[gaming_mouse]

[ QUOTE ]
You could get away with a single integral here, if I'm interpreting the assumptions right:

P(A<B,A<C,A<D)=E[P(A<B,A<C,A<D | A)]=E[f(A)]

f(x)=P(x<B)P(x<C)P(x<D)

[/ QUOTE ]

Jason, I don't follow this. Are you saying you can just multiply the individual probabilities together? I could be wrong, but for some reason that didn't seem correct to me.

Also, would you mind taking a look at that other thread I started, down below?

Thanks,
gm