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The necktie paradox

  1. Dec 15, 2009 #1
    There's discussion on the 'necktie paradox' on this blog, where I'm a regular visitor.

    I don't agree with the perspectives of those who have responded on that blog.

    In the wikipedia, The necktie paradox states that each stands to either win or lose an expensive tie, each at 50% probability, so the game has no advantage to either man.



    But, I look at this problem differently:

    Say, the cheaper necktie has value y, and
    the other one has the value (y + z) with z > 0.

    Let's assume that both men have an equal chance of being correct,
    the expected value in winnings for either man is,

    (0.50)(y + z) – (0.50)(y) = (0.50)z

    Both men are expected to make money if they bet.
    So both men are correct in choosing to bet.

    Is this correct?
     
    Last edited: Dec 15, 2009
  2. jcsd
  3. Dec 15, 2009 #2
    No, since you assume that z > 0, you have that y + z > y and will thus win by a probability 100% (not 50%)

    The correct way of calculating expected gain is:

    for player with necktie A (worth A dollars)
    P(B>A)*B - P(A>B)*A = 0.5*(B-A)

    for player with necktie B (worth B dollars)
    P(A>B)*A - P(B>A)*B = 0.5*(A-B)

    but since A>B or B>A with equal probability, both have equal chance of winning / losing the difference
     
  4. Dec 15, 2009 #3
    If you keep reading that wikipedia article, it answers your question.

     
  5. Dec 15, 2009 #4
    I read the whole wikipedia article, but i wasn't convinced... i needed to do the math, and see what others may say.

    I thought since we have 2 neckties of different $ values, the cheaper one is y and the other is (y+z), with z > 0, etc.

    Now, I see that my reasoning was flawed.

    I understand this:

     
  6. Dec 15, 2009 #5
    Only the more expensive tie is won or lost, so the expected winnings are zero:

    (0.50)(y + z) – (0.50)(y+z) = 0
     
  7. Dec 15, 2009 #6
    What do you, guys, think of the variation of the "necktie paradox": http://wapedia.mobi/en/Two_envelopes_problem [Broken]

    the expected value of the money in the other envelope is:
    (0.50) *1/2 * x + (0.50) * 2x = 5x/4

    Or

    the expected value in either of the envelopes 1.5 C, C being the lower of the 2 amounts

    Read about the new strategy for the two-envelope paradox:

    http://www.physorg.com/news169811689.html
     
    Last edited by a moderator: May 4, 2017
  8. Dec 15, 2009 #7

    CRGreathouse

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    In the paradox, the essential flaw is calculating relative gain rather than absolute gain. A $100 tie is 10 times more expensive than a $10 tie, but you can't say you expect to gain 900% or lose 90%, since the basis for the two is different. Better to say you stand to gain or lose $90 either way.
     
  9. Dec 16, 2009 #8
    Thanks to all of you for the feedback and help.

    Imagine if we had 3 people exchanging 3 neckties A1, A2 and A3

    Would you help me create an interesting problem on this basis?
     
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