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28 April, 2012 at 5:17 pm #494518
@ian52hants wrote:
Unfortunatley this problem cant be solved from the information given. There are variious answers on the internet but they all fail without the knowledge that there is a mixture of colours on the folks heads.
The basic logic is
“If either of the two sighted can see two reds then they know their hat is green.
So by deduction there is either – two green hats and one red – OR – all the hats are green.
The Internet answers go –
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Start with prisoner A. They don’t know which hat they have on. If B and C were both wearing red they would know their hat was green . That must mean either B or C is wearing a red hat or they are both green. A could be wearing a red or green hat in this case, so A doesn’t know.
So, let’s go to B. B realising what A is thinking, B realises that either themselves and C is wearing a red hat. If B sees a green hat on on C, then B must be wearing a red hat. Since B does not know which hat they have on that means that B MUST see a red hat on C. So, that leaves them wondering if they have a green or a red hat on.
That leaves C (the blind dude). C realises that he must have a red hat, since neither A or B could deduce the colors of their own hats.
That’s the solution to the problem. There is no way that C could have a green hat and produce the same results. If B also had a green hat, student A would know he had a red hat. Or if B had a red hat, A would not know his own hat color, but B would know that since C had a green hat, B must have a red hat.”
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Unfortunately the answer does not hold true for the three green hat scenario because there is the assumption for B that A is seeing two two different colours, and for C also that there are two different colours in play. Without them knowing that there are at least two different colours on the heads it is impossible to solve through logic.
The internet answers such as the one above are couched in such a way as to appear plausible but they rely upon the assumption of two different colours.
Probably one of those things lost in transcription from the original.
The original post did mention that the blind man solved the puzzle, so the only scenario was that they all had to be wearing green.
28 April, 2012 at 5:19 pm #494519@pepsi wrote:
Aha!
Well done ! Just one thing though what about the blue hat ? Lol :shock:
:shock: :lol:
28 April, 2012 at 7:07 pm #494520OK. I’m not one to let go of a problem. I obviously know who had done what or not because Sceptical told me. I needed to know why it worked. So I got my resident geniuses on the case and got an explanation I could take in. And it goes like this:
The sighted prisoners could each see either two green hats or one red and one green hat. If the blind prisoner was wearing a red hat then one of the others would either have seen two red hats or noticed that the second sighted person had not seen two reds and spoken up.
I thought I’d put this up on the basis that if I needed it explained like this, then someone else probably needs it too.
I’m sure I’d have got it if my sinuses hadn’t been playing up.
28 April, 2012 at 7:31 pm #494521Major edake now!
28 April, 2012 at 7:46 pm #494522The jailer must have been a kind man, because through his choice of hats only the blind man could win his freedom.
28 April, 2012 at 8:45 pm #494523Yeah, and what about the white hats that google mentioned eh?!! pmsl
29 April, 2012 at 8:30 am #494524@ian52hants wrote:
My favourite one of these is
Mr and Mrs Brown have 7 children -If exactly half of them are girls what sex are the others?
Brings this to the fore as the next one to be solved….. thanks to Ian.
Goes off to think about it…. I hope you know the answer Ian…. :D
29 April, 2012 at 8:39 am #494525I was trying to do the arithmetic on this one…..because obviously its not possible to cut a child in half but actually its not asking you to do that…. is the answer simply… boys?
29 April, 2012 at 9:47 am #494526well, I’ll shut up then! :lol:
29 April, 2012 at 9:48 am #494527……………not for long! :D
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