Maybe I'm missing something, but it seems that if player 2 lands away from one of the extremal tiles in less than $d$ turns, (s)he has lost (since, then player 1 can order player 2 to move n-1 places).

So wouldn't the winning strategy (for $n>2$ and $d>2$) be $n-1$, $1$, $n-1$?

Maybe I'm missing something, but it seems that if player 2 lands away from one of the extremal tiles in less than $d$ turns, (s)he has lost (since, then player 1 can order player 2 to move n-1 places).

So wouldn't the winning strategy (for $n>2$ and $d>2$) be $n-1$, $1$, $n-1$?

Edit: Steven Stadnicki points out that even when player 1 cannot use the same order twice, there is still a winning strategy: $n-1, 2, n-2$. This works if $n>4$. For $n=4$, player 1 should first order 0. If player 2 puts his piece on the edge, then $1,3$ wins, and if player 2 puts his piece off the edge, then 3 wins.