Let $ 2n$ distinct points on a circle be given. Arrange them into disjoint pairs in an arbitrary way and join the couples by chords. Determine the probability that no two of these $ n$ chords intersect. (All possible arrangement into pairs are supposed to have the same probability.)
1. **Identify the total number of pairings:**
   The total number of ways to pair \(2n\) distinct points on a circle is given by the double factorial \((2n-1)!!\). This can be expressed as:
   \[
   (2n-1)!! = \frac{(2n)!}{2^n n!}
   \]
   This formula counts the number of ways to pair \(2n\) points into \(n\) pairs.

2. **Identify the number of "good" pairings:**
   A "good" pairing is one where no two chords intersect. The number of such pairings is given by the \(n\)-th Catalan number \(C_n\). The \(n\)-th Catalan number is defined as:
   \[
   C_n = \frac{1}{n+1} \binom{2n}{n} = \frac{(2n)!}{n!(n+1)!}
   \]
   This formula counts the number of ways to pair \(2n\) points such that no two chords intersect.

3. **Calculate the probability:**
   The probability that a random pairing of \(2n\) points results in no intersecting chords is the ratio of the number of "good" pairings to the total number of pairings. Therefore, the probability \(P\) is:
   \[
   P = \frac{\text{Number of "good" pairings}}{\text{Total number of pairings}} = \frac{C_n}{(2n-1)!!}
   \]
   Substituting the expressions for \(C_n\) and \((2n-1)!!\), we get:
   \[
   P = \frac{\frac{(2n)!}{n!(n+1)!}}{\frac{(2n)!}{2^n n!}} = \frac{(2n)!}{n!(n+1)!} \cdot \frac{2^n n!}{(2n)!} = \frac{2^n}{(n+1)!}
   \]

Conclusion:
\[
\boxed{\frac{2^n}{(n+1)!}}
\]