What is the probability that at least two of the $n^{\rm th}$ biggest elements of $\mathbf{A}$ share the same column? I have a random matrix $\mathbf{A}=\left[a_{ij}\right]$ for all $i,j\in\{1,\ldots,n\}$. 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 probability that the $n^{\rm th}$ biggest elements of $\mathbf{A}$ belongs to different columns. Or, alternatively, the probability that at least two of the $n^{\rm th}$ biggest elements of $\mathbf{A}$ share the same column. That is,
$$\Pr\left[X_j\ge 2\right],$$
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 $X_3=2$.
The three biggest elements of $\mathbf{A}$ are $6,10$ and $11$ which are not in different columns because $X_3\ge 2$. Given $\mathbf{A}$, I would like to know the probability that the $n$ biggest elements of $\mathbf{A}$ are in different columns?

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
$$\Pr\left[X_j\ge 2\right]=1-\left(1-\dfrac{1}{n}\right)^{n-1}-\left(1-\dfrac{1}{n}\right)^{n},$$
which gives me the probability the the $n$ biggest elements of $\mathbf{A}$ are in different columns equals to:
$$\left(1-\dfrac{1}{n}\right)^{n-1}\left(2-\dfrac{1}{n}\right)\to\dfrac{2}{e}.$$
What is weird is that, in my calculation, I never used the fact that the $a_{ij}$ are exponential random variables.
 A: The exact distribution doesn't matter. The only important part is that the $a_{i, j}$ are independent and every row uses the same distribution. It follows from these assumptions that the $x_i$ are independent and uniformly distributed on $\{1, \ldots, n\}$. [*]
Now your question is asking the following: What is the probability that the maxima are all in different columens, i.e. $x_i \ne x_j$ for all $i \ne j$? This question can be answered by simple combinatorics. There are $n^n$ possibly ways to assign the numbers $\{1, \ldots, n\}$ to $\{x_1, \ldots, x_n\}$, but only $n!$ many of them satisfy $x_i \ne x_j$ for all $i \ne j$. This gives you a probability of $\frac{n!}{n^n}$.
[*] This is under the assumption that if you have multiple maxima in a row, you choose one of them randomly [i.e. uniformly, independent] to be the argmax. If you would i.e. always choose the smallest index at which a maximum is attained, you would get a bias towards smaller indices and a much, much, much more complicated solution that depends on the actual distribution.
A: There are $n^n$ different ways the $X_i$ can be assigned, and they're all equiprobable, independent of the details of the distribution of the $A_{ij}$ as long as they're i.i.d. There are $n!$ of these such that there's exactly one $1$ per column. Thus the probability for this is $n!/n^n$, and the probability that there's more than one $1$ in some column is the complement,
$$
1-\frac{n!}{n^n}\approx1-\sqrt{2\pi n}\mathrm e^{-n}\;,
$$
where the approximation is Stirling's approximation of the factorial for large $n$.
