Coin throw probability on a chessboard square

Question

I need some insights on the following problem. You are given a coin whose diameter is equivalent to half the length of a chessboard unit square. If you throw the coin on the board, what is the probability that it lands completely within a single square?

Here is what I need to know,

1. Does it matter here that the chessboard has 64 (8 X 8) squares? Or should I just concentrate on a single square?
2. Is this probability simply the coin's area (since it is a unit square)?

I am asking because it appears quite easy and odd! Please let me know if I am overthinking.

Thank you.

Here is what I did,

The area of a circle is given by the formula A = πr^2, where r is the radius. Since the coin's diameter is 0.5 units, the radius is half of that, which is 0.5/2 = 0.25 units.

The area of the coin is then A = π(0.25)^2 = 0.1963495.

The area of a single square on the chessboard is simply the side length squared, which is 1 square unit.

To find the probability that the coin lands completely within a single square, I divided the area of the nickel by the area of a square:

P = A of coin / A of square = 0.1963495 / 1

Calculating this probability:

P ≈ 0.1963

Is my approach correct?

Answer 1

I would try thinking of the problem in terms of the center point of the circle, and where this center point is permitted to fall if the coin is to be entirely within a square. Take an empty box, and a coin with an ink dot on its center point, placed in one of the box's corners. Then trace along the entire boundary of the box with the coin, the center point of the coin will draw out a rectangle within the box. If the diameter of the coin is half the square size, then the inner drawn square will have side lengths of 1/2 (relative to the side length of the squares).

So, this inner square has an area of 0.25 (0.5 * 0.5), while the outer square has an area of 1. The probability of your coin landing perfectly within a square is therefore 0.25.

Diagram for reference: