Let $ p>2 $ be a prime number. For any permutation $ \pi = ( \pi(1) , \pi(2) , \cdots , \pi(p) ) $ of the set $ S = \{ 1, 2, \cdots , p \} $, let $ f( \pi ) $ denote the number of multiples of $ p $ among the following $ p $ numbers:
\[ \pi(1) , \pi(1) + \pi(2) , \cdots , \pi(1) + \pi(2) + \cdots + \pi(p)  \]
Determine the average value of $ f( \pi) $ taken over all permutations $ \pi $ of $ S $.
To determine the average value of \( f(\pi) \) taken over all permutations \( \pi \) of \( S = \{1, 2, \cdots, p\} \), we will follow these steps:

1. **Understand the Problem:**
   We need to count the number of multiples of \( p \) among the partial sums:
   \[
   \pi(1), \pi(1) + \pi(2), \cdots, \pi(1) + \pi(2) + \cdots + \pi(p)
   \]
   for all permutations \( \pi \) of the set \( S \).

2. **Example for Clarity:**
   Let's take \( p = 7 \) and consider a permutation \( \pi = (4, 1, 3, 2, 6, 7, 5) \). We will generate \( p \) permutations by cyclically shifting the elements and then examine the partial sums.

3. **Generate Cyclic Permutations:**
   The cyclic permutations of \( \pi \) are:
   \[
   \begin{aligned}
   &4, 1, 3, 2, 6, 7, 5 \\
   &5, 2, 4, 3, 7, 1, 6 \\
   &6, 3, 5, 4, 1, 2, 7 \\
   &7, 4, 6, 5, 2, 3, 1 \\
   &1, 5, 7, 6, 3, 4, 2 \\
   &2, 6, 1, 7, 4, 5, 3 \\
   &3, 7, 2, 1, 5, 6, 4 \\
   \end{aligned}
   \]

4. **Partial Sums and Multiples of \( p \):**
   For each permutation, we calculate the partial sums and check for multiples of \( p \). The claim is that for each \( 1 \leq i < p \), exactly one of these permutations will have \( p \mid \pi(1) + \cdots + \pi(i) \).

5. **Summing Up:**
   Since \( p \mid \pi(1) + \cdots + \pi(p) \) is always true, we have \( p \) multiples of \( p \) from the full sums. For the other \( p-1 \) partial sums, each of the \( p \) cyclic permutations contributes exactly one multiple of \( p \).

6. **Total Multiples of \( p \):**
   Therefore, for each set of \( p \) cyclic permutations, we have \( (p-1) + p = 2p-1 \) multiples of \( p \).

7. **Average Value:**
   The average value of \( f(\pi) \) over all permutations \( \pi \) is:
   \[
   \frac{2p-1}{p} = 2 - \frac{1}{p}
   \]

The final answer is \(\boxed{2 - \frac{1}{p}}\)