Derangement probability distribution question: Why $\text{Var}[x] \approx E[X]$ is fulfilled? Suppose I have shuffled a deck of $N$ distinct cards, the probability that none of the cards stays in their original position is simply $ \frac{!N}{N!} \to \frac1e =0.3678...$ (by derangement).
Now let $X_N$ denote the probability distribution for the total number of cards that stay in their original position.
Is there a statistical interpretation of why for sufficiently large $N$, the values of $E[X_N]$ and $\text{Var}[X_N]$ are only slightly above one, and their difference is almost 0?
For example, $E[X_{100}] \approx 1.0000000000000002, \text{Var}[X_{100}] \approx 1.0000000000000004$.
And, $E[X_{101}] \approx 1.0000000000000002, \text{Var}[X_{101}] = 1$.
(The results above are obtained from a code I've written)
I know that $\text{Pr}[X=k] = \dfrac{\binom{50}k \cdot !(50-k)}{50!} $
Is there any statistical reasoning behind these "near to 1" values?
 A: They're near $1$ because they are $1$.
Let $X_i$ denote the indicator random variable that is $1$ if $i$ is fixed and $0$ if $i$ is moved. Then $X = \sum X_i$ counts the number of fixed points and the expected number of fixed points, by linearity of expectation, is
$$ \mathbf{E}X = \sum_{i = 1}^n \mathbf{E}X_i = \sum_{i = 1}^n \frac1n = \frac nn = 1. $$
Note that $i$ has an equal probability of moving to any position $j$ so the probability that $i$ is fixed is $1/n$.
Then for the variance we have $\operatorname{Var}(X) = \mathbf{E}X^2 - (\mathbf{E}X)^2$
\begin{align*}
\mathbf{E}X^2 &= \mathbf{E} \left(\sum X_i\right)^2  \\
&= \mathbf{E} \left[ \sum_i X_i^2 + \sum_{i \ne j} X_iX_j \right]
\end{align*}
Note that $X_i^2 = X_i$ and $X_iX_j$ is $1$ if $i$ and $j$ are both fixed and $0$ otherwise. So $\mathbf{E}[X_iX_j] = (n-2)!/n!$ (number of permutations fixing $i$ and $j$ divided by the total number of permutations). Therefore, we have
\begin{align*}
\mathbf{E}X^2 &= \sum_{i = 1}^n \mathbf{E} X_i + \sum_{i \ne j} \mathbf{E}[X_iX_j] \\
&= \sum_{i = 1}^n \frac1n + \sum_{i \ne j} \frac{1}{n(n-1)} \\
&= n \cdot \frac1n + n(n-1) \cdot \frac{1}{n(n-1)} \\
&= 1 + 1 = 2.
\end{align*}
And therefore the variance is $2 - 1 = 1$.
