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View Diary: Morning feature: The Monty Hall problem (with poll and statistics questions answered) (310 comments)

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  •  There is a very simple solution to the Monty Hall (1+ / 0-)
    Recommended by:
    nathguy

    problem, which is given as follows:

    Take the 10-door version of the MH problem. If all the doors are closed, your chosen door will have a 1/10 chance of being correct. The remaining 9 doors "collectively" will have a 9/10 of having the correct answer. If the host now opens 8 of the remaining 9 doors empty, then the remaining unchosen door will still retain a 90% of being correct, for two reasons,:

    (a) the host knew the correct answer and withheld the answer, or  

    (b) even if the host did not know the correct answer, by blind luck he did not choose the correct door...

    In either case, there is a 90% probability of choosing the correct door by switching from the original choice, which still remains at 10%.

    Thus, the 3-door Monty Hall problem is the simplest form of the 10, or 100, or 10,000 door problem. Remember, computer solutions of the 3-door MH problem are not programmed to know the correct answer. In other words, the original 1/3 odds for the original chosen door does not change, while there is a "collective" 2/3 chance for the two remaining doors to have the correct answer. By switching doors in the 3-door MH scenario, one simply shifts from the "one door 1/3 odds set" over to the "collective two door 2/3 odds set" (again, assuming that the door that is opened by the host is empty).  

    The key to understanding the Monty Hall problem is the concept of "group statistics", which current probability theory does not allow for. Only by using elementary set theory in probability statistics will this logical confusion among professional mathematicians get finally fixed. This is the main reason most statisticians (or pure number theorists, such as Paul Erdos) do not get (or even understand) the correct answer to the Monty Hall problem...

    "The blackbird whirled in the autumn winds. It was a small part of the pantomime." Wallace Stevens

    by mobiusein on Tue May 05, 2009 at 07:28:54 AM PDT

    [ Parent ]

    •  This is incorrect (1+ / 0-)
      Recommended by:
      Toon

      It makes a great deal of difference whether Monty knows which door has the car, and the computer programs WERE programmed to know that.

      If Monty does not know which door, then switching makes no difference

      If Monty does know which door, then switching improves your odds.

      •  the monty hall problem defines (1+ / 0-)
        Recommended by:
        plf515

        monty as knowing which door to open to reveal a goat.

        now you could define it as

        "Monty asks Jay to please open a door,"  but
        it doesn't matter what monty knows, merely that the act of opening
        a door, compresses the other door set to have a much higher probability
        of being a car..

        George Bush is Living proof of the axiom "Never send a boy to do a man's job" E -2.25 S -4.10

        by nathguy on Wed May 06, 2009 at 06:48:07 AM PDT

        [ Parent ]

        •  It does matter (1+ / 0-)
          Recommended by:
          plf515

          In the puzzle statement, Monty knows and always opens a goat door, telling you nothing about your own door.

          pick the car: 1/3
           - monty opens a door with a goat: 1/3
             - stay to win or change to lose
          pick a goat: 2/3
           - monty opens a door with a goat: 2/3
             - stay to lose or change to win

          Results: Stay wins 1 in 3. Change wins 2 in 3.

          If Monty didn't know and opened a non-chosen door at random, you get a goat revealed 2/3 of the time and a car 1/3 of the time. If you still get to change after that, you get a no-risk win 1/3 of the time.

          pick the car: 1/3
           - randomly open a door with a goat: 1/3
             - stay to win or change to lose
          pick a goat: 2/3
           - randomly open a door with a goat: 1/3
             - stay to lose or change to win
           - randomly open the door with the car: 1/3
             - change to win

          Results: No risk win 1 in 3. In the uncertain choice cases (2 in 3), stay wins 50% and change wins 50%.

          Deranged neoconservative militarism isn't the solution to nuclear proliferation; it's a cause. -- Glenn Greenwald

          by factbased on Wed May 06, 2009 at 09:27:20 AM PDT

          [ Parent ]

    •  It's stunning Paul Erdos gets it wrong (1+ / 0-)
      Recommended by:
      plf515

      I studied information theory as part of my undergrad, so,
      I understand how a little bit of information wildly varies outcomes for probabilistic events,  and that conveyed information should alter gaming strategies.  

      I have sat down with lots of smart guys and been the only one who has it
      right on the monty hall problem, so it is very counter intuitive.

      George Bush is Living proof of the axiom "Never send a boy to do a man's job" E -2.25 S -4.10

      by nathguy on Tue May 05, 2009 at 07:41:48 AM PDT

      [ Parent ]

    •  Not quite.... (1+ / 0-)
      Recommended by:
      plf515

      If there are 100 doors and the computer opens them totally at random, there's a very good chance the computer will open the "right" door long before you're down to only two doors.  Among the times when the computer hasn't already opened the right door, once it gets down to two doors, it's a 50/50 question as to which door is the winner.  The other probabilities are included in the times you never get to a two-door problem because the computer shows you the winning door earlier.

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