27.4. (GDR, 78). A circle is circumscribed around a regular $2 n$-gon. A triplet of its distinct vertices is called one-sided if there exists a semicircle on which these vertices lie (the ends of the semicircle belong to it). What is the probability that a randomly chosen triplet of vertices will be one-sided?
27.4. If two vertices out of three have already been chosen, the number of ways to choose the third vertex so that the resulting triplet is one-sided depends on the angular distance 1 between the two chosen vertices. (The angular distance between vertices $A$ and $B$ is defined as $l=\widehat{A O B} \cdot(n / \pi)$, where $O$ is the center of the $2n$-gon; here, $l \leqslant n$, and the angular distance between adjacent vertices is 1.) If $l < n$, the third vertex can be chosen in $(l-1) + 2(n-l) = 2n - 1 - l$ ways; if $l = n$, it can be chosen arbitrarily, i.e., in $2n - 2$ ways. Further, for each value of $l = 1, 2, \ldots, n-1$, there are exactly $2n \cdot 2 = 4n$ ways to choose the first and then the second vertex at an angular distance $l$ from the first. Finally, for $l = n$, there are only $2n$ such ways. Therefore, the total number of ways to sequentially choose three vertices is

$$
\begin{aligned}
& 4n \cdot \sum_{l=1}^{n-1}(2n-1-l) + 2n(2n-2) = \\
& = 4n(2n-1)(n-1) - 4n \frac{(n-1)n}{2} + 4n(n-1) = 6n^2(n-1)
\end{aligned}
$$

We have calculated the number of one-sided triplets under the condition that in each triplet, the first, second, and third vertices are distinguished. If the vertices are not ordered, the number of triplets will be 6 times smaller. The number of ways to arbitrarily choose three vertices is

$$
C_{2n}^{3} = \frac{1}{6} \cdot 2n \cdot (2n-1) \cdot (2n-2)
$$

Therefore, the desired probability is

$$
\frac{n^3 - n^2}{\frac{1}{6} \cdot 2n \cdot (2n-1) \cdot (2n-2)} = \frac{3n}{2(2n-1)}
$$