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P L _ C _ Clue
THE SHEEP WILL SURVIVE. IF THERE WERE 1 LION AND 1 SHEEP, THEN THE LION WOULD SIMPLY EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 2 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IF ONE OF THEM WOULD, IT WOULD SURELY BE EATEN BY THE OTHER LION AFTERWARDS. THE SHEEP WILL SURVIVE. IF THERE WERE 3 LIONS AND 1 SHEEP, THEN ONE OF THE LIONS COULD SAFELY EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 2 LIONS, WHERE NO ONE CAN EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 4 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 3 LIONS. THE SHEEP WILL SURVIVE. CONTINUING THIS ARGUMENT, THE CONCLUSION IS AS FOLLOWS: IF THERE IS AN EVEN NUMBER OF LIONS, THEN NOTHING HAPPENS AND THE SHEEP SURVIVES. IF THERE IS AN ODD NUMBER OF LIONS, THEN ANY LION COULD SAFELY EAT THE SHEEP AND THE SHEEP WILL NOT SURVIVE. THIS IS SIMILAR TO THE UNEXPECTED HANGING PARADOX LET'S SEE HOW IT WOULD PLAY OUT IF THE HATS WERE DISTRIBUTED LIKE THIS. THE TALLEST CAPTIVE SEES THREE BLACK HATS IN FRONT OF HIM, SO HE SAYS "BLACK," TELLING EVERYONE ELSE HE SEES AN ODD NUMBER OF BLACK HATS. HE GETS HIS OWN HAT COLOR WRONG, BUT THAT'S OKAY SINCE YOU'RE COLLECTIVELY ALLOWED TO HAVE ONE WRONG ANSWER. PRISONER TWO ALSO SEES AN ODD NUMBER OF BLACK HATS, SO SHE KNOWS HERS IS WHITE, AND ANSWERS CORRECTLY. PRISONER THREE SEES AN EVEN NUMBER OF BLACK HATS, SO HE KNOWS THAT HIS MUST BE ONE OF THE BLACK HATS THE FIRST TWO PRISONERS SAW. PRISONER FOUR HEARS THAT AND KNOWS THAT SHE SHOULD BE LOOKING FOR AN EVEN NUMBER OF BLACK HATS SINCE ONE WAS BEHIND HER. BUT SHE ONLY SEES ONE, SO SHE DEDUCES THAT HER HAT IS ALSO BLACK. PRISONERS FIVE THROUGH NINE ARE EACH LOOKING FOR AN ODD NUMBER OF BLACK HATS, WHICH THEY SEE, SO THEY FIGURE OUT THAT THEIR HATS ARE WHITE. NOW IT ALL COMES DOWN TO YOU AT THE FRONT OF THE LINE. IF THE NINTH PRISONER SAW AN ODD NUMBER OF BLACK HATS, THAT CAN ONLY MEAN ONE THING. YOU'LL FIND THAT THIS STRATEGY WORKS FOR ANY POSSIBLE ARRANGEMENT OF THE HATS. THE FIRST PRISONER HAS A 50% CHANCE OF GIVING A WRONG ANSWER ABOUT HIS OWN HAT, BUT THE PARITY INFORMATION HE CONVEYS ALLOWS EVERYONE ELSE TO GUESS THEIRS WITH ABSOLUTE CERTAINTY. EACH BEGINS BY EXPECTING TO SEE AN ODD OR EVEN NUMBER OF HATS OF THE SPECIFIED COLOR. IF WHAT THEY COUNT DOESN'T MATCH, THAT MEANS THEIR OWN HAT IS THAT COLOR. AND EVERY TIME THIS HAPPENS, THE NEXT PERSON IN LINE WILL SWITCH THE PARITY THEY EXPECT TO SEE. SO THAT'S IT, YOU'RE FREE TO GO. IT LOOKS LIKE THESE ALIENS WILL HAVE TO GO HUNGRY, OR FIND SOME LESS LOGICAL ORGANISMS TO ABDUCT LOCKER 1,000,000 WILL BE OPEN. THE NUMBER OF TIMES THAT EACH LOCKER IS FLIPPED IS EQUAL TO THE NUMBER OF FACTORS IT HAS. FOR EXAMPLE, LOCKER 12 HAS FACTORS 1, 2, 3, 4, 6, AND 12, AND WILL THUS BE FLIPPED 6 TIMES (IT WILL BE FLIPPED WHEN YOU FLIP EVERY ONE, EVERY 2ND, EVERY 3RD, EVERY 4TH, EVERY 6TH, AND EVERY 12TH LOCKER). IT WILL END UP CLOSED, SINCE FLIPPING AN EVEN NUMBER OF TIMES WILL RETURN IT TO ITS STARTING POSITION. SO IF A LOCKER NUMBER HAS AN EVEN NUMBER OF FACTORS, IT WILL END UP CLOSED. IF IT HAS AN ODD NUMBER OF FACTORS, IT WILL END UP OPEN. AS IT TURNS OUT, THE ONLY TYPES OF NUMBERS THAT HAVE AN ODD NUMBER OF FACTORS ARE SQUARES. THIS IS BECAUSE FACTORS COME IN PAIRS, AND FOR SQUARES, ONE OF THOSE PAIRS IS THE SQUARE ROOT, WHICH IS DUPLICATED AND THUS DOESN'T COUNT TWICE AS A FACTOR. FOR EXAMPLE, 12'S FACTORS ARE 1 X 12, 2 X 6, AND 3 X 4 (6 TOTAL FACTORS). ON THE OTHER HAND, 16'S FACTORS ARE 1 X 16, 2 X 8, AND 4 X 4 (5 TOTAL FACTORS). SO LOCKERS 1, 4, 9, 16, 25, ETC... WILL ALL BE OPEN. SINCE 1,000,000 IS A SQUARE NUMBER (1000 X 1000), IT WILL BE OPEN AS WELL IF YOU CHOSE TO GO TO THE CLEARING, YOU'RE RIGHT, BUT THE HARD PART IS CORRECTLY CALCULATING YOUR ODDS. THERE ARE TWO COMMON INCORRECT WAYS OF SOLVING THIS PROBLEM. WRONG ANSWER NUMBER ONE: ASSUMING THERE'S A ROUGHLY EQUAL NUMBER OF MALES AND FEMALES, THE PROBABILITY OF ANY ONE FROG BEING EITHER SEX IS ONE IN TWO, WHICH IS 0.5, OR 50%. AND SINCE ALL FROGS ARE INDEPENDENT OF EACH OTHER, THE CHANCE OF ANY ONE OF THEM BEING FEMALE SHOULD STILL BE 50% EACH TIME YOU CHOOSE. THIS LOGIC ACTUALLY IS CORRECT FOR THE TREE STUMP, BUT NOT FOR THE CLEARING. WRONG ANSWER TWO: FIRST, YOU SAW TWO FROGS IN THE CLEARING. NOW YOU'VE LEARNED THAT AT LEAST ONE OF THEM IS MALE, BUT WHAT ARE THE CHANCES THAT BOTH ARE? IF THE PROBABILITY OF EACH INDIVIDUAL FROG BEING MALE IS 0.5, THEN MULTIPLYING THE TWO TOGETHER WILL GIVE YOU 0.25, WHICH IS ONE IN FOUR, OR 25%. SO, YOU HAVE A 75% CHANCE OF GETTING AT LEAST ONE FEMALE AND RECEIVING THE ANTIDOTE. SO HERE'S THE RIGHT ANSWER. GOING FOR THE CLEARING GIVES YOU A TWO IN THREE CHANCE OF SURVIVAL, OR ABOUT 67%. IF YOU'RE WONDERING HOW THIS COULD POSSIBLY BE RIGHT, IT'S BECAUSE OF SOMETHING CALLED CONDITIONAL PROBABILITY. LET'S SEE HOW IT UNFOLDS. WHEN WE FIRST SEE THE TWO FROGS, THERE ARE SEVERAL POSSIBLE COMBINATIONS OF MALE AND FEMALE. IF WE WRITE OUT THE FULL LIST, WE HAVE WHAT MATHEMATICIANS CALL THE SAMPLE SPACE, AND AS WE CAN SEE, OUT OF THE FOUR POSSIBLE COMBINATIONS, ONLY ONE HAS TWO MALES. SO WHY WAS THE ANSWER OF 75% WRONG? BECAUSE THE CROAK GIVES US ADDITIONAL INFORMATION. AS SOON AS WE KNOW THAT ONE OF THE FROGS IS MALE, THAT TELLS US THERE CAN'T BE A PAIR OF FEMALES, WHICH MEANS WE CAN ELIMINATE THAT POSSIBILITY FROM THE SAMPLE SPACE, LEAVING US WITH THREE POSSIBLE COMBINATIONS. OF THEM, ONE STILL HAS TWO MALES, GIVING US OUR TWO IN THREE, OR 67% CHANCE OF GETTING A FEMALE. THIS IS HOW CONDITIONAL PROBABILITY WORKS. YOU START OFF WITH A LARGE SAMPLE SPACE THAT INCLUDES EVERY POSSIBILITY. BUT EVERY ADDITIONAL PIECE OF INFORMATION ALLOWS YOU TO ELIMINATE POSSIBILITIES, SHRINKING THE SAMPLE SPACE AND INCREASING THE PROBABILITY OF GETTING A PARTICULAR COMBINATION. THE POINT IS THAT INFORMATION AFFECTS PROBABILITY. AND CONDITIONAL PROBABILITY ISN'T JUST THE STUFF OF ABSTRACT MATHEMATICAL GAMES. IT POPS UP IN THE REAL WORLD, AS WELL. COMPUTERS AND OTHER DEVICES USE CONDITIONAL PROBABILITY TO DETECT LIKELY ERRORS IN THE STRINGS OF 1'S AND 0'S THAT ALL OUR DATA CONSISTS OF. AND IN MANY OF OUR OWN LIFE DECISIONS, WE USE INFORMATION GAINED FROM PAST EXPERIENCE AND OUR SURROUNDINGS TO NARROW DOWN OUR CHOICES TO THE BEST OPTIONS SO THAT MAYBE NEXT TIME, WE CAN AVOID EATING THAT POISONOUS MUSHROOM IN THE FIRST PLACE