Gambling problem/ standard deviation

1. May 23, 2007

misu200

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. May 23, 2007

EnumaElish

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
3. May 23, 2007

misu200

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
4. May 23, 2007

EnumaElish

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.