TITLE: Objects are shuffled. What is the probability that exactly one object remains in its original position?
QUESTION [7 upvotes]: We have a deck with $n$ cards enumerated $1,2,\ldots,n$. The deck is shuffled. What is the probability of exactly one card to remain on its original position? What is the limit as $n$ rises to infinity?
$$
\begin{array}{rcl}
\{1\} & : & \dfrac 11 \\[6pt]
\{12,21\} & : & \dfrac 02 \\[6pt]
\{123,132,213,231,312,321\} & : & \dfrac 36 \\[6pt]
& \vdots
\end{array}
$$
At $n = 100 0$ and $10 000$ trials:   
$$
\begin{array}{rcl}
\text{value of }n & & \text{probability} \\
\hline
1000 & & 0.3739 \\
1001 & & 0.3689 \\
1002 & & 0.3722 \\
1003 & & 0.3638 \\
1004 & & 0.3707 \\
1005 & & 0.3664 \\
1006 & & 0.3616 \\
1007 & & 0.3728 \\
1008 & & 0.3702 \\
1009 & & 0.3801
\end{array}
$$
At $n = 100 000$ and $10 000$ trials:  

$\text{value of } n \quad \text{probability}$  
$\quad 100000 \quad \quad 0.3659$
$\quad 100001 \quad \quad 0.3552$
$\quad 100002 \quad \quad 0.356$
$\quad 100003 \quad \quad 0.367$
$\quad 100004 \quad \quad 0.3738$
$\quad 100005 \quad \quad 0.3647$
$\quad 100006 \quad \quad 0.3654$
$\quad 100007 \quad \quad 0.3637$
$\quad 100008 \quad \quad 0.3718$
$\quad 100009 \quad \quad 0.3708$

Apparently, probability approaches $0.36-0.38$, but how can one derive it analytically?

REPLY [11 votes]: By the inclusion-exclusion principle, in the symmetric group $S_n$ there are
$$ n!\sum_{k=0}^{n}\frac{(-1)^k}{k!} $$
permutations without fixed points (see derangements). It follows that in the same group there are
$$ n\cdot (n-1)!\sum_{k=0}^{n-1}\frac{(-1)^k}{k!} $$
elements with exactly one fixed point, and the limit probability is $\color{red}{\large\frac{1}{e}}$ in both cases.