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   Easy: Three Hats
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   Author  Topic: Easy: Three Hats  (Read 8082 times)
ootte
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Easy: Three Hats  
« on: Jul 24th, 2002, 1:59am »
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First a notation thing.
 
A: {x, y} -> {z} means B wears a x hat, C wears a y hat, and that leaves z hat left to A.
 
A: {<B>, <C>} -> {<A>}
 
1. A: {b, w} -> {b, w}
2. A: {b, b} -> {b, w}
3. A: {w, b} -> {b, w}
4. A: {w, w} -> {b}   => doesn't count, since then A would know about the hat he wears
 
Since A and B see C's hat, we come to
 
B: {<C>} -> {<B>}
 
 
1. B: {w} -> {b} (Because if C wears a white hat, and B and B would also wear a white hat, A would know about his black hat)
2. B: {b} -> {b, w}
 
That would make ({A, B, C})
 
1. {{b,w}, b, w}
2. {{b,w}, {b, w}, b}
 
A states he cannot conclude his own hat color, that means B and C wear {b,b}, {b,w}, or {w,b}. B states he can also not conclude his hat color. If B would see C wearing a white hat, he would know, that his hat color is black, but this contradicts that he doesn't know the color. So C must wear a black hat.
 
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Oliver
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jmlyle
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Re: Easy: Three Hats  
« Reply #1 on: Jul 25th, 2002, 9:12am »
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Of course, we're assuming that Man A and Man B have gone through the reasoning. If it was my friends, they'd just say;
How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing? How the hell would I know what color hat I'm wearing?
which would not give me much help as Man C.
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Alex
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Re: Easy: Three Hats  
« Reply #2 on: Jul 25th, 2002, 5:35pm »
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No, no. Person C has to be wearing a white hat. It works like this:
 
 
A looks infront of him. If B and C were both wearing black hats, he would know that he was wearing a white hat. But, he says that he doesn't know his hat color, so we know that inbetween B and C they are wearing either two white hats, or one black hat and one white hat.
 
No it's B's turn. If C was wearing a black hat, he would know that he was wearing a white hate (cause if C had a black hat, and B also had a black hat A would have known he had a white hate). But, B says that he doesn't know what his hat color is, meaning C must be wearing a white hat.
 
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ootte
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Re: Easy: Three Hats  
« Reply #3 on: Jul 25th, 2002, 10:33pm »
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on Jul 25th, 2002, 5:35pm, Alex wrote:
A looks infront of him. If B and C were both wearing black hats, he would know that he was wearing a white hat.

 
That's wrong. That would rule out the chance that A also wears a black hat. The riddle doesn't state this, that there are only two black hats, it states there are THREE black hats and TWO white hats in the box.
 
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Alex
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Re: Easy: Three Hats  
« Reply #4 on: Jul 27th, 2002, 8:03pm »
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Ooops. I read it wrong. My reasoning was correct, but I misread the number of each type of hat. I thought 3 white, 2 black.
 
I appologize! Smiley
 
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Rico
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Re: Easy: Three Hats  
« Reply #5 on: Jul 30th, 2002, 9:41pm »
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Of course, you're assuming that they answer in order: A, then B, then C. Nowhere in the riddle is this stated, it just says "WHEN A is asked... WHEN B is asked... WHEN C is asked..." What if C was asked first? It also doesn't say that they could each hear the others' responses. It also doesn't say that there is a mirror C can see... So if you can add the ungiven givens of them being asked in order and that they hear the previous response(s), you can add the ungiven given of C having a mirror.
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Luni_B
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Re: Easy: Three Hats  
« Reply #6 on: Aug 2nd, 2002, 10:49pm »
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none of the posts make any sense - especially the one with all the formulas!
 
I want someone who knows the answer to comment on this line of thinking...
 
FACT: man A would ONLY know what color he had on if he saw 2 white hats on the others.  
 
Therefore there is AT LEAST 1 black hat on another man.  Man B realizes this.  So if he looks at C and sees a white hat he would KNOW he had black. (because there had to be AT LEAST one)
 
BUT since man B couldn't tell his hat color he MUST have seen black.
 
Therefore man C will answer Black and be correct.
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[Jett]-CRX
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Re: Easy: Three Hats  
« Reply #7 on: Aug 3rd, 2002, 6:38pm »
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Luni_B,  
 
That is the same conclusion I came to and the same reasoning process I used to get there.
 
Good job!
 
~[Jett]-CRX
http://www.jettclan.net
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oren
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Re: Easy: Three Hats  
« Reply #8 on: Aug 8th, 2002, 5:05pm »
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It's funny how while the formulae make things seem like straight-forward reasoning, it usually turns out that there's a simpler, clearer, more convincing answer, like your's Luni_B. You concluded the same thing, for the same reasons, just you managed to say it a lot more simply.
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enigma
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Re: Easy: Three Hats  
« Reply #9 on: Aug 22nd, 2002, 3:44pm »
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he's wearing a black hat.
 
Reason - from the information we're given we con only conclude that man C is making an educated guess. Therefore assuming he is intelligent he would have guessed black (given a 60% chance of success vs 40% for guessing white), and since we know he was right...
 
 
--e
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Ash
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Re: Easy: Three Hats  
« Reply #10 on: Aug 25th, 2002, 6:37pm »
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Or...Mr. C looked in the hat box after the hats were passed out and saw two white hats.
 
(This works out exactly the same as Mr. A seeing at least one black hat, and Mr. B seeing a black hat as well.)
 
ash
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Mary T.
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Re: Easy: Three Hats  
« Reply #11 on: Aug 28th, 2002, 2:20pm »
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C has to be wearing a black hat. This is why..
 
If A says that he doesn't know, there's 2 probabilities. Either B & C are wearing 2 black hats, or a white and a black hat.  
 
B understands this, but since he doesn't know either - that means C MUST be wearing a black hat, because black is the only way B wouldn't know what he is wearing himself. (If C was wearing white, he'd know that he himself was wearing black and that would be the end of it)
 
C knows all of this and comes to the conclusion that he is wearing a black hat.
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Icarus
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Re: Easy: Three Hats  
« Reply #12 on: Oct 8th, 2002, 9:08pm »
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There is no mirror. Seeing in a mirror is still seeing, and C cannot see any hats. Also, since we are told that C knows what hat he is wearing, he didn't guess. Since the only way he could have known (barring a peek in the box) is by the senario practically everyone described in their own way, we also can conclude that A and B answered first, in order, and that C knew that they were able to figure out their hat colors if it were possible. Case closed! Wink
« Last Edit: Oct 8th, 2002, 9:14pm by Icarus » IP Logged

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Re: Easy: Three Hats  
« Reply #13 on: Dec 3rd, 2002, 1:12am »
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This could be classified as a Microsoft question.  Not because its hard, its actually one of the easier easy questions, but because my friend actually got asked this question in a Microsoft interview.  My friend is not that bright and didn't get the answer or the job.
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tymalynj
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Re: Easy: Three Hats  
« Reply #14 on: Dec 9th, 2002, 6:05pm »
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LUNI B,  who answered in August, absolutely had it right. Nothing more to say.
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jollytall
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Re: Easy: Three Hats  
« Reply #15 on: Feb 17th, 2006, 12:48pm »
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Sorry to dig out one that is really old (and there is nothing more to say according to tymalynj), but I want to read all the old ones before I share with you some riddles I know. Few comments:
 
- Unfortunatelly in case of these old ones, I cannot see how the thread started. E.g. this one starts with the reply using the math description.
- From the answers it seems, that the question was 3 black and 2 white hats, three people (A,B,C), A and B do not know the answer, so C knows (and he has black as it was correctly analysed).
- Actually this was one of the, if not THE, first riddle I have ever heard, and already it was thousand year old then (or more).
- In that version, the test was to choose the smartest man from the three men, and they could all speak when they switched on the lights (probably lit a candle then). Who answered first right was the smartest.
- So the first (who saw two black ones) thought: if I had white, the two others would see one white and one black each. One of them would be smart enough to realise that he had black, because otherwise it would be obvious to the third. So mine is black too. (Basically, it is the same solution said earlier in the thread).
- Why I still wanted to share my view on this is this:
 
I never really liked this one. If I were one of the three, I could have named my black hat even before the lights were on, simply because I would know that such an important test cannot be non symmetrical (i.e. different conditions). So in my view this is a silly riddle, even if you need to think a bit to solveSmiley
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Icarus
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Re: Easy: Three Hats  
« Reply #16 on: Feb 17th, 2006, 4:03pm »
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There is nothing missing from this thread. It was started to address one of the riddles on the Main riddle page (click on "Return to Riddles" on the bar at the top of this page). ootte was relying on the name of the thread to indicate which riddle it was for.
 
Originally, the forums were just a discussion arena for the Riddles, but as time went on, people started posting riddles here, and William gave up on trying to update the Riddle pages. So the practice changed. But these really old threads don't conform to the newer standard.  
 
By the way, this is not quite the riddle you are describing. It is the 3 riddle down on the Easy page.
« Last Edit: Feb 17th, 2006, 4:08pm by Icarus » IP Logged

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Re: Easy: Three Hats  
« Reply #17 on: Feb 18th, 2006, 5:45am »
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Thanks Icarus,
 
I found that page now: this makes other old threads also more understandable.
Anyway, even if the riddle is different the solution of the one I mentioned is very similar to the one of this riddle (as I already gave the solution too).
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Xetheare
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Simplest by Doubts  
« Reply #18 on: Dec 8th, 2006, 12:16pm »
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B & C cannot both have white hats for A would not have a doubt.
C is certain that B sees a black hat for this is the only configuration that leaves doubt.
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