Probability that coin will fall into a square So the exercise is this:
We have and infinite chessboard and we have a coin. Every grid is of length and width $a$, whereas the coin has diameter $2 \cdot r<a$. We throw a coin into a chessboard and we want to know with what probability the coin will falll into the grid.
 
So let $S_{1}$=area of green rectangle, $S_{2}=S_{1}+$ area of the red border.
So the probability in my opinion is 
$ P(a)= \frac{S_{1}}{S_{2}} $.
Is this in any shape or form correct?
 A: The rounded corners of your red area are a mistake.
The best way to look at the problem is to reformulate

The coin falls completely inside a single square.

to

The center of the coin is at least one coin radius from any grid line.

You can then shade the area of a grid square where the coin center cannot fall -- this shaded area will look the same in every square. If the coin radius is $r$ and the grid square sides have length $d$, there's a small square of side length $d-2r$ in each square where the center can fall without the coin extending beyond the grid square.
So the probablity of staying within the square becomes
$$ \frac{(d-2r)^2}{d^2} $$
assuming that the position of the coin's center relative to the square it happens to land on is uniformly distributed.
A: The term completely inside the circle means that we have to exclude the areas where the coin may or may not lie within the square region. Hence the probability will be $$(d-4r)^{2}/d^{2}.$$ 
In addition to the above 
the probability that the coin will never be inside the square will be 
((d^2 -(d-2r)^2))/d^2)
