TITLE: What is the probability of losing in the Taiwainese IMO team's game?
QUESTION [6 upvotes]: Evan Chen's recount of the Taiwanese IMO team's journey recorded a game the team members played at their free time, which runs as the following:
There are $n$ team members (in the actual case $n=6$ but here we simply take $n\geq2$)Every team member points at another team member (whom must be different from him/herself) and thus we obtain a (directed) graph with $n$ vertices and $n$ edges. It has one more edge than a tree and therefore must contain a cycle. Any member who is a vertex of any cycle in this graph loses the game. (So it is possible that everyone loses but it's impossible that no one loses.)
Assume everyone chooses the person s/he points at randomly, what is the probability of a player losing the game? 
Reference: Evan Chen's recount

REPLY [3 votes]: Equivalently to the other answers so far, it's $\displaystyle\sum_{k=2}^n\frac{(k-1)!\binom{n-1}{k-1}}{(n-1)^k}$, or to make the sum work out nicely (adding the missing first term and shifting the index $k$ by one):
$$
-\frac{1}{n-1}+\displaystyle\sum_{k=0}^{n-1}\frac{k!\binom{n-1}{k}}{(n-1)^{k+1}}
$$
I'd just like to note that this can be simplified via using generating functions. We start with the simple generating function
$$
(1+x)^{n-1}=\displaystyle\sum_{k=0}^{n-1}\binom{n-1}{k}x^k.
$$
Note that for $y$ positive, $\displaystyle\int_0^{\infty}x^ke^{-x/y}dx=y^{k+1}\displaystyle\int_{0}^{\infty}z^ke^{-z}dz=k!y^{k+1}$ (using the substitution $x=yz$).
This is a convenient tool allowing us to get the desired $k!$ in our generating function. In particular, we find that
$$
\displaystyle\int_0^{\infty}(1+x)^{n-1}e^{-x/y}dx=\displaystyle\sum_{k=0}^{n-1}\binom{n-1}{k}\displaystyle\int_0^{\infty}x^ke^{-x/y}dx=\displaystyle\sum_{k=0}^{n-1}k!\binom{n-1}{k}y^{k+1}.
$$
Now the idea is clear, as we have that subsituting $y=\frac{1}{n-1}$,
$$
\displaystyle\sum_{k=0}^{n-1}\frac{k!\binom{n-1}{k}}{(n-1)^{k+1}}=\displaystyle\int_0^{\infty}(1+x)^{n-1}e^{-(n-1)x}dx=\displaystyle\int_0^{\infty}((1+x)e^{-x})^{n-1}dx,
$$
so we have that the probability with $n$ people is
$$
P_n=-\frac{1}{n-1}+\displaystyle\int_0^{\infty}((1+x)e^{-x})^{n-1}dx.
$$
This integral formulation is nice in that it allows us to evaluate this asymptotically, as the integral is
$$
\displaystyle\int_0^{\infty}e^{(n-1)(\ln(1+x)-x)}dx\approx\displaystyle\int_0^{\infty}e^{(n-1)(-x^2/2)}dx
$$
for large $n$, which is a Gaussian integral and evaluates to $\sqrt{\frac{\pi}{2(n-1)}}$. The $\frac{1}{n-1}$ term can thus be ignored and we get an asymptotic result of $\sqrt{\frac{\pi}{2(n-1)}}$.