A player throws a pair of fair dice and compares two totals: the two numbers added, and the two numbers multiplied. What is the probability that adding gives the strictly bigger result?

If either die shows 1 the sum wins: 1 + b is always bigger than 1 × b. That covers the 6 throws with the first die 1 and the 6 with the second die 1, but (1,1) sits in both lists, so it is 6 + 6 − 1 = 11 throws. With both dice at 2 or more the product never falls behind, since ab − (a + b) = (a − 1)(b − 1) − 1 ≥ 0. So 11 throws out of 36. Trap 1/3 = 12/36 also counts (2,2), where sum and product both equal 4 — 'strictly greater' rules a tie out. Trap 1/6 counts only the throws whose FIRST die is 1. Trap 5/18 = 10/36 subtracts the shared throw (1,1) twice.