Which is more likely: rolling a total of nine when two dice are rolled or when three dice are rolled

Solution:

When two fair six-sided dice are rolled

We have to find the probability that the sum is 9 or higher

About 36 different combos are present for the two dice i.e. 6 possibilities for the first dice and 6 possibilities for the second.

Among the 36, 10 have the sum 9 or higher

3 and 6

4 and 6

4 and 5

5 and 4

5 and 5

5 and 6

6 and 3

6 and 4

6 and 5

6 and 6

The favorable outcomes are {(3,6),(4,6),(4,5),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}

Number of favorable outcomes = 10

Probability = 10/36 = 5/18

Therefore, the probability that the sum is 9 or higher is 5/18.

Summary:

If you roll two fair six-sided dice, the probability that the sum is 9 or higher is 5/18.

Convince yourself that when you roll two dice you are more likely to score 9 than 10.

If I roll 3 dice, which is more likely, a total score of 9, or a total score of 10?

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The Question

Which is more likely: rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled?

My Work

First we have to determine the probability of two die summing to $9$. I began by calculating the sample space. There are $6*6 = 36$ different ways the two die could sum. Now to figure out how many of those ways sum to $9$. This is essentially the same question as how many integer solutions to the equality $x_1 + x_2 = 9$ where $1 \leq x\le6$ ($x$ represents the two dice in this case). The number of ways to do this is $\binom{8}{7} - 2$ (where $2$ represents the case where $x\geq6$. So the probability of rolling two dice which sum to $9$ is $\frac{6}{36}$

Figuring out how many ways $3$ dice sum to $9$ is a similar process. Find all solutions to $x_1 + x_2 + x_3 =9$ where $1\le x\le6$ which is $\binom{8}{6} - 3 = 25$ ways to do this (3 represents all the solutions where $x_i$ is greater than $6$). There are $6^3$ ways $3$ dice can be rolled therefore the probability $\frac{25}{6^3}$.

Computing the probabilities I found $\frac{6}{36} > \frac{25}{6^3}$ Therefore, it is more likely with 2 dice.

The Problem

My book says it is more likely with 3 dice. Can anyone point out where I went wrong in my approach, and how to correct it?