Stacking Semicircular Disks

The answer is $1-\frac{n}{2^{n-1}}$.

First, consider the boundary lines (diameters) of the $n$ semicircular disks. Placing a semicircle randomly is mathematically equivalent to first choosing a random diametrical line passing through the center, and then randomly choosing which of the two halves the disk will occupy. Let us fix these $n$ random lines. With probability $1$, no two lines coincide. These $n$ intersecting lines divide the full circle into exactly $2n$ distinct sectors.

For each of the $n$ fixed lines, the semicircle has exactly $2$ possible orientations (each occurring with probability $\frac{1}{2}$). Since the orientations are independent, there are $2^n$ equally likely configurations for any fixed set of $n$ lines.

We want to find the probability that the circle is not fully covered. The circle is not fully covered if and only if there is one sector missing. Because every disk is exactly a semicircle (spanning an angle of $\pi$), it is geometrically impossible to have more than one sector missing. Therefore, if the circle is incomplete, the empty space must perfectly correspond to exactly one of the $2n$ sectors.

For any specific sector to be entirely empty, every single one of the $n$ semicircles must be oriented in the direction pointing away from that sector. Out of the $2^n$ possible orientation configurations, there is exactly $1$ unique configuration that leaves a given sector empty.

Since there are $2n$ sectors, and the event of one sector being empty is mutually exclusive from another sector being empty (due to the $\pi$ span of the disks), there are exactly $2n$ configurations out of $2^n$ that result in an incomplete circle.

The probability of the circle remaining incomplete is the ratio of these failing configurations to the total configurations:

Thus, the probability that the $n$ semicircles completely cover the circle is the complement: