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So I got a new fun one from friend. Again, try your best, post answer in spoilers, and have fun!
I think most people probably can't wait too long to see the final answer so I'll just put the answer in spoiler of the day, as I don't have much time as before where I could post puzzle for a day, wait for answers, and post right answer the next day...
Question:
There are 3 people, one is blind.
There are 2 white hats, 3 black hats.
I put 3 of those hats on the 3 people. Each person can see other two's hat color but not his own, blind person cannot see anything.
They want to guess their own hat color
There is a long silence then, the blind person says "I know my own hat color"
How?
+ Show Spoiler [answer] ++ Show Spoiler [u sure? it'll spoil all the fun] +
Blind person is wearing black hat, here's why:
Let's call the 3 people p1,p2,b.
If b=white, then 2 cases:
b = white, p1 = white, p2 = black (1)
b = white, p1 = black, p2 = black (2)
b = white, p1 = black, p2 = white (symmetric to 1)
In case 1, p2 will see 2 white, and say that he himself is black.
In case 2, p1 will see a white and a black, and not know his own color and remain quiet. However, p2, noticing that p1 is quiet, will reason:
If I am wearing white, then p1 would've said something as in case 1. Since p1 is quiet, I must be wearing black.
Therefore p2 will say that he himself is black.
Therefore the blind person cannot be wearing white.
Therefore he must wear black
 
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you need to include the fact that they need to guess their own hat colour. at the moment, you give them the hats and they do not need to guess anything.
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+ Show Spoiler +I think in the question it should say that the people do say something if they know their own hat colour. And it's not really a math puzzle, it's more like a logic puzzle.
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It would help if you explained the people want to say their own hat color.
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On November 09 2009 13:12 betaben wrote:
you need to include the fact that they need to guess their own hat colour. at the moment, you give them the hats and they do not need to guess anything.
thx, added.
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Well, assuming that the men have some reason to announce their hat colors, he might reason
+ Show Spoiler +
1. If either sighted man sees two white hats, he will know his own hat is black. Therefore, the blind man's hat is black or both sighted men's hats are black.
2. If the blind man's hat is white, both sighted men's hats must be black. Therefore, since each sighted man can go through the same reasoning: "If both my hat and the blind man's hat are white, the other man would know his is black. Since he hasn't said anything, my hat must be black," but in fact, neither has figured this out, the blind man knows his hat must be black, thus giving the other men no useful information.
edit: problem was corrected as I typed
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+ Show Spoiler +
If he and one other are white then the remaining person would be able to determine themselves as a black hat.
If he was the only white hat, one of the other two would be able to determine themselves as a black hat. This is because they would realize that the other could not determine himself as a black hat following the above, so the other person must not see two white hats.
Therefore he realizes that if he has a white hat, one of the other two can solve the problem. Because they cannot solve it, he must have a black hat.
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Hong Kong20321 Posts
is it this
+ Show Spoiler +
the room is pretty damn dark and if everyone was wearing black hats no one would've been able to see anythnig and thus the long silence.
so thus the blind person is also wearing a black hat
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Hong Kong20321 Posts
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On November 09 2009 14:00 alffla wrote:is it this + Show Spoiler +
the room is pretty damn dark and if everyone was wearing black hats no one would've been able to see anythnig and thus the long silence.
no.
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Lol... the question is poorly phrased. It's less of a math problem, and more of a Mensa brainteaser. Like a guy in a scuba suit is found in a forest, how does he get there?
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Hong Kong20321 Posts
On November 09 2009 14:04 meeple wrote:
Lol... the question is poorly phrased. It's less of a math problem, and more of a Mensa brainteaser. Like a guy in a scuba suit is found in a forest, how does he get there?
puts on a scuba suit and walks to a forest???
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United States4796 Posts
Question was cool. Nice puzzle, thanks evan!
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+ Show Spoiler +No, in case 2, p2 was actually suspicious that p1 was trying to trick him by remaining quiet, and so he remained quiet himself. Therefore the blind guy wouldn't know his color.
Sure you can add depth to the problem to give an answer, but then you can add even more depth to make that answer invalid.
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On November 09 2009 13:13 dcberkeley wrote:+ Show Spoiler +I think in the question it should say that the people do say something if they know their own hat colour. And it's not really a math puzzle, it's more like a logic puzzle.
Math is logic.
I mean numbers are made of empty sets, come on.
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On November 09 2009 14:04 alffla wrote:Show nested quote +On November 09 2009 14:04 meeple wrote:
Lol... the question is poorly phrased. It's less of a math problem, and more of a Mensa brainteaser. Like a guy in a scuba suit is found in a forest, how does he get there? puts on a scuba suit and walks to a forest???
Nope sorry, his name is aquaman and he's an action figure that a little kid was playing with while camping. Good try though
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On November 09 2009 14:10 El.Divino wrote:
Question was cool. Nice puzzle, thanks evan!
no problem :3
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yeah question needs to be reworded to say "if a person knows his hat color then he will say something," I didn't get the long silence part as part of the problem. I deduced black and was right for the wrong reason.
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i didnt know they would say their own hat color if they knew it.....
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On November 09 2009 14:00 alffla wrote:is it this + Show Spoiler +
the room is pretty damn dark and if everyone was wearing black hats no one would've been able to see anythnig and thus the long silence.
so thus the blind person is also wearing a black hat
I solved it using trivial logic initially, but I find this solution much more clever 
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This is pretty much the same problem as what was posted recently. Different conditions but almost the almost identical deductions, you can do better than that...
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On November 09 2009 15:48 igotmyown wrote:
This is pretty much the same problem as what was posted recently. Different conditions but almost the almost identical deductions, you can do better than that...
link me
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I see. You can't expect me to know every blog in existence and command me to "do better than that" though :p
I got tons of other puzzles, just stay in tuned! 
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just saw this thread
these questions are not really a puzzle, they kind of solve themselves.. but anyway
here's my thought process
+ Show Spoiler +i'ma call em s1 s2 B (non-blind dude 1, nonblind-dude 2, blind dude) in that order
so what can we have? let's just go thru the possibilities
BWW
WBW
WWB
BBW
BWB
WBB
BBB
we can't obviously have a BWW or a WBW situation cos then s1 or s2 will instantly know they're B
so we're down to..
WWB
BBW
BWB
WBB
BBB
so.. hmm gets interesting
S1:W sees WB, S2:W sees WB
S1:B sees BW, S2:B sees BW
S1:B sees WB, S2:W sees BB
S1:W sees BB, S2:B sees WB
S1:B sees BB, S2:B sees BB
Now we can also see we can't have the second situation, because if they see a BW situation, they'd auto-know they were B
we're down to..
S1:W sees WB, S2:W sees WB
S1:B sees WB, S2:W sees BB
S1:W sees BB, S2:B sees WB
S1:B sees BB, S2:B sees BB
now they can't figure shit out
so the blind dude comes in, does this thinking somehow, and figures out "holy shit, i must have a black hat"
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Oh noes, I'm dumber than the blind person assumes everyone is =_=
And in the other problem. What if D saw 2 white, didn't trust the guards and just wanted to live 9 more minutes...
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On November 09 2009 14:25 Kwidowmaker wrote:Show nested quote +On November 09 2009 13:13 dcberkeley wrote:+ Show Spoiler +I think in the question it should say that the people do say something if they know their own hat colour. And it's not really a math puzzle, it's more like a logic puzzle. Math is logic. I mean numbers are made of empty sets, come on.
Yeah this is so right =D.
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EPT![[image loading]](http://i35.tinypic.com/24mx114.jpg)
