Let $x_i∼N(u_i,\sigma^2_i)$. For two normal distribution, I know that $P(x_1>x_2|u_1,u_2) = {\displaystyle \Phi }(\frac{u_1-u_2}{\sqrt{\sigma^2_1+\sigma^2_2}})$. But for more than two distributions, e.g. swimming, only 1 winner out of $n$ players, how to find $P(x_1>x_2,...,x_n)$ if the order of non-winners doesn't matter?

  • $\begingroup$ How can possibly negative $\sigma_1^2-\sigma_2^2$ appears under the square root? $\endgroup$ – NCh Jan 24 at 4:20
  • $\begingroup$ Sorry, just fixed it $\endgroup$ – chan ak Jan 24 at 4:30
  • $\begingroup$ Look at the @antkam's comment to the question math.stackexchange.com/q/3520600 $\endgroup$ – NCh Jan 24 at 7:58

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