1. **Problem statement:** Two players, A and B, each throw a pair of dice. Player A throws a total of 9. We need to find the probability that player B throws a higher total than 9. 2. **Possible sums when throwing two dice:** The sums range from 2 to 12. The number of ways to get each sum is: - 2: 1 way - 3: 2 ways - 4: 3 ways - 5: 4 ways - 6: 5 ways - 7: 6 ways - 8: 5 ways - 9: 4 ways - 10: 3 ways - 11: 2 ways - 12: 1 way 3. **Total possible outcomes:** Since each die has 6 faces, total outcomes for two dice are $6 \times 6 = 36$. 4. **Calculate the number of ways B can throw a higher number than 9:** - Sums higher than 9 are 10, 11, and 12. - Number of ways for 10: 3 - Number of ways for 11: 2 - Number of ways for 12: 1 - Total ways = $3 + 2 + 1 = 6$ 5. **Calculate the probability:** $$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}$$ 6. **Final answer:** The probability that B throws a higher number than A's 9 is $\boxed{\frac{1}{6}}$.