If we change the game so that the player to move can take up to 3 stones only, then with stones, the successor states have nimbers , giving a mex of 0. Since the nimber for 4 stones is 0, the first player loses. The second player's strategy is to respond to whatever move the first player makes by taking the rest of the stones. For stones, the nimbers of the successor states of 2, 3, and 4 stones are the nimbers 2, 3, and 0 (as we just calculated); the mex of the set of nimbers is the nimber 1, so starting with 5 stones in this game is a win for the first player.