Twenty-one pebbles lie in a heap. Two players take turns, and a player must take 1, 2 or 3 pebbles on a turn. Whoever takes the very last pebble wins. How many pebbles should the first player take on the opening turn to be sure of winning?

The heap sizes that are hopeless for the player about to move are the multiples of 4: whatever that player takes, the opponent takes the rest of a block of 4 and hands back another multiple of 4, ending with 0 left for the loser. So the winning plan is to leave a multiple of 4 every time. $21=4\times 5+1$, so the first player takes 1 pebble, leaving 20, and then simply keeps the total of each round at 4. The trap is 3: taking the largest allowed amount looks aggressive but leaves 18, which is not a multiple of 4 and hands the winning plan to the opponent.