Coins on a grid

Since the coin has a radius of $\frac{1}{2} \text{ cm}$ , if the center of the coin is within $\frac{1}{2} \text{ cm}$ from a grid line, then it will cross it, otherwise it will not cross it. The red area below shows where the center of the coin can land and cross a grid line, and the green area below shows where the center of the coin can land and not cross a grid line.

Therefore, the probability that the coin does not touch the grid lines is the area of the green square to each grid square, which is $\frac{1 \cdot 1}{2 \cdot 2} = \boxed{\frac{1}{4}}$ .

Instead of considering areas, consider the position of the center of the coin, first in the x-direction and then in the y-direction. Since the coin is randomly thrown, these are independent of each other. We can simplify to consider just one square.

X-direction : The center of the coin has to land between x = 0.5cm and x = 1.5cm to ensure the coin does not touch a vertical line. This is $\frac{(1.5-0.5)}{2}$ , i.e. $\frac{1}{2}$ .

Y-direction : Analogous to the x-direction, so the probability is the same, i.e. $\frac{1}{2}$ .

Overall : Multiple these together to get the overall probability that the coin does not touch a vertical or horizontal grid line: $\frac{1}{2}$ x $\frac{1}{2}$ = $\frac{1}{4}$ .

Consider the grid as the set of lines $\{x=2n | n \in \mathbb{Z}\} \cup \{y =2n | n \in \mathbb{Z}\}$ . Let $(a,b)$ be coordinates of the center of the coin. Then there are indepedent identical random variables. The coin does not intersect a vertical line if $a \in [1/2, 3/2] \mod 2$ and similarly it misses a horizontal line if $b\in [1/2, 3/2] \mod 2$ . Thus it misses the horizontal lines with probability 1/2 and the vertical lines also with probability 1/2. By independence, the probability of missing the grid is 1/4.

HOWEVER

I would like to point out that technically , the problem is not well-defined because there is no translation invariant probability measure on $\mathbb{R}^2$ , or for that matter on $\mathbb{R}$ . One way to get around this is to view the plane as a torus by identifying the grid lines. This is implicitly what I am doing in my solution by reducing the coordinates mod 2. Another way is to think of the coin as being fixed at the origin, and asking about a random placement of the grid lines. Since the space of placements of the grid lines is compact, it does have a uniform probability measure.

The total area the coin can cover with respect to one grid is 12 + π ( This includes area of the square + area the coin can cover while still touching the square from the outside.), while the area required will be such that the coin is completely inside the square, which is 3 + π/4. Hence, the probability is 1/4

The area of one square of the grid is $2*2=4 cm^2$ and the area of the square which is surrounding the circle is $1*1=1 cm^2$ . the ratio of their areas is $1:4$ .

• We know the area inside a square, in this grid, is 2 cm squared (= a 4 cm area.)

• And we know the diameter and area of a single coin (circle) is 1 cm squared.

• So according to though measurements, we can only fit 4 coins/circle in one square, without touching the lines.

-{ Ego, a single coin is 1/4 of the area inside a square. | In other words, there is 1/4 chance for the coin to stay inside the square and ego not to touch the grid lines. }-

1/2*1/2=1/4

First consider a square in a grid.The total area where the coin of radius 1/2 cm can move so that it does not actually touch the grid lines is pi(1)^2(which is the area of the inscribed circle in square). The actual area of coin is pi(1/2)^2. Probability=area of coin /area for free movement I.e inscribed circle =1/4 No matter there be any number of grids,the combined probability is same as considering only one square in this case.