Who wins the take-1-to-3 game from 1000 stones

A pile of $1000$ stones sits between two players. On each turn a player removes $1$, $2$ or $3$ stones, and whoever removes the final stone wins. Assuming perfect play, which player is favoured: the one who moves first or the one who moves second?

The second player. Facing a multiple of $4$ is a losing position: whatever $k$ stones the mover takes, the opponent takes $4-k$ and restores a multiple of 4. Since $1000$ is divisible by $4$, the second player keeps the count at multiples of 4 and takes the last stone.