sure.

W(i) is between 0 and 1

For example:

Imagine that you are in a casino and:

At t0 time you have a chance of 20% to win 100$.

This means W(0) = 0.2 and P(0)=100

At t1 time you have a chance of 70% to win 150$.

This means W(1) = 0.7 and P(1)=150

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,,,

At t(N-1) time you have a chance W(N-1) = 0.8 to win P(N-1) = 70

On average you expect to win E=W(0)P(0)+W(1)P(1) ........W(N-1)P(N-1)

In reality you win R

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.

I've tried to use "weighted standard deviation" (

http://www-minos.phyast.pitt.edu/disdocs/weightsd.pdf ) and "weighted mean" (

http://www-minos.phyast.pitt.edu/disdocs/weigmean.pdf )

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

}

I'm not sure if this is right.

Thanks,