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S _ _ Clue
THE KEY IS THAT THE PERSON AT THE BACK OF THE LINE WHO CAN SEE EVERYONE ELSE'S HATS CAN USE THE WORDS "BLACK" OR "WHITE" TO COMMUNICATE SOME CODED INFORMATION. SO WHAT MEANING CAN BE ASSIGNED TO THOSE WORDS THAT WILL ALLOW EVERYONE ELSE TO DEDUCE THEIR HAT COLORS? IT CAN'T BE THE TOTAL NUMBER OF BLACK OR WHITE HATS. THERE ARE MORE THAN TWO POSSIBLE VALUES, BUT WHAT DOES HAVE TWO POSSIBLE VALUES IS THAT NUMBER'S PARITY, THAT IS WHETHER IT'S ODD OR EVEN. SO THE SOLUTION IS TO AGREE THAT WHOEVER GOES FIRST WILL, FOR EXAMPLE, SAY "BLACK" IF HE SEES AN ODD NUMBER OF BLACK HATS AND "WHITE" IF HE SEES AN EVEN NUMBER OF BLACK HATS. 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 MIKE COOPER IS THE CULPRIT. WE CAN EXCLUDE MARTIN FROM OUR INVESTIGATIONS AS HE WAS THE VICTIM. TERRY SINGER POSSESSES AN ORANGE CAR. THE THIEF POSSESSES A BLUE CAR, THUS TERRY SINGER CANNOT BE THE CULPRIT. THE SUSPECT WHO OWNS A BLUE CAR WAS WEARING PURPLE SHOES. THE SUSPECT WHO WEIGHS 190 POUNDS WAS WEARING PURPLE SHOES. THE SUSPECT WHO WEIGHS 190 POUNDS IS NOT THE ONE WHO HAS BLACK HAIR. THEREFORE, ONE SUSPECT WAS WEARING PURPLE SHOES, POSSESSES A BLUE CAR, WEIGHS 190 POUNDS, AND IS NOT THE ONE WHO HAS BLACK HAIR. BUT JIM CONDON WAS WEARING BROWN SHOES, SYLVIAN BOGARD HAS BLACK HAIR, JONAH JAMESON WEIGHS 210 POUNDS, AND TERRY SINGER POSSESSES AN ORANGE CAR. THEREFORE, THE ONLY SUSPECT WHO COULD BE THIS SUSPECT IS MIKE COOPER. AS MIKE COOPER, OWNS A BLUE CAR AND THE THIEF OWNS A BLUE CAR, HE IS DEFINITELY THE THIEF 45 MINUTES. METHOD 1 LET US TALK ABOUT 24 HOURS AS IT WILL SIMPLIFY THE PROBLEM. IN 24 HOURS: THE COLD TAP WILL BE ABLE TO FILL THE BATHTUB 80 TIMES. THE HOT TAP WILL BE ABLE TO FILL THE BATHTUB 96 TIMES. THE DRAIN WILL BE ABLE TO DRAIN THE BATHTUB 144 TIMES. THUS IN 24 HOURS, THE BATHTUB WILL BE FILLED 80 + 96 - 144 = 32 TIMES (IF BOTH TAPS ARE TURNED ON WITH THE PLUG LEFT OUT) THUS IT WILL TAKE 45 MINUTES TO FILL THE TUB ONCE. METHOD 2 THE BATHTUB FILLING RATE IS: THE COLD TAP WILL FILL THE BATHTUB AT 1/18 TUB/MINUTE THE HOT TAP WILL FILL THE BATHTUB AT 1/15 TUB/MINUTE THE DRAIN WILL DRAIN THE BATHTUB AT 1/10 TUB/MINUTE THE NETT BATHTUB FILL RATE IS: 1/18 + 1/15 - 1/10 = 1/45 TUB/MINUTE THE WATER WILL FILL 1/45 OF THE TUB IN 1 MINUTE, SO THE TUB WILL BE FULL IN 45 MINUTES 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