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    Re: The Darn Old Cocked Hat - the sequel 1
    From: Don Seltzer
    Date: 2013 Mar 17, 13:12 -0700

    Tom Sult replied:
    >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.

    If the move from the three curtain choice to the two curtain choice was a random event then the probabilities may well change. In the event that it were a random choice occasionally the world cruise would have been revealed and eliminated. This would've placed the likelihood of winning at zero. But because that is never an option in this scenario they are actually two independent throws of the "dice".

    We have two strategies to consider:
    A. Always stick to the first pick
    B. Always switch to the other curtain that Monty did not reveal

    After my initial pick there are exactly nine possible cases:
    Prize My Pick
    1 1
    1 2
    1 3
    2 1
    2 2
    2 3
    3 1
    3 2
    3 3

    In Strategy A, regardless of what Monty subsequently reveals I stick to my guns and my initial pick will win in 3 out of 9 case.

    In Strategy B, I always switch my pick and I will win in the other six out of nine cases because of the extra information revealed.

    Don Seltzer

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