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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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An answer of balance | Comment 39 of 75 |
Many people have said that there are 2 of each colour hat and indeed this is correct (at least I believe so), however there has not been a valid explanation as to why this must be.
In this problem there are only 3 possible combinations of hats:
RYBRRR 4a, 1b, 1c
RYBRRY 3a, 2c, 1c
RYBRYB 2a, 2b, 2c
*Note : The colours can be interchanged but that is irrelevant to the solution.

The key to this problem is the timing, at first no one stood up and then all stood up at once.
Now if we look at the first case of RYBRYB. If this were to be how the hats were arranged then immediately the ones with Y and B hats would have stood up as they would see that their colours were not present while the four with the red hats would still be puzzled as to what their colour was and it would only be after the Y and B hats stood up they would realize that each of those two had not seen their own colour and then all the reds would know what they were wearing. This arrangement has no delay and everyone would not rise at once.
Secondly, if there were 3 red 2 yellow and 1 blue immediately the blue would stand up as there would be no other of their colour while the rest would remain seated. Once again there is no delay and all do not rise at the same time.
Finally the only possible arrangement is to have 2 of each colour. At first each would look around the room and see 2 of two colours and one of another thus not immediately being able to deduce the colour of their cap. Then after a while each would notice that no one has stood up making it clear that there is no one there who sees only 2 colours (this would mean that they wear the missing colour) if there is no one who sees only 2 colours (i.e. each see 2 of 2 colours and 1 of a third) each can deduce that as there is no odd one out then the only wat this can occur is if each wears the same colour hat. Realising this at the same time they all stand.
  Posted by Michael on 2004-11-29 02:02:55
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