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The Logician's Birthday (Posted on 2004-11-17) Difficulty: 2 of 5

A logician invites 6 of his logician friends to help him celebrate his birthday. Each of the 6 guests is wearing a hat which is either red, yellow, or blue, and the logician host informs them all that there is at least 1 of each color. After they eat the cake, the host stands up and exclaims there is a special party prize for the first person who can deduce what color hat is on their head. The party guests all looked around the room at each other but no one claimed the prize immediately. Suddenly all 6 guests stood up and correctly identify what color hat was on their head.

What were the colors of their hats and how did they know?

See The Solution Submitted by Erik O.    
Rating: 3.5417 (24 votes)

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Some Thoughts Another Aproach: | Comment 9 of 75 |

 

Now let's think as one of the logicians:I see two blue hats, two yellow hats and a red hat. I might have any of this colors.

But I know, that if there is a logician who doesn't see a color at the other 5 , he would immediatly know he has that color and he would say it.

If nobody say it, that means everybody see all the colors at least once. But because there are only 6 hats and 3 colors, the only possibility for this to happen is that there are two hats of each color,so that anyone see all the colors .In this moment I know that the hat I wear has a pair and that pair is the red hat I see alone.

In the same way must think all 6 logicians, when they all say the solution.

So a solution is (red,red,yellow,yellow,blue,blue)

Suppose I see three red hats, one blue and one yellow, and I wear red but I don't know it.The blue and yellow would see the solution and they would tell it.Because they tell which hat they wear it means they didn't see that color at the others including me.So, I know too, that I don't wear blue or yellow and so, I wear red for sure.The same thing is done by the other three red logicians.

So, another solution is (color1,color2,color3*4);

The big problem appears when is (color1,2*color2,3*color3);In this case the only one who can see the solution is color1, but that thing just doesn't help the others.This is not a solution.

Of, course there is a condition in the problem that says: "no one claimed the prize immediately but suddenly all 6 guests stood up and correctly identify what color hat was on their head " and in this case the second solution is eliminated because for the second the prize is claimed by the logicians who wear color1,color2 hat.Color3 waits for them to say the answer so that color3 logicians to tell for sure their color.

 But it is a nice solution too. :)

 

 


  Posted by cezar on 2004-11-18 23:38:12
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