I have a random matrix $\mathbf{A}=\left[a\_{ij}\right]$ for all $(i,j)\in\{1,\ldots,k\}\times\{1,\ldots,n\}$. Every entry $a\_{ij}$ of the matrix $\mathbf{A}$ is generated randomly with exponential distribution. The $a\_{ij}$ are i.i.d and have the same parameter $\lambda$.

Now, for each row $i$ of $\mathbf{A}$, I select the argument of the maximum element. That is,

$$x\_i=\arg\max\limits\_{j} a\_{ij}.$$

Let $X\_{ij}$ be the binary random variable that is equal $1$ if $x\_i=j$, and $0$ otherwise. Also, let $X\_j=\sum\_{i=1}^nX\_{ij}$.

I am interested in calculating the expected number of ones in column $j$. That is,

$$\mathbb{E}\left[X\_j\right],$$

for all $j\in\{1,\ldots,n\}$.

How can I solve this problem?

I will give an example to illustrate the problem: Let $n=3$ and $\mathbf{A}$ given by:

$$\mathbf{A}=\begin{bmatrix}
1 & 3 & 6\\
9 & 7 & 10\\
11 & 5 & 8
\end{bmatrix}.$$

Now, given $\mathbf{A}$, I can calculate $\mathbf{X}=[X\_{ij}]$ as:
$$\mathbf{A}=\begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 1\\
1 & 0 & 0
\end{bmatrix},$$
since $x\_1=3,x\_2=3$ and $x\_3=1$. Then, I get $X\_1=1,X\_2=0$ and $X\_3=2$.

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When I tried to solve the problem I find that 
$$\Pr\left[X\_{ij}=1\right]=\dfrac{1}{n}.$$

After my work, I find that
$$\begin{align}\mathbb{E}\left[X\_j\right]&=\sum\_{j=1}^n\mathbb{E}\left[X\_{ij}\right]\\&=1,\end{align}$$

What is weird is that, in my calculation, I never used the fact that the $a\_{ij}$ are exponential random variables. Also, why the expected number of ones is $1$?