Solution:
The probability of the coining landing in the correct position in any of the boxes is equal, since all of the boxes have an equal chance of containing the center of the coin.
Now, let's make the coin a square - specifcially, an 8 x 8 square. As long as the center of the square is at least \(\frac{8}{2} = 4\) away from the sides, and since each of the square have a side length of \(16\), we have a square of side length \(16-4-4 = 8\) that contains the "safe zone", that is, the area in which the center of the coin may land on. Therefore, the answer is \(\frac{8^2}{16^2} = \boxed{\frac{1}{4}}\).
-Bobinthius!
