Problem in logic

[Lindley]

The key to the whole problem is this:

IF there were only a single blue, THEN he would leave on the first day.

If you can't establish that, none of the other logic can follow. This statement can only be true with the guru's statement. It's that simple.

 

[Yoda]

And you can't establish that, because you can't say what the Guru would say if there was only a single blue.

 

[Scout101]

You can't say what the Guru would say if Godzilla suddenly showed up, either, but that's not part of the equation, so you don't account for it. You're given a set of conditions, it results in a solvable equation, i don't get the problem...

 

[Yoda]

Godzilla isn't part of the proposed solution. If he was, then that would be a flaw too.

 

[Scout101]

"What-if?" scenarios for what the Guru might have said in other circumstances aren't part of the solution either. You're given the sample populations and a condition created by the Guru, and that's all you need to solve the problem.

You're the one trying to change the initial conditions by assuming the Guru's statement would change based on the number of one subset of the population. if the guru makes the same statement with only 1 blue eyed person present, they know it's about them, and they walk away, problem over. If you want to assume he'd make a different statement, you're welcome to do so, but you're changing the conditions yourself, and it's not part of the supplied (or needed) conditions.

 

[Yoda]

I'm not assuming the Guru's statement changes, I'm saying that from the initial conditions we can't assume that it stays the same (and we can't assume that it changes either). The idea that the guru makes the same statement with 1 person is what's not part of the supplied conditions. If it was intended to be then the riddle was phrased very poorly.

 

[Bisz]

This is quite a tricky problem.

I originally made the same mistake as RoJoHen and assumed we had to assume that the guru was talking about someone different each day, but in that case the problem is nearly trivial.

Once I realized my mistake in understanding the problem's parameters my mind was truly boggled. I want to thank the people who have taken their time to explain things, it was quite helpful.



And since this horse has not entirely decayed quite yet, I'll take a whack at it. Assuming three blues and three browns.

Day 1:

The guru told you that they see someone with blue eyes. If you have blue eyes you see two people with blue and three with brown. The guru may have been talking about the other blue eyed people you see or about you - you don't know which therefore you don't know if you have blue eyes or not.

The other blue eyed people are thinking the exact same thing you are thinking, they also do know know if they have blue or not-blue eyes.

No one leaves the island.

Day 2:

You still see two blue eyed and three brown eyed people. The only way that anyone would have left the island on Day 1 was if they saw no blue eyed people, making them the one and only blue eyed person the guru was talking about.

Since no one left this tells you that there is more than one blue-eyed person on the island. The only possibilities are that there are two blue eyed people, the two that you can see, or three, those same two and yourself.

Your blue eyed colleagues are thinking the exact same thing, but they might be better off than you. You don't know if there are two or three blue eyed people on the island, they might - if you don't have blue eyes then each one of them sees only one other person with blue eyes, they conclude that there are two blue eyed people, the one they see and themselves, and they leave the island on Day 2.

But that is not the case, no one leaves the island.

Day 3:

Yesterday you knew that there were either two or three blue eyed people, the two you can see and maybe you. Since the two you can see did not leave you now know that there are three blue eyed people on the island and that you are one of them.

The other two are thinking the exact same thing.

All three of you get on the boat on Day 3 and sail off into the sunset.

 

[Scout101]

I'm not assuming the Guru's statement changes, I'm saying that from the initial conditions we can't assume that it stays the same (and we can't assume that it changes either). The idea that the guru makes the same statement with 1 person is what's not part of the supplied conditions. If it was intended to be then the riddle was phrased very poorly.

Complete non-sequitur, though, because that has nothing to do with the problem presented. You have 100 blue, 100 brown, plus the green-eyed Guru, and he makes a statement (one time) that he can see a blue. None of the 'what-if' scenarios apply, it's just math from there. You can see N number of people with blue eyes, so the answer is either N or N+1.

 

[Yoda]

But you can derive the same statement from the initial conditions even if the Guru doesn't say it, because of the everyone can see a common blue situation. If what the Guru says at N=1 is irrelevant then it's also irrelevant that 'common blue' doesn't work for N=1.

 

[boobatuba]

But you can derive the same statement from the initial conditions even if the Guru doesn't say it, because of the everyone can see a common blue situation. If what the Guru says at N=1 is irrelevant then it's also irrelevant that 'common blue' doesn't work for N=1.

200 people can't derive, all together at the same time, that we're only going to consider blue or not-blue (actually, they consider talking about me/not talking about me but that's the same thing).

 

[Yoda]

Why not? They can consider blue or not blue, green or not green, brown or not brown. They would consider all of those as a consequence of not knowing what color their eyes are. Why is the 'only blue' necessary?

 

[Lindley]

But you can derive the same statement from the initial conditions even if the Guru doesn't say it, because of the everyone can see a common blue situation. If what the Guru says at N=1 is irrelevant then it's also irrelevant that 'common blue' doesn't work for N=1.

Personally I don't really think the focus on "blue or not blue" is really helpful. It may be true but it's not the key issue.

Here's the thing you're not getting. There's a dichotomy at work. On the one hand, there are 100 blues. On the other hand, it's possible to consider the case when there is only one blue.

The one blue case does not exist, nor does it need to exist, in reality. The situation is what it is, and it won't change no matter what the people on the island are thinking in their heads.

Blue A thinks "Hey, maybe I'm not blue....that would mean there are 99 blues." This does not mean there are 99 blues, it is merely a thought. It does not change what the guru said.

Blue A further considers that in that case, Blue B might suppose he's actually not blue (not knowing any better, since Blue A can't tell him he is). In that case Blue B would think there are 98 blues. Important note: This is still in blue A's head. In reality, Blue B would see 99 other blues even if he thought he might not be blue, but Blue A cannot use that information. A has assumed he is not blue, so if B further assumes himself not to be blue, that would leave 98 others.

Blue A knows that Blue B thinks the same way he does, so he knows Blue B will try to look at things from Blue C's point of view. Blue C may also consider the I'm-not-blue assumption, in which case Blue C will suppose there are 97 blues. This is still within Blue A's head, and he's imagining it to also be within Blue B's head.

It continues in this manner. We are not changing the actual number of blues and therefore cannot be changing the guru's statement-----we are merely performing a thought experiment in the form of a proof by contradiction.

Blue A assumes that Blue B assumes that Blue C assumes that Blue D assumes that Blue E may be seeing only 95 blues. And so on.

 

[Yoda]

Ok, but I can do the same thought experiment with the derived common blue statement. Since it's a a thought experiment we're not changing the actual number of blues and therefore aren't invalidating that statement.

 

[Lindley]

Ok, but I can do the same thought experiment with the derived common blue statement. Since it's a a thought experiment we're not changing the actual number of blues and therefore aren't invalidating that statement.

No, you can't. The one-blue case may not be reality but if you are able to follow a chain of logic to that point (which you can), then once there you are not allowed to use facts which go against previous assumptions (including the recursive assumptions that no one else is blue).

 

[boobatuba]

I can see Yoda's point. Inductive reasoning does break down, logically, when n>2. And here's why.

Lindley, in your explanation you claim:

Blue A knows that Blue B thinks the same way he does, so he knows Blue B will try to look at things from Blue C's point of view. Blue C may also consider the I'm-not-blue assumption, in which case Blue C will suppose there are 97 blues. This is still within Blue A's head, and he's imagining it to also be within Blue B's head.

While this is correct for an inductive reasoning argument, the logical fact is that it is impossible for Blue C to suppose there are 97 blues because we know he can see at least 98. Everyone knows this.

Blue A assumes that Blue B assumes that Blue C assumes that Blue D assumes that Blue E may be seeing only 95 blues. And so on.

Again, it's illogical to assume that Blue E may be seeing only 95 blues because everyone knows he sees at least 98. Certainly when you get to Blue UUUU, who we have to imagine supposes there is only 1 blue, the argument falls apart. How can we possibly think anyone supposes there is only 1 blue?

In order for inductive reasoning to work, you have to consider cases that you know are not true. That's really not logical.

I'm still pretty convinced that the answer lies in determining the entire subset of people which include the one the guru was talking about. Everyone with blues eyes knows it's either 99 or 100 people. Everyone with brown eyes knows it's either 100 or 101 people.

What's hard to determine is why they wait until day 100 to figure it out and why everyone knows to wait the same number of days as number of people in the subset. Everyone thinking about the inductive reasoning method may explain that answer, but it's hard to put your mind around the fact the everyone can logically know that everyone else uses that method to wait for day 100.

 

[sojourner]

the fact the everyone can logically know that everyone else uses that method to wait for day 100.

we know that because we are told that everyone on the island is a "perfect logician". Basically, all of the people think exactly the same way. They are all following the same path of reasoning.

 

[Nociod]

Ok, I think I understand the riddle now.


There are 100 blue eyed people, 100 brown eyed people and 1 green eyed woman on the island.

Before the guru speaks, they were all aware of the fact that there was someone with blue eyes on the island, but that fact didn’t help them at all. Since they had no way of communicating, 201 people would show up every night at the ferry station, and all 201 would leave empty handed every night since they saw someone with a different eye color.

Let’s say that one night when all 201 are gathered at the ferry station, the guru suddenly utters the words “ I see someone with blue eyes”. Now everyone has a common reference of how they are going to get off the island.

The blue eyed person will see 99 other blue eyed persons, and know that if he/she also has blue eyes, they will have to show up on the 100th night.

The brown eyed person will see 100 other with blue eyes, and know that if he/she also has blue eyes, they will have to show up on the 101st night.

The green eyed guru will see 100 other with blue eyes, and know that if she also has blue eyes, she will have to show up on the 101st night.

On the 100th night, no blue eyed person will leave when they see each other, and therefore they can all go on the ferry together.

On the 101st night, a 100 brown eyed people and one green lady will show up, but the guru will screw them all over because she has a different eye color, and no one of them will leave.
They are back to the first scenario where they all go to the ferry every night, and they all have to go back empty handed. This will go on until the green eyed guru dies, or for some reason (illness, vacation on the other side of the island, given up on trying to leave etc) doesn’t show up. If that happens, the brown eyed people will be able to leave, since no one will leave.

One of the brown eyed people could also sacrifice himself so that the rest can leave.

One of them could find the guru one night and detain her. By doing this they would be dooming themselves to spend the rest of their life on the island, and forever live with the uncertainty of their own eye color. If one does that, all the other brown eyed people would routinely go to the ferry station, and no one would leave when they saw each other, and they would figure out that they are brown eyed.

However, the “sacrificial lamb” and the guru would be stuck on the island forever. When both of them show up the next day, they wouldn’t know if they should stick or leave, cause they have no one to reference themselves with

 

[Lindley]

I can see Yoda's point. Inductive reasoning does break down, logically, when n>2. And here's why.

Lindley, in your explanation you claim:

Blue A knows that Blue B thinks the same way he does, so he knows Blue B will try to look at things from Blue C's point of view. Blue C may also consider the I'm-not-blue assumption, in which case Blue C will suppose there are 97 blues. This is still within Blue A's head, and he's imagining it to also be within Blue B's head.

While this is correct for an inductive reasoning argument, the logical fact is that it is impossible for Blue C to suppose there are 97 blues because we know he can see at least 98. Everyone knows this.

Blue A certainly knows that Blue C will see at least 98 blues, because blue A sees 99. However, what Blue A sees is not relevant to Blue B's thought process within Blue A's head.

The real blue B will see 99 blues as well because A has blue eyes. But Blue A supposes that he is not blue, and within this context, Blue B would see 98 others.

Now, Blue B would certainly not know he is blue. So he also supposed, what if he isn't blue. In that case, Blue B (within Blue A's head) would suppose that there are only 98 total blues. In reality blue B can see 99, but blue A doesn't know that. So then A supposes that B supposes that perhaps Blue C can only see 97 blues.

 

[boobatuba]

Blue A certainly knows that Blue C will see at least 98 blues, because blue A sees 99. However, what Blue A sees is not relevant to Blue B's thought process within Blue A's head.

The real blue B will see 99 blues as well because A has blue eyes. But Blue A supposes that he is not blue, and within this context, Blue B would see 98 others.

Now, Blue B would certainly not know he is blue. So he also supposed, what if he isn't blue. In that case, Blue B (within Blue A's head) would suppose that there are only 98 total blues. In reality blue B can see 99, but blue A doesn't know that. So then A supposes that B supposes that perhaps Blue C can only see 97 blues.

I don't disagree with you. I understand the inductive reasoning process, and how by using it you can get down to 1 theoretical person assuming that they don't have blue eyes. I can also see how everyone understands that system and why they would wait for the 100th day to confirm that there are indeed 100 people with blue eyes.

It's still just not very logical in a "real-world" scenario.

 

[Lindley]

It's perfectly logical. It just isn't intuitive.