Solution:

For $i \geq 0$, let $f_{i}(x)$ denote the polynomial on the blackboard after Hobbes' $i$-th turn. We let Calvin decrease the coefficient of $x$ by $1$. Therefore $f_{i+1}(2) = f_{i}(2) - 1$ or $f_{i+1}(2) = f_{i}(2) - 3$ (depending on whether Hobbes increases or decreases the constant term). So for some $i$, we have $0 \leq f_{i}(2) \leq 2$. If $f_{i}(2) = 0$ then Calvin has won the game. If $f_{i}(2) = 2$ then Calvin wins the game by reducing the coefficient of $x$ by $1$. If $f_{i}(2) = 1$ then $f_{i+1}(2) = 0$ or $f_{i+1}(2) = -2$. In the former case, Calvin has won the game and in the latter case Calvin wins the game by increasing the coefficient of $x$ by $1$.