IMO Hk TST, 1997

The answer is $\sqrt{2}-1$.
Let $P(n)$ be the probability that there is eventually no alien on the Earth given that there are currently $n$ aliens, where $n$ is a positive integer. Let $P(1) = c$. Since different aliens act independently, we have $P(n) = c^n$.
By the given actions, we know that
$$P(1) = \frac{1}{4} + \frac{1}{4}P(2) + \frac{1}{4}P(3) + \frac{1}{4}P(1).$$
This becomes $4c = 1 + c^2 + c^3 + c$, which can be factorized as
$$(c-1)(c^2+2c-1) = 0.$$
Thus, we have $c = 1$ or $-1 \pm \sqrt{2}$. As $c$ is a probability, we have $0 \le c \le 1$, so that $c = -1 - \sqrt{2}$ is rejected. We shall also reject $c = 1$ in the following.
Indeed, note that the expected number of aliens arisen from one of the aliens is $\frac{0+2+3+1}{4} = \frac{3}{2} > 1$. Therefore, the expected number of aliens increases indefinitely. So it is impossible that all aliens must eventually be destroyed. This shows $c < 1$.
Therefore, the probability is $c = \sqrt{2}-1$.