What is the probability of losing in the Taiwainese IMO team's game? 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

It is: $$p_{1}+\cdots+p_{n-1}$$ where: $$p_{k}:=\frac{n-2}{n-1}\frac{n-3}{n-1}\cdots\frac{n-k}{n-1}\frac{1}{n-1}$$ is the probability of losing together with $k$ mates that are in the same (losing) cycle.

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addendum:

Let's start with denoting the player who starts with $P_0$ and the player he points at with $P_1$. Inductively denote the player pointed at by $P_{n-1}$ with $P_n$. Then $P_0$ will lose with exactly $k$ mates in the same cycle if $P_0,\dots,P_{k-1}$ are distinct and $P_k=P_0$.

If e.g. $P_0$ loses together with $3$ mates $P_1$ must point at $P_2\
eq P_0$ (probability $\frac{n-2}{n-1}$), $P_2$ must point at $P_3\
otin\\{P_0,P_1\\}$ (probability $\frac{n-3}{n-1}$) and finally $P_3$ must point at $P_0$ (probability $\frac1{n-1}$).

This leads to $p_3=\frac{n-2}{n-1}\frac{n-3}{n-1}\frac{1}{n-1}$.