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$ 6 . 6 6 Clue
A MUST NOT HAVE SEEN TWO WHITE HATS ON B AND C, OR HE WOULD HAVE KNOWN HIS OWN HAT MUST BE BLACK SINCE THERE ARE ONLY TWO WHITE HATS. SO A'S ANSWER ESTABLISHES THAT AT LEAST ONE OF B OR C'S HAT IS BLACK. BASED ON A'S ANSWER, B KNOWS THAT HE AND C ARE EITHER BOTH WEARING BLACK, OR ONE IS WEARING BLACK AND ONE IS WEARING A WHITE HAT. IF B SEES THAT C IS WEARING A WHITE HAT, THEN HE WOULD KNOW HIS OWN HAT HAD TO BE BLACK. BUT B DOES NOT KNOW WHAT COLOR HAT HE IS WEARING, WHICH MEAN'S C'S HAT IS NOT WHITE AND MUST BE BLACK. SINCE BOTH A AND B CANNOT DEDUCE THE COLOR OF THEIR OWN HATS, C WILL KNOW THAT HE IS WEARING A BLACK HAT THE ANSWER MIGHT SEEM A LITTLE COUNTER INTUITIVE AT FIRST BUT WE'LL SEE... THE SHORT ANSWER IS THAT IT IS IN YOUR ADVANTAGE TO EXCHANGE. BUT WHY? WELL INITIALLY THERE WAS A 1/3 CHANCE THAT YOU WERE HOLDING THE ENVELOPE WITH THE NOTE IN IT AND A 2/3 CHANCE THAT THE NOTE WAS ON THE TABLE. THIS IS STILL THE CASE AFTER ONE OF THE ENVELOPES ON THE TABLE HAS BEEN REMOVED, THERE IS STILL A 1/3 CHANCE THAT YOU HAVE THE NOTE AND A 2/3 CHANCE OF IT BEING ON THE TABLE. IF THIS IS CONFUSING THEN IT MAY HELP TO THINK THAT THE QUESTIONER KNOWS WHICH ENVELOPE THE $20 NOTE IS IN, THOUGH IN PRACTICE IT DOESN'T ACTUALLY MATTER. THE QUESTIONER WOULD ALWAYS BE ABLE TO DEMONSTRATE THAT THE NOTE WAS NOT IN ONE OF THE ENVELOPES ON THE TABLE REGARDLESS OF WHERE THE NOTE WAS, SO THE FACT THAT HE WAS ABLE TO DO THIS CHANGES NOTHING. CONSIDER A DIFFERENT EXAMPLE.... SAY THERE ARE A 1000 ENVELOPES ON THE TABLE, 1 WITH A NOTE INSIDE. YOU PICK 1 ENVELOPE, THE CHANCE THAT THIS HAS THE NOTE IN IT IS CLEARLY 1/1000, WHERE AS THE CHANCE THAT IT IS STILL ON THE TABLE IS 999/1000. ODDS ARE ITS ON THE TABLE. NOW THE QUESTIONER COULD, ASSUMING HE CAN REMEMBER WHERE THE NOTE IS DEMONSTRATE TO YOU THAT THE NOTE IS NOT IN 998 OF THE ENVELOPES ON THE TABLE. IN THIS CASE NOTHING WOULD HAVE HAPPENED TO CHANGE THE FACT THAT THERE IS ONLY A 1/1000 CHANCE OF YOU HAVING THE NOTE. THAT IS WHY YOU EXCHANGE. WHAT IS THE VALUE OF THE EXCHANGE? SIMPLY BEFORE THE EXCHANGE YOU HAVE 1/3 OF $20 AND AFTERWARDS YOU WILL HAVE 2/3 OF $20, IE THE ADVANTAGE TO YOU IS ABOUT $6.66 SPLIT INTO 4 GROUPS: 2 GROUPS OF 3, 1 GROUP OF 2 AND YOU. ALL GO IN SEPARATE DIRECTIONS AND RETURN TO THE STARTING POINT IN 16 HOURS. THE TRICK IS TO DETERMINE WHO ARE THE SPIES. REMEMBER THAT THEY WILL NOT ALWAYS LIE. IF YOU FIND THE VILLAGE, TAKE THE SAME PATH AGAIN. IF YOU DID NOT FIND THE VILLAGE AND EVERYONE IN THE 2 GROUPS OF 3 ALSO REPORT THAT THEY DID NOT FIND THE VILLAGE, THEN IT IS DOWN THE PATH WITH THE GROUP OF 2 MEN AND THEY ARE BOTH SPIES. IF ONLY ONE OF THE GROUPS OF 3 HAS ONE MAN WHO SAY THAT THE VILLAGE IS THERE, THEN HE IS RIGHT AND THE OTHER TWO MEN ARE SPIES. IF BOTH GROUPS OF 3 ARE IN DISAGREEMENT, THEN THE MAJORITY VOTE WINS. EACH GROUP OF 3 HAS 1 SPY EACH. IF THE GROUP OF 2 IS IN DISAGREEMENT AND OF THE GROUPS OF 3 IS ALSO IN DISAGREEMENT, THEN WE CAN TRUST THE MAJORITY IN THE GROUP OF 3. IF IT'S NOT IN THE PATH OF EITHER OF THE GROUPS OF 3 MEN, THEN TAKE THE PATH WITH THE TWO IN DISAGREEMENT Q1: ASK GOD B, "IF I ASKED YOU 'IS A RANDOM?', WOULD YOU SAY JA?". IF B ANSWERS JA, EITHER B IS RANDOM (AND IS ANSWERING RANDOMLY), OR B IS NOT RANDOM AND THE ANSWER INDICATES THAT A IS INDEED RANDOM. EITHER WAY, C IS NOT RANDOM. IF B ANSWERS DA, EITHER B IS RANDOM (AND IS ANSWERING RANDOMLY), OR B IS NOT RANDOM AND THE ANSWER INDICATES THAT A IS NOT RANDOM. EITHER WAY, YOU KNOW THE IDENTITY OF A GOD WHO IS NOT RANDOM. Q2: GO TO THE GOD WHO WAS IDENTIFIED AS NOT BEING RANDOM BY THE PREVIOUS QUESTION (EITHER A OR C), AND ASK HIM: "IF I ASKED YOU 'ARE YOU FALSE?', WOULD YOU SAY JA?". SINCE HE IS NOT RANDOM, AN ANSWER OF DA INDICATES THAT HE IS TRUE AND AN ANSWER OF JA INDICATES THAT HE IS FALSE. Q3: ASK THE SAME GOD THE QUESTION: "IF I ASKED YOU 'IS B RANDOM?', WOULD YOU SAY JA?". IF THE ANSWER IS JA, B IS RANDOM; IF THE ANSWER IS DA, THE GOD YOU HAVE NOT YET SPOKEN TO IS RANDOM. THE REMAINING GOD CAN BE IDENTIFIED BY ELIMINATION. SOURCE: WIKIPEDIA