pretty tough/hard riddle

[looseswing]

Friend showed me this, see if you guys can reason it out. I have not figured it out yet, he's going to tell me the answer tomorrow if I have not gotten it by then:

200 men and a queen are living on an island. Queen has brown eyes, 100 men have blue eyes and the other 100 have green eyes. Nobody knows the color of his own eyes but he can see the other people living on the island, so a green eyed man will see 99 other green eyes and 100 blue eyes. No communication is allowed between the men so no one can simply tell the others the color of their eyes using any kind of communication. Also, No one knows that there's 100 men with green eyes and 100 with blue eyes except the queen, so every man might think his eyes are brown, black or whatever. There's only one rule in the island:

-If someone is sure of the color of his eye, he has to leave the island in the same day!

The queen wakes up everyday and makes one announcement:

"There's at least one man with green eyes", if there's at least one man with green eyes on the island.
"There are no men with green eyes", if there are no green eyed men on the island.

The queen never lies.

The question here is: Who will leave the island and when ?

---------------------------------------------------------

Hint: The Answer is not "No one will ever leave". I promise!

Clarifications:

1- No one can use any kind of material to know the color of his eyes(mirros, water, etc)

2- Queen makes sure that everyone hears the announcment everyday.

3- No one can ask the queen anything.

 

[SuperFly]

I got it.






















The men execute a coup against the queen because she's making their lives harder by giving them a stupid riddle while they are stuck on a deserted island. The men then proceed to build shelters, gather food and make a rescue fire.

 

[cucio]

The queen. As soon as possible. No woman could endure sharing an island with 200 horny baastards.

 

[TheJRK]

Is this the finale from "Lost"? Don't spoil it for me, I'm still on Season 3.

 

[Kenny022593]

Is this the finale from "Lost"? Don't spoil it for me, I'm still on Season 3.

Magnets

10char

 

[sureshs]

Can any guy remember everyone and count to see how many have what eye color? If he counts only 99 of a color, that means he has that color?

 

[ProgressoR]

think i figured it out.....do you want me to post my guess on here? or let others have time to ponder on it?

 

[TheJRK]

I had two other guys at work helping me out with this and we were drawing it out on a dry-erase board.

I think we figured it out.

 

[looseswing]

Well I put this up so people coul guess and collaborate on it here, thought it would be a fun little activity for the board. And he can't be sure that he has a certain color if he counts 99 of it because remember he could have black/brown etc eyes and does not know that there 100 blue and 100 green.

 

[ProgressoR]

ok here are my thoughts. Nobody will leave on the first day because some will see 100 greens and some will see 99 greens.

Actually i might change my thoughts as i write this....

Imagine there are 2 greens (g1 and g2) and 100 blues. Bear with me.
The blues see 99 blues and 2 greens, and wouldnt leave as they have no idea of their own colour.

The greens will see 1 green and 100 blues. So will not leave as that doesnt tell them their own colour.

on the second day, again queen says there is at least one green. G1 can still only see 1 green, so can G2.

But G1 knows g2 didnt leave yesterday, but if g2 did not see any greens yesterday, he would have left as g2 would have known he was the green, so g1 knows g2 saw at least one other green. But g1 sees no other greens, so today g1 knows he must be the other green. He will leave on the second day. Similarly, G2 goes through the same process and he would also leave the second day. The 100 blues stay where they are. The next day queen says no greens and no one leaves because they know they are not green and see only 99 other blues, so cannot know their own colour.

To extrapolate, imagine there are N greens. if n=2 they both leave on second day. the same process tells us if n=3 then they all leave on the 3rd day. So if there n=100, then all 100 greens should leave on the 100th day.

Nobody will leave after that day.

that is what my initial thoughts are....

comments?

 

[fed_the_savior]

Theoretically, if all the men wanted to know how many had green eyes, and they were logical geniuses, they would know there was only one sure-fire way to find out without communicating, and that is to use a system where they wait as many days to leave as they see green eyed men. That way the men that actually have green eyes and would see the lowest number of green eyed men, would all leave together on the soonest possible day, which would be the 100th day in this case. The leftover men would never leave, because even though they all know the others are blue eyed, they would never know for sure what color they had themselves, they would just know "There are no men with green eyes." But this system requires that the men have a desire to find out how many green eyed there are. If they don't care, they will won't follow the plan, and they will never know.

 

[Jennifer]

Queen is the only one that leaves the island. She is the only one who knows her eye color.

-Jennifer

 

[looseswing]

ok here are my thoughts. Nobody will leave on the first day because some will see 100 greens and some will see 99 greens.

Actually i might change my thoughts as i write this....

Imagine there are 2 greens (g1 and g2) and 100 blues. Bear with me.
The blues see 99 blues and 2 greens, and wouldnt leave as they have no idea of their own colour.

The greens will see 1 green and 100 blues. So will not leave as that doesnt tell them their own colour.

on the second day, again queen says there is at least one green. G1 can still only see 1 green, so can G2.

But G1 knows g2 didnt leave yesterday, but if g2 did not see any greens yesterday, he would have left as g2 would have known he was the green, so g1 knows g2 saw at least one other green. But g1 sees no other greens, so today g1 knows he must be the other green. He will leave on the second day. Similarly, G2 goes through the same process and he would also leave the second day. The 100 blues stay where they are. The next day queen says no greens and no one leaves because they know they are not green and see only 99 other blues, so cannot know their own colour.

To extrapolate, imagine there are N greens. if n=2 they both leave on second day. the same process tells us if n=3 then they all leave on the 3rd day. So if there n=100, then all 100 greens should leave on the 100th day.

Nobody will leave after that day.

that is what my initial thoughts are....

comments?

I thought the same at first. However, think about it this way in the same scenario (g1 and g2 with 100 blue eyes).

Day 1: queen makes the announcement
g1: looks around and sees g2, and the blues.
g2: knows that g1 must have seen someone with green eyes because otherwise he would have left himself. He knows that he must be that someone because everyone besides him and g1 have blue eyes. He leaves.

Day 2: queen makes the announcement
g1 leaves as everyone else has blue eyes.

At this point I'm not really sure how to extrapolate out though.

 

[aceX]

I also came to the independent conclusion that all the greens will leave on the 100th day.

 

[aceX]

I thought the same at first. However, think about it this way in the same scenario (g1 and g2 with 100 blue eyes).

Day 1: queen makes the announcement
g1: looks around and sees g2, and the blues.
g2: knows that g1 must have seen someone with green eyes because otherwise he would have left himself. He knows that he must be that someone because everyone besides him and g1 have blue eyes. He leaves.

Day 2: queen makes the announcement
g1 leaves as everyone else has blue eyes.

At this point I'm not really sure how to extrapolate out though.

I don't think g2 will know that g1 doesn't know until the next day, but I'm not sure on this

 

[ProgressoR]

I thought the same at first. However, think about it this way in the same scenario (g1 and g2 with 100 blue eyes).

Day 1: queen makes the announcement
g1: looks around and sees g2, and the blues.
g2: knows that g1 must have seen someone with green eyes because otherwise he would have left himself. He knows that he must be that someone because everyone besides him and g1 have blue eyes. He leaves.

Day 2: queen makes the announcement
g1 leaves as everyone else has blue eyes.

At this point I'm not really sure how to extrapolate out though.

problem is the timing...if g2 leaves on day 1, then does g1 have time to process this and leave at the same time, because he will work out he must be green the second g2 leaves, he doesnt need to wait till next day. I am assuming that the leaving part happens or it doesnt, not that others can see them leave and then process something and leave immediately after them.

if that is the case, g1 has no time to process that g2 hasnt left, presumably he only finds out the next day - and i think that is a sensible approach, otherwise we get into a circular situation

 

[ProgressoR]

I don't think g2 will know that g1 doesn't know until the next day, but I'm not sure on this

yep that is the conclusion i came to also

 

[ProgressoR]

Queen is the only one that leaves the island. She is the only one who knows her eye color.

-Jennifer

lol smart answer, however being pedantic, and why not be pedantic, what makes you assume she knows her own eye colour?

and the question posited that if someone knows their eye colour, then HE must leave the island.

 

[aceX]

I thought the same at first. However, think about it this way in the same scenario (g1 and g2 with 100 blue eyes).

Day 1: queen makes the announcement
g1: looks around and sees g2, and the blues.
g2: knows that g1 must have seen someone with green eyes because otherwise he would have left himself. He knows that he must be that someone because everyone besides him and g1 have blue eyes. He leaves.

Day 2: queen makes the announcement
g1 leaves as everyone else has blue eyes.

At this point I'm not really sure how to extrapolate out though.

Day 1: queen makes the announcement (at least one green)
g1: looks around and sees g2, and the blues.
g2: looks around and sees g1, and the blues.
g1: knows that g2 must have seen someone with green eyes because otherwise he would have left himself.
g2: knows that g1 must have seen someone with green eyes because otherwise he would have left himself.

So they both leave as soon as the announcement is made.
But if they both leave as soon as the announcement is made they wouldn't know to leave because they wouldn't see the other person wait.

So they must only know if a person knows or not on the next day.

So for 2gs 100 bs the 2 gs leave on the 2nd day.
For the original, the 100gs leave on the 100th day

 

[aceX]

lol smart answer, however being pedantic, and why not be pedantic, what makes you assume she knows her own eye colour?

and the question posited that if someone knows their eye colour, then HE must leave the island.

The first two answers I wrote down were:
- Queen is a man and leaves the Island
- Queen tells them all their colour without them asking

But because this goes against the spirit of the riddle I decided there would be a more satisfying answer.

 

[TheJRK]

Basically we narrowed it down to 3 dudes (for simple math reasons) and we said 2 of us had green eyes and one dude had blue eyes.

Day 1

Queen says "I see at least one mofo with green eyes."

Me - I see (dude 1) with green eyes and (dude 2) with blue eyes

Nobody leaves the island as we don't know what color our eyes are.

Day 2

Me - I see (dude 1) with green eyes and (dude 2) with blue eyes.

They are still on this f***ing island with me! That means that each one of those dudes also saw someone with green eyes. Since (dude 1) can't see himself and I know he saw at least one other person with green eyes, it has to be me since (dude 2) has blue eyes.

I now know my eye color and therefore must leave the island. (dude 1) sees me leaving the island and thinks "that guy must have seen someone with green eyes otherwise he would have left on day 1". (dude 1) then realizes that he has to have green eyes since (dude 2) has blue eyes. (dude 1) then gets the sand out of his ass and follows me to leave the island as well.

Basically, (whatever) the number of people with green eyes will take an equal number of days to figure out what their eye color is. In this case, 100 days.

I guess the blue eyed people get to stay since they have no f***ing idea what color their eyes are.

 

[JohnnyCracker]

All the green guys will leave on the first day. Here's why. Each of the green guys looks around and see only blue and green eyes, except for the queen. Each will realize that she is the exception. Each will count and see that there are 200 men, self-included, with 100 blues and 99 greens. Each green guy will see the undeniable pattern and realize that he has green eyes himself to make it even 100 blues and 100 greens.

On second thought. They all leave the same day because all the blue guys would figure out their own eye color using the same observation and logic.

This is not a math problem. It is a rational problem. Once each man reasons it out, he will realize that his eyes can only be either blue or green and figure it out from there with confident. (that this queen has a fetish for men with blue or green eyes)

BTW, one just has to be "sure" (confident) about his eye color in order to leave the island, not necessarily "correct", right? It probably doesn't even matter either way.

 

[looseswing]

Well, just for anyone who wants to know the answer, just found it out:


100 green eyed men will leave on the 100th day.

Explanation:

Let's assume there's only one man with green eyes and 100 men with blue eyes:

The queen announces that there's at least one man with green eyes. But, the man with green eyes can only see 100 blue eyed men. He concludes that he has the green eyes and he will leave on the first day.

Fact #1: If there's only 1 man with green eyes, he will leave on the first day.

Now, Let's assume there are 2 men with green eyes and 100 men with blue eyes:

First day: The queen announces that there's at least one man with green eyes. Both of the green eyed men sees one other man with green eyes so they are not sure of their eye color and they will wait.

Second day: They will both think "If I didn't have green eyes, the other green eyed man would've left on the first day(Fact #1). Since he waited to the second day, this means he saw a green eyed man which made him wait. But, I only see 1 green eyed man so the other green eyed man must be me!" and they will both leave on the second day.

Fact #2: If there are only 2 men with green eyes, they will both leave on the second day.

This is where it gets harder, 3 men with green eyes and 100 men with blue eyes:

First day: The queen announces that there's at least one man with green eyes. all of the green eyed men sees 2 other green eyed men so all of them will wait to the second day.

Second day: Again, they all see 2 other green eyed men so they will wait.

Third day: All of them will think "If I didn't have green eyes, the other 2 green eyed men should've used Fact #2 and left on the second day. Since they waited to the third day, this means there's a third man with green eyes which made them wait. But, I can only see 2 men with green eyes so the third one must be me! and they will all leave on the third day.

Fact #3: If there are only 3 men with green eyes, all of them will leave on the third day.

Assume there are 4 green eyed men and 100 blue eyed men:

First 3 days: All of the green eyed men sees 3 other green eyed men, they will just hear the announcements and wait.

Day 4: All of the green eyed men will think "If I didn't have green eyes, the other 3 green eyed men should've left on the 3rd day(Fact #3). Since they waited to the 4th day, this means there's a 4th person with green eyes which made them wait. But, I can only see 3 green eyed men so the 4th green eyed man must be me!" and they will all leave on the 4th day.

Finally, Assume there are N green eyed men and 100 blue eyed men:

First N-1 Days they will all just hear the announcements and wait.

Day N: All of the green eyed men will think "If I didn't have green eyes, the other N-1 green eyed men should've left on the (N-1)th day(Fact #N-1). Since they waited to the Nth day, this means there's an Nth person with green eyes which made them wait. But, I can only see N-1 green eyed men so the Nth green eyed man must be me!" and they will all leave on the Nth day.

As for the blue eyed men, they will never leave, simply because they don't know if they have blue, red or brown eyes since the only information is given about green eyes.

Btw, only the first announcement was enough for all of the green eyed men to conclude their eye color, all the other announcements made from Day 2 to Day N are worthless.

 

[TheJRK]

Okay here's an easy one...

There is a man in prison. The warden tells him he will let him go free if can solve this puzzle.

On one end of the prison is a room (with no windows) with three light switches.

On the opposite end of the prison there is another room (with no windows) with three lights in it.

The prisoner must start in the first room (with the switches). He can flip as many switches as he wants. Then he must head over to the second room (with the lights). He must tell the warden which switch corresponds to which light.

-----
Rules:


The prisoner cannot return to the first room, once he has left.

This is not a "glass prison" i.e. the prisoner cannot see one room while he is in the other.

No he does not bribe the guards to tell him.

He does not tell anyone "to get busy living or get busy dying".

 

[athiker]

:idea: Do I have a ladder?

 

[Djlpenguin]

got it! There are 200 men so they all die of thirst. no water, right?
the queen then dies of lonelyness or thrirst

 

[Kenny022593]

What if this riddle was with women instead of men?

 

[cucio]

On one end of the prison is a room (with no windows) with three light switches.

On the opposite end of the prison there is another room (with no windows) with three lights in it.

This'd better be a small prison and the temperature in the second room not too low, otherwise the guy is screwed. ;-)

 

[statto]

Well, just for anyone who wants to know the answer, just found it out:


100 green eyed men will leave on the 100th day.

Explanation:

Let's assume there's only one man with green eyes and 100 men with blue eyes:

The queen announces that there's at least one man with green eyes. But, the man with green eyes can only see 100 blue eyed men. He concludes that he has the green eyes and he will leave on the first day.

Fact #1: If there's only 1 man with green eyes, he will leave on the first day.

Now, Let's assume there are 2 men with green eyes and 100 men with blue eyes:

First day: The queen announces that there's at least one man with green eyes. Both of the green eyed men sees one other man with green eyes so they are not sure of their eye color and they will wait.

Second day: They will both think "If I didn't have green eyes, the other green eyed man would've left on the first day(Fact #1). Since he waited to the second day, this means he saw a green eyed man which made him wait. But, I only see 1 green eyed man so the other green eyed man must be me!" and they will both leave on the second day.

Fact #2: If there are only 2 men with green eyes, they will both leave on the second day.

This is where it gets harder, 3 men with green eyes and 100 men with blue eyes:

First day: The queen announces that there's at least one man with green eyes. all of the green eyed men sees 2 other green eyed men so all of them will wait to the second day.

Second day: Again, they all see 2 other green eyed men so they will wait.

Third day: All of them will think "If I didn't have green eyes, the other 2 green eyed men should've used Fact #2 and left on the second day. Since they waited to the third day, this means there's a third man with green eyes which made them wait. But, I can only see 2 men with green eyes so the third one must be me! and they will all leave on the third day.

Fact #3: If there are only 3 men with green eyes, all of them will leave on the third day.

Assume there are 4 green eyed men and 100 blue eyed men:

First 3 days: All of the green eyed men sees 3 other green eyed men, they will just hear the announcements and wait.

Day 4: All of the green eyed men will think "If I didn't have green eyes, the other 3 green eyed men should've left on the 3rd day(Fact #3). Since they waited to the 4th day, this means there's a 4th person with green eyes which made them wait. But, I can only see 3 green eyed men so the 4th green eyed man must be me!" and they will all leave on the 4th day.

Finally, Assume there are N green eyed men and 100 blue eyed men:

First N-1 Days they will all just hear the announcements and wait.

Day N: All of the green eyed men will think "If I didn't have green eyes, the other N-1 green eyed men should've left on the (N-1)th day(Fact #N-1). Since they waited to the Nth day, this means there's an Nth person with green eyes which made them wait. But, I can only see N-1 green eyed men so the Nth green eyed man must be me!" and they will all leave on the Nth day.

As for the blue eyed men, they will never leave, simply because they don't know if they have blue, red or brown eyes since the only information is given about green eyes.

Btw, only the first announcement was enough for all of the green eyed men to conclude their eye color, all the other announcements made from Day 2 to Day N are worthless.

The problem with the solution is that it assumes all the men are logical geniuses. In a real world situation none of them would ever leave.

Okay here's an easy one...

There is a man in prison. The warden tells him he will let him go free if can solve this puzzle.

On one end of the prison is a room (with no windows) with three light switches.

On the opposite end of the prison there is another room (with no windows) with three lights in it.

The prisoner must start in the first room (with the switches). He can flip as many switches as he wants. Then he must head over to the second room (with the lights). He must tell the warden which switch corresponds to which light.

-----
Rules:


The prisoner cannot return to the first room, once he has left.

This is not a "glass prison" i.e. the prisoner cannot see one room while he is in the other.

No he does not bribe the guards to tell him.

He does not tell anyone "to get busy living or get busy dying".

Spoiler (Highlight to reveal answer):

If we call the switches A, B and C.

The prisoner switches on A and B, and leaves C off. After ten minutes he switches B off.

Then he goes to the three rooms. Whichever room has the light on is paired with A. Whichever room has a warm bulb is B. Whichever room has a cold bulb is C.

 

[xr3fgb]

^^^^^^^
2 problems with your answer:

(1) All three lights are in ONE room (@athiker, you are right, this is probably irrelevant); and
(2) The riddle did not state that the on/off positions are indicated on the switches (i.e., "up" can be on or off) (@athiker, but this is the crucial assumption).

 

[athiker]

Spoiler (Highlight to reveal answer):

Love the "Highlight to reveal answer" trick...I'll have to remember that!

Agree w/ your answer btw (and cucio's).

EDIT: It doesn't matter if they are in the same room or not, the solution still works. Most light switches are labeled on/off...if not then...

 

[LameTennisPlayer]

just a thought, if one of them is blind he would never know what color their eyes are, what if they are all blind lol

 

[looseswing]

The problem with the solution is that it assumes all the men are logical geniuses. In a real world situation none of them would ever leave.



Spoiler (Highlight to reveal answer):

If we call the switches A, B and C.

The prisoner switches on A and B, and leaves C off. After ten minutes he switches B off.

Then he goes to the three rooms. Whichever room has the light on is paired with A. Whichever room has a warm bulb is B. Whichever room has a cold bulb is C.

Was this right?

 

[looseswing]

The problem with the solution is that it assumes all the men are logical geniuses. In a real world situation none of them would ever leave.



Spoiler (Highlight to reveal answer):

If we call the switches A, B and C.

The prisoner switches on A and B, and leaves C off. After ten minutes he switches B off.

Then he goes to the three rooms. Whichever room has the light on is paired with A. Whichever room has a warm bulb is B. Whichever room has a cold bulb is C.

Was this right?

 

[TheJRK]

Yeah, that was the right answer.