We consider the set $B$ of the numbers on the blackboard at the end of the game. It is clear that $B \subseteq \{1, \dots, 2016\}$. Let $m = \min B$ and $n \in B$. We claim that $m \mid n$. Otherwise, write $n = qm + r$ with $0 < r < m$. By induction on $k$, we have $n - k m \in B$ for $0 \le k \le q$ (because no more moves are possible, these numbers must be on the blackboard). Thus $r = n - q m \in B$, which contradicts the minimality of $m$.

We conclude that $m \mid 1 = \gcd(2016, 47) \in B$. By induction on $c$, we have $n - c \in B$ for $0 \le c \le 2015$. This also implies that $B = \{1, \dots, 2016\}$.

As $2$ numbers had been on the blackboard at the beginning of the game, the game ends after $2014$ moves when all other numbers have been written. Therefore, Bob wins after move $2014$.