Probability that the sum of three dice is not greater than 9

Question

What is the probability that the sum of 3 indistinguishable dice is less than or equal to 9?

I have tried counting the pairs $X_1, X_2$ such that their sum is less than or equal to 8, but I seem to be overcounting because I get that for each pair of dice I have 46 possibilities for their sum to be less than 8.

Answer 1

You have $27$ combinations of the digits $(1,2,3)$ plus $15$ combinations that include $4$: $(1,1,4), (1,2,4), (2,2,4), (1,3,4)$ and $(2,3,4)$ plus $12$ combinations that include a $5$: $(1,1,5), (1,2,5), (2,2,5), (1,3,5)$ over the $216$ possible outcomes. So $54/216 = .25$

Answer 2

Consider the sum of the first two dice. If the sum is lower than or equal to 3, the third die can have any value. As the sum increases from 3 to 8, fewer possible values can be thrown by the third die. Since the number of ways to throw a sum of $n$ with two dice equals $6 - |n - 7|$, the number of ways to throw a total of 9 or less equals:

$$1 \cdot 6 + \sum_{i = 3}^{8} (6 - |i - 7|) (9 - i) = 1 \cdot 6 + 2 \cdot 6 + 3 \cdot 5 + 4 \cdot 4 + 5 \cdot 3 + 6 \cdot 2 + 5 \cdot 1 = 81$$

Since there are $6^3 = 216$ ways to throw the three dice, the probability of throwing 9 or less equals:

$$\frac{81}{216} = 0.375$$

Answer 3

The "indistinguishable" part is potentially misleading. If you want the probability of an event produced by rolling three dice, you should label the dice first, second, and third.

The number of ways the dice can add up to $9$ or less is the number of ways you can make a sum of $9$ or less from three positive integers in sequence provided that no integer is greater than $6.$

If you remove the "greater than $6$" condition then the number of ways to make a sum of $9$ or less from three positive integers is the same as the number of ways to make a sum of exactly $10$ from four positive integers. A well-known way to compute that number is the so-called "stars and bars" method, which tells us the answer is $\binom93 = 84.$

But the "greater than $6$" condition says the first die (for example) cannot be $7$ or greater. There is only one such case for the first die (the rolled numbers are $7,1,1$), and likewise for the other two dice, so we have to remove those three cases from the total. We are left with $84 - 3 = 81.$

There are $216$ possible rolls altogether, each one equally likely, so the probability is $\frac{81}{216} = \frac38.$