In a game of n people, each person is asked to select two other players. No player knows whether or not they've been selected. What is the probability that at least one person is not selected by anyone?

What we've got so far:

For n=3 players, it is impossible for anyone to be left out.

For n=4 players, the probability that one player is left out is $\displaystyle \frac{12}{81} = \frac{4}{27}$.

For n=5 players, the probability that one player is left out is $\displaystyle \frac{2340}{7776} = \frac{65}{216}$.

For n=5 players, the probability that two players are left out is $\displaystyle \frac{90}{7776} = \frac{5}{432}$.

For n=5 players, the overall probability that someone is left out is $\displaystyle \frac{2430}{7776} = \frac{5}{16}$.

For n=6 players, the probability that one player is left out is $\displaystyle \frac{380700}{1000000} = \frac{3807}{10000}$.

For n=6 players, the probability that two players are left out is $\displaystyle \frac{42120}{1000000} = \frac{1053}{25000}$.

For n=6 players, the probability that three players are left out is $\displaystyle \frac{540}{1000000} = \frac{27}{50000}$.

For n=6 players, the overall probability that someone is left out is $\displaystyle \frac{423360}{1000000} = \frac{1323}{3125}$.

For n players, the denominator will be $\displaystyle \binom{n-1}{2}^n$.

Any thoughts on how to generalize this, or any similar problems we could relate this back to? Thanks!