Take 1 to 3 matches where the last one loses

Two players alternately remove $1$, $2$ or $3$ matches from a pile, and the player forced to take the final match loses. What is the optimal strategy?

Always leave your opponent a number of matches that is one more than a multiple of 4. From $4m+1$ any move $k$ can be answered with $4-k$, eventually leaving the opponent the single last match. So move first unless the starting pile is already $\equiv1\pmod 4$, in which case you want to go second.