## Blue Eyes with Superrationality

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flewk
Posts: 12
Joined: Fri Dec 25, 2015 12:47 pm UTC

### Blue Eyes with Superrationality

I believe the solution might be incorrect since it is based on rationality as opposed to superrationality, at least based on my understanding of superrationality.
<EDIT 2> Refer to my second post for a clearer explanation. I will leave my original post as a spoiler here.</EDIT 2>
<EDIT 3>For simplicity sake, let's ignore the island and think about something other than highest possible minimum. If it can be shown that "at least 1" is known by everyone, then the solution would be wrong even without figuring out some higher minimum. If we can first agree that "at least 1" is not new information, then we will at least know that the original solution is wrong.

Let's say that I am 1 of 20 people in a circle. I see 19 people wearing pants. I have no idea if I am wearing pants and I have no way of determining this.
1) Can I say with certainty that everyone else in the circle sees at least 1 pair of pants?
2) Can I say with certainty that everyone else in the circle knows that I see at least 1 pair of pants?
3) If #1 and #2 are true, then this logic can be applied to any individual in the circle, so it can be concluded that everyone sees at least 1 of pants based solely on my seeing 19 pairs of pants.
</EDIT 3>

Spoiler:
<EDIT> Here's a clearer(?) explanation. I have no idea how to explain my idea clearly.
The solution relies on the fact that "at least 1 blue" is new information which triggers a cascade.
Wouldn't the entire population of the island be able to conclude that everyone else on the island knows there is at least 1 blue eyed individual already?
For example, every person on the island will see at least 99 blues and 99 browns. From this, they can assume that everyone else on the island can see at least 98 blues and 98 browns. Of course, the actual numbers will differ, but 98 is the lower limit for all perspectives.
A blue will see 99 blues and 100 browns, so he will assume that all other blues can see at least 98 and all browns can see at least 99 blues. Similar logic for a brown. </EDIT>

The rules of puzzle indicates that everyone is perfectly logical and is aware of the fact that everyone else is perfectly logical as well. Everyone also knows the same set of parameters, which means similar starting assumptions. This means that the choices made here will follow the superrational path rather than the rational path because the logic is shared while initial conditions differ by 1. Blue - 99 blue, 100 brown. Brown - 100 blue, 99 brown.

Short explanation -
Everyone already knows there is at least 1 blue-eyed individual on the island. Everyone also knows everyone else knows this because the minimum safe assumption will be 98/99 (depends on self color), based on superrationality. The guru stating there is at least 1 does not change anything.

Long explanation -
Even before the guru says there is at least one blue eyed person on the island, everyone on the island will have already made some logical conclusions. Ignoring guru for most of this.
1) A blue eyed person will have observed 99 blues, 100 browns and made some conclusions about what those people observe and what conclusions they can make.
2) A brown eyed person will have observed 100 blues, 99 browns and made some conclusions about what those people observe and what conclusions they can make.
3) All individuals know that they can be one of two states blue or not blue. They also know that everyone else thinks this as well. Everyone also knows everyone else knows everyone thinks this way. And so on. Works for any color.
4) Every person on the island can assume a lower bound for what every other person sees. It will be exactly the same thing but with 1 additional unknown. This is true regardless of color.

A brown knows that the 100 blue-eyed individuals he sees must see either 99 or 100 other blues(if he is blue). It would be impossible for any of the 100 blue-eyed individuals to see less than 99 or more than 100 blue eyed individuals.
Going with the lower bound, the brown knows that the 100 blues will all see at least 99 blue.
Those 100 blues will also see at least 99 browns as the original brown sees.
The 99 browns will see at least 100 blues.

Focusing on the blue seen. The brown can conclude that the 100 blues will conclude that there are at least 98 blues because the 100 see at least 99. Here is where the cascade begins for the rational solution. The brown can conclude that the 100 blues will conclude that there are at least 99 blues and will conclude that those 99 will see at least 98 and those 98 will see at least 97... all the way down to none.
However, being superrational beings, and knowing about each other's superrationality, they can conclude that the blues will see at least 98 and assume everyone else will see at least 98. This is because the brown will consider the blue's reference point and knows the blue will consider his as well.

5) Brown knows everyone he sees will see at least 98 browns and 99 blues with self unknown and the original brown unknown. Since we consider the least case, it will be 98 browns, 99 blues, self unknown, 1 not blue.

To illustrate the point, we must now consider what a blue would see and assume about everyone else which everyone on the island will do as well.
A blue knows that the 99 blue-eyed individuals he sees must see either 98 or 99 other blues(if he is blue).
The blue knows that those 99 blues see at least 98 blues. This will cascade to zero as well. However, the blue knows that all of the 99 blues he sees are superrational. They will consider themselves as stateless and perform the same analysis with 1 less blue and that every other blue will think exactly as he thinks and see at least 98 other blues.
The blue knows a brown will see 99 blues and assume exactly as he has.

6) Blue knows everyone he sees will see at least 99 browns and 98 blues with self unknown and the original blue unknown. Since we consider the least case, it will be 99 browns, 98 blues, self unknown, 1 not blue.\\

The minimum information required to trigger a cascade for blues would be 98 blues.
We can consider what happens if the guru says "at least 97 blues".
For the brown, he already knows this. He knows that of the 100 blues he sees, they also know this because they can't see less than 99. He also knows that the 100 blues knows that the people they see will see at least 99 blues as well. This is because of [4] which means between any two individuals, there is only one additional unknown.
The blue also knows that of the 99 blues he sees, they will see at least 98, and knows the browns he sees will see at least 99.

Conclusion -
I have no idea if this is logically sound or not. Superrationality on its own is already confusing because there is a giant loop of logical deductions. Instead of maximizing their gain, I am just using the fact that they know everyone else can reach the same conclusion as them to stop the cascade because no one on the island can see less than 98 blues or 98 browns from any perspective. This can be arrived at from every perspective as well.
Last edited by flewk on Wed Jan 06, 2016 4:50 am UTC, edited 2 times in total.

Vytron
Posts: 432
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### Re: Blue Eyes with Superrationality

Superrationality has already been tried on Rethinking Blue Eyes - A Logic Puzzle, where variations of the puzzle were tried, in which the people of the island were superrational, were allowed to come up with a plan before the experiment started, had changed their conditions so they could make guesses even if they didn't know things, and such.

And...

Spoiler:
NOTHING WORKS

The beauty of the original puzzle is that it has the least assumptions and has the people on the island "as weak as possible" as long as it still works, and it does, but increasing their power or knowledge or potential behavior doesn't help them any.

The problem is that you are trying to solve the puzzle from the point of view of an outside observer, and this is wrong. Yes, what you say it's factual, but those are only facts that work because you already know how many people are on the island. What if you were on the island? Well, it no longer works...

I think the crux of the issue is here:

Everyone also knows everyone else knows this because the minimum safe assumption will be 98/99 (depends on self color), based on superrationality.

Yes.

But that's only true for an island with 100 blue eyed colored people.

What would you say if the puzzle said "there's 99 blue eyed people on the island"?

Everyone also knows everyone else knows this because the minimum safe assumption will be 97/98 (depends on self color), based on superrationality.

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.

Unfortunately, solving the puzzle requires fictional worlds that you imagine (that was one: one where your eyes are blue in reality, but on the fiction they're brown) other people imagining (so in those worlds their blue eyes turn brown too) other people imagining... 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 even tried in that thread to pre-agree to some "minimum safe assumption" of x for when you see at least y blue eyed people on the island. Say, you see at least 97? Agree that the "minimum safe assumption" can't drop below 95. Now, this works. The problem is, this doesn't work for all cases. For any such strategy where you leave early, there's 1 case where it doesn't work, so even if you're on the case in which it does, you don't really know your eyes' color (which is required to leave), because you don't you that you're on a case that works.

This is the conclusion from the other thread (though people continued arguing afterwards):

It's impossible to skip any number of days even with superrationality, because if you managed to find a strategy that skips n number of days, then it should work for n+1 days, so the prisoners should be able to use it immediately and leave. This isn't the case because whenever they decide to leave shouldn't be confused by brown eyed people (i.e. if they don't leave immediately then blue eyed people leave Day 2, so brown eyed people would leave Day 3 if blue eyed people were there, because they were seeing them.)

The bottom looks like this:

1 Blue eyed person leaves D1 (they don't see anybody else)
2 Blue eyed people leave D2 (they see the only blue eyed guy they see didn't leave D1)
3 Blue eyed people leave D3 (they see the only 2 blue eyed guys they see didn't leave D2)
4 Blue eyed people leave D4 (they see the only 3 blue eyed guys they see didn't leave D3)
5 Blue eyed people leave D5 (they see the only 4 blue eyed guys they see didn't leave D5)
...

The next cases are identical, and there's no magic number that is different. It follows that:

The only way to leave earlier for n number of people should work for 3 number of people.

If you manage to find a way for 3 people to leave earlier than D3 then you have a solution that should work generally.

Alas, with 3 people they don't know if it's only 2 people, and they won't know it's more than 2 until D3 and because the others didn't leave.

flewk
Posts: 12
Joined: Fri Dec 25, 2015 12:47 pm UTC

### Re: Blue Eyes with Superrationality

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.

PeteP
What the peck?
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### Re: Blue Eyes with Superrationality

MOJr
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### Re: Blue Eyes with Superrationality

flewk wrote: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.

How is 7/A/II the same as 7/B/II, where in one I have trigger at 99 and in the second 98 blue eyes.....

The problem with this puzzle is that yes you can deduce the minimum of blue eyes everyone sees. But this minimum number will be deduced as two different numbers for the blue eyed and brown eyed groups. So you cannot set it as a trigger.

douglasm
Posts: 630
Joined: Mon Apr 21, 2008 4:53 am UTC

### Re: Blue Eyes with Superrationality

You are on an island with all the same rules but a different number of people. You see 73 people with blue eyes and 74 with brown. Everyone there is superrational, and this fact is common knowledge. What color are your eyes, and when and how do you figure this out?

flewk wrote:
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.

Superrationality and collusion are functionally equivalent.

drachefly
Posts: 197
Joined: Thu Apr 23, 2009 3:25 pm UTC

### Re: Blue Eyes with Superrationality

Going back to the previous thread for a moment -

stekp wrote:It's hard to define. I was in the middle of a post trying to justify my brilliant scheme. Also I was thinking about this post about iterating down to zero. I had got it in my mind that something changed at N=3, this being the case where everyone's model of everyone observations would have to contain at least on blue-eye, meaning the reasoning process would need to be different in this case due to the "extra knowledge". But then I realized this was introducing a artificial discontinuity and that in any case, the islanders themselves would not know beforehand that N was 3 and not 2. Suddenly I saw the foundation of my argument falling away. It's seems obvious now, but somehow I managed to completely convince myself!

It's not crazy to suppose that a discontinuity can kick in at 3 or 4 since 3 is a minimal cycle and 4 allows one to look at a minimal cycle. If that helped, it wouldn't be an *artificial* discontinuity, but a natural one.

But it doesn't help.

Base case: N = 1 blue-eyed person. He's screwed.

General: N blue-eyed people. Since the N-1 case is screwed, that they don't leave on day N-1 is not surprising and yields no information. Since we got no new information, we too are screwed.

Not even superrationality can get rid of the need for a guru, let alone allow them to get off quicker (short of fighting the hypothetical by speeding up the rate at which they gather information in some pre-coordinated fashion)

Vytron
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Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Blue Eyes with Superrationality

@flewk: Try it with five. You are on the island, and there's 5 blue eyed people, but you don't know it.

So you see 4 blue eyed people.

You go through your logic process and conclude:

Everyone on the island sees either 3 blue eyed people, or they see 4.

3 Is the minimum.

However, if your eyes ARE brown, the other superrational people aren't thinking like you. They're thinking:

Everyone on the island sees either 2 blue eyed people, or they see 3.

2 Is the minimum.

What you argue is that things stop here, but, take any of those blue eyes people that you see. If YOUR eyes are brown, they're wondering what would happen if THEIR eyes were brown, and they'd see the other guys thinking:

Everyone on the island sees either 1 blue eyed people, or they see 2.

1 Is the minimum.

The cascade can't be stopped, because, whenever there's some hypothetical scenario where your eyes are brown, you don't matter. You're out of the formula. The other people don't care about you. So there's a new scenario where you're excluded, and this forces a new scenario where some other random guy is excluded.

This never ends, until you end with 0 blue eyed people (everyone excluded) and this changes when the guru speaks (the "aha! I should have blue eyes!")

I think you're confusing superrationality with perfect logic, as what you said could be said by a perfect logician (as long as it's common knowledge) though it doesn't really help them.

flewk
Posts: 12
Joined: Fri Dec 25, 2015 12:47 pm UTC

### Re: Blue Eyes with Superrationality

General response to everyone: The important thing here is that between any two individuals, there is only two unknowns total. If we just ignore those unknowns, we can still deduce the other's conclusions about us, hence the -2. This logic applies to every individual that you see and you know they will apply the same logic to everyone they see as well.

MOJr wrote:
flewk wrote: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.

How is 7/A/II the same as 7/B/II, where in one I have trigger at 99 and in the second 98 blue eyes.....

The problem with this puzzle is that yes you can deduce the minimum of blue eyes everyone sees. But this minimum number will be deduced as two different numbers for the blue eyed and brown eyed groups. So you cannot set it as a trigger.

Even two separate numbers won't matter as long as you can determine the minimum.
Just imagine you are on a different island and you see 99 blues and 99 browns. It is logical to assume that everyone else on the island will see at least 1 brown and 1 blue. It is also logical to assume that every other person can logically deduce that everyone they see will see at least 1 blue and 1 brown. Just expand that logic all the way up to the minimum.

douglasm wrote:You are on an island with all the same rules but a different number of people. You see 73 people with blue eyes and 74 with brown. Everyone there is superrational, and this fact is common knowledge. What color are your eyes, and when and how do you figure this out?

flewk wrote:
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.

Superrationality and collusion are functionally equivalent.

Read my response above. I do not need to know my own eye color to know the minimum amount that everyone else sees on the island. I can also deduce what others will assume about everyone else on the island.
If I see 73 blues and 74 browns, in the simplest case, I can assume that everyone else sees at least 1 blue and 1 brown person. Does it not also make sense that every other islander will assume that I can see at least 1 blue and 1 brown as well?

I do not believe superrationality is functionally equivalent to collusion, but that is another debate.

Vytron wrote:@flewk: Try it with five. You are on the island, and there's 5 blue eyed people, but you don't know it.

So you see 4 blue eyed people.

You go through your logic process and conclude:

Everyone on the island sees either 3 blue eyed people, or they see 4.

3 Is the minimum.

However, if your eyes ARE brown, the other superrational people aren't thinking like you. They're thinking:

Everyone on the island sees either 2 blue eyed people, or they see 3.

2 Is the minimum.

What you argue is that things stop here, but, take any of those blue eyes people that you see. If YOUR eyes are brown, they're wondering what would happen if THEIR eyes were brown, and they'd see the other guys thinking:

Everyone on the island sees either 1 blue eyed people, or they see 2.

1 Is the minimum.

The cascade can't be stopped, because, whenever there's some hypothetical scenario where your eyes are brown, you don't matter. You're out of the formula. The other people don't care about you. So there's a new scenario where you're excluded, and this forces a new scenario where some other random guy is excluded.

This never ends, until you end with 0 blue eyed people (everyone excluded) and this changes when the guru speaks (the "aha! I should have blue eyes!")

I think you're confusing superrationality with perfect logic, as what you said could be said by a perfect logician (as long as it's common knowledge) though it doesn't really help them.

1) I am on an island. I know there are 5 people total, including me.
2) I see 4 blue eyed people.
3) I know everyone else can see at least 3 blue eyed people. This is my conclusion based on observation.
4) Since the others see at least 3 blue eyed people, I can assume that they will assume that everyone else will see at least 2 blue eyed people.
5) This minimum is unique to my perspective. It does not matter what anyone else actually sees because from my perspective, everyone will see at least 2 blue eyed people.
6) Super logical people will know that everyone else can figure out #5 specific to their perspective, so everyone's trigger will differ. My trigger will be at at least 3 blue eyed people as that is new information. It will be at least 2 people for people who see 3. This means that we have different triggers, which we would have to, if we see different number of blue people.

This is just repeating my logic from before. So maybe we can try to view it from a different angle. Let's label the people A,B,C,D,E.

2) The combined unknowns between A and B are the colors of A and B. This works for any two individuals.
3) If I am A, I know that B and C will both see D and E.
4) If I am A, I know that B knows between C and I(A), we will both see B and E.
5) #3 and #4 work for any pair's perspective regarding any other pair.
6) The combined unknown between myself and every other person is 2. Every other person's combined unknown between themselves and others will also be 2.
7) Everyone sees at least 2 blues if I see 4 blues.
8) I have no way of knowing what my color is or what they actually assume about me. I do not actually need to know what they see or assume about me. I just need to figure out the minimum information they will assume about everyone else.
9) I can deduce that if my eyes were not blue, then their trigger will be 1 less than mine. They would assume that everyone sees at least 1 blue. This would follow logically, as saying there are at least 2 blues will be new information for them, but not me. They will leave the island after one night instead of waiting two nights like I would. If they waited until the second night, then I know for sure that I am blue because the people who see 3 blues would have left on the first night.

For simplicity sake, let's ignore the island and think about something other than highest possible minimum. If it can be shown that "at least 1" is known by everyone, then the solution would be wrong even without an optimal minimum.
20 people are in a circle. I see 19 people wearing pants. I have no idea if I am wearing pants.
1) Can I say with certainty that everyone in the circle sees at least 1 pair of pants?
2) Can I say with certainty that everyone in the circle knows that I see at least 1 pair of pants?
3) If #1 and #2 are true, then this logic can be applied to any individual in the circle, so it can be concluded that everyone sees at least 1 of pants based solely on my seeing 19 pairs of pants.

jaap
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### Re: Blue Eyes with Superrationality

Even though they all can agree that there is at least one blue-eyed person on the island, they don't mean the same thing when they say that. It really means, "amongst the N-1 people that are not me, there is at least one blue-eyed person". That statement is self-referential, and refers to a different set of people for each person. It's like a group of people all agreeing that Tom is a great actor, but with one thinking of Tom Cruise, another of Tom Hanks, and a third of Tom Hardy.
There is no single person that they can all agree on that has blue eyes - because that person himself doesn't know he has blue eyes. When the guru says there is a person with blue eyes, then they all know that there is a single person that they can all agree has blue eyes, even if the guru doesn't actually reveal who that is at that moment.

MOJr
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### Re: Blue Eyes with Superrationality

flewk wrote:Even two separate numbers won't matter as long as you can determine the minimum.
Just imagine you are on a different island and you see 99 blues and 99 browns. It is logical to assume that everyone else on the island will see at least 1 brown and 1 blue. It is also logical to assume that every other person can logically deduce that everyone they see will see at least 1 blue and 1 brown. Just expand that logic all the way up to the minimum.

I see 2 flaws with your logic:

1. This approach will fail if a only a single person has particular color of eyes or small number of people have. (hence, the need for guru).
2. You cannot deduce any single minimum higher than 1, that would be common for the whole group (in other words, others can deduce it using your logic). You will have allways 2 different groups of people, who will see different minimums, and so they cannot use them as trigger. (i.e. if I used your logic with 10 blue eyes (and 100 brown), then blue eyed people would have minimum 7 for blue and brown eyed 8 for blue. (What this means is that you still have to wait 10 days to leave))

Gwydion
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### Re: Blue Eyes with Superrationality

flewk wrote:They will leave the island after one night instead of waiting two nights like I would. If they waited until the second night, then I know for sure that I am blue because the people who see 3 blues would have left on the first night.
Ok, given that statement, what would you need to see to leave on night 1 without a guru? If you saw 3 blue and 1 brown, how does that lead you to be certain your eyes are blue? If the answer is instead "that's not possible", what information has been gained to allow someone to leave night 2? What would you need to see in order to leave on night 2....

flewk
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### Re: Blue Eyes with Superrationality

jaap wrote:Even though they all can agree that there is at least one blue-eyed person on the island, they don't mean the same thing when they say that. It really means, "amongst the N-1 people that are not me, there is at least one blue-eyed person". That statement is self-referential, and refers to a different set of people for each person. It's like a group of people all agreeing that Tom is a great actor, but with one thinking of Tom Cruise, another of Tom Hanks, and a third of Tom Hardy.
There is no single person that they can all agree on that has blue eyes - because that person himself doesn't know he has blue eyes. When the guru says there is a person with blue eyes, then they all know that there is a single person that they can all agree has blue eyes, even if the guru doesn't actually reveal who that is at that moment.

The important thing is not an individual knowing there is at least 1 blue, but that the individual knows that everyone else sees at least 1 blue. Read the underlined portion from my previous post.

MOJr wrote:
flewk wrote:Even two separate numbers won't matter as long as you can determine the minimum.
Just imagine you are on a different island and you see 99 blues and 99 browns. It is logical to assume that everyone else on the island will see at least 1 brown and 1 blue. It is also logical to assume that every other person can logically deduce that everyone they see will see at least 1 blue and 1 brown. Just expand that logic all the way up to the minimum.

I see 2 flaws with your logic:

1. This approach will fail if a only a single person has particular color of eyes or small number of people have. (hence, the need for guru).
2. You cannot deduce any single minimum higher than 1, that would be common for the whole group (in other words, others can deduce it using your logic). You will have allways 2 different groups of people, who will see different minimums, and so they cannot use them as trigger. (i.e. if I used your logic with 10 blue eyes (and 100 brown), then blue eyed people would have minimum 7 for blue and brown eyed 8 for blue. (What this means is that you still have to wait 10 days to leave))

1. Yes. It fails if I see 2 blue people. Since the problem indicated 100 blue people, that is not an issue.
2. It seems like you are suggesting that a minimum of 1 is logically sound. This would mean that the original solution is invalid as the guru stating "there is at least 1 blue" is treated as new information.
If we use 10 blue and 100 browns, then telling everyone on the island that there is at least 1 blue will not be new information. Everyone on the island already knows that everyone else knows this.

Gwydion wrote:
flewk wrote:They will leave the island after one night instead of waiting two nights like I would. If they waited until the second night, then I know for sure that I am blue because the people who see 3 blues would have left on the first night.
Ok, given that statement, what would you need to see to leave on night 1 without a guru? If you saw 3 blue and 1 brown, how does that lead you to be certain your eyes are blue? If the answer is instead "that's not possible", what information has been gained to allow someone to leave night 2? What would you need to see in order to leave on night 2....

The introduction of a brown does not affect the minimum number of blues. If I see 3 blues out of 4 people or 3 blues out of 5 people, the minimum is still 1 blue. Telling me there is at least 1 blue will not be new information for me. It could be new information for one of the blues I see. If they leave, then I would know I am not blue.

Question: Can anyone find a flaw in the math from the second post? Does the minimum of x-2 and y-2 make sense? If not, which part?

PeteP
What the peck?
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### Re: Blue Eyes with Superrationality

"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." is wrong. You can't combine the minimums because people don't know the other minimum. Yes their own minimum is higher than the general minimum but that doesn't make the general minimum shared knowledge.
flewk wrote:1. Yes. It fails if I see 2 blue people. Since the problem indicated 100 blue people, that is not an issue.

Of course it is an issue. There are plenty of plans that work for specific numbers the problem are plans that work in general.

emlightened
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### Re: Blue Eyes with Superrationality

I think part of the issue is a misconception about what the guru's information is actually used for. Consider, for instance, the case where the guru says that there are at least 10 people with blue eyes. If you generalise this case, you can easily see that the guru actually tells the group when to leave based on this, but doesn't actually give any new information. With superrationality, if there is a 'initial number' that makes sense (1, probably, but this is the hard part), then everyone will have decided on it (superrationality), and the guru is not needed.

"Therefore it is in the interests not only of public safety but also public sanity if the buttered toast on cats idea is scrapped, to be replaced by a monorail powered by cats smeared with chicken tikka masala floating above a rail made from white shag pile carpet."

Gwydion
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### Re: Blue Eyes with Superrationality

flewk wrote:
Gwydion wrote:
flewk wrote:They will leave the island after one night instead of waiting two nights like I would. If they waited until the second night, then I know for sure that I am blue because the people who see 3 blues would have left on the first night.
Ok, given that statement, what would you need to see to leave on night 1 without a guru? If you saw 3 blue and 1 brown, how does that lead you to be certain your eyes are blue? If the answer is instead "that's not possible", what information has been gained to allow someone to leave night 2? What would you need to see in order to leave on night 2....

The introduction of a brown does not affect the minimum number of blues. If I see 3 blues out of 4 people or 3 blues out of 5 people, the minimum is still 1 blue. Telling me there is at least 1 blue will not be new information for me. It could be new information for one of the blues I see. If they leave, then I would know I am not blue.
I'm not sure how this game works then. Based on your logic, for a 5-person island, if all 5 have blue eyes you would leave on night 3? And in your earlier posts, you said this was because if there were only 4 (4 blue and you are brown) the others would have left night 2, and that is because they knew "the people who see 3 blues would have left on the first night". Why would they know this? If you had brown eyes, the others would all see 3 blues and a brown, though you don't know this. What about that scenario would enable them to say immediately, on the first night, "I know I also have blue eyes because it is impossible that there are 3 blues and 2 browns on this island." Put differently, what if you were on an island and saw 2 blues and 2 browns. What would you do?

I don't think the problem is with declaring a minimum. I think the issue comes up when you try to say there are at least n-2 people with a certain eye color, because n varies for different people. The way you've set up the game, everyone always leaves night 3, the night when (n-2) + 2 becomes known. Trouble is, nobody's learned anything - you can't draw inferences about your eye color from anyone's behavior in this game, because their behavior is always "leave night 3" but without a logical reason why.

Superrationality is an interesting concept in game theory, but don't let the name fool you: it is a special case of irrationality, not rationality. Following a strategy because you know your opponent is going to follow the same strategy, without any further guiding reasoning, is not rational at all. If you remove rationality from decision-making in a situation like this, the ability to draw valid inferences from others' behavior is much more limited.

emlightened
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### Re: Blue Eyes with Superrationality

Gwydion wrote:Superrationality is an interesting concept in game theory, but don't let the name fool you: it is a special case of irrationality, not rationality. Following a strategy because you know your opponent is going to follow the same strategy, without any further guiding reasoning, is not rational at all. If you remove rationality from decision-making in a situation like this, the ability to draw valid inferences from others' behavior is much more limited.

The way I understood it was that superrationality is only typically applied against superrational opponents. Namely, the assumption is that instead of the guru telling you about the number of blue-eyed people, she tells you that she knows, without any doubt, that everyone is superrational.

Also, an interesting fact is that, if the guru says "Without this information, none of you would leave on a Sunday," everyone is able to leave before next Sunday (the one in 8-14 days).

Also, I think that this thread has a lot more discussion on the subject.

"Therefore it is in the interests not only of public safety but also public sanity if the buttered toast on cats idea is scrapped, to be replaced by a monorail powered by cats smeared with chicken tikka masala floating above a rail made from white shag pile carpet."

Xias
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### Re: Blue Eyes with Superrationality

flewk wrote:Question: Can anyone find a flaw in the math from the second post? Does the minimum of x-2 and y-2 make sense? If not, which part?

The problem comes in that in 7)A, you acknowledge that each of the other islanders will have gone through the process of conclusions through 6. However, you do not take that further and understand that

8) each of the other islanders will also come to their own version of 7)A with one less minimum. Furthermore:

9) each of the other islanders will come to their own version of 8 with one less minimum.

etc.

I do not believe that superrationality negates this point. There are still going to be infinitely many nested hypotheticals, all of which eventually reach (and stay at) the possibility of 0 blue eyes, which only terminates when the guru speaks. There is also nothing preventing the superrational agents from using the same strategy as the rational ones, because even in the original problem the information gained from the guru isn't that there is a blue eyed person, it's the conditional statement "If there were only one blue eyed person, that person would leave tonight." It doesn't matter that the condition in that conditional statement is known by all of the agents to be false; it's the conditional statement itself being true that starts the chain and leads to the given solution.

MOJr
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### Re: Blue Eyes with Superrationality

flewk wrote:
MOJr wrote:1. This approach will fail if a only a single person has particular color of eyes or small number of people have. (hence, the need for guru).
2. You cannot deduce any single minimum higher than 1, that would be common for the whole group (in other words, others can deduce it using your logic). You will have allways 2 different groups of people, who will see different minimums, and so they cannot use them as trigger. (i.e. if I used your logic with 10 blue eyes (and 100 brown), then blue eyed people would have minimum 7 for blue and brown eyed 8 for blue. (What this means is that you still have to wait 10 days to leave))

1. Yes. It fails if I see 2 blue people. Since the problem indicated 100 blue people, that is not an issue.
2. It seems like you are suggesting that a minimum of 1 is logically sound. This would mean that the original solution is invalid as the guru stating "there is at least 1 blue" is treated as new information.
If we use 10 blue and 100 browns, then telling everyone on the island that there is at least 1 blue will not be new information. Everyone on the island already knows that everyone else knows this.

As far as I understand the superrationality puzzle there are 2 questions: Do we need guru and can we skip days. My answers were related to these questions resp..

1. As was said general strategy should work with any number not just 100. Saying it works with 100 blue is the equivalent of guru saying there are at least 100 blue eyed people...
2. No, I was just pointing that you cannot use X-2 as a minimum, and "if you could use something as a minimum", the lowest number you could use is 1. But you cannot use 1 as a minimum since the strategy doesn't work with 1 person.

Just to ilustrate the point (seen from my perspective of having blue eyes):
1. The strategy doesn't work with 1 blue. (obvious)
2. The strategy doesn't work with 2 blue: I see 1 person with blue eyes. Since that person could be the only one the strategy doesn't work.
3. The strategy doesn't work with 3 blue: I see 2 persons with blue eyes, so they can each see one persons so the strategy doesn't work due to 2.
4. The strategy doesn't work with 4 blue: I see 3 with blue eyes, so they can see only two, and since the strategy doesn't work for 2 it doesn't work.

and then you find that you cannot set any minimum number where it should work, since the two groups will see two minimums as described at start. So you wil get somthing like: Lets say you can start the process if you see at least 100 blue eyes. But then if you have green eyes, the blue eyed people see only 99 eyes, and won't start the countdown, so you shouldn't start the countdown.

Vytron
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### Re: Blue Eyes with Superrationality

Anyway, yeah. I think superrationality works best if we assume collusion. Say, infinite collusion in where everyone has already agreed to act in some way in a given scenario, and knows how everyone will act in that scenario.

This way, indeed, 100 islanders would be able to skip 70 days, with the following strategy:

- If you see at least 90 islanders, skip to day 70.

This works in the superrational case, because you know that if you skip to day 70 if you see 90 people, all other superrational beings are doing the same.

The problem is, if you are on the island and see 90 islanders. Should you skip 70 days? But what if your eyes aren't blue? Oh, it's possible those people are seeing just 89 blue eyed people, and won't skip any day, so they not leaving at some day doesn't tell you anything.

This breaks if you change the 100, 90 and the 70 I used for any number, and thus, there's no superrational strategy that works for the general case (where you're told you will be put on the island, but don't know how many blue eyed people will be there.)

flewk
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### Re: Blue Eyes with Superrationality

PeteP wrote:"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." is wrong. You can't combine the minimums because people don't know the other minimum. Yes their own minimum is higher than the general minimum but that doesn't make the general minimum shared knowledge.
flewk wrote:1. Yes. It fails if I see 2 blue people. Since the problem indicated 100 blue people, that is not an issue.

Of course it is an issue. There are plenty of plans that work for specific numbers the problem are plans that work in general.

General case is x-2. Was shown above.
The purpose is not to combine minimums, the purpose is to determine a minimum based on your perspective. The fact that the beings are superrational means that everyone else can determine their own minimum in the same manner. This minimum can be applied to any perspective.

People here are all trying to nest the minimums until 0. This is not needed because the largest possible unknowns between any two individuals is two. This means that A would only need to imagine what B sees about C. There would be no need for A => B => C => D... all the way down to 0.
Gwydion wrote:
flewk wrote:
Gwydion wrote:
flewk wrote:They will leave the island after one night instead of waiting two nights like I would. If they waited until the second night, then I know for sure that I am blue because the people who see 3 blues would have left on the first night.
Ok, given that statement, what would you need to see to leave on night 1 without a guru? If you saw 3 blue and 1 brown, how does that lead you to be certain your eyes are blue? If the answer is instead "that's not possible", what information has been gained to allow someone to leave night 2? What would you need to see in order to leave on night 2....

The introduction of a brown does not affect the minimum number of blues. If I see 3 blues out of 4 people or 3 blues out of 5 people, the minimum is still 1 blue. Telling me there is at least 1 blue will not be new information for me. It could be new information for one of the blues I see. If they leave, then I would know I am not blue.
I'm not sure how this game works then. Based on your logic, for a 5-person island, if all 5 have blue eyes you would leave on night 3? And in your earlier posts, you said this was because if there were only 4 (4 blue and you are brown) the others would have left night 2, and that is because they knew "the people who see 3 blues would have left on the first night". Why would they know this? If you had brown eyes, the others would all see 3 blues and a brown, though you don't know this. What about that scenario would enable them to say immediately, on the first night, "I know I also have blue eyes because it is impossible that there are 3 blues and 2 browns on this island." Put differently, what if you were on an island and saw 2 blues and 2 browns. What would you do?

I don't think the problem is with declaring a minimum. I think the issue comes up when you try to say there are at least n-2 people with a certain eye color, because n varies for different people. The way you've set up the game, everyone always leaves night 3, the night when (n-2) + 2 becomes known. Trouble is, nobody's learned anything - you can't draw inferences about your eye color from anyone's behavior in this game, because their behavior is always "leave night 3" but without a logical reason why.

Superrationality is an interesting concept in game theory, but don't let the name fool you: it is a special case of irrationality, not rationality. Following a strategy because you know your opponent is going to follow the same strategy, without any further guiding reasoning, is not rational at all. If you remove rationality from decision-making in a situation like this, the ability to draw valid inferences from others' behavior is much more limited.

It looks like we are talking about different problems. "If I see 3 blues out of 4 people or 3 blues out of 5 people, the minimum is still 1 blue." vs "Based on your logic, for a 5-person island, if all 5 have blue eyes you would leave on night 3?"
emlightened wrote:
Gwydion wrote:Superrationality is an interesting concept in game theory, but don't let the name fool you: it is a special case of irrationality, not rationality. Following a strategy because you know your opponent is going to follow the same strategy, without any further guiding reasoning, is not rational at all. If you remove rationality from decision-making in a situation like this, the ability to draw valid inferences from others' behavior is much more limited.

The way I understood it was that superrationality is only typically applied against superrational opponents. Namely, the assumption is that instead of the guru telling you about the number of blue-eyed people, she tells you that she knows, without any doubt, that everyone is superrational.

Also, an interesting fact is that, if the guru says "Without this information, none of you would leave on a Sunday," everyone is able to leave before next Sunday (the one in 8-14 days).

Also, I think that this thread has a lot more discussion on the subject.

Aren't all the islanders superrational? The rules specified at the beginning of the puzzle indicate that all the islanders are aware of all the rules and of each other's awareness. They are also perfectly rational beings. This means that all the islanders know exactly the same thing with the exception of two unknowns between any two individuals.

I actually read that thread before making this one. That one focuses mainly on collusion which is not required for superrationality. This would be the third time I have indicated as such.
Xias wrote:
flewk wrote:Question: Can anyone find a flaw in the math from the second post? Does the minimum of x-2 and y-2 make sense? If not, which part?

The problem comes in that in 7)A, you acknowledge that each of the other islanders will have gone through the process of conclusions through 6. However, you do not take that further and understand that

8) each of the other islanders will also come to their own version of 7)A with one less minimum. Furthermore:

9) each of the other islanders will come to their own version of 8 with one less minimum.

etc.

I do not believe that superrationality negates this point. There are still going to be infinitely many nested hypotheticals, all of which eventually reach (and stay at) the possibility of 0 blue eyes, which only terminates when the guru speaks. There is also nothing preventing the superrational agents from using the same strategy as the rational ones, because even in the original problem the information gained from the guru isn't that there is a blue eyed person, it's the conditional statement "If there were only one blue eyed person, that person would leave tonight." It doesn't matter that the condition in that conditional statement is known by all of the agents to be false; it's the conditional statement itself being true that starts the chain and leads to the given solution.

The nested structure is the basis for the original solution. It is basically saying A sees 100 people, those 100 people see 99, those 99 see 98, and so on. This is why I included #3. Between any two individuals on the island, the number of unknowns will be 2. A only needs to think about what B thinks A sees. B only needs to think about what A thinks B sees. Nesting any further is unneeded as A=>B=>A can be applied between any two individuals because of the 2 unknown combination. No need for A thinks about B thinks about C thinks about D...

I think I just thought of an even better way to show why subtracting two equalizes the perspectives, let any individual determine a minimum.
For some sufficiently populated island, we pick out two random individuals, A and B.

A sees X blues plus B's color.
B sees Y blues plus A's color.

A does not see A's color and B does not see B's color. Combined unknown would be A and B's color.

If we remove both A and B from both individual's perspective, we get the following.
A sees X blues. (A could not see A to begin with, so only remove B)
B sees Y blues. (similar to above)

X equals Y.

This is because the only difference between two individual perspectives is each other's color. Both individuals will see everything else exactly the same.

From A's perspective:
A sees X blues. (at least)
B sees X blues. (at least)
Island minimum safe assumption is X-1 (All the X blues will see at least X-1 blues).

From B's perspective:
A sees Y blues.
B sees Y blues.
Island minimum safe assumption is Y-1.

The only difference would be the consideration of B and A's colors by A and B respectively. This would shift the minimum by 1.
Since A knows B can figure all this out, and B knows A can figure all this out, it means that everyone on the island will have some minimum that will only differ slightly since the perspectives can only differ slightly (2) between any two individuals.

Is this math more clear? Normalization of perspectives allows for a logical determination of a minimum.

PeteP
What the peck?
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### Re: Blue Eyes with Superrationality

Two of them can indeed agree on a number, if you get three to agree to a number that would be impressive. Example case: A and B have brown eyes, C has blue eyes. A+B as pairing=1, A and C=B and C=0. Show how they can arrive at the same minimum. Otherwise the difference of one is what causes all plans that reduce time to fail at some number.

Edit: I can generalize the concept to three people of course but for two people it works by excluding themselves, to generalize it to more people works but if you include the whole group the result is always 0 because you don't take the eye color of anyone in the group into account. So generalizing it that way isn't very useful.
Last edited by PeteP on Fri Jan 08, 2016 9:15 am UTC, edited 1 time in total.

flewk
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### Re: Blue Eyes with Superrationality

PeteP wrote:Two of them can indeed agree on a number, if you get three to agree to a number that would be impressive. Example case: A and B have brown eyes, C has blue eyes. A+B as pairing=1, A and C=B and C=0. Show how they can arrive at the same minimum. Otherwise the difference of one is what causes all plans that reduce time to fail at some number.

The minimum is not shared, nor should it be. Each individual will have a different trigger. It would not make sense for a person who sees 99 blues and a person who sees 100 blues to have the same trigger.

So many people seem to think I am talking about a minimum shared by all, instead of a minimum that is applied to every other islander you see...

Let's just apply the above logic to A,B,C. A and B see X blues. C sees Y blues.
A plus B both see X blues ignoring each other's eye color.
A plus C and B plus C both see Y blues ignoring respective colors.

A is able to determine that B and C will both see at least X-1 blues.
B is able to determine that A and C will both see at least X-1 blues.
C is able to determine that A and B will both see at least Y-1 blues

This means that A and B will share a minimum trigger. A and B will leave or not leave together. Fates are bound.
C has a different minimum trigger. This has to be so, or else the original solution would not work either.

When A accounts for the colors of B and C, A knows that B sees the same amount of blues, but C will see 1||0 less, hence C (and all other blues) will have a trigger that will be 1 less than A's if A is not blue and the same as A if A is blue.
Similar logic for B.
When C accounts for the colors of A and B, C knows that A and B will see the same amount of blues. C has no idea what his color is, so C knows his trigger will be either the same or 1 less than A and B's.

The indeterminate part is where the waiting takes place. Everyone waits for the 1 less assumption to pass before knowing they are blue. For A and B, this will be 1 higher than C's, so C will leave with the rest of the blues 1 night earlier and A and B will know they are not blue.

PeteP
What the peck?
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### Re: Blue Eyes with Superrationality

So how is that not an needlessly convoluted way of saying that people know the blue people they see see one less blue than the brown people they see?

flewk
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### Re: Blue Eyes with Superrationality

PeteP wrote:So how is that not an needlessly convoluted way of saying that people know the blue people they see see one less blue than the brown people they see?

It doesn't only apply to blue eyed people. The whole point of using arbitrary individuals with indeterminate personal colors in my original example is to show that the normalization is applicable to all individuals within one's perspective.

Since it is applied to everyone, and the beings are superrational, they would know that everyone else has also determined some minimum based on their own perspective which can only differ by 2 as pointed out earlier. This means that "at least 1" is not new information to anyone on the island.

EDIT 1: Read the original post for determining x=y. Since it ignores the personal colors, this works for any person to any other person on the island. A's minimum will never differ by more than 2 with any person he sees. If he also applies that person's actual color to the situation, it could even be only a distance of 1.

PeteP
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### Re: Blue Eyes with Superrationality

Going back I'm noẁ unsure where you think you are using superrationality. There is no superrationality necessary in any scenario with more than 2 blues to determine that nobody sees less than 1 blue, thinking about what others see or what conclusions they arrive at does not require superrationality. The original solution already relies on everyone being perfect logicans and everyone else knowing that.

So I must apologize, I skimmed the first time and made assumptions and thought you were trying to do something with superrationality instead of just calling reasoning about each others logic superrationality and so I got confused.
Last edited by PeteP on Fri Jan 08, 2016 10:00 am UTC, edited 2 times in total.

flewk
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### Re: Blue Eyes with Superrationality

PeteP wrote:Going back I'm noẁ unsure where you think you are using superrationality. There is no superrationality necessary in any scenario with more than 2 blues to determine that nobody sees less than 1 blue, thinking about what others see or what conclusions they arrive at does not require superrationality. The original solution already relies on everyone being perfect logicans and everyone else knowing that.

So I must apologize, I skimmed the first time and made assumptions and thought you were trying to do something with superrationality instead of just calling reasoning about each others logic superrationality and so I got confused.

The superrational part is that the islanders know everyone else must arrive at the same conclusions with the same given information. Even the original solution required superrationality as the islanders moved en masse and even waited together based on the same piece of information.

"The original solution already relies on everyone being perfect logicans and everyone else knowing that." is basicallly superrationality. Add on the fact that all the islanders know the rules and know that the other islanders know the rules, and it is a superrational situation.

PeteP
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### Re: Blue Eyes with Superrationality

No that is just normal logic, reasoning about the thinking of other perfect logicians isn't outside of normal logic. Superrationality is taking getting the same answer into account in things like symmetrical games. Like concluding in the prisoner dilemma that you should cooperate. It isn't logical your actions have no way of influencing your opponent so defecting can't cause the other to defect so it remains the logical solution. But superrationality says the other superrational individual will arrive at the same choice no matter what coupling your choices and making cooperating the solution. That is what is special about superrationality. Assuming the other person is logical and that they know you are logical and reasoning about what they will reason about your reasoning is not sufficient to label it superrationality.

My explanation is likely to be horrible flawed but superrationality is a game theory concept which probably has no relevance to the blue eyes puzzle. It's not just the assumption that everyone makes all possible logical conclusions and behaves accordingly.

flewk
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### Re: Blue Eyes with Superrationality

PeteP wrote:No that is just normal logic, reasoning about the thinking of other perfect logicians isn't outside of normal logic. Superrationality is taking getting the same answer into account like in things like symmetrical games. Like concluding in the prisoner dilemma that you should cooperate. It isn't logical your actions have no way of influencing your opponent so defecting can't cause the other to defect so it remains the logical solution. But superrationality says the other superrational individual will arrive at the same choice no matter what coupling your choices and making cooperating the solution. That is what is special about superrationality. Assuming the other person is logical and that they know you are logical and reasoning about what they will reason about your reasoning is not sufficient to label it superrationality.

It seems like we are working with two similar definitions of superrationality, but we have arrived at different conclusions. I am going by the Hofstadter one.

For the prisoner's dilemma, being logical and cooperating is not enough for superrationality. The most important part of superrationality is that you must know that the other person is also superrational. He must know that you know. You must know that he knows you know. And so on. Hofstadter likes infinite recursions. The recursive part seems to be the difference between our definitions.

1) Logic differs between individuals, because given information and assumptions differ. However, in a thought experiment, we could assign the same set of parameters to multiple individuals to force superrationality.

2) Think about what you mean by "cooperating the solution". How do you think the original islanders were waiting together and leaving together?

3) Every islander knows
I) the parameters of the game
II) everyone else knows the parameters
III) everyone is a perfect logician

This is a superrational problem.

The problem with the original solution is that it did not consider other conclusions a superrational population might make. A cascade caused by the nested logic only works if the trigger is new information.

Regardless of the definition of superrationality, it seems we agree that they are perfect logicians, which means they will make every possible logical conclusion. Can you find a flaw in the normalized perspective argument? If not, then "at least 1" will not be new information.

PeteP
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### Re: Blue Eyes with Superrationality

I don't think the way superrationality is used can really be derived purely from that recursive definition "Superrational thinkers, by recursive definition, include in their calculations the fact that they are in a group of superrational thinkers." replace superrational with blublubs, blublubs do things purely to damage others and if they include others also being blublubs in their calculations they don't arrive at superrationality, so I don't think the recursive definition alone actually provides a proper definition of superrationality. Also perfect logicians knowing that the other is a perfect logician wouldn't automatically arrive at the cooperation solution of the prisoners dilemma. Because being a perfect logician does not mean all perfect logicians automatically act the same, just that they know everything that can be derived from their knowledge. => There is more to superrationality than that.

But yeah it is sufficient to know what you mean by superrationality and it would derail the thread to talk about it more.

Anyway that they are perfect logicans and everyone knows is part of the problem statement so yes we agree. Iirc why the guru information starts the chain despite everyone knowing that everyone knows there is at least 1 (namely because it starts the collapse of the nested hypothetical) has been discussed at length in the original 35 pages thread and I don't think I could add anything new to that, thus I will leave and wait for others to rehash that discussion.

SirGabriel
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### Re: Blue Eyes with Superrationality

flewk, you keep talking about minimums and ignoring the real problem: when will the blue-eyed people leave, and why will they leave then?

Gwydion
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### Re: Blue Eyes with Superrationality

I want to expand on SirGabriel's point further. Flewk, what happens in the following scenarios (with superrational islanders etc)? Who leaves, and on what day?

1) There are 5 blue eyed islanders.
2) There are 4 blue eyed islanders and 1 brown eyed islander.
3) You are on an island and see 4 blue eyed islanders.

Alternatively,
1a) There are 5 blue eyed islanders and 4 brown eyed islanders.
2a) There are 4 blue eyed islanders and 5 brown eyed islanders.
3a) You are on an island and see 4 blue eyed and 4 brown eyed islanders.

Xias
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### Re: Blue Eyes with Superrationality

flewk wrote:The nested structure is the basis for the original solution. It is basically saying A sees 100 people, those 100 people see 99, those 99 see 98, and so on. This is why I included #3. Between any two individuals on the island, the number of unknowns will be 2. A only needs to think about what B thinks A sees. B only needs to think about what A thinks B sees. Nesting any further is unneeded as A=>B=>A can be applied between any two individuals because of the 2 unknown combination. No need for A thinks about B thinks about C thinks about D...

The bolded portion is the point that you keep making, and the point that I think is wrong. None of your math demonstrates to me that it is true.

If I see k blue eyed people, then for any conclusion that I come to, any minimum that I calculate, any "cascade trigger" that I come up with, any number at all that I calculate that depends on k, each other blue eyed person will calculate either the same number (if I have blue eyes) or one based on k-1.

This means that if you decide that the minimum number the Guru must say for me to figure out my eye color is X, then the minimum for each blue eyed person is either X or X-1. If the minimum is 98, then theirs is 98 or 97. If the minimum is 40, theirs is 40 or 39. But if the Guru says X-1, and none of the blue eyed people leave before the day that k blue people would leave, then we have terminated the possibility that my eyes aren't blue. This is a contradiction, though, because you said that the minimum for you is X.

1. Your conclusion is that the minimum number of Blues the Guru has to say for you to leave is 98.
2. This means that the minimum number for each of the other blue eyes is either 97 (if you do not have blue eyes) or 98 (if you do).
3. The Guru says "There are 97 blues." This is below your minimum.
4. On the day that you have determined that the other blues would leave if you were not blue, they do not leave. This means that they all see exactly the same as you, and their minimum is 98, and your eyes are blue.
5. You leave on the next day. Your minimum is 97.

Which step do you think is wrong, there?

Again, the information the Guru says is not merely "There is someone with blue eyes." The information the Guru gives is "If there were only one blue eyes, that person would leave tonight."

SirGabriel wrote:flewk, you keep talking about minimums and ignoring the real problem: when will the blue-eyed people leave, and why will they leave then?

Gwydion wrote:I want to expand on SirGabriel's point further. Flewk, what happens in the following scenarios (with superrational islanders etc)? Who leaves, and on what day?

I don't think these questions, as worded, are very helpful, because Flewk's claim is that the islanders can only leave if the Guru says a number greater than the minimum that Flewk has calcuated. What the Guru says is important to the solution, and important to Flewk's reasoning. So I'll further expand on Gwydion's questions:

What happens in the following scenarios (with superrational islanders etc)? Who leaves, and on what day?

1) There are 5 blue eyed islanders, and
- a) The Guru says "I see 3 blue eyed islanders."
- b) The Guru says "I see 2 blue eyed islanders."
- c) The Guru says "I see 1 blue eyed islanders."
2) There are 4 blue eyed islanders and 1 brown eyed islander, and
- a) The Guru says "I see 3 blue eyed islanders."
- b) The Guru says "I see 2 blue eyed islanders."
- c) The Guru says "I see 1 blue eyed islanders."

PeteP
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### Re: Blue Eyes with Superrationality

I think I will try an explanation anyway. Everyone knows that the number of blue eyes is greater than 1. And it tells you that the last case of the stacked hypothetical is wrong (there can't be zero) but of course everyone knows that already, everyone knows that most of the levels of the hypothetical can't exist. So what does it tell us? Well it tells us that all level of the hypothetical can use that information. If there are n levels where level n is (no people with blue eyes) then after the first day level n-2 knows than level n-1 is wrong. After two days level n-3 can conclude n-2 is wrong etc.. And yes this level are entirely fictional but eventually the chain will reach the level where you are not sure if it's a hypothetical or the real situation. Yes the whole benefit is that it gives information everyone knows to hypothetical constructs in a logicians mind that don't have that information. But it does cause a chain reaction that removes the hypothetical layer by layer. That can't start without the Guru.

Btw about one sentence that Xias quoted "No need for A thinks about B thinks about C thinks about D...". I hope you understand that the nested hypothetical is not between a chain of specific persons but a chain of imaginary groups? Because your reasoning about pairs and the naming scheme makes it sound like you don't?

Xias
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### Re: Blue Eyes with Superrationality

PeteP wrote:I think I will try an explanation anyway. Everyone knows that the number of blue eyes is greater than 1. And it tells you that the last case of the stacked hypothetical is wrong (there can't be zero) but of course everyone knows that already, everyone knows that most of the levels of the hypothetical can't exist. So what does it tell us? Well it tells us that all level of the hypothetical can use that information. If there are n levels where level n is (no people with blue eyes) then after the first day level n-2 knows than level n-1 is wrong. After two days level n-3 can conclude n-2 is wrong etc.. And yes this level are entirely fictional but eventually the chain will reach the level where you are not sure if it's a hypothetical or the real situation. Yes the whole benefit is that it gives information everyone knows to hypothetical constructs in a logicians mind that don't have that information. But it does cause a chain reaction that removes the hypothetical layer by layer. That can't start without the Guru.

Yeah, the whole thing about nested hypotheticals is an imperfect way to understand the abstraction of why the Guru's statement does anything at all. Talking about hypotheticals and what everyone knows that everyone knows that everyone knows... it's not very rigorous. Where it is most helpful is in understanding why everyone just doesn't accept that there are 98 or 97 or 96 or whatever the arbitrary minimum number of blue eyes you come up with.

The most rigorous proof of the solution itself is the proof by induction. Flewk's process tries to show that the induction doesn't work, for some reason.

Btw about one sentence that Xias quoted "No need for A thinks about B thinks about C thinks about D...". I hope you understand that the nested hypothetical is not between a chain of specific persons but a chain of imaginary groups? Because your reasoning about pairs and the naming scheme makes it sound like you don't?

I had that feeling as well. The chain of nested hypotheticals says just as much for what A knows that B knows that A knows that B knows... as it does for A>B>C>D or A>B>A>D or any other chain.

douglasm
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### Re: Blue Eyes with Superrationality

Suppose you have numbers X and Y for how many blue and brown-eyed people a given person on the island sees. You give me an algorithm for determining what this person's eye color is and when he figures it out and leaves. Skip explaining the logic behind your algorithm and just give me the algorithm. If you are unable to do this, then by definition you do not in fact have a solution to the problem.

I personally guarantee you that, no matter the logic behind it, if your algorithm results in anyone leaving earlier than in the original solution then there are numbers X and Y where it fails, having people get their eye color wrong.

Try this a few times, and maybe it will help you understand why your reasoning is wrong. Give me an algorithm - just the algorithm (e.g. "wait X-3 days, if no one's left yet then my eyes are blue") - and I'll post an X and Y where it fails. I suspect this practical exercise, trying to plug the hole I point out only for me to point out a new one you just created, may illustrate the problem for you better than all the abstract reasoning people have been trying so far.

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### Re: Blue Eyes with Superrationality

Superrationality is irrelevant when you're not playing against an opponent.

After the Guru speaks, there is a deterministic sequence of events that progress in the following days. These events are dependent on the eye color distribution of the population. You are allowed to observe the sequence of events and deduce what eye color distribution based on your observations. It just happens that these events involve other cognitive people, but it doesn't have to be.

The events could just as well be LEDs lighting up on a big billboard. If you see G blue-eyed people and Y brown-eyed people, You know you can leave the island when there are Y+1 Yellow LEDs or X+1 Green LEDs. The Guru speaking is equivalent to her flipping a timed switch that turns on the first LED. Once that LED lights, it will trigger the next LED in 24 hours, which will then trigger the next in another 24 hours, etc.
Superrationality is equivalent to knowing exactly how each LED work. It doesn't help you deduce their color.
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Vytron
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### Re: Blue Eyes with Superrationality

In any case, this is the original problem for n people as n increases, with perfectly logical people that have common knowledge about everything.

n=0
The guru can't speak, because she'd be lying.

n=1
After the guru speaks, the person that doesn't see any other blue eyed people realizes it must be him, and leaves D1.

n=2
After the guru speaks, the person that doesn't see the other blue eyed people leave D1 realizes their eyes must be blue as well, and leaves D2. And the other person goes through the same reasoning, and leaves as well at the same time.

n=3
After the guru speaks, the third person looks at 2 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D2, or their eyes are blue, and if those people are still here D3, it's because they are seeing them, and all of them leave D3.

n=4
After the guru speaks, the next person looks at 3 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D3, or their eyes are blue, and if those people are still here D4, it's because they are seeing them, and all of them leave D4.

Etc.
Spoiler:
n=5
After the guru speaks, the next person looks at 4 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D4, or their eyes are blue, and if those people are still here D5, it's because they are seeing them, and all of them leave D5.

n=6
After the guru speaks, the next person looks at 4 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D5, or their eyes are blue, and if those people are still here D6, it's because they are seeing them, and all of them leave D6.

n=7
After the guru speaks, the next person looks at 4 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D6, or their eyes are blue, and if those people are still here D7, it's because they are seeing them, and all of them leave D7.

n=8
After the guru speaks, the next person looks at 4 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D7, or their eyes are blue, and if those people are still here D8, it's because they are seeing them, and all of them leave D8.

n=9
After the guru speaks, the next person looks at 4 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D8, or their eyes are blue, and if those people are still here D9, it's because they are seeing them, and all of them leave D9.

n=10
After the guru speaks, the next person looks at 4 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D9, or their eyes are blue, and if those people are still here D10, it's because they are seeing them, and all of them leave D10.

...

n=100
After the guru speaks, the next person looks at 4 blue eyed people. They know that either their own eyes aren't blue, and they'll leave D99, or their eyes are blue, and if those people are still here D100, it's because they are seeing them, and all of them leave D100.

What flewk would need to tell us, is at what point his superrationality makes a difference and allows people to leave earlier than this.

Xias
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### Re: Blue Eyes with Superrationality

Vytron wrote:What flewk would need to tell us, is at what point his superrationality makes a difference and allows people to leave earlier than this.

It's my understanding that Flewk's argument is not that they can leave sooner, it's that they would not leave if the Guru only says she can see one blue eyed person. That the typical "The Guru doesn't tell them anything they don't already know!" argument applies to superrational islanders.

Vytron
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### Re: Blue Eyes with Superrationality

My question still applies. If superrational beings don't need the guru at all, then, at what point does it make a difference?

Mainly:

How can a single islander tell between n=0 and n=1? It can't its eyes might be blue but she needs the guru.

How can 2 islanders tell between n=1 and n=2? They can't, their eyes might be blue but they need the guru to know that the other islander is in the above case or not.

How can 3 islanders tell between n=2 and n=3? They can't, their eyes might be blue but they need the guru to know that the other islanders are in the above case or not.

And so on.

Or I don't get the argument at all? Is it that superrational beings don't leave at all because the guru speaking doesn't change anything? But how does that compare to my case posted? Because it's clear a single islander still leaves at D1 and that forces 2 islanders to leave D2 and such. The guru clearly changes the base case and superrationality doesn't change anything.