Five. (25 points) A math game is played between two very smart students, A and B. The referee first writes the positive integers $2,3, \cdots, 2006$ on the blackboard, then randomly erases one number. Next, B and A take turns erasing one of the numbers (i.e., B erases one number 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).

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Five, the key to winning is to see which number the referee erases.
Notice 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 numbers will definitely be even, thus the remaining two numbers will not be coprime, so 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:
$(2,3), \cdots,(2m-2,2m-1),(2m+1,2m+2)$, $\cdots,(2005,2006)$.

This way, no matter which number player B erases, player A will erase the other number in the paired set, ensuring that the last two remaining numbers are coprime. Hence, player A wins.
Therefore, the probability of player A winning is $\frac{1003}{2005}$.