Factorizing $\displaystyle (4r^2) - \frac{\pi r^2}{2}$ gives me the following:
$\displaystyle \frac {2(4r^2)}{2} - \frac {\pi r^2}{2} \longleftrightarrow \frac {2(4r^2)-\pi r^2}{2} \longleftrightarrow \frac {8r^2-\pi r^2}{2}$.
We know that the area of the half-circle divided by the area of the whole square multiplied by 100 gives us the percentage that the half-circles occupies in the square. Then, 100% minus the percentage found gives us the percentage of the rest of the square, which is the probability we're looking for. But I've tried to find a concrete percentage but I couldn't, I'm a bit rusty tonight and it's getting late
Tell me if you get anything new...
Those are the calculations for what I've stated above:
$\displaystyle \frac {2\pi r^2}{8r^2 - \pi r^2}$ x 100
$\displaystyle \frac {2\pi r^2}{r^2(8-\pi)}$ x 100
$\displaystyle \frac {2\pi}{(8-\pi)}$ x 100
$\displaystyle \frac {200\pi}{(8-\pi)}$% = The percentage that the half-square occupies
100% - $\displaystyle \frac {200\pi}{(8-\pi)}$%= The probability we're looking for...
$\displaystyle \frac {200\pi}{(8-\pi)} \approx $ 129%
100% - 129% = -29%