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Bertrand's Box Paradox and Monty Hall Problem

  1. Sep 27, 2008 #1
    This thing is making me pull my hair out: http://en.wikipedia.org/wiki/Bertrand's_box_paradox Can someone give me a good explanation of this?

    As for the Monty Hall Problem, I think I understand it. This is how I think of it: There is a 1/3 possibilty of picking the correct door and 2/3 possibility of picking the wrong door. And after one of the empty doors has been revealed, then by switching your choice, that 2/3 possibility of picking the wrong door becomes a 2/3 possibility of picking the correct door. -- Is this a legitimate way of understanding it?
     
    Last edited: Sep 27, 2008
  2. jcsd
  3. Sep 27, 2008 #2
    All the explanation was trying to say was that when you drew a gold coin, there is a 1/3 chance of drawing each gold coin of the three gold coins. 2 of these gold coins when checked with its partner will give another gold since both are in the same drawer whereas the last gold coin is with a silver coin in a drawer. Therefore, the chance that the partner coin is gold is 2/3.
     
  4. Oct 6, 2008 #3

    LAF

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    Monty Hall Problem: it´s easy to understand it with 10 doors. There is a 1/10 possibilty of picking the correct door and 9/10 possibility of picking the wrong door. You make your choice. And after 8 of the empty doors has been revealed, you choice still 1/10 possibilty of picking the correct door and the other closed door is 9/10 possibility of picking the correct door.
     
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