Lethal russian roulette (modifed) $21$ players stand in the circle. Every player chooses aim for the shot - another player. No one can shoot themselves or into the air. 
What the probability that there are at least $2$ players that have taken aim at each other? 

The solution is
$$1+\sum_{s=1}^{\lfloor n/1 \rfloor}(-1)^s\frac{n!}{2^ss!(n-2s)!(n-1)^{2s}}$$
with $n=21$
The problem is explained in
http://msor.victoria.ac.nz/twiki/pub/Courses/MATH214_2009T1/Enumeration/214_2009_notes3.pdf
page 20, where it is called Zen stares.
Credits go to the people from www.wiskundeforum.nl (a Dutch math forum). 
