6. Let $m, n$ be given positive integers. In each square of an $m \times n$ chessboard, fill a number according to the following rule: first, fill the numbers in the 1st row and the 1st column arbitrarily, then, for any other square, let the number filled in it be $x$, and the number in the 1st row in the same column as $x$ be $y$, the number in the 1st column in the same row as $x$ be $z$, and the number in the 1st row and 1st column be $a$, then $a + x = y + z$. Thus, a unique number can be filled in each square. After all squares are filled with numbers, take any rectangle on the chessboard, then the probability that the sum of the numbers in the two diagonal corners of the rectangle is equal is $\qquad$
6.1.

Let the cell at the $i$-th row and $j$-th column of the chessboard be denoted as $a_{i j}$, and the number filled in cell $a_{i j}$ is also represented by $a_{i j}$.

Consider any rectangle on the chessboard, and let the four corner cells of this rectangle be $a_{i j}, a_{i t}, a_{s j}, a_{s t} (i<s, j<t)$.
By the rule of filling numbers, we have
$$
\begin{array}{l}
a_{i j}=a_{1 j}+a_{i 1}-a_{11}, a_{i t}=a_{1 t}+a_{i 1}-a_{11}, \\
a_{s j}=a_{1 j}+a_{s 1}-a_{11}, a_{s t}=a_{1 t}+a_{s 1}-a_{11} . \\
\text { Then } a_{i j}+a_{s t}=\left(a_{1 j}+a_{i 1}-a_{11}\right)+\left(a_{1 t}+a_{s 1}-a_{11}\right) \\
=a_{1 j}+a_{i 1}+a_{1 t}+a_{s 1}-2 a_{11}, \\
a_{i t}+a_{s j}=\left(a_{1 t}+a_{i 1}-a_{11}\right)+\left(a_{1 j}+a_{s 1}-a_{11}\right) \\
=a_{1 j}+a_{i 1}+a_{1 t}+a_{s 1}-2 a_{11} .
\end{array}
$$

Therefore, $a_{i j}+a_{s t}=a_{i t}+a_{s j}$
Thus, the required probability is 1.