Consider $n$ independent, normally distributed random variables $X_i$ with mean $\mu$ and variance $\sigma$. Suppose the maximum value of the realizations of each of these random variables is $\text{max}\left(X_i\right)$, while the second largest value is $\text{smax}\left(X_i\right)$.
How would one calculate an expression, even if only asymptotically, for how the expected value of the difference between the two largest realizations of the $X_i$s depends on $n$? In other words, can we write an expression (or just an approximation) for
$$\mathbb{E}\left[\text{max}\left(X_i\right) - \text{smax}\left(X_i\right)\right]$$
as a function of $n$?
Having spent a bit of time working on the problem, I think I've found an answer and concluded that the problem is relatively trivial. I will let the votes (if any) decide if the question should be deleted or not.
Using a corollary from the Fisher-Tipett-Gnedenko theorem, we have that
$$\mathbb{E}\left[\text{max}(X_i)\right]\sim\sqrt{2\ln n}$$ as $n \to \infty$.
We can consider finding the expectation for $\mathbb{E}\left[\text{smax}(X_i)\right]$ through conditioning. For example,
$$\mathbb{E}\left[\text{smax}(X_i)\right] = \sum_j\mathbb{E}\left[\text{max}(X_{i\neq j})|X_j=\text{max}(X_i)\right]P(X_j = \text{max}(X_i))$$
This is saying that the expected value of the second-largest realization can be found by conditioning over the result that the $j$th random variable is the largest one, finding the expectation of the maximum value of the remaining random variables, and adding up over $j$ while weighing the sum over the probability that the $j$th random variable is the largest.
Because each of the $X_i$s is independent from each other and they all have the same distribution, one can find from symmetry and an application of the previous asymptotic estimate that:
$$\mathbb{E}\left[\text{smax}(X_i)\right] \sim \sqrt{2\ln (n-1)}$$
Therefore, we can use linearity of expectation and this asymptotic estimate to determine that:
$$\boxed{\mathbb{E}\left[\text{max}(X_i) - \text{smax}(X_i)\right] \sim \sqrt{2\ln n} - \sqrt{2\ln(n-1)}}$$