Given an ordered $n$-tuple $A=(a_1,a_2,\cdots ,a_n)$ of real numbers, where $n\ge 2$, we define $b_k=\max{a_1,\ldots a_k}$ for each k. We define $B=(b_1,b_2,\cdots ,b_n)$ to be the “[i]innovated tuple[/i]” of $A$. The number of distinct elements in $B$ is called the “[i]innovated degree[/i]” of $A$.
Consider all permutations of $1,2,\ldots ,n$ as an ordered $n$-tuple. Find the arithmetic mean of the first term of the permutations whose innovated degrees are all equal to $2$
1. **Understanding the Innovated Degree**:
   - The "innovated degree" of a permutation is the number of distinct elements in the tuple $B = (b_1, b_2, \ldots, b_n)$, where $b_k = \max(a_1, a_2, \ldots, a_k)$.
   - For a permutation of $\{1, 2, \ldots, n\}$ to have an innovated degree of 2, it must have exactly two left-to-right maxima.

2. **Characterizing Permutations with Innovated Degree 2**:
   - A permutation has exactly two left-to-right maxima if and only if its first entry $k$ satisfies $k < n$ and all entries in the permutation between (in position) $k$ and $n$ have values less than $k$.
   - This means the permutation starts with $k$, followed by a sequence of numbers all less than $k$, and ends with $n$.

3. **Counting Permutations with Innovated Degree 2**:
   - For a given $k$, the number of permutations with exactly two left-to-right maxima and first entry $k$ can be calculated as follows:
     \[
     \sum_{j=0}^{k-1} \binom{k-1}{j} j! (n-j-2)!
     \]
     - Here, $\binom{k-1}{j}$ chooses $j$ positions out of the first $k-1$ positions for the elements less than $k$.
     - $j!$ is the number of ways to arrange these $j$ elements.
     - $(n-j-2)!$ is the number of ways to arrange the remaining $n-j-2$ elements.

4. **Simplifying the Count**:
   - The sum can be simplified using the binomial theorem and properties of factorials:
     \[
     \sum_{j=0}^{k-1} \binom{k-1}{j} j! (n-j-2)! = (k-1)! (n-k-1)! \sum_{j=0}^{k-1} \binom{n-j-2}{n-k-1}
     \]
     - The sum $\sum_{j=0}^{k-1} \binom{n-j-2}{n-k-1}$ simplifies to $\binom{n-1}{n-k}$.
     - Therefore, the number of such permutations is:
       \[
       (k-1)! (n-k-1)! \binom{n-1}{n-k} = \frac{(n-1)!}{n-k}
       \]

5. **Calculating the Arithmetic Mean**:
   - The arithmetic mean of the first term of the permutations whose innovated degrees are all equal to 2 is given by:
     \[
     \frac{\sum_{k=1}^{n-1} \frac{k}{n-k}}{\sum_{k=1}^{n-1} \frac{1}{n-k}}
     \]
   - The numerator is:
     \[
     \sum_{k=1}^{n-1} \frac{k}{n-k}
     \]
   - The denominator is:
     \[
     \sum_{k=1}^{n-1} \frac{1}{n-k}
     \]

6. **Simplifying the Expression**:
   - The denominator is the harmonic number $H_{n-1}$:
     \[
     H_{n-1} = \sum_{k=1}^{n-1} \frac{1}{k}
     \]
   - The numerator can be rewritten using the properties of harmonic numbers:
     \[
     \sum_{k=1}^{n-1} \frac{k}{n-k} = n \sum_{k=1}^{n-1} \frac{1}{k} - (n-1)
     \]
   - Therefore, the arithmetic mean is:
     \[
     \frac{n H_{n-1} - (n-1)}{H_{n-1}} = n - \frac{n-1}{H_{n-1}}
     \]

The final answer is $\boxed{n - \frac{n-1}{H_{n-1}}}$