In a game, Jimmy and Jacob each randomly choose to either roll a fair six-sided die or to automatically roll a $1$ on their die. If the product of the two numbers face up on their dice is even, Jimmy wins the game. Otherwise, Jacob wins. The probability Jimmy wins $3$ games before Jacob wins $3$ games can be written as $\frac{p}{2^q}$, where $p$ and $q$ are positive integers, and $p$ is odd. Find the remainder when $p+q$ is divided by $1000$.

Let's think step by step and output the final answer within \boxed{}.
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