AHA! I think I managed the deriviation of your result with the help of the article above and of course your end result..

A moment please, and I shall post it here and you can let me know if I did it right

...

Given a set of nonidentically distributed random varialbes \(X_0, X_1, X_2,\ldots,X_n\) what is the probability that \(X_i\) is the maximum value.

\(P[X_i>Y_i]\) where \(Y_i = max(X_{i'}|_{\forall i'\neq i})\)

\(P[X_i>Y_i]=P[X_i-Y_i>0]=1-F_{X_i-Y_i}(0)\) where \(F_{X_i-Y_i}(0)\) is the CDF of \(X_i-Y_i\)

Take the convolution.. \(F_{X_i-Y_i}(0)=\int_{-\infty}^{\infty} F_i(0-(-y))\cdot f_y(y)dy=\int_{-\infty}^{\infty} F_i(y)\cdot f_y(y)dy\) where \(F_i\) is the CDF of \(X_i\)

\(f_y(y)\) is the PDF of Y so we can calculate Y's CDF and differentiate.

\(P[Y<a] = P[X_{i'}<a]\cdot P[X_{i'+1}<a]\cdots = \prod_{\forall i'\neq i}P[X_i'<a] = F_y(a)\)

\(F_y(y)=\prod_{\forall i'\neq i}F_i(y)\)

\(f_y(y)=\frac{d}{dy} F_y(y)= \frac{d}{dy}F_i \cdot F_{i+1} \cdots = f_i \cdot F_{i+1} \cdot F_{i+2} \cdots + F_i \cdot f_{i+1} \cdot F_{i+2} \cdots + \cdots\)

ok so my notation is a bit crazy... but I think it turns out the same as yours, Dason.

I'm just wondering now if there is a sneakier way to do it. I suppose it would be different depending on the distributions used for the random variables. I could take a hint from the article posted for the normal distributions..