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Direct simple solution of the envelopes paradox

  1. Dec 19, 2006 #1


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    The proposed (in)equality is A >=< p2A 2A + pA/2 A/2. Since the two amounts are either "A" and "2A," or "A" and "A/2," p2A > 0 implies pA/2 = 0, and vice versa. Therefore p2ApA/2 = 0. If pA/2 > 0 then

    pA/2 A >=< pA/2p2A 2A + pA/22 A/2

    pA/2 A >=< pA/22 A/2

    A >=< pA/2 A/2

    Clearly A > pA/2 A/2, and one would not switch.

    Alternatively, if p2A > 0 then

    p2A A >=< p2A2 2A + p2ApA/2 A/2

    p2A A >=< p2A2 2A

    A >=< p2A 2A

    1/2 >=< p2A, and one would switch only if 1/2 < p2A. Taking p2A = 1/2 in the original problem as a given, one would not switch.

    [Edit: this does not work in the general case where the values are kA and A/k with k > 2.]
    Last edited: Dec 20, 2006
  2. jcsd
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