Three Dice Rolls: Achieving a Sum of Ten

Solution & Explanation

Problem Statement: Calculate the probability of achieving a total sum of 10 when rolling a six-sided die three times.

Step-by-Step Solution:

1. Understanding the Problem:

   • Each die has 6 faces, numbered from 1 to 6.
   • We roll the die three times and want the sum of the numbers on the top faces to be 10.

2. Total Possible Outcomes:

   • Each roll is independent, and there are 6 possible outcomes per roll.
   • Total possible outcomes when rolling three dice = $6 \times 6 \times 6 = 216$.

3. Desired Outcomes:

   • We need to find all combinations of three numbers ($x, y, z$) such that $x + y + z = 10$ where each number is between 1 and 6.
   • Possible combinations are:
     • $(1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4), (3, 3, 4)$

4. Calculating Permutations for Each Combination:

   • $(1, 3, 6)$: All numbers are different, so there are $3! = 6$ permutations.
   • $(1, 4, 5)$: All numbers are different, so there are $3! = 6$ permutations.
   • $(2, 2, 6)$: Two numbers are the same, so there are $\frac{3!}{2!} = 3$ permutations.
   • $(2, 3, 5)$: All numbers are different, so there are $3! = 6$ permutations.
   • $(2, 4, 4)$: Two numbers are the same, so there are $\frac{3!}{2!} = 3$ permutations.
   • $(3, 3, 4)$: Two numbers are the same, so there are $\frac{3!}{2!} = 3$ permutations.

5. Total Favorable Outcomes:

   • Summing all permutations gives us the total number of favorable outcomes:
     • $6 + 6 + 3 + 6 + 3 + 3 = 27$

6. Probability Calculation:

   • Probability of achieving a sum of 10 = $\frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}}$
   • $\frac{27}{216} = \frac{1}{8}$

7. Conclusion:

   • The probability of rolling three six-sided dice and getting a sum of 10 is $\frac{1}{8}$ or 12.5%.