# Probability for a coin to be cut by exactly 5 lines

Assume you have a sheet of paper made of horizontal and vertical lines, and the distance between parallel lines is $8\ mm$. A circular coin with radius $1\ cm$ is being tossed on the paper. What is the chance that the coin is being intersected by exactly 5 lines (doesn't matter how many are horizontal or vertical).

I found out that if the coin is tangent to exactly one line, it can move perpendicular to that line for $4\ mm$ until the next line is tangent, and it can move parallel to both sides until the lines are tangent to the coin. In other words, the center of the coin can be in a square of area $16\ mm^2$, out of an area of $5.76\ cm^2$.

The problem I am facing that prevents me approving this answer logically, is the total sampling space area.

• We need to make the probability model explicit. One can for example assume that whatever square the center falls in, the location of the center is uniformly distributed in that square. – André Nicolas Feb 23 '16 at 17:48

If $x$ denotes the distance in millimeters of the left end of the coin from the preceding vertical line then for $0<x<4$ we have two lines and for $4<x<8$ we have three lines. In other words, the probability for two or three vertical lines is $0.5$ each. The same argument applies to the number of horizontal lines. To get five lines in total we must have either two plus three or three plus two lines. Hence the final answer is $$p=0.5\cdot0.5+0.5\cdot 0.5=0.5$$
The coin covers $2$ cm in both directions. We can assume that its distance to the next line at the left (or top) is uniformly distributed between $0$ and $8$ mm. If it is below $4$ mm, the coin intersects two lines in that direction, and if above, three. Thus in both directions chances are $50$/$50$ that the coin intersects $2$ or $3$ lines, and since we need either $2+3$ or $3+2$, the probability is
$$\frac12\cdot\frac12+\frac12\cdot\frac12=\frac12\;.$$