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

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    – Kuba
    Commented Oct 15, 2015 at 18:42

3 Answers 3


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


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 ($6^3=216$).


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


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.


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