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Gambling problem/ standard deviation

  1. May 23, 2007 #1
    Read gambling problem/Standard deviation
    Posted: May 23, 2007 4:57 AM
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    I'm analyzing some data from online gambling sites.I'm trying to prove they are stealing the customers.
    I'm not an expert in math ....just having "common sense" knowledge.

    Here is the problem I need to solve:

    I have a series of N independent event where I have a chance ( W ) to win some money ( P )

    At the end of day my mathematical expectation is :
    E= sum after i (W*P)

    In reality I will have won R dollars after these N independe events.

    If N is big then R and E should converge somehow.


    Is it possible to apply here some Standard deviation/Chebyshev's inequality/Weighted standard deviation to get some statistical interpretation about this?

    I would really like a formula that will say to me something like this:
    There is a 80% chance that the possible values of R to be in (E-x,E+x) range.
     
  2. jcsd
  3. May 23, 2007 #2

    EnumaElish

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    Assume R is distributed F (e.g. the normal distribution, even as an approximation). Normalize E = 0. Then you can calculate the probability R in (E-x, E+x) = (-x, x) as prob(-x,x) = 1 - 2F(-x). You can also solve for x* such that 1 - 2F(-x*) = 0.8.

    F doesn't have to be the normal. This approach will work for other distributions as well.
     
    Last edited: May 23, 2007
  4. May 23, 2007 #3
    Thanks. I will try to apply what you said.

    Until now I've tried to use weighted standard deviation / weighted mean

    with the weights being W(i):
    w(i)=W(i)

    and x(i) to be the outcome at the moment t(i)
    x(i) = {
    0, if you loose
    P(i),if you won
    }


    but probable that's not good.
     
    Last edited: May 23, 2007
  5. May 23, 2007 #4

    EnumaElish

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    Example:
    F is the uniform dist. over [-1,1]. Then F(x) = (x+1)/2.

    1 - 2F(-x*) = 1 - 2(-x*+1)/2 = 1 - (-x*+1) = 0.8 ===> x* = 0.8.
     
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