A red die and a green die, both fair, are rolled together. What is the probability that the number on the green die is at least 3 more than the number on the red die?

List the throws with green ≥ red + 3. Red 1 allows green 4, 5 or 6; red 2 allows green 5 or 6; red 3 allows green 6; red 4, 5 or 6 allows nothing. That is 3 + 2 + 1 = 6 throws out of 36, so the probability is 6/36 = 1/6. Trap 5/12 = 15/36 answers the much looser 'the green die simply beats the red one'. Trap 1/3 = 12/36 counts a gap of 3 or more in EITHER direction, but the question names which die has to be the larger. Trap 1/12 = 3/36 demands a gap of more than 3 and drops the throws where the gap is exactly 3.