Eye color riddle

Miscellaneous and Personal Stuff I Must Share
61

This is only true at the first meta level. In other words, on its face the statement does not give anyone any information that they did not already have. However, each blue-eyed person is, on some level, expecting that either it will add new information for someone, OR that they themselves have blue eyes. When it turns out that the statement did not add new information in the way they expected, the first option becomes a logical contradiction and they are forced to conclude the second option (i.e. that they are blue-eyed) is correct. Thus, it’s true that the statement adds no new information, but the fact that that is so is a new piece of meta-information for each blue-eyed person.

The fact that the statement does not provide any new information, by itself, does not prove anything, as you (FBG) have already acknowledged by stipulating to the fact that the logic works for the case of 3 blues. In the case of 3, it is true that the statement adds no information, and that everyone knows this. Yet you agree that the logic still works.

So, yes, the statement does not add information. However, it does add meta-information, in that it shapes each person’s expectations of how the others will react, and each person realizes that those reactions have implications for their own eye color.

62

Wait, wait, wait … you’re both right. I mean you’re both wrong. The good news is, nobody has to die here.

ncarbtpmo has de-lurked to give us the concept of meta-information, but I wouldn’t exactly say it’s not new information – it really is. Well, it is in the lower end case; not in the case of N= 100.

I’d seen this riddle before but hadn’t thought about it until I saw this thread, and then played around with brown sugar and equal packets at lunch. Since I’d already poured a ton of sugar into my tea, I think they expected me to go crazy. I also only took three packets of each, which worried me when I saw I needed more than two but three works, so I’ll kill 97 or so blue-eyed folks myself. (Wait, I said nobody gets hurt. Let’s just ignore them for now).

I think it was Libertarian who once introduced some of the symbols to describe it and if he were around he’d probably do a clearer job of this. But I’ll try my best. The key to understanding this is the information that A knows that B knows, and what everybody knows. First, some terminology: Let P be some statement, and A§ mean ‘A knows P’, A->B § mean ‘A knows that B knows P’, E[P] to mean “Everybody knows P”. The meta- step takes us higher, to E[E[P]], which means “Everybody knows that everybody knows P” of course. Think about this : Just because E[P], does that logically imply E[E[P]]?

In terms of the island riddle, let P=“there is somebody with blue eyes”. With a sufficient number of blue eyes, it’s true that E[P] before the guru shows up. Then the guru shows up, and with his declaration, it is now true that E[E[P]]. That information can lead to somebody going away when it didn’t before.

However it is NOT true that for larger numbers that this is new information. In fact, it does not help any group with more than three of an eye color.

To clarify how this works, let’s look at various populations. B means an Equal packet, I mean a blue-eyed person and b means a brown-eyed person. What is seen (?) is shown in each column; note that B->B (?) means ‘a blue-eyed person knows that a different blue-eyed person sees ?’, and x is the unknown of themselves. (-- means not possible, 0 means no information). Here’s a nice big, format-ruining chart :



Population     B(?)    b(?)  B->B(?)  B->b(?)  b->B(?)  b->b(?)   E[?]  E[E[?]]

     Bb          b       B      --        x        x       --       0      0
    Bbb         bb      Bb      --       bx       bx       Bx       b      0
   BBbb        Bbb     BBb     bbx      Bbx      Bbx      BBx      Bb      0
 BBBbbb      BBbbb   BBBbb   Bbbbx    BBbbx    BBbbx    BBBbx     BBbb    Bb
 BBBBbb      BBBbb   BBBBb   BBbbx    BBBbx    BBBbx    BBBBx     BBBb    BB
 BBbbbb      Bbbbb   BBbbb   bbbbx    Bbbbx    Bbbbx    BBbbx     Bbbb   bbb

Here’s what can be seen from the chart (hopefully): Nobody knows for any population set what color their eyes are, since they can see different things depending on what the true population is. They need more information if they are going to know. So the population starts stable.

For either the B’s or the b’s (let’s use B’s), if there are at least N>=3 of them, then E[E[there are at least N-2 B’s on the island]]. If N < 3, then there’s no information about that group that everybody knows everybody knows. So when the guru makes his statement, it does effect the small groups (with N < 3), but this is not true for the larger groups, since he really is telling them something they already know.

There are bits of information that could be given by the guru that lets people know what they are, but I’m too tired to find that out.

To extend an already overlong post, consider the point that’s been brought up ( I think by Bob Cos) when they try to work backwards. They have to say “if there were 97 blue-eyed people” but there aren’t, they know that, so any conclusion is invalid.

(6 hours ago, when I thought the solution was 100 days, I realized that anyone would know what day they were supposed to kill themselves - just count the number of blue people they see, and add one. Of course, since the solution doesn’t work out that’s not that exciting; though I wonder if it was legal for them to talk about that, since it’s not revealing their eye color but it would let others figure it out immediately.)

63

It is very late at night and I’ll be able to focus on this more tomorrow. However, here are some random thoughts on the very interesting chart by panamajack:

  1. I understand why the chart applies on the first day. I understand why it does not help any group of more than 3 blues. However, the inaction of the other blue-eyes each day up through day 99 is the equivalent of an amendment to the guru’s original pronouncement. Like a message from Holmes’ famous dog that doesn’t bark, the non-action is a new statement telling the blue eyes to alter the guru’s original words. The non-suicide of a blue-eye on the first day has the same result as if the guru were to have said on the first day: “I see 2 blues eyes.” And non-action on the third day = an amended first day statement to: “I see 3 blue eyes.” etc.

  2. We are concerned not just with E[P] and E[E[P]], but also E[E[E[P]]], E[E[E[E[P]]]]…** ALL THE WAY UP TO 99 E’S**. This seems absurd, but that’s where you go if you look from 99 blue eyes down to 1 instead of working it up.

  3. For me now the best way of looking at it backwards is:

Say I’m a blue eye, but I don’t know it. I know there are either 99 or 100 blue eyes. I’m now going to assume I am a brown eye. If true, then that means my blue-eyed friend, Joe, is going to be thinking this:

“There are either 99 or 98 blue eyes. I’m going to assume my eyes are brown. That means my blue-eyed friend Sue is going to be thinking…”

Thinking this:“There are either 97 or 98 blue eyes. Now I (Sue) assume my eyes are brown, that means my blue-eyed friend Bob is going to be thinking there are either 96 or 97 blue eyeds, and Bob blue-eyed friend Fred is going to be postulating 95 or 96 blue eyes…”

and so on.

We know these assumptions are wrong, but as the primary questioning blue-eyed person, I’m not saying that Sue is going to think anything. I’m saying that Joe is going to think that Sue is going to think that Bob is going to think that Fred is going to think that Mary is going to think that James is going to think that …right on down to the last blue-eye (whom we’ll call Bart.) When Bart doesn’t kill himself, I have to rethink the whole thing. Or to be more accurate, Morris (who was the second to last blue-eye), is going to have to rethink), I think.

Now I am going to bed.

64

Ok I have a challenge to the dissenters.

It works for n=0, n=1, n=2 and n=3. It provably works for n and n+1 (seperate from the specific agreed upon chain). At what number does it break down? Why? That is explain why the assumption of generalizability that undergirds most numerical logic based upon proving for the n and n+1 case? What is it that makes the numbers you accept it up to so special?

I think every explanation that can be given has been given with various clarities. The objections are fairly valid when thinking reasonably. However these aren’t reasonable people, they are logical people. Reasonable people don’t kill themselves over eye-color. So what specific logical objection can you give to prove that some sub-set of cases (the low numbers) is different?

I think that producing a formal disagreement will help clarify exactly where the objectors are objecting and will hopefully win a few over as they attempt it.

65

Gah stupid sleep deprivation. Obviously the question about the assumption for generalization is asking why doesn’t it hold in this case.

Stupid final in the middle of the semester destroying my mind.

66

As an example, no one can allow for the possibility of anyone postulating a possible 95 sets of eyes, correct? Not when everyone on the island knows that everyone else can see at least 98 sets of blue eyes. Sorry, this must be frustrating, but I just can’t see how this point isn’t indisuptable (and I’m trying to see).

67

How about this?
Each and every blue eyed person is thinking “I can see 99 people with blue eyes. If 99 people commit suicide on the 99th day, then I will know that each of them are seeing only 98 blue eyed people, therefore I have brown eyes. If not, then I will know that each of the blue eyed people I am seeing are also seeing 99 others, in which case I must be one of them.”
Nobody is expecting anything to happen until either the 99th or 100th day.

68

Everyone lisen to Elret. His explanation rocks :slight_smile:

69

I’m not positing that Adam can posit that Bill can posit there are 97 blue eyed people. I’m positing that he can posit that he can posit that Colin (and 96 or so other people) posit this.

And then he might assume that someone supposes that someone supposes that someone supposes that someone supposes that someone supposes that someone supposes that someone supposes that someone supposes that…

But that’s not a good way to imagine it

70

Apparently I’ve had one too many sugar packets.

Never mind, I was wrong about all that. The very fact that E[E**] isn’t new information means that N must be greater than or equal to 3 … so if I see only two blue-eyed folk, I know I’m one of them, etc., etc.

Nevertheless, I’m not so sure about the working backwards thing. Since I know that Joe->Sue’s assumptions are false (even though Joe doesn’t know this), I can’t draw anything from this.

So I only had one interesting point (which I already mentioned, ) – with two colors, anyone knows what day they will die – it’s (number of blue-eyed people +1). So does telling someone the immediately false statement “I’m dying on day 100” violate the rules?

71

I got a wicked bad headache trying to figure this one out. I’m going to stick to surfing porn . . .

72

Thanks to everyone for the explanations. I drift in and out of understanding it, and generally come down on the side of Biotop. However, the one concern I had has already been submitted by dil when he pointed out:

The only thing the guru is doing is starting the clock for the counting of the days in figuring out how many blue eyed people there are on the island. The fact remains, however, that the scenario in the riddle is impossible to meet, because it would have already been met at X number of days after the rules were enacted. You cannot possibly say these rules existed for any days longer than there are a number of people on the island. It is a nice thought experiment, but it is a logically inconsistent. You would have to change the scenario and make rules of the riddle were voted on and set to take effect on a certain day, and then X number of days later is when the mass suicide would happen. The rules could not just have been always in effect.

73

Hamlet:

No, “Ready, set, go!” will not work.

Note that if there are 100 blue eyes, and 100 brown eyes then the statement “I see a blue-eye” only sets off the blues, not the browns. Unless the browns know there are only two eye colors, they remain safe. Or, even if the browns are aware that there are only two eye colors, they get an extra day more of life than the blues.

74

Hamlet,

I’m wrong. If the islanders know beforehand that there are only two eye colors, then “Ready, Set, Start thinking about your eye color, Go!” does set them all off! The eye-color segment with the fewest members will do itself in first.

However, if we postulate the existence of more than two eye colors, the guru/stranger must designate one to go, while un-named colors are spared.

75

But the riddle still requires this island to just plop into existance with all these rules already in effect. Otherwise, the mass suicides would have occurred X days after these rules take effect, where x is the number of blue eyed people. It would be impossible, and illogical, for it to be otherwise. You could never have 100 people on the island with blue eyes, unless the island/rules came into effect less than 100 days ago.

Say the island/rules were created 500 days before the guru shows up. Then, 400 days before the guru showed up, all the blue eyed people (100) would have killed themselves, because of the same logic you use to explain the guru’s situation. If the guru shows up on say the 50th day of the island/rules existance, and says “I see someone with blue eyes”, then 100 people would kill themselves on the 50th day after the guru’s comment, not the 100th.

I guess what I’m getting at it is impossible for there to be any more than one blue eyed people on the island, otherwise they would have killed themselves already. Only if the blue eyed people had the belief that there were NO blue eyed people on the island, would any of them still be alive. As soon as they know there is at least one blue eye on the island, then the suicides begin. So you could never have more than 1 person on the island with blue eyes. And that one person would have to think that NOBODY on the island had blue eyes, or else he’d off himself.

I’m not sure if I’m making this clear, but the island could not logically exist unless there was only one blue eyed person, and he didn’t know there were any blue eyed people on the island.

Brain lie, hurt now down I.

76

Final correction:

If all the Islanders are somehow previously told (but not at the same time) that there are only two eye colors, *and, that they are both represented, then the statement “Ready, Set…Go” works. Otherwise, not.

However, if the natives are somehow collectively aware that only blue and brown-eyed people can be on the Island, without saying for sure that both actually do exist, then the guru/stranger must make the statement for any demises to occur.

77

gonna bump this evil grin :smiley:

the ‘2 blue eyed’ logic holds*. my objection is that the guru doesn’t impart any new information, but i’ve already accepted that his role is more towards starting the thoughts rolling**, as it were.

*however, at some point (assuming the natives are humans and not pure logical robots) it gets really ridiculous. because if there are 100 blue eyes and everybody can see 99, then everyone will do nothing for at least 98 days. that doesn’t make common sense, and some 50+ days along the way, some natives will get really confused and ask someone ‘what day is today??’

**as Hamlet mentioned that doesn’t make sense too, and it would be much better if the scenario states that the guru says ‘from today onwards, i henceforth pass down the law that whoever discovers his eye color must go into exile and, oh btw, i see at least one person with blue eyes (hehehe)’

78

But if you have more than one islander with blue eyes, they will know that at least one exists, because they will see one. Maybe that is why in the second riddle, there is a person with green eyes. If you add another possibility exists, then nobody will be assured that they have blue eyes.

79

It’s not necessary, though, for green or any third eye color. Because even though you know everyone can see blue and brown eyes in the 2-color version, and you know everyone knows everyone knows, and you know everyone knows that everyone knows that everyone knows etc., you eventually reach a level (see the E[E[E[E…[P]]]] example) where you get new information.

As long as no one exactly notes, for everyone to hear, that at least one of a certain eye color exists, it doesn’t matter if there are two or fifty different eye shades.

** Coincidentally enough, the May issue of Games Magazine, which arrived in my mail yesterday, contains still another version of this riddle. Check it out.

80

I understand the logic. It took many re-readings of people’s answers, and the great wisdom of Elret, but I understand it.

I also agree with everyone who points out that this scenario is impossible (i.e., the rules coming into existence, someone’s eye color changing, etc.) however, I am going to set that aside for now.

Imagine this. The stranger lands, looks around. He sees 100 people with blue-eyes. Instead of saying “I see a person with blue eyes,” he actually says, “I see two blue eyes.”

It is essentially the same thing, because each person has two eyes. So everyone looks around to check. They each see many blue eyes. As a result, everyone realizes the stranger was omitting a detail. They can do nothing.

Now. Am I right or wrong in this logic?