Select $ n$ points on a circle independently with uniform distribution. Let $ P_n$ be the probability that the center of the
circle is in the interior of the convex hull of these $ n$ points. Calculate the probabilities $ P_3$ and $ P_4$. [A. Renyi]
1. **Calculate \( P_3 \):**

   We start by considering three points \( Z_1, Z_2, Z_3 \) on the unit circle. Without loss of generality, we can assume that \( Z_1 \) is at \( \alpha_1 = \pi \) (using polar coordinates) and \( Z_2 \) is in the upper half-plane, i.e., \( 0 < \alpha_2 < \pi \).

   The center of the circle \( O \) will be inside the convex hull of \( Z_1, Z_2, Z_3 \) if and only if the argument of \( Z_3 \) satisfies \( \alpha_2 - \pi < \alpha_3 < 0 \). This is because \( Z_3 \) must be in the lower half-plane and between \( Z_1 \) and \( Z_2 \) in terms of the argument.

   Given \( Z_1 \) and \( Z_2 \), the conditional probability that \( O \) is inside the convex hull of \( Z_1, Z_2, Z_3 \) is:
   \[
   \frac{\pi - \alpha_2}{2\pi}
   \]

   To find the overall probability \( P_3 \), we integrate this conditional probability over all possible positions of \( \alpha_2 \) from 0 to \( \pi \):
   \[
   P_3 = \int_{0}^{\pi} \frac{\pi - \alpha_2}{2\pi} \cdot \frac{d\alpha_2}{\pi}
   \]

   Simplifying the integral:
   \[
   P_3 = \int_{0}^{\pi} \frac{\pi - \alpha_2}{2\pi^2} \, d\alpha_2
   \]

   \[
   P_3 = \frac{1}{2\pi^2} \int_{0}^{\pi} (\pi - \alpha_2) \, d\alpha_2
   \]

   \[
   P_3 = \frac{1}{2\pi^2} \left[ \pi \alpha_2 - \frac{\alpha_2^2}{2} \right]_{0}^{\pi}
   \]

   \[
   P_3 = \frac{1}{2\pi^2} \left( \pi^2 - \frac{\pi^2}{2} \right)
   \]

   \[
   P_3 = \frac{1}{2\pi^2} \cdot \frac{\pi^2}{2}
   \]

   \[
   P_3 = \frac{1}{4}
   \]

2. **Calculate \( P_4 \):**

   For four points \( Z_1, Z_2, Z_3, Z_4 \) on the unit circle, the center \( O \) will be inside the convex hull if and only if all four points are not on the same semicircle. This is equivalent to saying that the points must be in such a configuration that no three points are on the same semicircle.

   The probability that four points are on the same semicircle is given by:
   \[
   \frac{4}{2^4} = \frac{1}{4}
   \]

   Therefore, the probability that the center is inside the convex hull of the four points is:
   \[
   P_4 = 1 - \frac{1}{4} = \frac{3}{4}
   \]

The final answer is \( \boxed{ P_3 = \frac{1}{4} } \) and \( P_4 = \frac{3}{4} \)