Question: We have a uniformly sampled permutation matrix $\Pi\in\{0,1\}^{n\times n}$. What is the probability that the $i$th row in $\Pi$ is in the "correct" position, which means that the $i$th row of $\Pi$ is equal to the $i$th row of an identity matrix $I$ with the same dimension as $\Pi$. For example, the 2nd row in the following permutation matrix is correct: $$ \Pi= \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}. $$
My attempt: If the matrix had no constraints on the row weight and column weight being one, then the answer would simply be $1/n$. However, the presence of the constrains of a permutation matrix makes this tricky. We can instead calculate this differently: Let us think of the random permutation matrix being generated in a random row-by-row fashion, starting with row $i$. Then the number of matrices with row $i$ being "correct" is $$ 1\cdot{n-1\choose 1}\dots{1\choose 1}=(n-1)!, $$ and the total number of matrices is $$ {n\choose 1}{n-1\choose 1}\dots{1\choose 1}=n!. $$ Hence, the probability of row $i$ being correct is $$ \frac{(n-1)!}{n!}=\frac{1}{n}, $$ which surprisingly is the same as our naive calculation... Is this correct? Thanks.