For a permutation $P = (p_1, p_2, ... , p_n)$ of $(1, 2, ... , n)$ define $X(P)$ as the number of $j$ such that $p_i < p_j$ for every $i < j$. What is the expected value of $X(P)$ if each permutation is equally likely?
1. **Define the problem and notation:**
   We are given a permutation \( P = (p_1, p_2, \ldots, p_n) \) of the set \(\{1, 2, \ldots, n\}\). We need to find the expected value of \( X(P) \), where \( X(P) \) is the number of indices \( j \) such that \( p_i < p_j \) for every \( i < j \).

2. **Define the event \( A_j \):**
   Let \( A_j \) be the event that \( p_1, p_2, \ldots, p_{j-1} \) are all less than \( p_j \). This event is equivalent to \( p_j \) being the maximum of the first \( j \) elements \( p_1, p_2, \ldots, p_j \).

3. **Calculate the probability of \( A_j \):**
   By symmetry, the probability that \( p_j \) is the maximum of \( p_1, p_2, \ldots, p_j \) is \(\frac{1}{j}\). This is because each of the \( j \) elements is equally likely to be the maximum.

4. **Use linearity of expectation:**
   The expected value of \( X(P) \) can be found by summing the probabilities of the events \( A_j \) for \( j = 1, 2, \ldots, n \). By the linearity of expectation, we have:
   \[
   E(X) = \sum_{j=1}^n \Pr(A_j)
   \]

5. **Sum the probabilities:**
   Since \(\Pr(A_j) = \frac{1}{j}\), we get:
   \[
   E(X) = \sum_{j=1}^n \frac{1}{j}
   \]

6. **Identify the result as the \( n \)-th harmonic number:**
   The sum \(\sum_{j=1}^n \frac{1}{j}\) is known as the \( n \)-th harmonic number, denoted by \( H_n \).

Conclusion:
\[
E(X) = H_n
\]

The final answer is \( \boxed{ H_n } \)