Logic

[CrazyIslander]

What I dont get is why did it take 100 days for everyone to figure out that they are the 100th member of a colour? Surely once they realise that no one knows the colour of their eyes then they look at the number of one colour is 100 and one is 99 not including you.

[Raggs]

What I dont get is why did it take 100 days for everyone to figure out that they are the 100th member of a colour? Surely once they realise that no one knows the colour of their eyes then they look at the number of one colour is 100 and one is 99 not including you.

How do they know that there's 100 people with brown eyes and 100 with blue? They're never told that.

[CrazyIslander]

What I dont get is why did it take 100 days for everyone to figure out that they are the 100th member of a colour? Surely once they realise that no one knows the colour of their eyes then they look at the number of one colour is 100 and one is 99 not including you.

How do they know that there's 100 people with brown eyes and 100 with blue? They're never told that.

They can see them. Why else would they realise how things are after 100 days?

[Woddy]

One hundred prisoners are lined up single file. A blue or red hat is placed on each of their heads randomly. The prisoners cannot see the colour of the hat on their own head, but they can see the colours of all the hats in front of them. The prisoner in the back can clearly see all 99 hats in front of him. The 50th prisoner in line can see the 49 hats in front of him, and the prisoner in the front of the line cannot see anything but the forest before him. Also, the prisoners don't know the proportion of red hats to blue hats in advance—it could be 50/50, but it could also be any combination that adds to 100.

A guard is going to walk down the line, starting in the back, and ask each prisoner what colour hat they have on. They can only answer "blue" or "red." If they answer incorrectly, or say anything else, they will be shot dead on the spot. If they answer correctly, they will be set free. Each prisoner can hear all of the other prisoners' responses, as well as any gunshots that indicate an incorrect response. They can remember all of this information.

Before the executions begin, the prisoners get to huddle up and make a plan. How can the prisoners ensure that the most people possible survive?

[Ymx]

What I dont get is why did it take 100 days for everyone to figure out that they are the 100th member of a colour? Surely once they realise that no one knows the colour of their eyes then they look at the number of one colour is 100 and one is 99 not including you.

How do they know that there's 100 people with brown eyes and 100 with blue? They're never told that.

They can see them. Why else would they realise how things are after 100 days?

They can’t see their own.

They only know based on what they see of each eye colour there might be either that number or that number plus one of each eye colour ie

99/100 of one colour
100/101 of the other colour

But they can’t deduce their own colour until time has played out.

[Ymx]

The interesting part of the puzzle for me is
All the islanders could already see there were at least one islander with blue eyes, so what difference did it make that the guru said something they already knew? Or wouldn’t it have mattered.

Presumably on day 101 the brown eyes would go too, seeing that the blue eyes had figured out who they were.

The guru, by naming only a single colour, confirms that colour exists, which is vital for the process. If he didn't exist, I couldn't know if my eyes were blue, just because someone else's were and they hadn't left.

The final 100 brown eyed people cannot use the same logic, they can be sure their eyes aren't blue, since the blues left too early for them to be, but they then can't know if they brown or green (or purple, or red) eyes. Only the confirmed colour can exist, and I believe, only confirming one colour allows it to work. Saying I see brown and blue eyes, wouldn't allow you to know if your eyes were blue or brown, and so the key day 1 logic, doesn't work.

EDIT - Hmm, need to think more about what would happen if the Guru said I see brown eyes and blue eyes. Maybe the logic could still hold up. However, for the original riddle, I'm not sure you could be sure of your own eye colour with the facts given.


EDIT EDIT - Spitballing. I see brown and blue eyes.

1 brown, 1 blue 100 red 1 green.

Brown leaves since he knows he's brown. Blue leaves since he knows he's blue.

1 brown, 2 blue, 100 red, 1 green.

Brown leaves night 1. 2 blue leave on night 2 following original logic.

2 brown, 2 blue, 100 red, 1 green.

All 4 leave on night 2.

3 brown, 2 blue, 100 red, 1 green

2 blue leave night 2, 3 brown leave night 3.

I think the Guru's a plum. If he'd said I see brown and blue eyes. Everyone else could have gotten off the Island (but for the Guru).

As these guys are perfectly logical, after seeing all blue eyes vanish you’d know they’d figured out their eye colour. So you’d be able to deduce you are not blue but brown.

Edit: actually that logic falls through because they’re not told they are only brown/blue, and could be red?

[Ymx]

One hundred prisoners are lined up single file. A blue or red hat is placed on each of their heads randomly. The prisoners cannot see the colour of the hat on their own head, but they can see the colours of all the hats in front of them. The prisoner in the back can clearly see all 99 hats in front of him. The 50th prisoner in line can see the 49 hats in front of him, and the prisoner in the front of the line cannot see anything but the forest before him. Also, the prisoners don't know the proportion of red hats to blue hats in advance—it could be 50/50, but it could also be any combination that adds to 100.

A guard is going to walk down the line, starting in the back, and ask each prisoner what colour hat they have on. They can only answer "blue" or "red." If they answer incorrectly, or say anything else, they will be shot dead on the spot. If they answer correctly, they will be set free. Each prisoner can hear all of the other prisoners' responses, as well as any gunshots that indicate an incorrect response. They can remember all of this information.

Before the executions begin, the prisoners get to huddle up and make a plan. How can the prisoners ensure that the most people possible survive?

Ok so the first guy should definitely just say the colour of the hat in front of him.

The next guy should hear this and save his own life.

Then same as the first guy the third guy should say the colour of the 4th guy.

Can’t think of another way. Would save at least 50%, but quite the sacrifice on the evens.

Am I close?

[Raggs]

One hundred prisoners are lined up single file. A blue or red hat is placed on each of their heads randomly. The prisoners cannot see the colour of the hat on their own head, but they can see the colours of all the hats in front of them. The prisoner in the back can clearly see all 99 hats in front of him. The 50th prisoner in line can see the 49 hats in front of him, and the prisoner in the front of the line cannot see anything but the forest before him. Also, the prisoners don't know the proportion of red hats to blue hats in advance—it could be 50/50, but it could also be any combination that adds to 100.

A guard is going to walk down the line, starting in the back, and ask each prisoner what colour hat they have on. They can only answer "blue" or "red." If they answer incorrectly, or say anything else, they will be shot dead on the spot. If they answer correctly, they will be set free. Each prisoner can hear all of the other prisoners' responses, as well as any gunshots that indicate an incorrect response. They can remember all of this information.

Before the executions begin, the prisoners get to huddle up and make a plan. How can the prisoners ensure that the most people possible survive?

Ok so the first guy should definitely just say the colour of the hat in front of him.

The next guy should hear this and save his own life.

Then same as the first guy the third guy should say the colour of the 4th guy.

Can’t think of another way. Would save at least 50%, but quite the sacrifice on the evens.

Am I close?

Going to spoiler my answer. I swear I haven't looked this up, and maybe there's a way to do better, but I reckon it'd be hard to do better than this:

Spoiler

Show

If they say that Blue = Odd and Red = Even, and they're counting the number of blue hats, then the first prisoner merely says blue/red based on what he counts in front of him.

Let's say 75 red and 25 blue, with the first prisoner wearing red.

He shouts Blue. Get's shot. No worries.

2nd prisoner has a red hat too, he can still see an odd number of blue hats, and therefore knows he must be wearing red. He survives.

3rd prisoner has a blue hat. He can see an even number of blue hats, and therefore knows he must be wearing a blue hat himself, says "Blue", survives.

4th prisoner with a red hat, can also see an Even number, therefore knows he must be red (remember, now one has shouted blue, the blues have switched to evens, not odds).

The man at the front can count whether or not Blue has been shouted an odd or even number of times (not counting the first man), to determine what his hat colour is. If the shout reaches him after an even number of people have shouted blue (not including the first man), then he is wearing blue

99 survivors guaranteed, 1st man has a 50/50.

[Ymx]

One hundred prisoners are lined up single file. A blue or red hat is placed on each of their heads randomly. The prisoners cannot see the colour of the hat on their own head, but they can see the colours of all the hats in front of them. The prisoner in the back can clearly see all 99 hats in front of him. The 50th prisoner in line can see the 49 hats in front of him, and the prisoner in the front of the line cannot see anything but the forest before him. Also, the prisoners don't know the proportion of red hats to blue hats in advance—it could be 50/50, but it could also be any combination that adds to 100.

A guard is going to walk down the line, starting in the back, and ask each prisoner what colour hat they have on. They can only answer "blue" or "red." If they answer incorrectly, or say anything else, they will be shot dead on the spot. If they answer correctly, they will be set free. Each prisoner can hear all of the other prisoners' responses, as well as any gunshots that indicate an incorrect response. They can remember all of this information.

Before the executions begin, the prisoners get to huddle up and make a plan. How can the prisoners ensure that the most people possible survive?

Ok so the first guy should definitely just say the colour of the hat in front of him.

The next guy should hear this and save his own life.

Then same as the first guy the third guy should say the colour of the 4th guy.

Can’t think of another way. Would save at least 50%, but quite the sacrifice on the evens.

Am I close?

Going to spoiler my answer. I swear I haven't looked this up, and maybe there's a way to do better, but I reckon it'd be hard to do better than this:

Spoiler

Show

If they say that Blue = Odd and Red = Even, and they're counting the number of blue hats, then the first prisoner merely says blue/red based on what he counts in front of him.

Let's say 75 red and 25 blue, with the first prisoner wearing red.

He shouts Blue. Get's shot. No worries.

2nd prisoner has a red hat too, he can still see an odd number of blue hats, and therefore knows he must be wearing red. He survives.

3rd prisoner has a blue hat. He can see an even number of blue hats, and therefore knows he must be wearing a blue hat himself, says "Blue", survives.

4th prisoner with a red hat, can also see an Even number, therefore knows he must be red (remember, now one has shouted blue, the blues have switched to evens, not odds).

The man at the front can count whether or not Blue has been shouted an odd or even number of times (not counting the first man), to determine what his hat colour is. If the shout reaches him after an even number of people have shouted blue (not including the first man), then he is wearing blue

99 survivors guaranteed, 1st man has a 50/50.

That is genius, I think.

[CrazyIslander]

Ok so the first guy should definitely just say the colour of the hat in front of him.

The next guy should hear this and save his own life.

Then same as the first guy the third guy should say the colour of the 4th guy.

Can’t think of another way. Would save at least 50%, but quite the sacrifice on the evens.

Am I close?

Going to spoiler my answer. I swear I haven't looked this up, and maybe there's a way to do better, but I reckon it'd be hard to do better than this:

Spoiler

Show

If they say that Blue = Odd and Red = Even, and they're counting the number of blue hats, then the first prisoner merely says blue/red based on what he counts in front of him.

Let's say 75 red and 25 blue, with the first prisoner wearing red.

He shouts Blue. Get's shot. No worries.

2nd prisoner has a red hat too, he can still see an odd number of blue hats, and therefore knows he must be wearing red. He survives.

3rd prisoner has a blue hat. He can see an even number of blue hats, and therefore knows he must be wearing a blue hat himself, says "Blue", survives.

4th prisoner with a red hat, can also see an Even number, therefore knows he must be red (remember, now one has shouted blue, the blues have switched to evens, not odds).

The man at the front can count whether or not Blue has been shouted an odd or even number of times (not counting the first man), to determine what his hat colour is. If the shout reaches him after an even number of people have shouted blue (not including the first man), then he is wearing blue

99 survivors guaranteed, 1st man has a 50/50.

That is genius, I think.