\section*{Problem 6B - 121236B}

If \(n\) is a natural number greater than 1, then on a line segment \(A B\), points \(P_{1}, P_{2}, P_{3}, \ldots, P_{2 n-1}\) are placed in this order such that they divide the line segment \(A B\) into \(2 n\) equal parts.

a) Give (as a function of \(n\)) the probability that two points \(P_{k}, P_{m}\) chosen from the points \(P_{1}, P_{2}, P_{3}, \ldots, P_{2 n-1}\) with \(0 < k < m < 2 n\) divide the line segment \(A B\) in such a way that a triangle can be constructed from the three segments \(A P_{k}, P_{k} P_{m}, P_{m} B\).

b) Investigate whether this probability converges to a limit as \(n \rightarrow \infty\), and determine this limit if it exists.

Note: The probability sought in a) is defined as follows: Each selection of two points \(P_{k}, P_{m}\) with \(0 < k < m < 2 n\) is referred to as a "case."

A "case" is called a "favorable case" if \(P_{k}\) and \(P_{m}\) are chosen such that a triangle can be formed from the segments \(A P_{k}, P_{k} P_{m}\), and \(P_{m} B\).

If \(z\) is the total number of possible "cases" and \(z_{1}\) is the number of "favorable cases," then the probability is defined as the quotient \(\frac{z_{1}}{z}\).
}

a) O.B.d.A. let the segment \(A B\) have the length \(2 n\), so that the three segments to be considered \(A P_{k}, P_{k} P_{m}\) and \(P_{m} B\) have the lengths \(k, m-k\) and \(2 n-m\) respectively. A triangle can be constructed from these segments exactly when the three triangle inequalities are satisfied:

The first triangle inequality \(\left|A P_{k}\right|+\left|P_{k} P_{m}\right|>\left|P_{m} B\right|\) is thus equivalent to \(k+(m-k)>2 n-m\) or \(2 m>2 n\), i.e., \(m>n\).

The second triangle inequality \(\left|P_{k} P_{m}\right|+\left|P_{m} B\right|>\left|A P_{k}\right|\) is equivalent to \((m-k)+(2 n-m)>k\) or \(2 n>2 k\), i.e., \(n>k\).

The third triangle inequality \(\left|A P_{k}\right|+\left|P_{m} B\right|>\left|P_{k} P_{m}\right|\) is equivalent to \(k+(2 n-m)>m-k\) or \(2 m<2 n+2 k\), i.e., \(m<n+k\).

Thus, for each \(1 \leq k \leq n-1\) there are exactly the \(k-1\) possibilities for \(m\) with \(n+1 \leq m \leq n+k-1\), such that a triangle can be constructed from the three resulting subsegments. Therefore, there are

\[
\sum_{k=1}^{n-1}(k-1)=\sum_{k=0}^{n-2} k=\frac{(n-2)(n-1)}{2}
\]

favorable selections of \(k\) and \(m\), while there are \(\binom{2 n-1}{2}=\frac{(2 n-1)(2 n-2)}{2}\) equally possible, resulting in a probability of

\[
P=\frac{(n-2)(n-1)}{(2 n-1)(2 n-2)}=\frac{n-2}{2(2 n-1)}=\frac{1}{4} \cdot \frac{4 n-8}{4 n-2}=\frac{1}{4} \cdot\left(1-\frac{6}{4 n-2}\right)
\]

b) Clearly, for \(n \rightarrow \infty\) the fraction \(\frac{6}{4 n-2}\) approaches 0, so \(\lim _{n \rightarrow \infty} P=\frac{1}{4}\) follows.