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Apr 00

Riddle Ye! VI & Riddle Ye! V Solution

Courtesy of Bonhomme De Neige.

Riddle Ye V Solution
Let us label the 3 men A, B, and C. We shall designate a liar with the letter L, a truth-teller with T, and the one who tells both lies and thruth with M. There are only 6 permutations of the letters L,T, and M. Arrange them thus:

  A B C
1:T L M
2:T M L
3:L M T
4:L T M
5:M T L
6:M L T

Then we ask A “Is it true that the probability of getting a truthful answer from B is greater than that of getting one from C?”

Case 1: A says yes. This eliminates permutations 1 and 4. So C is not an M. Ask him “Are you the man who sometimes lies and sometimes tells the truth?” — depending on his answer you know whether he’s a liar or a truth-teller. Then ask him whether B is a the man who lies and tells truth.

Case 2: A says no. This eliminates permutations 2 and 3. So then B can’t be an M. From there same as above, really.

Riddle Ye VI
You have 12 coins, and you know one (and only one) of them is counterfeit. You know the counterfeit coin weighs differently from a real one, but you don’t know whether it weighs less or more. You have a balance (not a set of electronic scales =P), and you are allowed 3 weighing to determine which coin is the counterfeit. How do you do it?