Problem 9.14 Easy Free 1 variation

Coin Inside a Square

A chessboard is tiled by squares of side $2$ inches. A coin of diameter $1$ inch is tossed so that it lands with its centre uniformly distributed over the board. Find the probability that the coin lies entirely within a single square (and does not cross into a neighbouring one).

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. A disc lies inside a square exactly when its centre is at least one radius from every edge. The admissible centres therefore form a smaller, concentric square.

Given which square the centre lands in, the centre is uniform over that square. This holds for every square of the board, including those on the outer edge: a coin hanging over the board’s edge is also not inside a single square.

The coin has radius $\tfrac12$ . It lies entirely within its square if and only if its centre is at distance at least $\tfrac12$ from all four sides. These centres form a concentric square of side $2-2\cdot\tfrac12 = 1$ . Hence

\mathbb{P}(\text{coin inside one square}) = \frac{1^2}{2^2} = \frac14 .

The conditional probability is the same for every square, so it is also the unconditional answer.

\boxed{\tfrac14}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 9.14a

A square of side length $20$ is drawn, and a point is chosen uniformly at random inside it. Independently, a radius $R \sim \mathrm{Unif}(0,10)$ is chosen and a circle of that radius is drawn centred at the chosen point. Find the probability that this circle lies entirely inside the square, i.e. never crosses the square’s boundary.

Work it out, then compare with the answer.

Solution

Condition on $R=r$ . The circle stays inside the square exactly when the centre is at least $r$ from every side. Those centres form a concentric square of side $20-2r$ , so

\mathbb{P}(E\mid R=r)=\frac{(20-2r)^2}{20^2}=\Bigl(1-\frac{r}{10}\Bigr)^2 .

Write $R/10=U\sim\mathrm{Unif}(0,1)$ . Since $1-U$ is also uniform on $(0,1)$ ,

\mathbb{P}(E)=\mathbb{E}\bigl[(1-U)^2\bigr]=\int_0^1 (1-u)^2\,du=\frac13 .

\boxed{\tfrac13}

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