A game uses one ordinary die with faces 1 to 6 together with a special die whose six faces read 1, 1, 2, 2, 3 and 3. Both dice are rolled once. What is the probability that the number on the ordinary die is strictly greater than the number on the special die?

The special die shows 1, 2 or 3, each with probability 2/6 = 1/3. Take the cases in turn: against a 1 the ordinary die wins with probability 5/6; against a 2, with probability 4/6; against a 3, with probability 3/6. So the answer is (1/3) × (5/6 + 4/6 + 3/6) = (1/3) × 12/6 = 12/18 = 2/3. Counting instead over the 18 equally likely (special, ordinary) pairs gives 12 winning pairs. Trap 1/2 assumes the two dice must be evenly matched by symmetry, but the special die can never beat 3 while the ordinary one reaches 6. Trap 5/6 = 15/18 answers 'at least as large', quietly including the 3 tied pairs. Trap 1/6 = 3/18 is the probability of a tie itself.