Mid-Week Brain Teaser

[estebanrey]

There is *not enough information* given to the first classroom to decide that that rule is true. We can only see that it's true objectively - a viewpoint that the kids do not have.

It clearly says ...

"The children can all see each other’s hats, but not their own."

In which case the first group of children can see there is "at least one red hat".

Put yourself in the shoes of a child in the first test. You can physically see all your classmates and can see a number of them are wearing red hats. So given you can see a number of children wearing red hats how do you not already know that "at least one hat is red"?

 

[estebanrey]

Because the first group could all have 'black' hats, so why could you ever be sure yours is 'red'. See my post above for a better explanation.

They could, but every child in the first case knows that isn't true because they can already see either 6 or 7 other kids wearing a red hat.

As I keep saying, I understand why the rule needs to know there is at least one red hat to work, what I don't understand is why the children in the first scenario can't rule out the possibility of the "all black hats" scenario immediately as soon as they see numerous children wearing red hats at the start.

 

[King Damager]

Probably been solved several time over by now, but surely it's simple?

Spoiler

As soon as you see 'all black hats', then you must be 'red'.

The reason for this, is that, if you weren't 'red' then the person before you would have solved it already.'

Start at the beginning, if person 1 sees all 'blacks' then they know they are the one 'red' hat. If they are unsure, then this indicates that they can still see at least 1 'red' hat. People keep being eliminated while they see 1 or more 'red' hats. As soon as someone can't see a 'red' hat, then they must be the last 'red' hat that was keeping the previous players unsure.

After that the remaining players know they are 'black'.

Yeah, that's right.

AJK, here you go. This one, I actually had more issue understanding, but it builds on from the red hat concept, and once you understand that, you should be able to understand this. (We were actually asked this one first, then told the red hat one to help us understand this one)

In an isolated village of cannibals, the following social customs are firmly established.

1) Each woman tells every other woman about the men who have flirted with her, except that (if a man is married) they never tell the man’s own wife. Nobody ever tells a woman that her husband has been in any way unfaithful.

2) A woman who discovers that her husband has been in any way unfaithful will eat him (the next day, before sunrise) for breakfast.

One day a missionary arrives and comments disapprovingly, in a sermon to all, that at least one male villager is unfaithful to his partner.

7 days later, before sunrise, 7 unfaithful husbands are eaten.

Why? What did the cannibals - specifically, the 7 wives - learn from the missionary?

kd

 

[wesimmo]

Probably been solved several time over by now, but surely it's simple?

Spoiler

As soon as you see 'all black hats', then you must be 'red'.

The reason for this, is that, if you weren't 'red' then the person before you would have solved it already.'

Start at the beginning, if person 1 sees all 'blacks' then they know they are the one 'red' hat. If they are unsure, then this indicates that they can still see at least 1 'red' hat. People keep being eliminated while they see 1 or more 'red' hats. As soon as someone can't see a 'red' hat, then they must be the last 'red' hat that was keeping the previous players unsure.

After that the remaining players know they are 'black'.

But nobody ever sees all black hats, the scenario states there are 7 red hats, so everybody will see at least 6 red hats.

But at the same time the kids don't know that there are only 7, just that there is at least 1.

 

[Bloomfield]

Yeah, you can't just ignore the one thing that's different...


Does this not make sense:

No, because in both classrooms the children know that there's atleast 6 red hats, whether they're told or not doesn't matter.

The only situation this would work is if there was only one red hat, or if the children couldn't see each other until they were asked AND ones already asked were removed from who they could see (so the last person with a red hat would see everyone in black hats, but not even know if a red hat exists).

As far as I can work out anyway.

I'm just going to go with classroom 1 being the "special" group .

 

[Bloomfield]

But nobody ever sees all black hats, the scenario states there are 7 red hats, so everybody will see at least 6 red hats.

But at the same time the kids don't know that there are only 7, just that there is at least 1.

You would ignore the hat that anyone who has already answered was wearing. Imagine that after answering they went out the room and then you checked everyone's hats again .

 

[wesimmo]

Sorry i don't understand your point there?

There must be at least 9 kids as the scenario states "some are black" meaning at least 2, and 7 are red.

So 9 or more kids are standing there and can see all the other hats, and all can see at least 6 red ones.

Unless the kids know that there are 7 red hats, which they don't, then the only condition that they know must be met, there is at least 1 red hat, is already met no matter what anybody answers.

 

[estebanrey]

The only situation this would work is if there was only one red hat

Even then this rule...

Spoiler

If I can see other children with red hats who haven't been asked the question by teacher yet I will always say "I don't know" otherwise I will say "red"

...still works because the children wearing black hats who are asked before they get to the one kid with the red one can see there is one red hatted kid that hasn't been asked so they'd say "I don't know". The sole kid with a red hat can only see all black hats so he says "red" under the rule and is correct.

The only time the above rule fails is if all the kids are wearing black hats, as the rule would dictate the first child would say "red" an be wrong. That's why I think King Damager and AJK think that "extra" piece of information in the second test is needed. But what they keep failing to grasp is that the "No red hats" possibility can be immediately eliminated by both groups as soon as they see at least one other kid wearing a red hat (which is true of all the children taking part).

 

[div0]

But nobody ever sees all black hats, the scenario states there are 7 red hats, so everybody will see at least 6 red hats.

But at the same time the kids don't know that there are only 7, just that there is at least 1.

As soon as you see 'all black hats' (left in the game). Not as the starting condition.

 

[AJK]

That's why I think King Damager and AJK think that "extra" piece of information in the second test is needed. But what they keep failing to grasp is that the "No red hats" possibility can be immediately eliminated by both groups as soon as they see at least one other kid wearing a red hat (which is true of all the children taking part).

We're not failing to grasp anything. The puzzle requires inductive logic, it cannot be solved based only on looking at the other hats. I keep saying that the "at least one red hat" is not observational, it is a piece of logical information required to begin the inductive reasoning process.

You keep insisting that you have a "rule" which both classrooms can apply - but where did it come from? It wasn't given in the statement of the puzzle, so you have created that rule from some kind of reasoning. What is your reasoning? No person in the first classroom can logically produce that rule.

 

[div0]

Even then this rule...

Spoiler

If I can see other children with red hats who haven't been asked the question by teacher yet I will always say "I don't know" otherwise I will say "red"

...still works because the children wearing black hats who are asked before they get to the one kid with the red one can see there is one red hatted kid that hasn't been asked so they'd say "I don't know". The sole kid with a red hat can only see all black hats so he says "red" under the rule and is correct.

The only time the above rule fails is if all the kids are wearing black hats. That's why I think King Damager and AJK think that "extra" piece of information in the second test is needed. But what they keep failing to grasp is that the "No red hats" possibility can be immediately eliminated by bothe groups as soon as they see at least one other kid wearing a red hat (which is true of all the children taking part).

Assuming all kids can see all hats, then I think you're right. Group 1 should be able to solve the puzzle too, assuming that the group starts with at least 2 'red' hats, or 2 'black' hats (ie everyone can visibly deduce that there was at least 1 'other' hat.)

I think the 'extra' bit of information is only required to guarantee that there can be a winner in all cases.

Or perhaps the OP meant that each person can only see the players 'in front of them', ie only the first player sees all hats. So players are in a line, looking forwards and you start at the back of the line. Game 2 would still be solvable, game 1 wouldn't be.

 

[Bloomfield]

Even then this rule...

Spoiler

If I can see other children with red hats who haven't been asked the question by teacher yet I will always say "I don't know" otherwise I will say "red"

...still works because the children wearing black hats who are asked before they get to the one kid with the red one can see there is one red hatted kid that hasn't been asked so they'd say "I don't know". The sole kid with a red hat can only see all black hats so he says "red" under the rule and is correct.

The only time the above rule fails is if all the kids are wearing black hats, as the rule would dictate the first child would say "red" an be wrong. That's why I think King Damager and AJK think that "extra" piece of information in the second test is needed. But what they keep failing to grasp is that the "No red hats" possibility can be immediately eliminated by both groups as soon as they see at least one other kid wearing a red hat (which is true of all the children taking part).

But the kid with the red hat wouldn't even know that there are red hats to begin with in that situation, so whilst he'd be right, he wouldn't KNOW he was right, which is the point . I meant that it'd work as in classroom 1 can't work it out and classroom 2 could, not that neither could, if you were referring to the second being able to work it out.

 

[div0]

The only time the above rule fails is if all the kids are wearing black hats, as the rule would dictate the first child would say "red" an be wrong.

Not quite true. If you're the only 'red' hat, then you only see all 'blacks' from the start. So from your point of view, the game could be 'all blacks'. So saying 'red' would be a guess.

However, if the game starts with 2 or more 'reds', then everyone sees that there is at least 1 red, gaining the same information as game 2, and so should be solvable.

I'm not sure if the OP intended the game to be a 'line', where you only see the players standing in front of you and the player at the back of the line starts. In this sort of game, the 'extra' information is crucial, as you don't see all hats at the start.

 

[wesimmo]

As soon as you see 'all black hats' (left in the game). Not as the starting condition.

Still doesn't help me!

Just for clarity, i understand the 1st classroom cannot answer.

I am also making the assumption that all kids have perfect knowledge in that they know what all the other kids are wearing, irrespective of the order in which they answer, or indeed who has already answered.

There are 7 red hats, but that is unknown to the kids, BUT they all know there are at least 6, they can see that.

In classroom 2, they are told at least 1 is red, but that isn't any additional information to what classroom 1 knew because in both rooms they can see at least 6, so the "additional information" doesn't actually increase their knowledge.

 

[rubberduck]

div0 gave by far the best explanation of the solution which was succinct and clear.

OP comes over as a bit of a tool in #86 with his "OMG why cant you thickos understand tone". Its not really suprising that if you are doing a maths degree you might be better equiped to understand such problems than the lay-person.. Still I suppose some people think that a couple of years at uni suddenly elevates them to the upper echelons of human intellect.

 

[div0]

Not quite true. If you're the only 'red' hat, then you only see all 'blacks' from the start. So from your point of view, the game could be 'all blacks'. So saying 'red' would be a guess.

However, if the game starts with 2 or more 'reds', then everyone sees that there is at least 1 red, gaining the same information as game 2, and so should be solvable.

I'm not sure if the OP intended the game to be a 'line', where you only see the players standing in front of you and the player at the back of the line starts. In this sort of game, the 'extra' information is crucial, as you don't see all hats at the start.

In fact, you might need at least '3 reds' at the start of game 1 - because the game relies on all players knowing that there is at least one red. If there are only '2 reds', then each of the two red players are unsure of their own colour, which means that from their point of view the other 'red' might be looking at 'all blacks'.

 

[div0]

There are 7 red hats, but that is unknown to the kids, BUT they all know there are at least 6, they can see that.

In classroom 2, they are told at least 1 is red, but that isn't any additional information to what classroom 1 knew because in both rooms they can see at least 6, so the "additional information" doesn't actually increase their knowledge.

You're correct here and I actually think that game 1 is solvable too. The difficulty is that in game 2 the teachers bit of information is actually more subtle than it may first appear.

By saying that there is at least 1 red, then it provides a situation not just where all players know there is a red, but also where they know that all other players know there is a red. A subtle but important difference.

In order to achieve this knowledge in game 1, I think there needs to be at least 3 'reds' (and at least 3 'blacks) in the starting condition. Which I believe there is in the OP's example and so it should be solvable using the same logic as game 2.

 

[estebanrey]

You keep insisting that you have a "rule" which both classrooms can apply - but where did it come from? It wasn't given in the statement of the puzzle, so you have created that rule from some kind of reasoning. What is your reasoning? No person in the first classroom can logically produce that rule.

If no person in the first classroom could deduce the rule then no person in the second class could either as there has been no new 'information' introduced to them.

The first class can see there is "at least one red hat" so it is a redundant statement. It's like me showing you this picture...

And afterwards telling you "there is at least one £1 coin there". You'd be right to think "well I can see that already", I've provided you with no new information with that statement.

But if you want me to go through this step by step using the first class and their logic I will. So for my explanation let's say this is the set up....

Child 1 = RED HAT
Child 2 = BLACK HAT
Child 3 = RED HAT
Child 4 = RED HAT
Child 5 = BLACK HAT
Child 6 = BLACK HAT
Child 7 = BLACK HAT
Child 8 = RED HAT
Child 9 = RED HAT
Child 10 = RED HAT
Child 11 = BLACK HAT
Child 12 = BLACK HAT
Child 13 = RED HAT
Child 14 = BLACK HAT
Child 15 = BLACK HAT

Now without any further information from teacher (i.e. the 'at least one red hat' statement) let's look at what each child knows.

Obviously they do not know their own hat colour but they do know....

* Children 1, 3, 4, 8, 9, 10 & 13 know there are at least 6 red hats (and don't know if they are the 7th)
* Children 2, 5, 6, 7, 11, 12, 14 & 15 know there are at least 7 red hats (and don't know if they are the 8th)

When the game plays Child 1 has no previous information when it comes to them being asked. Child 2 knows what Child 1 answered and that's all. Child 3 knows what Child 2 and Child 1 answered...and so on.

When asked, Child 1 knows there are 6 children left to asked wearing red hats and 8 children with black hats left to be asked and by definition this means they also have no information about their own hat. So working our way through the children and their options...

Child 1 = 6 Red Hats Left, 8 Black
Child 2 = 6 Red Hats Left, 7 Black
Child 3 = 5 Red Hats Left, 7 Black
Child 4 = 4 Red Hats Left, 7 Black
Child 5 = 4 Red Hats Left, 6 Black
Child 6 = 4 Red Hats Left, 5 Black
Child 7 = 4 Red Hats Left, 4 Black
Child 8 = 3 Red Hats Left, 4 Black
Child 9 = 2 Red Hats Left, 4 Black
Child 10 = 1 Red Hat Left, 4 Black
Child 11 = 1 Red Hat Left, 3 Black
Child 12 = 1 Red Hat Left, 2 Black
Child 13 = 0 Red Hats Left, 2 Black
Child 14 = N/A
Child 15 = N/A

So child 13 knows that there are no more children left to ask with a red hat on but there are still 2 kids left wearing black hats. Because child 13 knows that child 10 said "I don't know" and that was the last person wearing a red hat who didn't have enough information to answer with certainty then they must be wearing a red hat.

But I'll say it again, in both scenarios both sets of children know from the start there is at least one red hat, telling them this before the game starts adds no new information and therefore cannot make or break the inductive logic that has to be used for Child 13 to guess correctly. It is only necessary to an example where you have 0 or (as Bloomfield correctly pointed out) 1 red hat. The minute you introduce 2 or more children wearing a red hat, all then of the children know there is 'at least 1 red hat' at the start of the game by definition regardless of the one they are wearing.

 

[BunnyKillBot]

Probably been solved several time over by now, but surely it's simple?

Spoiler

As soon as you see 'all black hats', then you must be 'red'.

The reason for this, is that, if you weren't 'red' then the person before you would have solved it already.'

Start at the beginning, if person 1 sees all 'blacks' then they know they are the one 'red' hat. If they are unsure, then this indicates that they can still see at least 1 'red' hat. People keep being eliminated while they see 1 or more 'red' hats. As soon as someone can't see a 'red' hat, then they must be the last 'red' hat that was keeping the previous players unsure.

After that the remaining players know they are 'black'.

That is what I thought at first, but on reflection I don't think it's a valid solution, because it assumes elimination. A response of 'I don't know' does not cause an elimination because it is the same, implied, information state prior to the asking of the question. The pool of red hat wearers does not tend to 1 and the pool is not aware the total sum of red hats.

Your given solution only works if there is exactly one red hat, and the wearer can only see all black hats.

 

[zenf]

In an isolated village of cannibals, the following social customs are firmly established.

1) Each woman tells every other woman about the men who have flirted with her, except that (if a man is married) they never tell the man’s own wife. Nobody ever tells a woman that her husband has been in any way unfaithful.

2) A woman who discovers that her husband has been in any way unfaithful will eat him (the next day, before sunrise) for breakfast.

One day a missionary arrives and comments disapprovingly, in a sermon to all, that at least one male villager is unfaithful to his partner.

7 days later, before sunrise, 7 unfaithful husbands are eaten.

Why? What did the cannibals - specifically, the 7 wives - learn from the missionary?

If there was only one unfaithful husband, all of the women of the village would be aware of him other than his wife. As soon as the missionary says that someone is unfaithful, she'll know it's her husband and will eat him.

If there are two unfaithful husbands, all of the women in the village will be aware of them other than their two wives, who each believe there to be only one unfaithful husband (the other one). So when the next morning arrives and the other wife hasn't eaten her husband, they will realise the above case was incorrect and that there must be more than one unfaithful husband (their own). So, on the second day, both of the wives eat their husbands.

If there are three unfaithful husbands, their three wives will believe there to be only two unfaithful husbands. So if on the second day the two other wives don't eat their husbands, they'll know that the above case was incorrect and that their husband must be the third unfaithful husband.

If there are seven unfaithful husbands, their wives will believe there to be only six unfaithful husbands on the village. If, on the sixth day, they haven't been eaten, the wives know that their husband is the seventh.