Mid-Week Brain Teaser

RDM · 7 Feb 2013 at 13:30

div0 said:
No it doesn't, the last red hat doesn't know that thev first red hat didn't say don't know, because the first red hat might have seen 'all blacks'. Remember that from the second reds point of view, the first red could think that if he doesn't know there's at least 1 red.

Yes, you are right, it does need at least three red hats. I stand corrected.

AJK · 7 Feb 2013 at 13:33

Let's try a PDF from Yale, if anyone fancies a bit of light reading:
http://cowles.econ.yale.edu/~gean/art/p0882.pdf

You're looking for section 2 "Puzzles about reasoning based on the reasoning of others", starting on page 3 of the PDF.

Quoting briefly:

Imagine three girls sitting in a circle, each wearing either a red or white hat. Suppose that all the hats are red. When the teacher asks if any student can identify the colour of her own hat, the answer is always negative, since nobody can see her own hat. But if the teacher happens to remark that there is at least one red hat in the room, a fact that is well-known to every child (who can see two red hats in the room) then the answers change. The first student who is asked cannot tell, nor can the second. But the third will be able to answer with confidence that she is indeed wearing a red hat.

Knock yourselves out.

estebanrey · 7 Feb 2013 at 13:36

I'm not with you either on this one div0. I don't agree with your statements above.

div0 said:
No it doesn't, the last red hat doesn't know that thev first red hat didn't say don't know, because the first red hat might have seen 'all blacks'. Remember that from the second reds point of view, the first red could think that if he doesn't know there's at least 1 red.

But they didn't because they could see the second person with a red hat exists.

In the situation where there are only 2 red hats, both red hat wearers know for a fact there is "at least one red hat" and neither can see "all black hats". The only thing they don't know is if there are 1 or 2 because the only think they don't know is their own hat colour.

The only time any child can see "all blacks" is if there are 0 or 1 red hats in the game. In the former all participants will see all black hats and in the latter all the children know there is at least 1 red hat except the person wearing it who can only see all black hats. Once you have two red hat wearers then all participants cans see at least one red hat and no one can claim anyone else can see no red hats.

Bloomfield · 7 Feb 2013 at 13:36

AJK said:
Let's try a PDF from Yale, if anyone fancies a bit of light reading:
http://cowles.econ.yale.edu/~gean/art/p0882.pdf

You're looking for section 2 "Puzzles about reasoning based on the reasoning of others", starting on page 3 of the PDF.

Quoting briefly:

Knock yourselves out.

And as you'll have already read (I assume) we agree that it isn't solvable with 2 or less red hats.

With more though everyone knows that there's atleast one red hat, and everyone knows that everyone else knows that there's atleast one red hat.

Bloomfield · 7 Feb 2013 at 13:39

estebanrey said:
I'm not with you either on this one div0. I don't agree with your statements above.

"the last red hat doesn't know that the first red hat didn't say don't know"

Yes they do unless we're introducing a new condition where the kids can't hear each others answers (in which case the whole things breaks down anyway).

"because the first red hat might have seen 'all blacks'. Remember that from the second reds point of view, the first red could think that if he doesn't know there's at least 1 red."

But they didn't because they could see the second person with a red hat exists.

In the situation where there are 2 red hats, both red hat wearers know for a fact there is "at least one red hat". The only thing they don't know is if there are 1 or 2 because the only think they don't know is their own hat colour.

You're supposed to know 100% that there's a red hat, and that everyone else knows it too. Whilst it's true that with 2 hats everyone knows there's a red hat, you can't be 100% sure that the other person wearing a red hat knows this, as your hat could be black

.

RDM · 7 Feb 2013 at 13:39

AJK said:
What do you mean "accounted for"? What about your scenario is "accounting for" the hats? Can you explain?

You are whilst wearing your red hat.

But now I think about it, I could be wrong, as the black hats saying I don't know will be referring to their own hat as being possibly red and not your hat.

But if that is the case then in neither scenario can you deduce the colour of your hat.

AJK · 7 Feb 2013 at 13:39

Bloomfield said:
And as you'll have already read (I assume) we agree that it isn't solvable with 2 or less red hats.

I don't agree that. I think it's perfectly solvable with 1 red hat, 2 red hats, or 48,962 red hats. That last one might take a while though.

AJK · 7 Feb 2013 at 13:41

RDM said:
You are whilst wearing your red hat.

But now I think about it, I could be wrong, as the black hats saying I don't know will be referring to their own hat as being possibly red and not your hat.

But if that is the case then in neither scenario can you deduce the colour of your hat.

Sure you can, if you're in classroom 2.

AJK said:
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 that all remaining (unspoken) hats are black, 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 that all remaining (unspoken) hats are black, 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 only see black hats remaining, then their hat is the last red hat.
Once that person speaks, all following people know that their hat is black.

RDM · 7 Feb 2013 at 13:44

AJK said:
Sure you can, if you're in classroom 2.

But surely that only works if only one hat is red? As 7 hats are red everyone is seeing a mix of red and black hats.

estebanrey · 7 Feb 2013 at 13:45

Bloomfield said:
You're supposed to know 100% that there's a red hat, and that everyone else knows it too. Whilst it's true that with 2 hats everyone knows there's a red hat, you can't be 100% sure that the other person wearing a red hat knows this, as your hat could be black .

Well explained. I get that point now.

But then it does seem strange that this rule stops at 2 though, maybe I was right earlier when I said the person who wrote it made the false assumption this logic continues to scale up forever (or maybe it does like AJK thinks and I'm just not able to carry on getting my head around it past 2).

EDIT: Whilst I understand your logic Bloomfield/div0 I still don't see how telling the children there is "at least one red hat" at the start solves any problems in the 2 hat scenario because you aren't introducing any new information to any of the children. The ones wearing black know there's either 2 or 3 and the 2 wearing red know there's either 1 or 2.

AJK · 7 Feb 2013 at 13:48

RDM said:
But surely that only works if only one hat is red? As 7 hats are red everyone is seeing a mix of red and black hats.

I will amend the above - in each case when I refer to someone "seeing" something I mean "hats whose owners have not yet spoken".

estebanrey · 7 Feb 2013 at 14:09

AJK said:
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 that all remaining (unspoken) hats are black, 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 that all remaining (unspoken) hats are black, 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 only see black hats remaining, then their hat is the last red hat.
Once that person speaks, all following people know that their hat is black.

I understand what you are trying to say AJK, you are basing it off the fact that the next person asked can establish what the previous person saw. I totally get that logic.

But where it falls down for me is when you introduce the the OP's specifics into it because for example...

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......

That dichotomy simply can't exist under the conditions of the riddle. Person 1 cannot see all black hats and person 2 knows that even before person 1's answer thus the logic then can't be continued as you've explained because they have no new information.

In reality, Person 1 sees a mixture of red and black hats so says I don't know (or doesn't speak as in your explanation). Now it's Person's 2 turn, they cannot deduce that the reason Person 1 didn't speak was because they saw a mixture of hats and that they would have said "red" if they saw all blacks because they already know person 1 can see a mixture of hats and not all blacks. Nothing can be deduced by person 1's answer, so the process of elimination which your explanation relies on can't hold.

Your explanation relies on dichotomies being created whereby the next person to answer has absolutely no clue about whether the person before them sees all black hats left or a mixture until they give their answer, but because person 2 knows for a fact that person 1 saw a mixture before the game starts, they don't need to know person 1's response to know that person 1 did not see a mixture of colours.

Bloomfield · 7 Feb 2013 at 14:44

AJK said:
I don't agree that. I think it's perfectly solvable with 1 red hat, 2 red hats, or 48,962 red hats. That last one might take a while though.

By we I meant me, div0 etc. And I was referring to when you haven't been told that there's atleast one hat as to whether it's solvable or not.

The Princeton example isn't the same as the one in the OP, which is what we're talking about.

fez · 7 Feb 2013 at 14:53

It doesn't make any sense at all between the two groups. You claim that the kids are all really smart and that they obviously catch on to the solution in the second group.

Thats fine. I understand how they do it. If you are the 7th red hat kid and everyone before you has said "I don't know" including the last red hat child to say "I don't know" then you can be sure that you have a red hat based on the assumption that the previous red hat to say "I don't know" could see at least one red and one black. Seeing as you can only see the blacks, you must also be a red.

Why didn't the first group get it though. Telling them that at least one hat is red does absolutely nothing in this situation. Its just to throw people off the read issue. You are also making the assumption that all of the children are perfect and act in a very particular way.

The whole thing is worded and put into a very odd situation.

aln · 7 Feb 2013 at 14:59

fez said:
It doesn't make any sense at all between the two groups. You claim that the kids are all really smart and that they obviously catch on to the solution in the second group.

Thats fine. I understand how they do it. If you are the 7th red hat kid and everyone before you has said "I don't know" including the last red hat child to say "I don't know" then you can be sure that you have a red hat based on the assumption that the previous red hat to say "I don't know" could see at least one red and one black. Seeing as you can only see the blacks, you must also be a red.

Why didn't the first group get it though. Telling them that at least one hat is red does absolutely nothing in this situation. Its just to throw people off the read issue. You are also making the assumption that all of the children are perfect and act in a very particular way.

The whole thing is worded and put into a very odd situation.

It actually does work.

Write it down and work through each of them. I've never read game theory before so it took writing it down to process it for me.

fez · 7 Feb 2013 at 15:06

There are 7 red hats in each class. Its impossible for the kids not to deduce that there is "at least one red hat" as every child would see at least 6 red hats.

I understand what people are saying but I simple cannot fathom how it applies to the wording given. If there was only a single hat then it would be fine but the whole thing is contradictory.

We are supposed to assume that the kids are super smart but can't tell that there is at least one red hat in the room by looking and seeing at least 6.

aln · 7 Feb 2013 at 15:17

fez said:
There are 7 red hats in each class. Its impossible for the kids not to deduce that there is "at least one red hat" as every child would see at least 6 red hats.

I understand what people are saying but I simple cannot fathom how it applies to the wording given. If there was only a single hat then it would be fine but the whole thing is contradictory.

We are supposed to assume that the kids are super smart but can't tell that there is at least one red hat in the room by looking and seeing at least 6.

Its because its recursive. You need to know what kids would be thinking if there was only 1-2 hats in order to figure out what they would be thinking when theres more. It gets increasingly complicated every hat you had, although the logic doesn't change beyond 4 afaik.

Now when you start with 1-2 hats, if you don't know theres at least 1, as the #2 red you cannot decide whether #1 red said he wasn't a red because he didn't see any reds and though there was really 0 reds or because he saw another red.

If you then state there is positively at least one red, then #2 red knows #1 red said he wasn't red because he saw another red and since that means theres two reds and #2 can only see one, he must be red.

The info is only useful at the start, but you need that start to start the recursion.

chriscubed · 7 Feb 2013 at 15:30

Having spent all day thinking and reading about this, I have come to 2 decisions
1. That I understand.
2. I don't like it.

estebanrey · 7 Feb 2013 at 15:31

But aln, in the first scenario in the OP all children in the class positively know there is more than one red hat because they can see that is the case. The teacher saying "there is at least one red hat" provides no new information to any of the children because every single one of them knows that already.

At the moment this is where I stand. If all the children 'just know' to play by this rule....

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"

Then the last kid wearing a red hat will guess correctly and everyone else will say "I don't know before him". That works but it works in both cases provided by the OP, the second class don't need the extra line that the teacher tells them.

If the children don't 'just know' that rule and they have work everything out on the fly using inductive logic then neither scenario can be solved because you cannot infer anything from the previous answers given.

I'd like to see the 'Princeton Example' Bloomfield referred to because it seems to me that this version just doesn't work properly and I'm convinced at the moment it's a rehash of a working puzzle that makes more sense.

Psiko · 7 Feb 2013 at 15:40

aln said:
Its because its recursive. You need to know what kids would be thinking if there was only 1-2 hats in order to figure out what they would be thinking when theres more. It gets increasingly complicated every hat you had, although the logic doesn't change beyond 4 afaik.

Now when you start with 1-2 hats, if you don't know theres at least 1, as the #2 red you cannot decide whether #1 red said he wasn't a red because he didn't see any reds and though there was really 0 reds or because he saw another red.

If you then state there is positively at least one red, then #2 red knows #1 red said he wasn't red because he saw another red and since that means theres two reds and #2 can only see one, he must be red.

The info is only useful at the start, but you need that start to start the recursion.

This is exactly right.

Let's rephrase the problem: Each child must determine how many red hats there are.

Now this is actually the same question as "what colour is my hat". This is because, as soon as I work out my own hat colour, I can see everybody else's, so I can deduce the total number of red hats.

When going through the logical process of determining the number of red hats, each person is really having to think to themselves, "I can see 6 (or 7 if they have a black hat, it doesn't matter) red hats, but what would have happened at this stage of the process if I could only see 5 redhats, which begs the question what would have happened if I could only see 4 red hats, ...3 red hats...2 red hats...1 red hat." Once you are asking what would have happened in the scenario had there been 1 red hat, in the first problem, you cannot answer the question. But in the second problem, you are told there is at least 1 red hat, so if you can't see someone else wearing it, it must be you!

This is where the extra information comes in to play. As aln said, it starts the recursion process. It may add nothing to what you see, but it does add something to the "what if" scenarios that you have to ask yourself.