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$ 6 . 6 6 Clue
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 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 BOTH WILL HAVE THE SAME AMOUNT. LET'S SAY THERE'S 100ML OF TEA AND 100ML OF COFFEE INITIALLY AND THE TEASPOON HOLDS 10ML OF LIQUID. AFTER TRANSFERRING 10ML OF LIQUID FROM THE TEA TO THE COFFEE CUP AND ANOTHER 10ML FROM THE COFFEE TO THE TEA CUP, THE QUANTITY WILL REMAIN THE SAME - 100ML OF COFFEE AND TEA MIX IN THE TEA CUP AND 100ML OF COFFEE AND TEA MIX IN THE COFFEE CUP. SO LET'S SAY AFTER THE MIXING IS DONE THERE IS 99ML OF TEA AND 1ML OF COFFEE IN THE TEA CUP. THE 1ML OF TEA HAS TO BE IN THE COFFEE CUP. THAT MEANS THAT THE COFFEE CUP HAS 99ML OF COFFEE AND 1ML OF TEA. IN OTHER WORDS, IF THERE'S (X)ML OF TEA MISSING FROM THE TEA CUP, THEN THERE HAS TO BE AN EQUAL AMOUNT OF OF COFFEE MISSING FROM THE COFFEE CUP