browns.6C is the superrational part, where everyone can make the same assumption about everyone else regardless of what they see.

So far, everything has been arbitrary. All the conclusions above can be made by any islander that can see at least two color groups with x and y members respectively. Let's say I am an islander for easier pronoun usage.

Based on #5, the people that I see will see at least x-1 blues and y-1 browns.

Based on #6, the people that I see will assume that everyone else will see at least x-2 blues and y-2 browns.

This means that I know for a fact, that everyone on the island knows there are at least x-2 blues and y-2 browns.

Below, I apply it to the actual Blue Eyes problem for more clarity.

7) Now, we apply all the previous conclusions to every possible viewpoint on the island based on what any islander can observe. Ignoring the guru, every islander will see one of two possibilities. They will see 99/100 blues and 100/99 browns. This means they will come to different minimums, which is why they will have different triggers for cascades.

A) I see 99 blues, 100 browns. I can assume everyone I see will see

at least 99-1 and 100-1 using #5. 98 blues and 99 browns. We also need to consider what those people can conclude about me using #6.

I) Everyone I see will assume everyone else sees at least 99-2 and 100-2.

I can assume everyone on the island knows there are at least 97 blues and 98 browns. II) My cascade trigger (new information) will be there are at least 98 blues or 99 browns. Recall #6. If the guru says 98 blues, then I will wait one night. If they do not leave, it means I am blue. Same for 99 browns, I will be brown.

III)

B) I see 100 blues, 99 browns. I can assume everyone I see will see

at least 100-1 and 99-1 using #5. 99 blues and 98 browns. We also need to consider what those people can conclude about me using #6.

I) Everyone I see will assume everyone else sees at least 100-2 and 99-2.

I can assume everyone on the island knows there are at least 98 blues and 97 browns. II) My cascade trigger (new information) will be there are at least 99 blues or 98 browns. Recall #6. Same as #7AII.

C)

It is important to note that a blue and a brown have different cascade triggers. This must be or they would leave on the same night based on the same information.Does this make for a better explanation? I think I should have opened with this. My first post is a jumbled mess.

Maybe an easier way to consider this is to focus on an islander who see 99 blues and 100 browns, is it logical to assume that 1) every islander knows there is at least 1 blue on the island and 2) every other islander can logically assume everyone else knows this as well?

If so, then "at least 1 blue" is not new information. My proof above is trying to show that even "at least 97 blues" is not new information.