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$ 6 . 6 6 Clue
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 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 THEIR HATS ARE ALL BLUE IN COLOR. THERE ARE 3 POSSIBLE HAT COLOR COMBINATIONS: [A] 1 BLUE, 2 WHITE [B] 2 BLUE, 1 WHITE [C] 3 BLUE THE COLOR COMBINATION OF 3 WHITE HATS IS NOT POSSIBLE SINCE THE KING HAS ALREADY SAID THAT AT LEAST ONE OF THE WISE MEN HAS A BLUE HAT. SO, LET'S START OUR ANALYSIS. WHAT IF THERE WERE ONE BLUE HAT AND TWO WHITE HATS? THEN THE WISE MAN WITH THE BLUE HAT WOULD HAVE SEEN TWO WHITE HATS AND IMMEDIATELY CALLED OUT THAT HIS OWN HAT WAS BLUE, SINCE HE KNEW THERE IS AT LEAST ONE BLUE HAT. THIS DIDN'T HAPPEN, AND THUS THE HAT COLOR COMBINATION [A] IS RULED OUT. NOW, THE WISE MEN KNEW THAT ONLY 2 HAT COLOR COMBINATIONS ARE POSSIBLE - COMBINATION [B] OR [C]. WHAT IF THERE WERE TWO BLUE HATS AND ONE WHITE HAT? THE MEN WITH THE BLUE HATS WILL SEE ONE WHITE HAT AND ONE BLUE HAT. THEY WILL CONCLUDE THAT COLOR COMBINATION [B] IS THE CASE AND WOULD CALL OUT BLUE AS THEIR HAT COLOR. THIS ALSO DID NOT HAPPEN, AND THUS THE HAT COLOR COMBINATION [B] IS ALSO RULED OUT. AFTER SOME TIME, WHEN NONE OF THE WISE MEN ARE ABLE TO IDENTIFY THE COLOR OF THEIR OWN HATS, COMBINATION [C] (OF 3 BLUE HATS) BECOMES THE ONLY POSSIBLE OPTION 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