The four aces are taken out of a standard 52-card pack and set aside. Two cards are then drawn together at random from the 48 cards that remain. What is the probability that both of them are face cards (jack, queen or king)?

Removing the aces leaves 48 cards which still hold all 12 face cards. Total draws = $C(48,2)=1128$, and both-face draws = $C(12,2)=66$, so the probability is 66/1128 = 11/188. Trap 11/221 = 66/1326 divides by $C$(52,2) and forgets that the pack has shrunk to 48. Trap $1/16=(12/48{)}^2$ assumes the first card is put back before the second is drawn, but the two cards come out together. Trap 1/4 = 12/48 is the chance for a SINGLE card and answers a one-card question.