Probability of normal random variable yielding highest value among other normal variables Let's say I have a competition with $N$ participants. Each participant yields a score that is normality distributed with unique means and unique variations. Each participant gets to post one score (we sample everyone once) and then we determine the winner. I want to know the probability of each participant posting the highest score.
I have seen a similar question when the means and the variations are all equal. I understand this and can write down a Bayes tree (which I did thinking it should help my problem) but I want too allow for each participants to have unique parameters. 
If I can only arrive at a solution via simulation, that's fine if there is no analytical solution, I would just like to know that is the case.
I started by thinking about how between two participants, the probability of participant $X_i$ beating $X_{j\neq i}$ is given by integrating over the cross-correlation (pdf$_{X_i} *\,$pdf$_{X_{j\neq i}}$) from 0 to infinity, where pdf$_i$ is the probability density function of participant $i$. So I claim to be able to write a probability where I compare two participants, $P(x_i > x_{j \neq i})$. I thought I could write down a solution in terms of these probabilities and attached them to branches of a Bayes tree... but I'm stumbling. It's writing down the conditional probabilities on the branches of the tree that I don't know how to do.
Some things I think are true:
$1 = \sum_{i=1}^{N}P(x_i$ is the largest value$)$
If I write a Bayes tree, the number of branches is $N!$
The number of branches where $x_i$ is the highest value is $(N-1)!$
Appreciate any help, thanks!
 A: Suppose $X_1, \ldots, X_n$ are independent absolutely continuous random variables with probability densities $f_{X_1}, \ldots, f_{X_n}$ and cumulative distribution functions $F_{X_1}, \ldots, F_{X_n}$.
Then by conditioning on $X_1$, we can see that the probability that $X_1$ is the biggest among $X_1, \ldots, X_n$ is
$$
\begin{aligned}
P(X_1 \geq X_2, \ldots, X_1 \geq X_n)
&= E[P(X_1 \geq X_2, \ldots, X_1 \geq X_n \mid X_1)] &&\text{(*)} \\
&= \int_{\mathbb{R}} P(X_1 \geq X_2, \ldots, X_1 \geq X_n \mid X_1 = x) f_{X_1}(x) \, dx \\
&= \int_{\mathbb{R}} P(X_2 \leq x, \ldots, X_n \leq x \mid X_1 = x) f_{X_1}(x) \, dx \\
&= \int_{\mathbb{R}} P(X_2 \leq x, \ldots, X_n \leq x) f_{X_1}(x) \, dx &&\text{(**)} \\
&= \int_{\mathbb{R}} P(X_2 \leq x) \cdots P(X_n \leq x) f_{X_1}(x) \, dx \\
&= \int_{\mathbb{R}} f_{X_1}(x) F_{X_2}(x) \cdots F_{X_n}(x) \, dx.
\end{aligned}
$$
In (*) we used the law of total expectation, and in (**) we used the independence of $X_1$ from $X_2, \ldots, X_n$.
I'll leave it to you to plug in whatever densities and cumulative distribution functions the individual random variables have.
Of course, there are analogous formulas to compute the probability that any of the $X_i$'s is the biggest.
