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  1. #1
    5 Star Lounger Ruff_Hi's Avatar
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    Anyone want to trade

    There are 2 sealed envelopes on the desk, one of which contains exactly twice as much money as the other. You select one of the envelopes and open it to reveal $2n. Therefore the other envelope contains either $n or $4n with equal probability.

    Your expected gain, if you now trade envelopes, is $n/2 (5n/2 less 2n) and hence it is advantageous to do so, regardless of the value of n. However, by the same reasoning, had you originally selected the other envelope, it would have been advantageous to trade it for the envelope you are already holding.

    How can this be?
    (Location Australia, then UK, but now USA. Heart, outlook, attitude, etc always Australian)
    Quote: "All Happiness is the release of internal pressure"

  2. #2
    Plutonium Lounger
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    Re: Anyone want to trade

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">It's not fair to say that the amount in the other envelope is $n or $4n.

    Say that the amounts in the envelopes are n and 2n.
    If you picked the envelope with $n, you'll get $2n if you trade.
    If you picked the envelope with $2n, you'll get $n if you trade.
    So both if you keep the first envelope, and if you trade, the expected amount is $1.5n.</span hide>

  3. #3
    Platinum Lounger
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    Re: Anyone want to trade

    The folly is that you're using "n" to represent different amounts in each case. Rephrasing the problem, with n in one envelope and 2n in the other, your expected gain, by swapping envelopes is:
    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">

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