I accidentally asked my students a question that reduced to the following: find $P(X>Y, X>Z)$ where $X, Y$, and $Z$ are independent, $X$ is standard normal, and $Y$ and $Z$ are both normal with mean $1$ and variance $1$.
Then I sat down to write a solution to the question. Of course this probability can be rewritten as $P(X>\max(Y,Z))$. The CDF of $\max(Y,Z)$ is $\Phi(y-1)^2$ where $\Phi$ is the standard normal cdf. The joint density of $X$ and $\max(Y,Z)$ is therefore $\phi(x) {d \over dx} \Phi(y-1)^2$ or $2 \phi(x) \phi(y-1) \Phi(y-1)$. (Here $\phi$ is the standard normal density.) Integrating this joint density over the half-plane $x > y$ gives the expression $ \int_{-\infty}^\infty \phi(x) \Phi(x-1)^2 \: dx $ which can be evaluated numerically -- it's about $0.113202$ - but I don't recognize this number.
Is there some way to write this number in terms of, say, values of $\phi$ and $\Phi$ (without integrating)?
Also, the context here was as follows: let $X_1, \ldots, X_n$ be standard normal and let $Y_1, \ldots, Y_n$ be normal(1,1). Find the variance of the number of pairs $(i, j)$ such that $X_i > Y_j$. My method is to write that variance as a sum of $n^2$ indicators and find covariances of those indicators, leading to the probability I opened the problem with. Is there a solution to this problem that doesn't go via $P(X>Y, X>Z)$, which seems annoyingly hard to find?