Let $p_0$ be a fixed point at $(0, 0)$. Let $p_1$ be a uniformly distributed random point inside the square $(-9, -9), (9, 9)$, and let $p_2$ be another random point inside the square $(11, -9), (29, 9)$.
Let $d_1$ be the distance between $p_0$ and $p_1$, and $d_2$ be the distance between $p_0$ and $p_2$. What is the probability of $d_2 < d_1$?
Using a montecarlo simulation I was able to estimate this probability to be arround $0.000394379$, but I would like to know where it comes from.
My attempts to solve this geometrically have failed, but here is a plot for intuition:
The two points must lay outside of the small circle, but inside of the bigger one, and still there is a chance that $p_1$ will be closer.
The condition $d_2\lt d_1$ can be fulfilled for $11\le d_2\lt d_1\le9\sqrt2$. For $d_1,d_2\in[11,9\sqrt2]$, the measures of the angle ranges that lie within the respective squares are $4(\frac\pi2-2\arccos\frac9{d_1})$ for the first square and $2\arccos\frac{11}{d_2}$ for the second square. Thus the desired probability is
\begin{eqnarray*} &&\frac8{18^4}\int_{11}^{9\sqrt2}\mathrm dd_1d_1\left(\frac\pi2-2\arccos\frac9{d_1}\right)\int_{11}^{d_1}\mathrm dd_2d_2\arccos\frac{11}{d_2} \\ &=& \frac8{2\cdot18^4}\int_{11}^{9\sqrt2}\mathrm dd_1d_1\left(\frac\pi2-2\arccos\frac9{d_1}\right)\left(d_1^2\arccos\frac{11}{d_1}-11d_1\sqrt{1-\left(\frac{11}{d_1}\right)^2}\right) \\ &\approx& 0.000394374 \end{eqnarray*}
(Wolfram|Alpha computation). I don't see how to analytically integrate the term involving both $\arccos\frac9{d_1}$ and $\arccos\frac{11}{d_1}$, and neither does Wolfram|Alpha.