Re: Probability question...
Looks to me like Jason is right.
He's *NOT* saying P(A beats B and C)=P(A beats B)*P(A beats C).
He's saying P(min{B,C,D}>x) = P(B>x)P(C>x)P(D>x), so that if we want P(A wins a 4-way race), we integrate f(x)P(B>x)P(C>x)P(D>x) over all x.
This still isn't a pretty integral to do - since for each x, we need need the CDF of a normal distribution 3 times. Just a matter of notation to write it in the above form instead of a quadruple integral, Integrate[f[w,x,y,z], {z,w,Infty}, {y,w,Infty}, {x,w,Infty}, {w,-Infty,Infry}] in Mathematica-ese.
|