, 2025

Because the endpoints of the segments are chosen at random from a uniform distribution, the probability that two segments share an endpoint is 0, so that possibility can be ignored. For the same reason, it can be assumed that no more than two of the segments intersect at the same point. Before any of the 27 line segments are drawn, the disk consists of 1 region. As each new line segment is drawn, 1 more region is created for each region that the new line segment enters. This is equal to 1 plus the number of times the new line segment intersects one of the previously drawn line segments. It follows that the total number of regions created by the 27 line segments is $1 + 27 + I = 28 + I$, where $I$ is the number of intersections of the line segments in the interior of the disk.

Label the four quadrants of the disk in order as 1, 2, 3, and 4. Then label each of the 25 randomly chosen line segments that Alex draws with the quadrants containing its endpoints; that is, each segment can be labeled 12, 13, 14, 23, 24, or 34, and each of these labels occurs with equal probability. The line segments labeled 12, 23, 34, and 14 connect adjacent quadrants and intersect one of the two diameters, but the line segments labeled 13 and 24 intersect both diameters. Thus the expected number of intersections of the 25 randomly drawn segments with the 2 diameters is $25 \cdot (1 \cdot \frac{2}{3} + 2 \cdot \frac{1}{3}) = \frac{100}{3}$. There is also 1 intersection at the center of the disk where the two diameters intersect.

It remains to find the expected number of intersections between pairs of the 25 randomly drawn line segments. There are $\binom{25}{2} = 300$ pairs of line segments. Each pair consists of two segments, and each segment is equally likely to be labeled 12, 13, 14, 23, 24, or 34. Of the $6^2 = 36$ possible combinations of two labels, 30 of them consist of labels that share at least one quadrant; that is, there is a probability of $\frac{30}{36} = \frac{5}{6}$ that the two line segments will have at least one endpoint in the same quadrant. These pairs of segments will intersect with probability $\frac{1}{2}$. Of the 6 remaining possible combinations of labels, 4 consist of pairs that cannot intersect: (12, 34), (14, 23), (23, 14), and (34, 12). The other 2 consist of pairs that must intersect: (13, 24) and (24, 13). Thus the expected number of intersections of pairs is

$$300 \left( \frac{5}{6} \cdot \frac{1}{2} + \frac{2}{36} \right) = \frac{425}{3}.$$

The expected value of $I$ is therefore $\frac{100}{3} + 1 + \frac{425}{3} = 176$. The expected number of regions is $28 + 176 = 204$.