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Monty hall problem

  1. May 25, 2012 #1
    I suppose you all know this famous problem. It is pretty clear to me why switching doors is beneficial, but I'm however unable to counter this argument from my friend:

    What is the difference between having picked a door and then the host revealing a goat, compared to not having picked one and then the host revealing the goat. Clearly there is difference but can someone elaborate in understandable terms because I tend to get very confused.
     
  2. jcsd
  3. May 25, 2012 #2
    The difference is that when you pick the door, the host must open ANOTHER door. Therefore if you had the goat in your door (which is 2/3 likely) then the door left closed will have the car 2/3 likely. This gives you information.

    If you hadn't picked a door, then the host could choose any of the doors with goats at random, thus not giving any other information about the closed doors.
     
  4. May 25, 2012 #3

    D H

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    To hammer this point home, imagine if there were 1,000 doors, one of which hides a car and each of the the remaining 999 doors hides a goat. You pick randomly a door. The probability you picked the right door is a paltry 1/1000. Monty then opens 998 doors, each of which shows a goat. Do you switch? Of course you do. The probability that the car is behind the unchosen, unopened door is 999/1000.

    Another way to look at it: Monty has just given you information, a whole lot of information in this case.
     
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