Mid-Week Brain Teaser

[Bloomfield]

Estebanrey, the Princeton one can't be solved because no one can confirm the existance (and therefore confirm that the others know this too) of a white hat.

That one is basically the same as if there were only black hats in the OP, you wouldn't be able to solve it, because you don't know if the previous people are saying "I don't know" because they can see red (yours) and black (everyone elses) hats, or because they only see black hats (and therefore your hat is black) and don't know whether any red ones exist.

 

[estebanrey]

Estebanrey, the Princeton one can't be solved because no one can confirm the existance (and therefore confirm that the others know this too) of a white hat.

That one is basically the same as if there were only black hats in the OP, you wouldn't be able to solve it, because you don't know if the previous people are saying "I don't know" because they can see red (yours) and black (everyone elses) hats, or because they only see black hats (and therefore your hat is black) and don't know whether any red ones exist.

If the group is all wearing white/black hats then the "at least one red hat" information allows them to solve it. I'm saying I agree with AJK that in that given scenario that 'extra' information is needed. It is also needed if there was 1 or 2 red hats (as you and div0 agree from previously).

However, once you introduce three red hats I believe the common knowledge bit can be deduced by each child making being told it redundant.

 

[aln]

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

I don't need to work it out for 2 hats, it's just the best way I had of explaining it because you use the same thought process for every amount, repeated to figure out what people are thinking.

You only need to think about 6 or 7 hats and whilst I don't entirely agree the difference between both groups is essential in reality, for the point of what it's trying to teach you it does matter.

I'm sorta bored of this now so take it or leave it.

 

[Bloomfield]

If the group is all wearing white/black hats then the "at least one red hat" information allows them to solve it. I'm saying I agree with AJK that in that given scenario that 'extra' information is needed. It is also needed if there was 1 or 2 red hats (as you and div0 agree from previously).

However, once you introduce three red hats I believe the common knowledge bit can be deduced by each child making being told it redundant.

The 3 hats in the last bit, are you saying you think the Princeton one is solvable without the extra information?

Edit: Heh, same aln . No one seems to be able to explain to me why Classroom 1 in the OP can't solve it, whether that's because I'm stupid, they're bad at explaining stuff or they're just wrong and Classroom 1 is full of special kids I don't know, but I think I'm just gonna assume it's the latter option .

 

[estebanrey]

I don't need to work it out for 2 hats, it's just the best way I had of explaining it because you use the same thought process for every amount, repeated to figure out what people are thinking.

You only need to think about 6 or 7 hats and whilst I don't entirely agree the difference between both groups is essential in reality, for the point of what it's trying to teach you it does matter.

I'm sorta bored of this now so take it or leave it.

Forget about that post and concentrate on my last one (#199) using the Princeton example which only uses three red hats for simplicity.

I understand the logic of common knowledge and why it's needed but now I'm stuck on why if at least 3 people are wearing the same colour hat why that common knowledge can't be deduced without specifically being told it by teacher.

The 3 hats in the last bit, are you saying you think the Princeton one is solvable without the extra information?

Yes, as above forget about my previous posts (because the more I think about it the more I change my mind on what is and what isn't possible ).

Princeton's explanation makes sense to me when you have 0, 1 or 2 red hats because I now understand the point being made between the difference between individual and common knowledge. I don't understand why that common knowledge can't be deduced when there are 3 or more red hats (I'm tempted to e-mail John Geanakoplos to see if he can show me where I'm going wrong )

 

[fez]

Forget about that post and concentrate on my last one (#199) using the Princeton example which only uses three red hats for simplicity.

I understand the logic of common knowledge and why it's needed but now I'm stuck on why if at least 3 people are wearing the same colour hat why that common knowledge can't be deduced without specifically being told it by teacher.



Yes, as above forget about my previous posts (because the more I think about it the more I change my mind on what is and what isn't possible ).

Princeton's explanation makes sense to me when you have 0, 1 or 2 red hats because I now understand the point being made between the difference between individual and common knowledge. I don't understand why that common knowledge can't be deduced when there are 3 or more red hats (I'm tempted to e-mail John Geanakoplos to see if he can show me where I'm going wrong )

This is what I am struggling with as well. Someone explains the theory or a different example and I understand it perfectly. It all works out. Then I go back to the OP and apply the same rationale and it all falls to pieces.

Whatever way I look at it. That last kid with the red hat always has exactly the same knowledge and position regardless of class.

I can see how this works if we do the "kid can only see the child in fronts hat" because that removes the source of info that tells all the kids that there are either 6 or 7 red hats. When they can all see each others then when all the reds are gone from class then the next kid knows they must have a red hat on. (when there are 7 like in the OP).

 

[estebanrey]

Where's King Damager to explain his riddle!!

 

[Houd]

I love these kind of things, hadn't heard this one before so thank you King Damager.

I'm going to try to explain the value of the teacher saying "at least one hat is red" using an example, which might take some time so hopefully someone might have explained it to everyone's satisfaction in a more succint manner in the time it takes to write this.

As everyone seems to accept the information is important when there are 2 or less red hats, but that it breaks down once there are 3 red hats, as every child then knows "there must be at least 1 red hat", I will use the simplest such example with both red and black hats, 4 kids, 3 with red hats, 1 with a black hat.

Let's number them 1-4, with 1-3 wearing a red hat, and number 4 wearing a black hat.

They have no other information other than what they can see and their knowledge of what each other can see.

They are then asked in turn, 1 to 4, what hat they are wearing.

I think again everyone agrees that kid 1 and kid 2 will both say "I don't know".

We must now step into kid 3's head. Kid 3 thinks "I may be wearing a red hat, or I may be wearing a black hat". Kid 3 can see that kid 1 and 2 are wearing red hats and that kid 4 is wearing a black hat.

Kid 3 thinks to himself, "what was kid 2 thinking when he said 'I don't know'". We must now step into kid 2's head with only kid 3's knowledge (ie. not knowing if kid 3 has a red or black hat).

With that restrction, kid 2 can be seeing one of two things:
A) He can see kid 1 wearing a red hat (and kid 1 has already said "I don't know"), kid 3 wearing a red hat and kid 4 wearing a black hat OR
B) He can see kid 1 wearing a red hat (and kid 1 has already said "I don't know"), kid 3 wearing a black hat and kid 4 wearing a black hat.

In scenario A, he sees someone (kid 3) wearing a red hat who hasn't spoken yet and again, I think everyone agrees that in this case he must say "I don't know".

In scenario B, only kid 1 is wearing a red hat and kid 1 has said "I don't know", BUT AT THIS POINT HE CAN'T JUST SAY HE HAS A RED HAT ON, HE MUST STEP INSIDE KID 1'S HEAD (sorry for the caps but I feel this point needs emphasis).

So, still in scenario B, kid 2 thinks to himself "what was kid 1 thinking when he said "I don't know".

In scenario B, from kid 2's perspective, kid 1 can be seeing one of 2 things:
B1) He can see kid 2 wearing a red hat, kid 3 wearing a black hat and kid 4 wearing a black hat OR
B2) He can see kid 2 wearing a black hat, kid 3 wearing a black hat and kid 4 wearing a black hat

In scenario B1, kid 1 sees kid 2 wearing a red hat and kid 2 hasn't spoken yet, so he must say "I don't know".

In scenario B2, kid 1 sees only black hats (and here's the kicker) BUT CAN'T MAKE ANY ASSUMPTION ABOUT THEIR OWN HAT, BECAUSE THERE MIGHT ONLY BE BLACK HATS. Therefore he still says "I don't know".

So in either scenario B1 or B2, kid 1 has to say "I don't know". Then again in either scenario A or B, kid 2 will say "I don't know". So by the time we get to kid 3, he thinks "regardless of anything that happened before, kid 2 will say "I don't know", so I have no new information, my hat might still be red or black, so I have to say "I don't know". I hope we are all in agreement on that.


Now let's step back to scenarios B1 and B2, but in this case the teacher has said "at least one hat is red".

In scenario B1, nothing changes, kid 1 sees kid 2 wearing a red hat and kid 2 hasn't spoken yet, so he must say "I don't know".

But in scenario B2, where he sees only black hats, he knows his must be red, so he says "red". Now, as he didn't say "red", kid 2 knows that kid 1 did not see only black hats, so given that kid 2 knows that kid 1 said "I don't know", it can't have been scenario B2. Therefore kid 2 (and kid 3) can rule out scenario B2. But, had it been scenario B1, kid 2 would been in a situation where he sees kid 3 and 4 with black hats and kid 1 with a red hat and kid 1 has said "I don't know", so kid 2 would have said "red". But kid 3 knows that kid 2 said "I don't know", so it can't have been scenario B1 or B2, therefore kid 2 can't have seen scenario B.

So kid 3 now knows that kid 2 saw scenario A) kid 1 wearing a red hat, kid 3 wearing a red hat and kid 4 wearing a black hat.

Therefore, kid 3 knows they are wearing a red hat.

That's pretty convuluted and might take a few reads but I think it holds. Also there might be people thinking "well ok, maybe that holds for 3 red hats, but not for 4 or 5 or 17!". What I've tried to point out is that for the final kids (be it 3, or 253) logic to hold up, he must consider every permutation of every previous kid, up to and including the fact that first kid might have seen all black hats and needs to know that there is at least one red hat to make the "red" conclusion.

If you're still unconvinced for cases higher than 3, just try from this starting point and follow my logic from above. 5 kids, 4 red, 1 black. Kid 4 thinks what kid 3 is thinking when saying "I don't know". Kid 3 could be seeing:
A) Kid 1 red, kid 2 red, kid 4 red, kid 5 black
B) Kid 1 red, kid 2 red, kid 4 black, kid 5 black


Hope that helps, apologies if I'm talking rubbish!

 

[estebanrey]

I've e-mailed John Geanakoplos [author of the Princeton document] with my post (#199) to see if he can explain where I'm going wrong.

I doubt he'll reply but hey ho I'll post it if he does.

 

[estebanrey]

snip...

Thanks for taking the time to write all that and you've explained the process and chain of logic well.

Personally I understand all that now but my issue is why can't the "at least one red hat" bit being common knowledge (not just individual) be deduced when there are three or more red hats?

I don't know if you saw my post #199 above that uses an even simpler scenario (three kids all wearing a red hat)n and brilliantly explains the difference between the individual knowledge of knowing there is 'at least one red hat' and the common knowledge of it.

The explanation given there is that individually knowing there is 'at least one red hat' isn't enough and the reason the teacher telling them works is because it then becomes common knowledge amongst the group which has a small but important effect on the inductive logic and I now understand that point. But again, I can't see why the fact this common knowledge can't be deduced itself once you have 3 of the same colour hat (which I explained why in my post above #199).

The way I look at, for any given colour hat these are the cases for the numbers that exist without any additional information from teacher

1 = All of the group individually know at least one 'red' exists bar the person wearing it who has no knowledge it exists. There is no common knowledge and incomplete individual knowledge. Teacher Must Inform Them To Solve

2 = All of the group individually knows at least one 'red' exists but not all of the group can conclude the rest of the group know at least red hat exists (so no common knowledge). Teacher Must Inform Them To Solve

3+ = All of the group individually know at least one 'red' exists but also they all know each other knows at least one red hat exists. Kids can deduce common knowledge exists without being told

So again, I'm not questioning the logic chain, the consideration of permutations of what the previous people were thinking etc, or the importance between the difference of individual vs. common knowledge, I get all that. I want to know why a group of three hat wearers of the same colour can't deduce that the common knowledge of "at least one red hat" exists without being told it by teacher?

 

[Psiko]

Thanks for taking the time to write all that and you've explained the process and chain of logic well.

Personally I understand all that now but my issue is why can't the "at least one red hat" bit being common knowledge (not just individual) be deduced when there are three or more red hats?

I don't know if you saw my post #199 above that uses an even simpler scenario (three kids all wearing a red hat)n and brilliantly explains the difference between the individual knowledge of knowing there is 'at least one red hat' and the common knowledge of it.

The explanation given there is that individually knowing there is 'at least one red hat' isn't enough and the reason the teacher telling them works is because it then becomes common knowledge amongst the group which has a small but important effect on the inductive logic and I now understand that point. But again, I can't see why the fact this common knowledge can't be deduced itself once you have 3 of the same colour hat (which I explained why in my post above #199).

The way I look at, for any given colour hat these are the cases for the numbers that exist without any additional information from teacher



So again, I'm not questioning the logic chain, the consideration of permutations of what the previous people were thinking etc, or the importance between the difference of individual vs. common knowledge, I get all that. I want to know why a group of three hat wearers of the same colour can't deduce that the common knowledge of "at least one red hat" exists without being told it by teacher?

In your scenario 3+, consider the following. There are three people each with red hats. I (call me person 3) have a red hat, but I don't know it. I am the third person to guess and the two people before me said they did not know (call them persons 1 and 2). My logic has to go as follows: if I have a black hat, what would the guy before me (person 2) have said? He would first have wondered what the case would be if he had a black hat. If he had a black hat (and I had a black hat because we are already assuming that in this thought process) then what would the first guy have said? The answer: the first guy wouldn't have a clue because in this particular thought process (but not in reality) he can only see two black hats, and does not have the additional information that there is at least 1 red hat. So it is not possible to determine what one's hat colour is in this situation.

EDIT: Sorry, just realised I've posted pretty much the same as Houd.

 

[div0]

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....

http://i45.tinypic.com/2cbxbq.png

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).

Have been away travelling and just thought I'd have a quick look through this again. Your post above got me thinking, so I'll try to explain how I see it.

Ok let's take the example in your picture.

3 players, all red hats.

Player 1 sees: P2=R, P3=R
And from his point of view:
Player 2 sees: P1=?, P3=R
Player 3 sees: P1=?, P2=R

BUT, P1 is thinking (if I'm white), then P2 is looking at 1 red and 1 white (P3=R, P1=W).

So from P1's perspective, P2 could be thinking that he (P2) is also a white and so P2 (from P1's perspective) could be thinking that P3 (the red hat) is looking at 2 whites. So P3 (from P1's perspective of P2), would therefore not know that there is at least 1 red.

This has now got me wondering whether I was right to say that game 1 (in the OP) was solvable (due to there being at least 3 reds). I'll have to go back and think about it some more. I think the difference in the OP, is that there are more hats in total, creating a slightly different puzzle given the right/wrong circumstances.

 

[Psiko]

Have been away travelling and just thought I'd have a quick look through this again. Your post above got me thinking, so I'll try to explain how I see it.

Ok let's take the example in your picture.

3 players, all red hats.

Player 1 sees: P2=R, P3=R
And from his point of view:
Player 2 sees: P1=?, P3=R
Player 3 sees: P1=?, P2=R

BUT, P1 is thinking (if I'm white), then P2 is looking at 1 red and 1 white (P3=R, P1=W).

So from P1's perspective, P2 could be thinking that he (P2) is also a white and so P2 (from P1's perspective) could be thinking that P3 (the red hat) is looking at 2 whites. So P3 (from P1's perspective of P2), would therefore not know that there is at least 1 red.

This has now got me wondering whether I was right to say that game 1 (in the OP) was solvable (due to there being at least 3 reds). I'll have to go back and think about it some more. I think the difference in the OP, is that there are more hats in total, creating a slightly different puzzle given the right/wrong circumstances.

I have been thinking along similar lines to you div0. I don't think game 1 is solvable.

I think the point here is the teacher is not saying "In the current situation there is at least one red hat" because everyone can see this. What they are saying is, before anyone has even put a hat on, "we are going to play a game where you all wear either red or white hats and at least one of you will be wearing a red hat". This is why when the final person (in your notation P3) thinks through which hypothetical permutations of hats could lead to me having a white hat, they eventually get to a contradiction of no-one having a red hat.

 

[estebanrey]

I understand how you both are thinking but I don't agree, I've quoted div0's post and edited it into stages to help with my explanation of why the thought process div0 describes is flawed.....

Have been away travelling and just thought I'd have a quick look through this again. Your post above got me thinking, so I'll try to explain how I see it.

Ok let's take the example in your picture.

3 players, all red hats.

STAGE 1
Player 1 sees: P2=R, P3=R
And from his point of view:
Player 2 sees: P1=?, P3=R
Player 3 sees: P1=?, P2=R

BUT, P1 is thinking (if I'm white), then P2 is looking at 1 red and 1 white (P3=R, P1=W).

STAGE 2
So from P1's perspective, P2 could be thinking that he (P2) is also a white

STAGE 3
....and so P2 (from P1's perspective) could be thinking that P3 (the red hat) is looking at 2 whites.

YOUR CONCLUSION
So P3 (from P1's perspective of P2), would therefore not know that there is at least 1 red.

In short, the possibility P1 considers in 'STAGE 3' is invalidated as a possibility by the information P1 learned in 'STAGE 1'.

You are using recursive logic BUT....[and this is the thing I think hasn't been considered by the author of that riddle or the OP's lecturer]...any recursive logic must STOP once you reach the limit of what you know is physical possible (which is at the end of 'STAGE 2' where the red text begins above) from EVERYTHING you've learned before and not only on what you learnt could be true in the last stage. Put another way, you can only consider possibilities that are possible from everything you've learned before.

P1 knows P3 can't physically be looking at two whites as he learnt this in 'STAGE 1' (He learnt in 'STAGE 1' that P3 can see P2's red hat), so at no point can he ever think P3 is looking at two whites ('STAGE 3') , which is the end point of your recursive logic and thus invalidates the conclusion stated above. P1 knows the end point of your recursion chain isn't true at the start so you can't get there and must stop where the white text does (after 'STAGE 2'. At that point it he still knows that the whole group knows there is at least one red hat).

The reason I think recursive logic fails here is because at at each stage you getting to the next stage by only using the knowledge established in the last and forgetting what has been learnt in the stage(s) before that. You must consider everything you've learned from the whole process when going to the next stage in the recursion and not just the specifics of the last which is how your explanation works.

To explain it in an anology, it's like you are writing down what you know ('STAGE 1'). Then writing a second line down what you can deduce about what the next person can know from what you know ('STAGE 2'). Then rubbing out the first line ('STAGE 1'), forgetting/disregarding it and then writing a new line about what what the third person can know only from what Person 2 knows ('STAGE 3').

You are only checking each stage against the last and not all the previous stages. You can't get to 'STAGE 3' if it contradicts what you learnt in 'STAGE 1' (which was that each player knows there is 1 red hat).

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

Also, and this is a completely separate topic, there is another flaw in the OP's example in the wording. As we agree you can only consider things that are 100% certain, well in Scenario 2 you are discounting the possibility that the teacher is lying when she says "There is at least one red hat".

OK so that's a bit flippant that second one

 

[Bloomfield]

Estebanrey, why does all (3) red hats work, but no red hats doesn't?

 

[Houd]

I understand how you both are thinking but I don't agree, I've quoted div0's post and edited it into stages to help with my explanation of why the thought process div0 describes is flawed.....




In short, the possibility considered in 'STAGE 3' is invalidated as a possibility by the information you learned in 'STAGE 1'.

You are using recursive logic BUT....[and this is the thing I think hasn't been considered by the author of that riddle or the OP's lecturer]...any recursive logic must STOP once you reach the limit of what you know is physical possible (which is at the end of 'STAGE 2' where the red text begins above) from EVERYTHING you've learned before and not only on what you learnt could be true in the last stage. Put another way, you can only consider possibilities that are possible from everything you've learned before.

P1 knows P3 can't physically be looking at two whites as he learnt this in 'STAGE 1' (He learnt in 'STAGE 1' that P3 can see P2's red hat), so at no point can he ever think P3 is looking at two whites ('STAGE 3') , which is the end point of your recursive logic and thus invalidates the conclusion stated above. P1 knows the end point of your recursion chain isn't true at the start so you can't get there and must stop where the white text does (after 'STAGE 2'. At that point it he still knows that the whole group knows there is at least one red hat).

The reason I think recursive logic fails here is because at at each stage you getting to the next stage by only using the knowledge established in the last and forgetting what has been learnt in the stage(s) before that. You must consider everything you've learned from the whole process when going to the next stage in the recursion and not just the specifics of the last which is how your explanation works.

To explain it in an anology, it's like you are writing down what you know ('STAGE 1'). Then writing a second line down what you can deduce about what the next person can know from what you know ('STAGE 2'). Then rubbing out the first line ('STAGE 1'), forgetting/disregarding it and then writing a new line about what what the third person can know only from what Person 2 knows ('STAGE 3').

You are only checking each stage against the last and not all the previous stages. You can't get to 'STAGE 3' if it contradicts what you learnt in 'STAGE 1' (which was that each player knows there is 1 red hat).

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

But, and this is a completely separate topic, there is another flaw in the OP's example in the wording. As we agree you can only consider things that are 100% certain, well in Scenario 2 you are discounting the possibility that the teacher is lying when she says "There is at least one red hat".

OK so that's a bit flippant that second one

I feel like div0 is on the right track but not quite there (or possibly just phrased it a bit clumsily), so you are right, 'STAGE 3' isn't right, P3 doesn't suddenly think there's a possibility that P1 is looking at 2 whites, that's not possible. In fact, as you point out, no player can possibly think any other player is looking at 2 whites. That isn't the point though.

Let's start again. All 3 players wear red hats.

Put yourself in player 3's shoes. You see 2 red hats. You know that P1 sees at least 1 red hat (P1 sees P2 is red). You know P2 sees at least 1 red hat (P2 sees P1 is red). In fact, P3 is not in a unique position this way, all 3 players know that each other player sees at least 1 red hat. This seemingly is the same as being told beforehand that "there is at least one red hat".

So you are P3. You see 2 red hats. You think "what could P2 possibly think". He could see that I'm (P3) wearing a black hat and P1 is wearing a red hat. He then thinks "what could P1 possibly think". Well, P2 FROM P3's perspective (not actually P2 or P3's perspective), thinks that P1 might be seeing that P3 is wearing a black hat (because P3 thinks that P2 could be seeing P3 in a black hat) and that P2 is wearing a black hat (because P3 knows that P2 doesn't know his own hat colour, so might think it is black).

It isn't the fact that P3 suddenly forgets the fact that he knows P1 can see at least 1 red hat as you state (which is not possible). It is the fact that P3 can think that P2 thinks that P1 can see 2 black hats and also think that P2 thinks that P1 thinks his own hat might be black, which makes the logic fall down UNLESS PRIOR KNOWLEDGE IS GIVEN TO P1 THAT AT LEAST ONE HAT IS RED (so that P1 would know his hat is red if he sees 2 black hats).

Let me reiterate the key points. Every player knows at least one hat is red. Every player knows that every other player knows that at least one hat is red. From player 3's perspective, player 2 CANNOT know that player 1 is seeing at least one red hat and therefore cannot make a conclusion about player 1's answer, without player 1 having prior knowledge that if he sees only black hats, his own hat is red.

If you can explain how from player 3's perspective (thinking his own hat may be black), player 2 (thinking his own hat may be black) knows that player 1 sees at least 1 red hat, then I'm willing to accept the logic is flawed.

 

[estebanrey]

If you can explain how from player 3's perspective (thinking his own hat may be black), player 2 (thinking his own hat may be black) knows that player 1 sees at least 1 red hat, then I'm willing to accept the logic is flawed.

Because she can physically obverse it! Player 3 knows Player 1 can see 1 red hat (on top of P2's head).

Besides you admit the common knowledge already exists without being told it specifically here...

Put yourself in player 3's shoes. You see 2 red hats. You know that P1 sees at least 1 red hat (P1 sees P2 is red). You know P2 sees at least 1 red hat (P2 sees P1 is red). In fact, P3 is not in a unique position this way, all 3 players know that each other player sees at least 1 red hat. This seemingly is the same as being told beforehand that "there is at least one red hat".

You are making the same error as div0 (IMO).......

So you are P3. You see 2 red hats. You think "what could P2 possibly think". He could see that I'm (P3) wearing a black hat and P1 is wearing a red hat. He then thinks "what could P1 possibly think". Well, P2 FROM P3's perspective (not actually P2 or P3's perspective), thinks that P1 might be seeing that P3 is wearing a black hat (because P3 thinks that P2 could be seeing P3 in a black hat) and that P2 is wearing a black hat (because P3 knows that P2 doesn't know his own hat colour, so might think it is black).

It isn't the fact that P3 suddenly forgets the fact that he knows P1 can see at least 1 red hat as you state (which is not possible). It is the fact that P3 can think that P2 thinks that P1 can see 2 black hats.

No can't, P3 knows that isn't a possibility because he already knows for a fact P1 can see at least one red hat (on top of P2's head).

Your explanation as to why the 'extra information' is needed even above 3 hats & beyond is no different to div0's in essence. You are doing recursion one stage and a time and not checking the information known from the stages previously against the new possibilities.

For P3 to ever consider P1 might be looking at two white hats (however you want to get there) he has to ignore the established fact that he knows that is physically impossible (because P3 knows P1 can see P2's red hat). That is by definition illogical.

 

[Houd]

I fear we're going round in circles slightly so this will be my last attempt.

Player 3 does not (and indeed cannot) think that Player 1 sees 2 black hats.

Player 2 does not think that Player 1 sees 2 black hats.

Player 3 CAN think that Player 2 can think that Player 1 sees 2 black hats, if he couldn't, he couldn't correctly identify his own hat as red.

I can't state it any plainer than I have above and in previous posts, so if you disagree with this point then you're right, the extra knowledge doesn't mean anything, but Player 3 also couldn't say with certainty that they are wearing a red hat.

 

[div0]

I understand how you both are thinking but I don't agree, I've quoted div0's post and edited it into stages to help with my explanation of why the thought process div0 describes is flawed.....

Late back to the topic again.

What I'm trying to say (which I realise is different from my original stance), is:

Player x knows there is at least 1 red hat
Player x knows that Player y knows there is at least 1 red hat
Player x knows that Player z knows there is at least 1 red hat
Player x does not know that Player y knows that Player z knows there is at least 1 red hat.

I believe that this is what Houd is also saying.

 

[rubberduck]

I believe one of the points being overlooked is that the solution must be reachable before the children observe the game. Furthermore, a principle assumption IMO that has also been overlooked is that all of the children are perfectly rationale and this is common knowledge between them i.e. every child knows every other child knows every other child is rationale. Perhaps this is ultimately a moot point, I will have to think about it some more.

In the second case of the OP. Once the children have been given the information 'at least 1 red' we can assume that the strategy for playing the game simulataneously forms within the head of each child. Moreover, each child can know that the strategy has formed within the head of every other child (etc etc) as they are all rationale and all have the same information.

Because the strategy/solution is formed BEFORE the children observe the game it is vital that the strategy will work with EVERY POSSIBLE variation of the game i.e. upon opening their eyes, they observe 99 red hats and 0 black hats or 99 black hats and no red hats.

Once the children open their eyes and see the game they know that assumed rationality combined with common knowledge and therefore common strategy will mean the game can be played out with each having 100% certainty in the 'correct' (rationale) answers being given.

Regarding the first case. If the children assume the others are completely rationale but do not have the common knowledge of 'at least one red' it is impossible for them to reach a solution before they are able to observe the game. We know this because because (using the above example) one possible game would have them opening their eyes and seeing no red hats in which case they could not rationaly determine the colour of their hat without the common knowledge.

If they did use the above 'solution' from the second case they are acting irrationally. This particular strategy would therefore become self-defeating as it is derived under he assumption of common knowledge and that assumed common rationality but as soon as a player decides to assume common knowledge he/she is acting irrationally.

I believe what estebanrey is doing is ripping a solution that was reached under common knowledge in one game, and, as an observer, applying it to a game that is equivalent for his as an observer but not equivalent for the players who are the ones that ultimately need to develop a rationale solution. Unfortuantely, once the game has been defined i.e. the total number of red hats is know to us we are now observers of the game. It is therefore vital that any 'solution' works with every possible variation of the game. In the first case of the game there is an extra variation which cannot occur in the second case i.e. 0 red hats.