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Topic: Yet another Monty Hall variant... (Read 654 times) |
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JocK
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Yet another Monty Hall variant...
« on: Aug 21st, 2005, 3:48am » |
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You're on a game show, and to select your prize you're given the choice of three doors. Behind one door is a car, and behind the other two doors is a goat. Once you've picked a door, the host - who knows what is behind the doors - will first open another door to reveal the goat behind it. He than will ask you "Do you want to switch?". Now, you have prepared very well for this moment supreme: you have watched the gameshow for years and kept statistics of which door hides the car. As a result you know that the three doors are not equally likely to hide the car. Rather, 3 out of 6 times the car is behing door three, 2 out of 6 times it is behind door two, and 1 out of 6 times it is behind door one. What door do you pick initially?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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towr
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Re: Yet another Monty Hall variant...
« Reply #1 on: Aug 21st, 2005, 7:55am » |
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First instinct: I pick the least likely door, because then Monty opening one of the other two doors should give me the most information.
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« Last Edit: Aug 21st, 2005, 7:58am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Icarus
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Re: Yet another Monty Hall variant...
« Reply #2 on: Aug 22nd, 2005, 4:21pm » |
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Good Instinct: The probabilities of winning are: Choose door #1: switch = 5/6; don't switch = 1/6. Choose door #2: switch = 2/3; don't switch = 1/3. Choose door #3: switch = 1/2; don't switch = 1/2.
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« Last Edit: Aug 22nd, 2005, 6:04pm by Icarus » |
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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