Mid-Week Brain Teaser

[estebanrey]

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

kd

I did, I sort of understand it to a point but the more kids you add the more complex it gets and I'm still not convinced it works with more than a very small number of children. Tell me if I'm on the right lines....

If there was 1 Girl_(1 permutation)
(P1) Girl_1 has a red hat
So she knows obviously and I can see why the 'extra information' is vital here because if she knows there is 1 red hat and she is the only one with a hat then clearly she is wearing a red hat.

If there was 2 kids (3 permutations)
(P1) Girl_1 has a red hat and Girl_2 has a black hat
Girl_1 sees Girl_2 has a black hat and because she knows there is at least 1 red hat then she knows she must be wearing a red one and thus answers "red" (and knows the one left is black because she can see it).

(P2) Girl_1 has a black hat and Girl_2 has a red hat
Girl_1 sees Girl_2 has a red hat but this doesn't help her because she only knows there is 'at least 1' red hat which means there could be 2 so she has to say "I don't know". Girl_2 sees Girl_1 with a black hat so knows she must have the "at least 1" red hat and answers "red"

(P3) Girl_1 & Girl_2 both have a red hats
Once again as above Girl_1 sees Girl_2 has a red hat this doesn't help her because she only knows there is 'at least 1' red hat which means there could be 2 so she has to say "I don't know". Girl_2 sees Girl_1 with a red hat and can deduce that because Girl_1 said "I don't know" then P1 isn't possible (as under P1 Girl_1 knows her hat colour and answers 'red') thus leaving P2 and P3 as the only possibilities, both of which she is wearing a red hat so she answers "red".

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

So up to that point I understand Psiko's logic but I'm struggling to take it to the next stage because each kid you add to the group adds an exponential amount of permutations. 3 kids gives 7 possibilities, 4 gives you 15 and so on. I'm still struggling to reconcile what 'new' information is presented in scenario 2 as well even though the logic I've said above uses the "at least 1 hat" rule the first group of kids would also know there was at least 1 red hat, in fact they each would know there were at least 6 red hats (or 7 if the child is wearing a black one).


Argggh this is making my head hurt!

 

[Bloomfield]

Stuff

Spoiler

After someone has said "I don't know" you immediately ignore what hat they're wearing, as it's now irrelevant. The only relevant part of it is that each one saw atleast 1 red hat, meaning they didn't know what colour their hat is.

Pretend that after someone guesses they leave the room, so all you can see is people that haven't guessed yet.

Then when it comes to your turn to guess and all you can see is black hats, whilst the person before you could see a red hat, you know that you have the red hat.

 

[Jono8]

There is a major flaw in the question though.

Were the first lot of kids told that the hats were going to be a mixture of black or red?

 

[Jono8]

Sorry, just to make my self more clear.

For the first class, did the teacher say " i am going to put hats on you now, some will be red, some will be black?" Or did she say " I am going to put hats on you now, they might be all black, all red, or a mixture of the two". Or did she not even mention the colour?

Were there any other colours mentioned?

For this to work in the second class, the kids must have known that the only two colours were red and black.

 

[estebanrey]

Also given grown adults are struggling to understand this, what's the odds that young children would and all the kids would have to know the system and for it to work. For example if the last child asked wearing a red hat says I don't know because they don't know the system, then the next kid who knows the system would incorrectly say 'red' and they'd be wrong

 

[Bloomfield]

For this to work in the second class, the kids must have known that the only two colours were red and black.

Spoiler

How?

If the person before you said "I don't know" it means they could see atleast one red hat. If you can now only see black hats then you must have a red hat. I don't see how thinking there may be other colours makes any difference?

 

[estebanrey]

I get what you're saying Bloomfield but I still don't see why it needed two goes and the first group couldn't use your system and do it right in the first place.

Nothing new has been introduced in the second trial.

 

[Jono8]

I get what you're saying Bloomfield but I still don't see why it needed two goes and the first group couldn't use your system and do it right in the first place.

Nothing new has been introduced in the second trial.

This was my point ( i think )

We need more specifics on what the first class were told.

 

[Bloomfield]

I get what you're saying Bloomfield but I still don't see why it needed two goes and the first group couldn't use your system and do it right in the first place.

Nothing new has been introduced in the second trial.

Neither do I tbh .

 

[FoxEye]

I think the question is wrong.

The 6 red people who already answered "don't know" can base that answer off seeing each other's hats, not the 7th person with the red hat.

Ie Red1 sees Red2 and says "don't know".
Red2 sees Red1, and says "don't know".

These people aren't then removed from the pool. Red7 can't conclude that Red6 saw his hat and answered "don't know". Red6 might equally have been looking at Red1-Red5. Those people aren't removed from the pool after they answer.

Therefore Red7 is not "only seeing black hats". He still sees Red1 - Red6 *unless they are removed from the pool after answering*.

 

[Bloomfield]

I think the question is wrong.

The 6 red people who already answered "don't know" can base that answer off seeing each other's hats, not the 7th person with the red hat.

Ie Red1 sees Red2 and says "don't know".
Red2 sees Red1, and says "don't know".

These people aren't then removed from the pool. Red7 can't conclude that Red6 saw his hat and answered "don't know". Red6 might equally have been looking at Red1-Red5. Those people aren't removed from the pool after they answer.

Therefore Red7 is not "only seeing black hats". He still sees Red1 - Red6 *unless they are removed from the pool after answering*.

Spoiler

They don't have to be specifically removed from the pool, like I explained above you would ignore whatever hat the person before you is wearing since that person sees atleast one red hat. It doesn't matter what hat they are wearing.

IE Red1 sees Red2 and says "don't know".
Red2 then knows that Red1 saw a red hat and answers "red".

 

[Voltar]

**** it. It's latin class. No one wants to do that. Therefore you chance your hand and say you've got a red hat: if you're right you get to have fun and if you're wrong you're stuck doing the latin anyway.

/arts student's interpretation of scientific theory

 

[estebanrey]

I think the question is wrong.

The 6 red people who already answered "don't know" can base that answer off seeing each other's hats, not the 7th person with the red hat.

Ie Red1 sees Red2 and says "don't know".
Red2 sees Red1, and says "don't know".

These people aren't then removed from the pool. Red7 can't conclude that Red6 saw his hat and answered "don't know". Red6 might equally have been looking at Red1-Red5. Those people aren't removed from the pool after they answer.

Therefore Red7 is not "only seeing black hats". He still sees Red1 - Red6 *unless they are removed from the pool after answering*.

Spoiler

...As long as each child answers "I don't know" so long as they can see other children with red hats who haven't been asked yet and "red" when that isn't the case that last child will be correct.

So the first kid with a red hat that is asked can see 6 other children with red hats that haven't been asked yet so says "I don't know".

The second kid with a red hat that is asked can see 5 other children with red hats that haven't been asked yet so again says "I don't know"

The third kid with a red hat that is asked can see 4 other children with red hats that haven't been asked yet so again says "I don't know"

..and so on until 6 children with red hats have said 'I don't know'.

At that point any subsequent children that are asked who have black hats can still see the last kid wearing a red hat that hasn't been asked yet so they say "I don't know" as well.

When it finally gets to the 7th child with a red hat on, she knows to say "red" because she can no longer see any more kids with a red hat on that haven't already been asked yet.

But I agree it's badly worded because their is no new information being introduced in the second trial. The OP should have presented as one test or said the difference the between the first and the second test was that in the first the children can't see the other kids hats and in the second they can or something.

 

[platinum87]

The 6th person with the red hat says "i dont know"

Why? = Because they see the 7th person has a red hat

The 7th person sees only black hats left and therefore he/she must have a red hat.

However the question of how do the children in the 2nd class figure it out, = 1) they haven't 2) Only one person got it right

 

[aln]

I think OP messed up the riddle.

 

[mame]

The 7th person sees only black hats left and therefore he/she must have a red hat.

What do you mean by left? there are still 6 or 7 red hats to look at

 

[Bloomfield]

What do you mean by left? there are still 6 or 7 red hats to look at

The hat colour of people that have already guessed is irrelevant, you would just ignore anyone who has gone before you when working it out.

 

[estebanrey]

What do you mean by left? there are still 6 or 7 red hats to look at

See my (or rather my re-worded Bloomfield) explanation...

Spoiler

...As long as each child answers "I don't know" so long as they can see other children with red hats who haven't been asked yet and "red" when that isn't the case that last child will be correct.

So when the teacher asks the first kid with a red hat, she can see 6 other children with red hats that haven't been asked yet so says "I don't know".

When the next kid with a red hat is asked, she can see 5 other children with red hats that haven't been asked yet so again says "I don't know"

The third kid with a red hat that is asked can see 4 other children with red hats that haven't been asked yet so again says "I don't know"

..and so on until 6 children with red hats have said 'I don't know'.

At that point any subsequent children that are asked who have black hats can still see the last kid wearing a red hat that hasn't been asked yet so they say "I don't know" as well.

When it finally gets to the 7th child with a red hat on, she knows to say "red" because she can no longer see any more kids with a red hat on that haven't already been asked yet.

So the ones 'left' (at any point in the game) are Kids wearing red hats who haven't been asked the question yet.

 

[AJK]

But I agree it's badly worded because their is no new information being introduced in the second trial.

It isn't badly worded. There is exactly one new piece of information introduced in the second trial, which is the statement "there is at least one red hat". Observationally, this doesn't tell you anything (every child can see at least 6 red hats), but logically it does.

Spoiler

Consider the first classroom, who do NOT have the statement "there is at least one red hat":

Person 1 will ALWAYS say "I don't know", regardless of whether they see red hats, black hats, or a mixture of colours. They do not have enough information to deduce what colour their hat is.

Since person 1 will always answer "I don't know", person 2 can't draw any information from that answer, and thus can only base their own guess on the hats they can see - which, as for person 1, is not enough information to draw a conclusion and so person 2 will also always answer "I don't know".

This continues for persons 3, 4, 5, etc. It's impossible to reach a point where you can logically conclude "I see only black hats remaining, and nobody else knew the colour of theirs, so my hat must be red". It's not logically true.

 

[Bloomfield]

Someone still needs to explain why the first kids couldn't work it out .

The only difference between the situations is that the teacher in the second says "at least one hat is red", but that's completely irrelevant since the kids can already see that there are at least 6 (if they have a red hat) or 7 (if they have a black hat) red hats, so why even tell them that? .