# Probability that one random number is larger than other random numbers

Suppose you have several sets of numbers, each with a normal distribution. For each set, you know the standard deviation and mean. Now, if you independently choose one random number from each set, what is the probability that the number chosen from a given set will be greater than each of the numbers chosen from the other sets?

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I'd prefer to express this in terms of random variables. Suppose $X_1, \ldots, X_n$ are independent normal random variables with means $\mu_1,\ldots, \mu_n$ and standard deviations $\sigma_1,\ldots,\sigma_n$ respectively. The conditional probability that $X_1$ is the greatest given that $X_1 = x_1$ is the probability that $X_2, \ldots, X_n$ all $\le x_1$, i.e. $\prod_{j=2}^n \Phi \left( \dfrac{x_1 - \mu_j}{\sigma_j}\right)$, where $\Phi$ is the standard normal cumulative distribution function. Thus the probability that $X_1$ is the greatest is $$\frac{1}{\sqrt{2\pi} \sigma_1} \int_{-\infty}^\infty e^{-(x_1 - \mu_1)^2/(2 \sigma_1^2)} \prod_{j=2}^n \Phi \left( \dfrac{x_1 - \mu_j}{\sigma_j}\right) \ dx_1$$ I don't expect this to be expressible in closed form in general.