You've a coin of diameter 2cm, and you throw it on to the chessboard. The center of the coin falls somewhere on the chessboard.

What is the likelihood that the coin is completely within a white square?

Well, the center of the coin definitely must land on a white square if the whole coin is to have any chance of being completely in a white square. Since the board is 50% white and 50% black, this means that the center of the coin has a 50:50 chance of landing on a white square.

So what I’ll do is calculate the likelihood of the coin being completely in a white square IF the center lands in a white square (call this W), and then "calculate" the likelihood of the coin being completely in a white square IF the center lands in a black square (call this B). Then the total chances of the coin being completely in a white square is T = 0.5*W + 0.5*B.

I said I’d "calculate" B because there is no calculation needed. If the center lands in a black square, then there is no chance for the coin to be completely in a white square. So B = 0.

For W, here is what I’ll do. I’ll draw a square that is 10cm on a side (80cm/8squares). Next, if I move the coin around inside this square, along the edges, I’ll see that the center of the coin traces out another square. This one is 8cm on a side, because the center of the 2cm in diameter coin will always be at least 1cm from each edge of the square.

So as long as the center of the coin lands inside the 8cm square, the coin will be entirely in the white square. So W = area of 8cm square / area of 10 cm square = (8cm)^2 / (10cm)^2 = 64/100 = 0.64.

So the answer is
T = 0.5*0.64 + 0.6 *0 = 0.32 = 32%

In general, if you call the side length of the square S, and the diameter of the coin D, W = (S-D)^2 / S^2.
Then T = (S-D)^2 / (2*S^2)