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Probability: What are the odds?

  1. Feb 5, 2009 #1
    My company split in two.
    Two owners, Fred and Jeff. Each got half the clients.
    Since then, Fred lost 235 clients and Jeff lost 285.
    My co-worker says it's just random variability.
    I says no way, that's one in a million.
    What are the odds? What's the equation?
    p.s. true story
     
    Last edited: Feb 5, 2009
  2. jcsd
  3. Feb 5, 2009 #2

    HallsofIvy

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    Randomness would assume this is a binomial distribution with probability p1 Fred will lose a client, probability p2 Jeff will lose a client. For large numbers those can be approximated by normal distributions but to do that you would need to know the total number of clients each one started with and I don't see how you can decide whether it was "random variability" or not without knowing how many clients there were to start with. If the original number of clients was not much more than 235+ 285= 520, this is probably not "random variability". If the original number of clients was much more, then it probably was.
     
  4. Feb 6, 2009 #3
    Oops, you're right. I thought at first it was just a coin toss x 520.
    In round numbers, the total was 2,000. So each began with 1,000 clients.
     
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