Example 8 A math game is played between two very smart students, A and B. The referee first writes the integers $2,3, \cdots, 2006$ on the blackboard, then randomly erases one number. Next, A and B take turns erasing one number (i.e., B erases one of the numbers first, then A erases one number, and so on). If the last two remaining numbers are coprime, A wins; otherwise, B wins. According to this game rule, find the probability of A winning (answer with a specific number).
Explanation: The key to winning is to see which number the referee erases. Note that $2,3, \cdots, 2006$, contains 1002 odd numbers and 1003 even numbers.
(1) If the referee erases an odd number, at this point, player B will definitely win.

No matter what number player A takes, as long as there are still odd numbers, player B will erase an odd number. This way, the last two remaining numbers will definitely be even, thus, the two remaining numbers will not be coprime. Hence, player B wins.
(2) If the referee erases an even number, at this point, player A will definitely win.

Suppose the number erased by the referee is $2m$. Then the remaining numbers can be paired into 1002 pairs:
$$
\begin{array}{l}
(2,3), \cdots,(2 m-2,2 m-1), \\
(2 m+1,2 m+2), \cdots,(2005,2006) .
\end{array}
$$

This way, no matter which number player B erases, player A will erase the other number in the pair. The last two remaining numbers will definitely be coprime. Hence, player A wins.
Therefore, the probability of player A winning is $\frac{1003}{2005}$.
Note: The key to solving this problem lies in, first, discussing based on whether the number erased by the referee is odd or even; second, effectively categorizing the numbers.