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Thread: A two-envelopes puzzle (probability)

  1. #1
    Sep 2008

    A two-envelopes puzzle (probability)

    You are handed two envelopes, and you know that each contains a positive integer dollar amount and that the two amounts are different. The values of these two amounts are modeled as constants that are unknown. Without knowing what the amounts are, you select at random one of the two envelopes, and after looking at the amount inside, you may switch the envelopes if you wish, A friend claims that the following strategy will increase above 1/2 your probability of ending up with the envelope with the larger amount: toss a coin repeatedly, let X be equal to 1/2 plus the number of tosses required to obtain heads for the first time, and switch if the amount in the envelope you selected is less than the value of X. Is your friend correct?
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  2. #2
    MHF Contributor Matt Westwood's Avatar
    Jul 2008
    Reading, UK

    If your envelope contains 1 dollar, then tossing the coin once gives you 1.5 which is more than 1 so switching will give you an envelope with a positive number different from 1, which is by def. greater than 1. So it works for the smallest case.

    If your envelope contains 2 dollars, then there is a chance that the other one contains 1 dollar, but (on the face of it) a bigger chance that there is more in the other envelope. However, there is a 50% chance of a head on the first throw and therefore there's a 50% chance you'll not switch.

    Goodness this is making my head hurt ... I'd need to think about it harder and dig out some of the probability maths that I've forgotten.
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