Let us assume the elements in the array are the multi-set $M=\{1,2,\cdots,n-k, d, d, \cdots, d\}$, where $d$, an element that is different from $1,2,\cdots, n-k$, appears $k\ge 0$ imes. The situation raised by OP is simply the special case when we let $n=2k$.
In order to discuss the average time-complexity of a linear search, we must specify the distribution of the array as well as the element or the distribution of the element we are searching for.
The distribution of the array assumed
The most common distribution discussed in various articles and in people's mindset is probably the following simplest one.
$$\text{All orderings of the multi-set are equally likely.}$$
That distribution is assumed throughout this answer.
The average number of comparisons needed when we search for $d$
The following answer is given in the wikipedia analysis.
$$\begin{cases}
n & \mbox{if } k = 0 \\
\displaystyle\frac{n + 1}{k + 1} & \mbox{if } 1 \le k \le n.
\end{cases}$$
The above answer is easy to prove by induction on $n$.
The average number of comparisons needed
Assume $k\ge 1$ from now on. Two cases will be shown.
the case with one extra element
Let $o$ stands for an element not in the set $S=\{1,2,\cdots,n-k, d\}$.
If we are searching for a specified element in $S\cup\{o\}$, where each element is equally likely to be specified, here is the computation.
$$\begin{align}
E[\text{search for } x] &= \sum_{x!=d \text { and } x!=0}\frac 1{n-k+2}E[\text{search for }x] + \frac1{n-k+2}E[\text{search for }d] + \frac1{n-k+2}E[\text{search for }o]\\
&=\frac{n-k}{n-k+2}\frac{n+1}2 + \frac1{n-k+2}\frac{n + 1}{k + 1} + \frac1{n-k+2}n\\
&=\frac{n+1}2 + \frac1{k+1} - \frac{2}{(k+1)(n-k+2)}
\end{align}$$
In particular, if we let $n=2k$, then
$$E[\text{search for } x] =\frac {n+1}2 + \frac2{n+2}-\frac{8}{(n+2)(n+4)}$$
the case with $M$ alone
What is the average number of comparisons need among the cases when the element we are search for turned out to be an element in $M$, assuming that we are searching for an element in the union of the multi-set $M$ and another multi-set $M'$ that is disjoint with $M$, where all elements in $M$ will be equally likely to be searched? The inclusion of the $M'$ is to ensure that we do NOT know whether the element to be found belongs to $M$ before we start the search.
Here is the computation.
$$\begin{align}
E[\text{search for } x] &= \sum_{x!=d}\frac 1{n}E[\text{search for }x] + \frac k{n}E[\text{search for }d]\\
&=\frac{n-k}n\frac{n+1}2 + \frac k{n}\frac{n + 1}{k + 1}\\
&=\left(1-\frac{k-2}n -\frac2{n(k+1)}\right)\frac {n+1}2
\end{align}$$
In particular, if we let $n=2k$, then
$$E[\text{search for } x] =\frac14\left(n+5-\frac 4{n+2}\right)$$
Please note that we have not computed the situations when we know the element to be searched is an element of array before we start the linear search, which might be the more common situations. Readers are encouraged to dive into those situations.
The wikipedia item is a good place to start your journey of finding, reading, and playing with the many variations and generalization of this question.
