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    Re: The Darn Old Cocked Hat - the sequel 1
    From: Luc Van den Borre
    Date: 2013 Mar 17, 21:12 +0100

    On 17/03/2013 20:17, Tom Sult wrote:
    > ------------------------------------------------------------------------
    > This explanation is a commonly held view but I think it is in error.
    > These are really two independent events. The first was a choice of
    > three. The second was a nonrandom elimination  of one of the choices.
    > This makes the second event an independent choice of one out of two.
    > Therefore keeping your current curtain or changing to the other curtain
    > results in the same 50-50 probability.
    
    I still remember the shame of getting this wrong in front of a hundred
    people at university, so I'll take a stab at explaining it.
    
    Imagine you do this experiment many times. You always choose door A as
    your initial choice. How often will you be correct if you stick with it?
    
    Obviously, since the prize is placed randomly, over many iterations door
    A will be the correct 1/3 of the time. So not switching doors has a 1/3
    chance of winning the prize. Monty doesn't move the prize around!
    
    Now, imagine you choose door A at first, but switch to the remaining
    door after Monty opens B or C.
    - One third of the time, the prize is behind A, Monty opens an empty
    door and you switch to the other empty door. Bad luck.
    - One third of the time, the prize is behind B, Monty opens C, and you
    switch to B to win the prize.
    - One third of the time, the prize is behind C, Monty opens B, and you
    switch to C to win the prize.
    
    So in conclusion, by switching you win 2/3 of the time, whereas by
    sticking with your initial choice you win 1/3 of the time.
    
    
    By the way, hello everyone! I recently got interested in celestial
    navigation on land. Most of the discussion so far is slightly over my
    head (badoom tish!).
    
    Luc Van den Borre
    
    

       
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