# Probability of more than three normal distributions

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?

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