Friday night Brain teaser

[Pundit]

<< Then an ultimate
> being comes to the island
> and tells them that when each blue-eyed person finds
> out he/she has blue
> eyes, he/she must kill themselves that night at
> exactly midnight. >>


Doh!!! From this sentence, I gather "that night" to refer to the night that the being came to them, not the night that one discovers his blue eyes. I was trying to solve this thinking that they must discover themselves in one day. Now I wish I hadn't read the solution, which does work!

p.s. I still think they can all rip one eye out of its socket and look at it with the other

 

[woolmilk]

Ok, i ve got it now. Following assumptions needed:
- Everyone meets everyone every day and counts the number of GFD
- Everyone counts the days and checks if someone leaves
- at least one with GFD

From logic, every monk can make up the equotion:
"If I see N cases of GFD, i will leave after N days - except some others leave before"

If I see no GFD, it must be me with GFD and I m leaving immediately.
If I see one GFD, i will leave tomorrow - except the other one is gone today (because he himself counted 0 GFD).
If I see two GFD, i will leave in two days - except the other two are gone before (because they counted 1 themself).

 

[MisterNi]

<< Ok, i ve got it now. Following assumptions needed:
- Everyone meets everyone every day and counts the number of GFD
- Everyone counts the days and checks if someone leaves
- at least one with GFD

From logic, every monk can make up the equotion:
"If I see N cases of GFD, i will leave after N days - except some others leave before"

If I see no GFD, it must be me with GFD and I m leaving immediately.
If I see one GFD, i will leave tomorrow - except the other one is gone today (because he himself counted 0 GFD).
If I see two GFD, i will leave in two days - except the other two are gone before (because they counted 1 themself). >>


That's exactly right. But saying, "I" is being an impartial third party observer.

 

[alrocky]

<< 100 people on an island. These 100 people can't speak or communicate with each other in any way. 10 of them have blue eyes and 90 of them have brown eyes. But there are no mirrors or anything so no one knows if he/she is blue or brown-eyed. Then an ultimate being comes to the island and tells them that when each blue-eyed person finds out he/she has blue eyes, he/she must kill themselves that night at exactly midnight. The 100 people are also very obedient so they will do as they are told The ultimate being doesn't tell them how many people on the island have blue eyes though. He just tells them that each blue-eyed person will kill himself when he finds out he has blue eyes. So what happens? >>



Shooters's solution is wrong and his answer does not apply. This riddle is bogus since the conditions set are ambiguous: "an ultimate being comes ... tells them that when each blue-eyed person finds out he/she has blue eyes, he/she must kill themselves that night." This suggests the night of the Being's arrival, not any subsequent night. Even if it's open to interpretation then the Islanders can do the same and follow the instructions to mean that first night only. Thus noone dies.

Even if Shooters's scenario was suppose to be right, the last person with blue eyes would never kill himself. Scenario (1) in Shooters's reply is incorrect since Islanders don't know the Blue / Brown mix.

Assuming Islanders don't know the Blue / Brown mix:

1 Blue 99 Brown. Blue cannot see own eyes, goes to sleep and wakes up and all 100 alive.
2 Blue 98 Brown. Both Blues wake up and see other Blue still alive. Each thinks the other Blue doesn't know he's Blue. Blue 1 thinks, "Islander B2 has Blue eyes and doesn't know it." Blue 2 thinks, "IslanderB1 has Blue eyes and doesn't know it." So neither kill themselves. (If Islanders knew the Blue / Brown ratio, ie 2 Blues, then both Blues only see 98 Brown and 1 Blue, which makes them Blue, and so they'd kill themselves that first night, not <U>n</U> nights as suggested by Shooters.)
...etc

With the riddle as given, each Blue would see 9 Blues and 90 Brown. Each Blue doesn't know if he is Blue or Brown. No Islander knows the Blue / Brown ratio. No Blue can make an assumption of his own eye color that night, and 100 Islanders wake up the next day. Since no additional information is discovered and the scenario has not changed save the passing of one day, Islanders cannot drawn any conclusion about their eye color and 100 Islanders wake up day 3. Repeat...

Even if we take Shooters's solution as correct that every night a Blue concludes he is in fact Blue, why does only one Blue take his life each night, and not 2 or more or the whole lot of 10?

Pundits answer on another thread about this riddle is correct. Noone dies or each Islander removes an eye to determine if he lives or dies that night.

edit italic format...

 

[MereMortal]

<< Even if we take Shooters's solution as correct that every night a Blue concludes he is in fact Blue, why does only one Blue take his life each night, and not 2 or more or the whole lot of 10? >>



You misunderstand the solution. A blue does not kill himself/herself each night. All n blues kill themselves on the nth night.

 

[calbear2000]

I see a couple people still do not understand the solution...

For the last time, the ultimate being does not need to say there is at least one blue eyed. The original problem states that there's 10 blue eyes. All the information is there. If you can't understand, please stop trying to reword the problem to fit your misconstrued concept or make lame statements trying to blame your misunderstanding on the wording of the problem.

Next thing you'll say is "you need to assume people on the island have 2 eyes and not 1. Thus the NUMBER of blue eyes is 20, not 10" Thus the riddle is bogus.


I'm glad most people understand... including you pundit

 

[alrocky]

"Take the simple example of 2 blues and 98 browns... you and I are blue and everyone else is brown. We see each other and I think you're the only blue and I wait for you to commit suicide on the 1st night. You're thinking the same thing. We both wait and nothing happens. That means 1 other person is definitely blue." No. What would be if you see the only one with blue eyes ? He has no proof that he has blue eyes and would not commit suicide. Just because superior says "if you recognize yourself as blueeyed you have to die" does not imply that any blueeyed must die unless there is a proof for the individual.
quote by calbear2000: Ok, I'll explain the base case one more time.... If you and I are blue eyes. I will count 1 blue and 98 brown. I do not know about myself yet. I assume I am brown, and I wait for you to commit suicide on the first night. You are thinking the same thing. The first night passes, and no one commits suicide. That is my PROOF that there has to be 1 more person thats blue eyes, and that person is me. I think I see the point you are bringing up though. I think your point is this: If you committed suicide on the first night, then I know that you saw 99 brown eyes and realized that you are the only one. But you cannot arrive at that conclusion if the superior being didn't say there was at least 1 blue. BUT, the initial condition was that there are 10 blues. So you would see > 1 blue. And thus, that negates your scenario.

"If you and I are blue eyes. I will count 1 blue and 98 brown... I assume I am brown, and I wait for you to commit suicide on the first night. You are thinking the same thing. The first night passes, and no one commits suicide. That is my PROOF that there has to be 1 more person thats blue eyes, and that person is me." This is no PROOF; the other Blue thinks you are the only Blue and that you just don't know that. Thus noone dies.

Your PROOF doesn't work as per items A) and B) below. Item C) fits your "see >1 Blue" and the Blues can still rationalize not dying.

A) 1 Blue 99 Brown. Blue will never know if he is Blue and noone dies.
B) 2 Blue 98 Brown. Blue1 and Blue2 only see (1) Blue and both think the same thing, "Hey, he's Blue and doesn't know it." Noone dies.
C) 3 Blue 97 Brown. Blue1, Blue2, and Blue3 each see (2) Blues and all (3) Blues think the same thing, "Hey, those (2) Blues think, " Hey, he's Blue and doesn't know it."" Noone dies.

<U>Blue = die. You = unknown eye color.</U> The total number of Browns is not important. Given an Island population of 1), 2), and 3) below can calbear2000 or anyone else show me how you determine who dies? Item 3) is a simplified version of the original 10 and is a vaild test that must precede the result for 10 Blues.

1) You
2) Blue and You
3) Blue, Blue, and You


Here's a question: If for some reason the original 10 Blues did not die on the 10th night, what would happen on the 11th night?
Thanks to MereMortal for the nth night clarification. Please, no copout like in the last post's 2nd paragraph. If you cannot explain a solution, just say so.

 

[calbear2000]

Ok you're taking this way too seriously...

I don't know why you can't understand... the initial problem says there are 10 people that are blue-eyed. Thus, your scenario goes out the window.

Are other people as confused as this guy is?

I admit the puzzle is challenging, but the solution isn't THAT difficult is it?

 

[her209]

Do they know how many blue eyed folks have killed themselves already?

Are each of the individuals distinguishable (i.e. do they all look different besides the eye color)?

 

[lizardboy]

answer please

 

[AzNmAnJLH]

ok 100 total
90 = brown
10 = blue

superior being told them to kill themselves after discovering that they are blue
time of death of established at midnight the same day that they discovered that they are blue
can't communicate also doesn't have knowledge of blue versus brown

so they are isolated to figure out if they are blue or brown by observing the others
the others are doing the exact same thing in trying to determine if they are blue or brown, they have to find out because they are obedient
the brown eye person will notice 10 blue eyes ont he island while the blue eye person will only see 9
yet the superior being didn't state that their life's goal was to find out if they were blue or brown but in discovering if they are blue to commit suicide

through observation from each individual they'll noticed after the first night that no one kill themselves because the day before they all saw that someone besides themselves had blue eyes
ok so no one killed themselves doesn't mean i'm blue eyes it just mean that the blue eye person saw 9 or possibily 10 other person with blue eyes be you yourself didn't know if you are blue eyes

all the brown saw that 10 are blue eyes yet they could assume that they are also blue bringing the total to 11

so blue eye people after the first midnight had pass could assume 9-10 person with themselves as blue eyes
but brown will assume after the first midnight had passed that 10-11 person with themselves as blue eyes

90 of the brown eye people could believe they are blue eye and 10 of the blue eye could believe they are brown

the inability to community will either cause them to commit mass suicide or to go on living assuming the other will commit suicide while the other will assume you will commit suicide if you happened to be blue


i know there is a logical solution....

 

[dfi]

Hrm, I've heard a similar riddle, where the anwer was based on the observations of others. But that riddle was flawed in the same way that this riddle seems to be...

dfi

 

[dfi]

<< Solution: Don't read if you want to figure it out for yourself.



1. If only one person on the island has blue eyes then he can look around and see 99 pairs of brown eyes. Therefore, he knows that he must be the one with the blue eyes and kills himself that night.

2. If two people have blue eyes, then each can look around and see the other person's blue eyes along with the remaining 98 pairs of brown eyes. They both go to sleep that night thinking that the person they see with blue eyes will kill himself like in situation 1 above. However, when both wake up the next morning and see that the other person with blue eyes didn't kill himself, then they know that they themselves must have blue eyes also and therefore both blue-eyed people kill themselves on the second night.

3. If three people have blue eyes then the above reasoning can be used again except now two nights will pass and the three blue-eyed people will kill themselves on the third night.


So in general, all the blue-eyed people will kill themselves on the nth night where n is the number of blue-eyed people on the island. >>



That doesn't work for a number of reasons.

First, the riddle is poorly stated. No one knows for sure that there are blues, so as far as they know, it's possible that no one is blue. Let's say there are 2 blues; let's call them #1 and #2. After a night of rest, they awake to see each other, and everyone else, to be still alive. #1 can easily assume that #2 thinks no one is blue, and #2 can assume the same. So then no one dies.

So, your logic is flawed. Say there are 2 blues and they kill each other (even though they shouldn't. But let's just assume they do, as the answer states). Then, there are 98 ppl left. But after another night, they awake to see all 98 other ppl alive. By your logic (that ppl kill themselves cuz they see no one else has died), what's to prevent the 98th person from assuming that he has blue eyes when he doesn't? Then everyone will keep killing themselves until no one is left.

Also, with 3 blue-eyed people, it is again screwed up. All 100 ppl go to bed, including the 3 blues. Next morning, they wake up. Let's call the blues A, B, and C. A sees that B and C are still alive. He can easily come to the conclusion that B didn't kill himself because B sees C, and C didn't kill himself because C sees B. B can come to the conclusion that A and C are alive because A sees C, and C sees A. While C thinks that B and A are alive because A sees B, and B sees A.

Like I posted earlier, this riddle and its asnwer is flawed in the exact same way that the other similar riddle (involving hats) is flawed.

dfi

EDIT: just read Alrocky's posts, completely agree.

Calbear2000, I'm sorry, but the riddle is flawed. I've heard a variation of this before, came up with the exact same answer you have presented, and realized the answer was flawed.

Btw, I think you'll feel quite ridiculous for questioning Alrocky's intelligence, once you realize that you were the one in error. It's not that the question/answer is difficult to understand; it's that they are simply wrong.

 

[AzNmAnJLH]

dude,,, if i saw you i could look at you and instantly know of you are blue or brown easily

 

[skace]

I first thought it was flawed also, but then realized some things:

Q: After all the blues have died, what stops the browns from looking around and thinking they are blue and killing themselves?

A: Well, you have to remember how the blues died. They won't kill themselves until they come to the realization that they themselves are blue. This is done by waiting for another blue to die, when that one doesn't die they come to the only other conclusion, they themselves are blue as well. Why is that the only other conclusion? Because they are obediant and the only reason for 1 not dying is because he saw another and thought he wasn't it. So if he saw someone else, and you can't see any other blues, that blue is YOU. So if the Blues were comfortable killing themselves, then that is your proof that you are not blue and can safely continue living.

Q: In a scenario where only 1 person has blue eyes and the rest have brown, why would he want to kill himself?

A: Well, why would your superior make a law if no one fits the rules. That would be like a god coming down to earth and saying, if you breathe water, walk on 2 legs, and have purple skin, you may go to heaven. Doesn't my superior know their arn't any people like this? Thinking that he must know better, an obediant person would think at least 1 exists. So if at least 1 exists, and you look at everyone else and they have brown eyes, then you're the unlucky sap. In the rare case that no one had blue eyes, everyone would kill themselves that night. But since these cases weren't setup in the condition, they don't really matter.

Q: What would stop a brown person from thinking there is 11 blues and he is one of them?

A: I can't figure this one out!! but maybe that is because I don't understand why it would take 10 nights for all 10 to kill themselves. I would think that each blue would see 9 people the first day, go to sleep and wake up seeing the same 9 people, he would then assume there must be 1 more, since he can't see 1 more he would know it is him and then all 10 would die the second night. So I must be missing something in the end process...

Nice teaser.

 

[dfi]

<< I first thought it was flawed also, but then realized some things:

Q: After all the blues have died, what stops the browns from looking around and thinking they are blue and killing themselves?

A: Well, you have to remember how the blues died. They won't kill themselves until they come to the realization that they themselves are blue. This is done by waiting for another blue to die, when that one doesn't die they come to the only other conclusion, they themselves are blue as well. Why is that the only other conclusion? Because they are obediant and the only reason for 1 not dying is because he saw another and thought he wasn't it. So if he saw someone else, and you can't see any other blues, that blue is YOU. So if the Blues were comfortable killing themselves, then that is your proof that you are not blue and can safely continue living. >>



That doesn't answer why the brown won't kill themselves. When 2 blues are alive, blue #1 looks at blue #2, and can just as easily come to the conclusion that "I am brown, but #2 doesnt know that he is blue." Just because you see another blue that's still alive, you can't rightly assume that you yourself are blue. It is just as logical to assume that the blue you see simply doesn't know that he is blue, and that you are brown. In other words, there is just as much justification to think you are blue as to think you are brown. So, either no one dies by one logic, or everyone dies by the other.




<< Q: In a scenario where only 1 person has blue eyes and the rest have brown, why would he want to kill himself?

A: Well, why would your superior make a law if no one fits the rules. That would be like a god coming down to earth and saying, if you breathe water, walk on 2 legs, and have purple skin, you may go to heaven. Doesn't my superior know their arn't any people like this? Thinking that he must know better, an obediant person would think at least 1 exists. So if at least 1 exists, and you look at everyone else and they have brown eyes, then you're the unlucky sap. In the rare case that no one had blue eyes, everyone would kill themselves that night. But since these cases weren't setup in the condition, they don't really matter. >>



Why would your superior make a rule if no one fits the rules? Someone asked Calbear if the superior had to explicitly state there was at least 1 blue, and Calbear said no.

But regardless, if they assumed there was at least 1 blue. And say there are actually 2 blues. So, (by your logic), one blue kills himself. We are left with 1 blue. That remaining blue can easily come to the conclusion that "there was only 1 blue, and he is dead now."



<< Q: What would stop a brown person from thinking there is 11 blues and he is one of them?

A: I can't figure this one out!! but maybe that is because I don't understand why it would take 10 nights for all 10 to kill themselves. I would think that each blue would see 9 people the first day, go to sleep and wake up seeing the same 9 people, he would then assume there must be 1 more, since he can't see 1 more he would know it is him and then all 10 would die the second night. So I must be missing something in the end process...

Nice teaser. >>



Reason you can't figure this out is because there are 2 equally logical ways for the brown/blues to think. One is the logic presented by the answer; but that would mean everyone dies. The other is the one I have presented, which means no one dies.

dfi

 

[alrocky]

Ha! My thanks, congratulations and apologies to MisterNi. I didn't before bother to read his post since it was a variant of the riddle, but he explained the answer a little bit better than Shooters or Calbear2000. I was a bit peeved at Calbear2000 since he didn't specifically explain what was wrong with my A), B), C), items 1), 2) and 3) scenarios or what it was I was missing. I think my "problem" with this riddle was that I assumed that Blue <U>might equal zero</U> , but Blue is greater than 1 - that is all Islanders see at least ONE BLUE Islander. Shooters's and Calbear2000's simplified explainations (that I had quoted before) were poorly stated, misleading and threw me of fthe track.

So can someone answer the question I posted last?: If for some reason the original 10 Blues did not die on the 10th night what would happen on the 11th night?

CalBear2000 you know the answer to that one, right?

 

[MereMortal]

<< So can someone answer the question I posted last?: If for some reason the original 10 Blues did not die on the 10th night what would happen on the 11th night? >>




Well, if we assume that there actually is an ultimate being, and it is not just a contrivance of the Browns to justify the elimination of the Blues because they are 'different', then the answer is simple.

The ultimate being will see their actions as seditious and will systematically expunge the lot of them.

 

[calbear2000]

<<

<< Solution: Don't read if you want to figure it out for yourself.



1. If only one person on the island has blue eyes then he can look around and see 99 pairs of brown eyes. Therefore, he knows that he must be the one with the blue eyes and kills himself that night.

2. If two people have blue eyes, then each can look around and see the other person's blue eyes along with the remaining 98 pairs of brown eyes. They both go to sleep that night thinking that the person they see with blue eyes will kill himself like in situation 1 above. However, when both wake up the next morning and see that the other person with blue eyes didn't kill himself, then they know that they themselves must have blue eyes also and therefore both blue-eyed people kill themselves on the second night.

3. If three people have blue eyes then the above reasoning can be used again except now two nights will pass and the three blue-eyed people will kill themselves on the third night.


So in general, all the blue-eyed people will kill themselves on the nth night where n is the number of blue-eyed people on the island. >>



That doesn't work for a number of reasons.

First, the riddle is poorly stated. No one knows for sure that there are blues, so as far as they know, it's possible that no one is blue. Let's say there are 2 blues; let's call them #1 and #2. After a night of rest, they awake to see each other, and everyone else, to be still alive. #1 can easily assume that #2 thinks no one is blue, and #2 can assume the same. So then no one dies.

So, your logic is flawed. Say there are 2 blues and they kill each other (even though they shouldn't. But let's just assume they do, as the answer states). Then, there are 98 ppl left. But after another night, they awake to see all 98 other ppl alive. By your logic (that ppl kill themselves cuz they see no one else has died), what's to prevent the 98th person from assuming that he has blue eyes when he doesn't? Then everyone will keep killing themselves until no one is left.

Also, with 3 blue-eyed people, it is again screwed up. All 100 ppl go to bed, including the 3 blues. Next morning, they wake up. Let's call the blues A, B, and C. A sees that B and C are still alive. He can easily come to the conclusion that B didn't kill himself because B sees C, and C didn't kill himself because C sees B. B can come to the conclusion that A and C are alive because A sees C, and C sees A. While C thinks that B and A are alive because A sees B, and B sees A.

Like I posted earlier, this riddle and its asnwer is flawed in the exact same way that the other similar riddle (involving hats) is flawed.

dfi

EDIT: just read Alrocky's posts, completely agree.

Calbear2000, I'm sorry, but the riddle is flawed. I've heard a variation of this before, came up with the exact same answer you have presented, and realized the answer was flawed.

Btw, I think you'll feel quite ridiculous for questioning Alrocky's intelligence, once you realize that you were the one in error. It's not that the question/answer is difficult to understand; it's that they are simply wrong. >>




Wow this thread is still alive...

Ok, don't get offended if you feel your intelligence is insulted... that is not my intent. I am just questioning your inability to understand the solution, as well as your need to blame your misunderstanding on the wording of the problem.

The solution is correct... on the 10th night, the 10 blue eyes commit suicide. You and the others before you (alrocky and woolmilk) bring up a valid point about the scenario of only 1, 2, or 3 blue eyes, which I agree with. BUT, as clearly stated in the original problem (which you seem to casually and conveniently disregard), there are 10 blues on the island. And I *guess* since shooter's solution describes these exact scenarios of 1, 2, and 3 blues, his answer is technically wrong. But you're being mighty anal if you refuse to realize he was using the case of 1, 2, and 3 as base cases to illustrate the recursive nature of the riddle when it involves 10 blues.

dfi, the only people that look ridiculous here is someone sticking to this reasoning to try and invalidate the answer.

 

[calbear2000]

<< dude,,, if i saw you i could look at you and instantly know of you are blue or brown easily >>



Yes but the problem didn't say that these people can come to this realization. For all we know, they can't use powers of observation.

Its a poorly worded riddle... we need to make all sorts of wild assumptions.

 

[dfi]

<<

<<

<< Solution: Don't read if you want to figure it out for yourself.



1. If only one person on the island has blue eyes then he can look around and see 99 pairs of brown eyes. Therefore, he knows that he must be the one with the blue eyes and kills himself that night.

2. If two people have blue eyes, then each can look around and see the other person's blue eyes along with the remaining 98 pairs of brown eyes. They both go to sleep that night thinking that the person they see with blue eyes will kill himself like in situation 1 above. However, when both wake up the next morning and see that the other person with blue eyes didn't kill himself, then they know that they themselves must have blue eyes also and therefore both blue-eyed people kill themselves on the second night.

3. If three people have blue eyes then the above reasoning can be used again except now two nights will pass and the three blue-eyed people will kill themselves on the third night.


So in general, all the blue-eyed people will kill themselves on the nth night where n is the number of blue-eyed people on the island. >>



That doesn't work for a number of reasons.

First, the riddle is poorly stated. No one knows for sure that there are blues, so as far as they know, it's possible that no one is blue. Let's say there are 2 blues; let's call them #1 and #2. After a night of rest, they awake to see each other, and everyone else, to be still alive. #1 can easily assume that #2 thinks no one is blue, and #2 can assume the same. So then no one dies.

So, your logic is flawed. Say there are 2 blues and they kill each other (even though they shouldn't. But let's just assume they do, as the answer states). Then, there are 98 ppl left. But after another night, they awake to see all 98 other ppl alive. By your logic (that ppl kill themselves cuz they see no one else has died), what's to prevent the 98th person from assuming that he has blue eyes when he doesn't? Then everyone will keep killing themselves until no one is left.

Also, with 3 blue-eyed people, it is again screwed up. All 100 ppl go to bed, including the 3 blues. Next morning, they wake up. Let's call the blues A, B, and C. A sees that B and C are still alive. He can easily come to the conclusion that B didn't kill himself because B sees C, and C didn't kill himself because C sees B. B can come to the conclusion that A and C are alive because A sees C, and C sees A. While C thinks that B and A are alive because A sees B, and B sees A.

Like I posted earlier, this riddle and its asnwer is flawed in the exact same way that the other similar riddle (involving hats) is flawed.

dfi

EDIT: just read Alrocky's posts, completely agree.

Calbear2000, I'm sorry, but the riddle is flawed. I've heard a variation of this before, came up with the exact same answer you have presented, and realized the answer was flawed.

Btw, I think you'll feel quite ridiculous for questioning Alrocky's intelligence, once you realize that you were the one in error. It's not that the question/answer is difficult to understand; it's that they are simply wrong. >>




Wow this thread is still alive...

Ok, don't get offended if you feel your intelligence is insulted... that is not my intent. I am just questioning your inability to understand the solution, as well as your need to blame your misunderstanding on the wording of the problem.

The solution is correct... on the 10th night, the 10 blue eyes commit suicide. You and the others before you (alrocky and woolmilk) bring up a valid point about the scenario of only 1, 2, or 3 blue eyes, which I agree with. BUT, as clearly stated in the original problem (which you seem to casually and conveniently disregard), there are 10 blues on the island. And I *guess* since shooter's solution describes these exact scenarios of 1, 2, and 3 blues, his answer is technically wrong. But you're being mighty anal if you refuse to realize he was using the case of 1, 2, and 3 as base cases to illustrate the recursive nature of the riddle when it involves 10 blues.

dfi, the only people that look ridiculous here is someone sticking to this reasoning to try and invalidate the answer. >>



It's not the wording of the problem that is the main concern. It is the answer that concerns me.

It's starting to get confusing go back and forth between everyone's post. So instead, I'm going to post how I see this riddle.

This is how I see the riddle:

One possible solution: There are 10 blues. The next morning, a blue awakes to see 9 blues and 90 browns. So, the blue can assume that because the 9 blues are still alive, he himself must be a blue. Therefore, this blue kills himself. HOWEVER, the only way the 10th blue can figure out that he himself is blue, is if he knows that there are 10 blues. Else, a brown can also come to the exact same conclusion as the blue, and therefore, all the browns will end up killing themselves as well.

Another possible solution: After the first night of rest, all 10 blues awake. A blue awakes to see 9 blues and 90 browns. However, this blue assumes that the 9 blues are alive because they don't know they are blue. The 10th blue cannot make a judgment on whether he himself is blue or brown on the ignorance of the blues. Since the 10th blue doesn't know there are 10 blues, he can easily assume that there are 9 blues, and that he is looking at them. Now, each blue can think the same thing, so no one dies.

Would someone care to explain where I've gone wrong?

dfi

Btw, Calbear: I didn't feel my intelligence was insulted. Your original attack was on Alrocky. It seemed to me that you were trying to belittle Alrocky, in a condescending "I can't believe this guy is so stupid that he doesn't understand the problem/answer" attitude. So I merely came to his defense. If that wasn't your intention, fine. That's just the impression I was left with.

 

[calbear2000]

Rather than analyzing your incorrect solutions, why don't you tell me what you don't understand about shooter's base cases bearing in mind that everyone on the island is well aware that there are multiple blues?

 

[Murphyrulez]

<< So instead, I'm going to post how I see this riddle.

This is how I see the riddle:

One possible solution: There are 10 blues. The next morning, a blue awakes to see 9 blues and 90 browns. So, the blue can assume that because the 9 blues are still alive, he himself must be a blue. Therefore, this blue kills himself. HOWEVER, the only way the 10th blue can figure out that he himself is blue, is if he knows that there are 10 blues. Else, a brown can also come to the exact same conclusion as the blue, and therefore, all the browns will end up killing themselves as well. >>



A brown will NEVER kill himself, because the day before it comes to that point, all the blues will have committed mass suicide. So the browns will know that they are brown.


<<

Another possible solution: After the first night of rest, all 10 blues awake. A blue awakes to see 9 blues and 90 browns. However, this blue assumes that the 9 blues are alive because they don't know they are blue. The 10th blue cannot make a judgment on whether he himself is blue or brown on the ignorance of the blues. Since the 10th blue doesn't know there are 10 blues, he can easily assume that there are 9 blues, and that he is looking at them. Now, each blue can think the same thing, so no one dies.
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After the 9th night and the 10th blue dude still sees 90 browns and 9 blues, he will KNOW that he is blue because otherwise the 9 blues will have killed themselves the night before.
You need to realize that all 100 people follow the same train of thought. They all think alike. They know that the other dudes are all going to kill themselves following the nth pattern.

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Would someone care to explain where I've gone wrong? >>




My pleasure....


<edit to remove some quoted previous message>

 

[Pundit]

Ok, aside from the "that night" part of the question being vague, the solution is totally workable.

In order for the blue-eyes to start killing themselves, they must know that the other blue-eyes have seen other blue eyes. That is why nobody dies if there is only 1 or 2 blue-eyed people.

Let's say there are 3 blue-eyed people now. You are one of them. You know that one has seen the other. You now think that he thinks there is only blue-eyed.

The next day, they are still alive. You now think that he thinks there are 2 blue-eyes and you think that he thinks himself and the other make up those 2.

The next day, they are still alive. You now know that he was expecting you and the other blue-eyed to kill yourselves, but you didn't. He now knows that he is blue-eyed as well, as you and the other blue-eyed do as well.

You all kill yourselves that night.

This works for 10 blue-eyed people as well.

 

[b0mbrman]

What threw me off was that the people on the island know how many people of each color eye there is on the island...