Mid-Week Brain Teaser

[King Damager]

I've thought it over and over and I just don't think it's possible unless they know that there's 7 red hats or they cheat such as the other class telling them (which isn't really a puzzle) If they do know there is 7 red hats then it's quite simple

It is, there is a small, but important difference in the first instance compared to the second. It is this that allows the students in the second class to know

Is the teacher wearing a hat? Or are all the kids girls?

It doesn't really matter

Mirror.
Are the children in the second class also told that there are seven hats? The brackets around the word seven in the first paragraph doesn't make it clear (to me) if they are aware that there are seven.

Assuming the second group knows that there are seven the the girl counts all of the other hats which she can see and therefore how many of each colour?

How many hats per child : p

No, neither class knows that there are seven hats.

kd

 

[estebanrey]

I don't understand what extra information is in the second example when the teacher says "at least one hat is red". If the kids can all see each other's hats then by definition they know that already (as did the first group, thus there is no new information in the second trial).

I'm going with the 'missing piece if information from your question' hypothesis for the time being.

 

[Robert]

I don't understand what extra information is in the second example when the teacher says "at least one hat is red". If the kids can all see each other's hats then by definition they know that already (as did the first group, thus there is no new information in the second trial).

I'm going with the 'missing piece if information from your question' hypothesis for the time being.

This.

 

[TurnipHead]

Can they communicate non verbally?

They could all group together, and one child could push all the black kids away. Then the remaining kids can decide if the original kid can stay.

 

[Psiko]

This is very similar to the green eyes/blue eyes problem (you can search the forums for it).

Here are some hints:

Spoiler

First think what happens if there is only 1 kid with 1 red hat and no black hats. Then think what if there are 2 kids with 2 red hats and no black hats. Then what if there were 3 kids with red hats and no black hats. This continues for as many red hats as you like.

Then think about what happens if there some black hats as well.

Here is my solution:

Spoiler

Assume no black hats.
1 red hat: kid knows straight away that he must have a red hat because he is told there is at least 1.
2 red hats: first kid doesn't know because he sees another kid with a red hat. If the first kid had seen the other kid with a black hat he would have known he had a red hat. Because the first kid didn't know, the second now does know that he has a red hat and says so. Because the 2nd said he has a red hat, the first now knows he must also have a red hat. (otherwise the second kid would have said he had a black hat)
...
n red hats: this continues by iteration so that it takes n-1 kids with red hats to say they don't know before the nth knows.

Now introducing black hats, the same thing happens, but you can ignore every time a kid with a black hat says he doesn't know because it gives you no additional information. Once the nth kid with a red hat says he has a red hat, everyone looks around and if they see n-1 red hats then they have a red hat, and if they see n red hats they must have a black hat.

I hope I explained that vaguely clearly...

 

[aa6972]

I am going to go with the second class are make up of the same children,

meaning they all know the class size, and each children will know there is at least 6 Red (cos they might be having wear one red in the first class), but this child saw 7 red in the first class and 6 red in the second class and she can see the reminder is black, meaning she can conclude she is wearing red. same with the reminder, they saw 7 red first class and 7 red have gone pass.


OK scrap that, just saw no children from either class know there is 7 hats.

 

[The Pat]

Answer

Spoiler

The kids with red hats on can see 6 reds hats, the ones wearing black can see 7 red hats. After the sixth child wearing a red hat says they don't know the one left wearing a red hat can only see black hats but the others still see a red one. As none of the black hat wearers know if they are red or black they will not answer, so the one wearing the red hat (and can only see black hats) will know that they must different. After the red hat is declared the others know they must have a black hat for this to happen.

Edit: actually this makes no sense as to why the first class couldn't do this.

 

[ChocolateStarfishRammer]

Why not just guess?

You can see everyone elses hats. You can see some are Red.

If you stand up and say I dont know then your in detention.

If you stand up and say its Red and your wrong, your in detention.

Either way, your in detention, so take the chance you could be right.

Seems logical to me.

 

[mame]

Answer

Spoiler

The kids with red hats on can see 6 reds hats, the ones wearing black can see 7 red hats. After the sixth child wearing a red hat says they don't know the one left wearing a red hat can only see black hats but the others still see a red one. As none of the black hat wearers know if they are red or black they will not answer, so the one wearing the red hat (and can only see black hats) will know that they must different. After the red hat is declared the others know they must have a black hat for this to happen.

Edit: actually this makes no sense as to why the first class couldn't do this.

even though I think that's the answer, and pretty much what I put, I keep coming back to the fact that every child, at every point can see numerous red hats and only knows that there is 'at least one red hat' and doesn't know that there are 7

 

[Bennie-Mac]

I think the riddle is wrong, there are several variants of the hat riddle extremely similar to the op, but the people wearing the hats can only see the person infront of them, where the solution is derived from what hat the person in front of them is wearing and the answers they give, based on what they can deduce from the same information.

 

[Sudden]

My answer above Is it right?

 

[King Damager]

My answer above Is it right?

No

Anyone who believes the riddle is incorrect, I suggest you look at Psiko's hint.

kd

 

[The Pat]

even though I think that's the answer, and pretty much what I put, I keep coming back to the fact that every child, at every point can see numerous red hats and only knows that there is 'at least one red hat' and doesn't know that there are 7

I don't get the 'at least one red hat' bit. In both situations the class would be able to see there are at least 6 or 7 red hats depending on if they were wearing one or not.

 

[estebanrey]

ahhhh

 

[Sudden]

No

 

[AJK]

The "at least one red hat" clue isn't about what each child can see, it's about what you can logically conclude that the previous child did NOT see.

My soln:

Spoiler

Given the fact that "at least one hat is red", and referring to persons 1-X in the order that they are asked "is your hat red":

If person 1 sees all black hats, he must be wearing a red hat (he can only conclude this because he knows that at least one hat is red).
Since person 1 doesn't speak, he must see a mix of red and black hats.

From above, person 2 knows that person 1 saw a mixture of hats.
If person 2 now sees all black hats, he knows that his hat must be red.
Since person 2 doesn't speak, he must see a mixture of red and black hats.

From above, person 3 knows that person 2 saw a mixture of hats.
If person 3 now sees all black hats, he knows that his hat must be red.
Since person 3 doesn't speak, he must see a mixture of red and black hats.

Continuing the above, each following person can conclude that if they see only black hats, then their hat is the last remaining red hat.
Once that person speaks, all following people know that their hat is black.

 

[estebanrey]

I don't get the 'at least one red hat' bit. In both situations the class would be able to see there are at least 6 or 7 red hats depending on if they were wearing one or not.

This is what I don't get either. I think I understand the logic of how a solution can be reached (just writing it out now to make sure it works out correctly but I don't see what 'new' information is introduced in the second trial and why the same logic can't be applied in the first scenario.

 

[estebanrey]

The "at least one red hat" clue isn't about what each child can see, it's about what you can logically conclude that the previous child did NOT see.

My soln (possibly badly explained!)

Spoiler

Given that "at least one red hat":

If person 1 sees all black hats, he must be wearing a red hat (he can only conclude this because he knows that at least one hat is red).
Since person 1 doesn't speak, he must see a mix of red and black hats.

From above, person 2 knows that person 1 saw a mixture of hats.
If person 2 now sees all black hats, he knows that his hat must be red.
Since person 2 doesn't speak, he must see a mixture of red and black hats.

From above, person 3 knows that person 2 saw a mixture of hats.
If person 3 now sees all black hats, he knows that his hat must be red.
Since person 3 doesn't speak, he must see a mixture of red and black hats.

Continuing the above, each following person can conclude that if they see only black hats, then their hat is the last remaining red hat.
Once that person speaks, all following people know that their hat is black.

Sorry but your explanation doesn't solve mine or Pat's question because we know there are seven red hats. Therefore given each child can see each other then no child can see "all black hats" like you explain as an example. Each child in either scenario can either see 6 red hats or 7 red hats depending on the one they are wearing.

 

[Irish_Tom]

Sorry but your explanation doesn't solve mine or Pat's question because we know there are seven red hats. Therefore given each child can see each other then no child can see "all black hats" like you explain as an example. Each child in either scenario can either see 6 red hats or 7 red hats depending on the one they are wearing.

The number of red hats is a red herring [sic].

If you can see a number of red hats in front of you and a number of black and you've been told at least one is red but non of them know they are red, the only logical conclusion is that yours is red because everyone else's is black.

All the people left can then determine that they have black hats.

 

[King Damager]

Sorry but your explanation doesn't solve mine or Pat's question because we know there are seven red hats. Therefore given each child can see each other then no child can see "all black hats" like you explain as an example. Each child in either scenario can either see 6 red hats or 7 red hats depending on the one they are wearing.

Read, Psiko's solution. It's probably the best of those posted so far.

kd