Question
Alice and Bob pick alternate turns in the following game: Initially, there are $n \geq 1$ stones in the middle. Either player can choose to take $1$ to $r$ stones from the pile. Assume that $n \equiv 1 \hspace{3pt} (\text{mod} \hspace{3pt} r+1)$. The player who draws the last stone loses. Alice gets to determine if she goes first or second. Assuming Alice selects her position in the game optimally and both Alice and Bob play optimally, in how many turns (total actions by the players) does Alice win the game, inclusive of the draw that makes Bob lose, when $n = 101$ and $r = 4$?