Soldato
- Joined
- 21 Oct 2009
- Posts
- 2,742
That's the one I was referring to. It's basically the same situation as there being no red hats (in the OPs question), which we already know isn't solvable without the hint.
Mid-Week Brain Teaser
- Thread starter King Damager
- Start date
That's the one I was referring to. It's basically the same situation as there being no red hats (in the OPs question), which we already know isn't solvable without the hint.
Soldato
- Joined
- 21 Oct 2009
- Posts
- 2,742
But how does thinking about what would happen if you could only see 1 red hat matter when you can see 6, and you know that everyone else can see atleast 5?
You use recursion to figure out what people should be thinking, therefore what comes at the start of that recursion is important. It's a bit of a brainfart because you're essentially making up a scenario to prove your current one, but when you do it you realise the math works, where it wouldn't if you assumed they didn't know.
So let me get this right. Unless there is only 1 red hat then it works for both classes?
Otherwise, to take the actual OP. There are 7 red hats. When you see that 6 red hatted "I don't know" kids have gone and you can't see any more, you know that you are a red too otherwise the 6th red hatted kid to go would not have said "I don't know".
Last edited:
Soldato
- Joined
- 8 Mar 2007
- Posts
- 10,938
It doesn't work though in the reality of the OP's example. No new information has been introduced in the second example. Everyone single child knows there is "at least one red hat" in both scenarios because of all them can see it before the game starts.
The flaw in the logic with the way this specific puzzle works with your/Psiko's explanation method is that you are working it out from 1 up (what would happen if only 1 red hat existed, then what if it was 2, then what if it was 3 etc) - which works that way, but then assuming that logic works backwards if you start with any number of red hats.
It doesn't, at least not in the set up of the OP's question.
Last edited:
The problem is you can't start the recursion if you assume the kids may be thinking they have 0 red hats.
With several on the table you can assume there will always be one and work from there, without the teacher needing to say anything, but if you start the recursion based on exactly what the teacher says, 0 being a possible value screws you over. If you take what the teacher says as a literal rule, you cannot solve it.
In reality, you probably could solve it assuming everyone else in the room knew to assume there was always 1. Then again in reality even if you knew the theory, you couldn't rely on anyone else knowing it, so it wouldn't work anyway.
Last edited:
Will you please stop saying this. Read the PDF I posted, it is exactly the information you need to understand what's happening.
This is what I am trying to get at. In certain cases this work perfectly. The OP just seems to have picked one that doesn't or has worded it in a way that allows loopholes
Let' assume the following:
1) I have a red hat (but I don't know that)
2) I can see six red hats
3) Six red hats have already said "I don't know"
There are two possibilities (from my POV):
1) I have a red hat so there are 7 red hats in total and I say "I have a red hat" and everyone's happy
2) I have a black hat and there are 6 red hats in total so I say "I don't know"
To prove 1 we must disprove 2. So I need to ask the question: what would the last guy with a red hat have said if:
1) He has a red hat (but doesn't know it)
2) He can see 5 red hats
3) 5 red hats have already said "I don't know"
If the answer is he would have said "I have a red hat", then we know possibility 2) is not possible, because assumption 3) above was that he didn't say "I have a red hate" (i.e proof by contradiction).
You see, we are now asking exactly the same questions as above but with 1 less hat. This is why we need to look at all the scenarios down to 1 red hat.
The only children that really matter are the last and second last, right.
In the first class. Every child knows that there are 6/7 red hats depending on their own colour.
All children say "I don't know" based on seeing at least 1 red still remaining including the second last.
The last kid then comes along and says. I have no idea how many reds there are total but the kid before me said "I don't know" so he must have seen a red somewhere. Because I don't see any red, I therefore must be a red.
Please explain how that reasoning is incorrect for the situation given.
I thought that at first too, it's wrong. The spoiler is how I worked that out. Every kid needs to be thinking about what the kid before him was thinking for this to work.
For 2 hats
=======
If theres 2 red hats, when #1 looks around and sees another red hat, he doesn't know if theres 1 or 2 hats so he can't say.
#2 red knows #1 red saw another red, can't see any more reds, so assumes it's himself.
For 3
=======
#1 red sees 2 red, but doesn't know if theres 2 or 3.
#2 red sees 2 red, knows theres at least 2 and accepts that #3 might be what #1 say to say "I don't know".
#3 sees 2. Knows that #1 sees at least 1 but #2 didn't solve it because he must see 2 and since #3 can only see 2, he's the 3rd.
For 4
====
#1 red sees 3 red, but doesn't know if theres 3 or 4.
#2 red sees 3 red, knows theres 3 or 4 but can only know #1 sees at least up to 2.
#3 red sees 3, he knows theres either 3 or 4, but only knows #1 can see 2 and #2 can see 2 but not whether either of them can see 3, so doesn't guess.
#4 can see 3. He knows that if there was 3 hats, #3 should have been able to guess he was red, so there must be another red and that'll have to be him.
=======
If theres 2 red hats, when #1 looks around and sees another red hat, he doesn't know if theres 1 or 2 hats so he can't say.
#2 red knows #1 red saw another red, can't see any more reds, so assumes it's himself.
For 3
=======
#1 red sees 2 red, but doesn't know if theres 2 or 3.
#2 red sees 2 red, knows theres at least 2 and accepts that #3 might be what #1 say to say "I don't know".
#3 sees 2. Knows that #1 sees at least 1 but #2 didn't solve it because he must see 2 and since #3 can only see 2, he's the 3rd.
For 4
====
#1 red sees 3 red, but doesn't know if theres 3 or 4.
#2 red sees 3 red, knows theres 3 or 4 but can only know #1 sees at least up to 2.
#3 red sees 3, he knows theres either 3 or 4, but only knows #1 can see 2 and #2 can see 2 but not whether either of them can see 3, so doesn't guess.
#4 can see 3. He knows that if there was 3 hats, #3 should have been able to guess he was red, so there must be another red and that'll have to be him.
You can follow this all the way up to 7, but it'd be a waste of time. You won't get it until you do the same in the spoiler while assuming there can be 0 hats in the kids answers.
Last edited:
Soldato
- Joined
- 8 Mar 2007
- Posts
- 10,938
Why should I stop saying something that is factually correct. The PDF poses a different scenario and it this specific one that we are looking at.
Please see post 172 where I point out the flaw with your explanation.
But when you get to 1 hat it only works if you are assuming a random number of hats can exist and everyone that follows has no idea about what the previous person saw until you get their answer, that is untrue.
Every single child knows that the person before them saw at least 1 unchecked hat wearer left and at most 6 unchecked left hat wearers. Only if there was only one hat could you ever get to the point where someone knows there are 0 unchecked hats in play, this can't happen under the OP's scenario.
You are working it out by changing the number of hats in the game. This doesn't work here because the every child from the first to the last knows there have to be 5 or 6.
Thinking what is there was only 1 hat in the game or 2 hats in the game is pointless because they know that isn't the case. They can only think what if there are 6 hats in the game or 7 hats in the game as it's a physical impossibility that any it can be any less.
AJK explains the method the same and you and I explain why that logic is flawed here...
http://forums.overclockers.co.uk/showpost.php?p=23713959&postcount=172
Last edited:
Soldato
- Joined
- 8 Mar 2007
- Posts
- 10,938
dup
But they do have new information - this is the whole point of the logical concept of common knowledge. I appreciate that this is confusing and not entirely intuitive, but the PDF I posted does explain why this works.
Stop right there. This is just WRONG. Person 1's reasoning remains the same regardless of what person 2 can see. Just because person 2 knows that person 1 didn't see all black hats DOES NOT CHANGE person 1's process of reasoning, and knowing what person 1's reasoning was is crucial to person 2 working out what he does or does not know.
Last edited:
I think I should have left well alone about 120 posts ago...
Soldato
- Joined
- 8 Mar 2007
- Posts
- 10,938
dup
Last edited:
Soldato
- Joined
- 8 Mar 2007
- Posts
- 10,938
Ok I've read it an maybe we should refer to this from this point forward for simplicity as it only deals with three hats so here it is for everyone else....
Right so I get the distinction between all of the children knowing there is at least one red hat individually and them all knowing that each other also knows there is at least one red hat (the crux of what you are saying) and how the process works from there.
But I would argue that even before they are given the "common knowledge" by the teacher they can deduce this common knowledge exists for themselves. So now I'm not just saying they all individually know there is more than one red hat but that they ALSO know that the other two children must know there is at least one red hat, or rather they can deduce that common knowledge exists without being told it, it seems to me.
Why? Well let's look at each child's point of view and what they know each other knows (in other words what common knowledge they can deduce).
Child 1 individually knows that at least two hats are red. But Child 1 also knows that Child 2 knows there is at least one red hat because she knows that Child 2 can see Child 3's red hat but doesn't know what she sees when he looks at her (Child 1). This is the same for Child 1's knowledge of what Child 3 knows. So Child 1 not only knows there is at least one red hat (individually she knows there are at least 2 of course) but she also knows that both other children also know there is at least 1 red hat.
Because they are all wearing red hats this same logic is true for all three children so not only do they all know individually there is more than 1 red hat but they also all know that each other already knows there is at least one red hat. So before the game starts, without being told anything each child individually knows that...
* There are at least 2 red hats in the game.
* That the other children in the game know for sure at least one red hat is in the game.
The common knowledge (bullet point 2) is thus already there or rather it can be deduced by each of the children without being told.
I want to make it clear that I'm not saying I don't understand the 'chain of logic' they explain and how the common knowledge of "at least 1 red hat" changes the scenario, what I'm now stuck on is why this common knowledge itself can't be deduced from the scenario without having to be told it specifically.
So I'm back to div0's point earlier that it 'breaks' after more than 2 red hats are introduced. Let's look at if only two red hats were involved in the same scenario above, then I totally get why that 'extra' information from the teacher is needed. So let's say Child 1 is wearing a white hat and Children 2 and 3 are wearing red. In this scenario the common knowledge that at least one red hat exists can't be deduced by Child 2 or 3.
Why? Well once again, individually they all know there is at least one red hat. However Child 2 sees Child 1 wearing white and Child 3 wearing red, therefore she doesn't know if Child 3 is looking at 1 red and 1 white, or 2 whites. Thus she can't say that Child 3 knows there is at least 1 red hat as in her mind from Child 3's P.O.V they could still all be white. This is the same for Child 3's view of Child 2 of course.
I hope I've explained that clearly
Last edited:
I agree. Group 1 should also have been able to figure it out.
The only time Group 1 wouldn't be able to solve it is if there were all red, all black, or 2 or less of any one colour.
In the case of 7 reds and > 2 blacks, it is solvable by group 1, isn't it.