No need to invoke Chebychev, you can use the

central limit theorem.

Let $\displaystyle X_1, X_2, ..., X_n$ be the results from $\displaystyle n$ tournaments, which we'll assume are independent and identically distributed. The operator $\displaystyle \mbox E$ is the expected value and $\displaystyle \mbox{Var}$ is the variance. Then obviously:

$\displaystyle \mbox E X_1 = 20.50 x + 9.7 y + 4.3 z - 6.50 (1 - x - y - z)$

and

$\displaystyle \mbox E X_1 ^ 2 = 420.25 x + 94.09 y + 18.49 z + 42.25 (1 - x - y - z)$

so $\displaystyle \mbox{Var}X_1 = \mbox E X_1 ^ 2 - \mbox E X_1$, which we'll denote $\displaystyle \sigma^2$. Your (empirical) PPT is given by $\displaystyle \bar X = n^{-1} \sum X_i$ and your (empirical) ROI is PPT / 6.5. It follows that

$\displaystyle

\mbox{Var PPT} = \frac {\sigma^2}{n}

$

and

$\displaystyle

\mbox{Var ROI} = \frac{\sigma^2}{(6.5^2) n}.

$

From the central limit theorem, for suitably large $\displaystyle n$ we have that both PPT and ROI are approximately normal, with the correct mean and variance as given above. So, you now have the asymptotic distribution of both of these statistics.

With this you can answer a few somewhat interesting questions, like "what is the probability that I will be a losing player over my next 1000 games" for various values of x, y, and z.