# Three fair six-sided dice are rolled. [closed]

Three fair six-sided dice are rolled. Find the probability that the total on all three dice is five or less.

## closed as off-topic by vonbrand, mrf, jameselmore, drhab, hardmathOct 16 '15 at 0:46

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• Hello. Welcome to Math SE. On this forum it's usually nice to explain what you have tried doing in order to solve the problem. We're not machines that solve random homeworks, we wanna make you understand the whole topic better. – I want to make games Oct 15 '15 at 18:42

## 3 Answers

Say $(D_1,D_2,D_3)$ denote the outcomes of the 3 dice respectively.
The possible ways are as follows:

$(1,1,1),(1,1,2),(2,1,1),(1,2,1),(1,2,2),(2,1,2),(2,2,1),(1,1,3),(1,3,1),(3,1,1)$

i.e. $10$ outcomes.

The total number of outcomes is

$$6\times 6\times 6 = 216$$

Hence probability is

$$\frac{10}{216} = \frac{5}{108}$$

If you want to approach the problem more analytically, consider that the probability is the number of ways of choosing $x_1,x_2,x_3$ (call this $N$) such that each is a positive integer (and the sum is five or less) divided by the the total number of different outcomes for the three dice ($6^3=216$).

$$x_1+x_2+x_3\le5$$

Introduce a slack variable $x_4\ge1$, and find the total number of solutions to $$x_1+x_2+x_3+x_4=6$$, where each of the $x_i\ge1$. By stars and bars, this is

$$N=\binom{6-1}{4-1}=\binom{5}{3}=10$$

so the probability is $\frac{10}{216}=\frac{5}{108}\approx0.0463$.

So, because dice re dice, we know that the minimum roll is a 1. This means that no die can be larger than a three as well, because $n=3$ in the equation $5-1(n-1)=3$.

Pretend die A is rolled and is a $3$ (a $\frac{1}{6}$ chance). At this point, both die B and die C need to be $1$s in order for the sum to be less than or equal to $5$. The combination of $[3,1,1]$ in that specific order has a $\frac{1*1*1}{6*6*6} = \frac{1}{216}$ chance of concurring.

What happens if die A is rolled and is a two? How many combinations of the last two dice exist where the sum is less than or equal to $3$? $[2,1,1],[2,2,1],[2,1,2]$ are all possibilities, and those are three more $\frac{1}{216}$s that can occur.

There are more combinations when die A is a $1$, but I'll leave that up to you.