What is the probability of throwing darts sufficiently far away? Suppose I have a dart board with radius 1 (1 foot, 1 meter, it don't matter). Suppose I throw darts one at a time and have decent aim, so they always land on the board, though they do so randomly.
What is the probability that the first two darts I throw are one unit of distance or greater away from on another?
I've set up my problem in terms of complex numbers polar coordinates.  If there is an easier interpretation, please let me know.
Let $r_1 \sim U(0,1)$,$r_2 \sim U(0,1)$, $\theta_1 \sim U(0,2\pi)$, $\theta_2 \sim U(0,2\pi)$.
If my two darts are $z_1 = r_1\exp(i \theta_1)$, $z_2 = r_2\exp(i \theta_2)$ The distance between two darts is
$$ \lvert z_1 - z_2 \rvert^2 = r_1^2 + r_2^2 -2r_1r_2\cos(\theta_1-\theta_2)$$
Performing a quadruple integral seems hopeless.  Is there a more efficient way to think of this?  The square of a uniform random variable has a known distribution, and so do the difference and product of iid uniform random variables.  Can I somehow leverage that?
EDIT:
I suppose I should parameterize my darts as $\sqrt{r_j }\exp(i\theta_j)$.
 A: Let's use the polar system and let $R,\theta$ denote the coordinates of the first hit.
If we may say that "uniform distribution over the unit disk" can be modeled by saying that (1) the distance form the origin, $R$, is uniform over $[0,1]$ and (2) the related angle $\theta$ is also uniform over $[0,2\pi]$ and  (3) $\theta$ and $R$ are independent then the distribution of $R,\theta$ is uniform over $[0,1]\times[0,2\pi]$.
Assume that $\theta$ is fixed. Choose a new coordinate system so that the angle be $0$. Note that $R$ remains the same. So we can follow with $\theta=0$ and $R$. 
Let's consider the following figure. Here $R=r$. The second hit will be further than one unit from the first one if it is in the grey area.

The surface area of the gray domain is
$$A_{\text{grey}}(r,0)=\pi-2\left(\color{red}{\cos^{-1}\left(\frac r2\right)}-\color{blue}{\frac r2\sqrt{1-\frac{r^2}4}}\right).$$
That is, conditioned on a given angle and a given $R=r$, the probability we are after is $\frac{A_{\text{grey}}(r,0)}{\pi}.$ In order to get the unconditioned probability, we have to integrate the conditional probability over $[0,1]$ with respect to $r$. (Note that the conditional probability does not depend on $\theta$ so we may omit the integration with respect to the angle.) Now, with the help of Wolfram Alpha, the result of the integration is easy to get:
$$\int_0^1 \frac{A_{\text{grey}}(r,0)}{\pi} \ dr\approx 0.31.$$
