What is the probability of rolling three dice and obtaining a sum between 5 to 10(inclusively)?

Question

For two dice, it is easy to create a list containing the outcomes of each dice since there are only 36 cases. ({1,1}, {1,2}, {1,3}, {1,4}, ... , {6,5}, {6,6}).

But for three dice, there are $6^3$ cases. Therefore, how can I calculate the probability of the given question without writing down every possible combinations ?

Stars and bars, also known as sticks and stones, dots and dividers, etc. If not, what topics have you learnt so far ? — true blue anil, Dec 05 '22 at 09:36
Suppose the dice are blue, red, and green. For each possible value on the blue die, in how many ways can the red and green die give you a sum in the correct range? — N. F. Taussig, Dec 05 '22 at 10:28
@trueblueanil Now I can understand the statement. I have learned the concept using '0' and '1', which is the same as stars and bars. I'm sorry for misunderstanding. — Woshi, Dec 05 '22 at 12:38

user2661923 · Answer 1 · 2022-12-05T11:42:52.583

This is not an answer.

Instead, this is a long-winded comment.

I am not allowed to answer because your posting has defects.

For an explanation of the defects, see this article on MathSE protocol. As onerous as the article may appear to you, it provides a defense mechanism against the MathSE forum being used as a do my homework forum. In particular, please see the Edit-Tools section of the article, and the portion of the article that discusses showing work. Homework problems are allowed, as long as the protocol is observed.

Most of the theory that you will need is contained in this answer.

There are two additional pieces of theory needed.

$\underline{\text{Additional Theory-1}}$

Consider the following problem:

$x_1 + x_2 + \cdots + x_k \leq n.$
$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}.$

The easy way to handle this is to create the variable $a$ and change the problem to

$x_1 + x_2 + \cdots + x_k + a = n.$
$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}.$
Note that by the constraint $x_1 + x_2 + \cdots + x_k \leq n,$
$a$ is forced to (also) be a non-negative integer.

$\underline{\text{Additional Theory-2}}$

The standard method of handling a problem like

$x_1 + x_2 + \cdots + x_k = n$
$x_1, x_2, \cdots, x_k \in \Bbb{Z_\geq {1}}.$

is via the change of variables:
$y_i = x_i - 1 ~: ~i \in \{1,2,\cdots,k\}.$

This leads to

$y_1 + y_2 + \cdots + y_k = (n-k)$
$y_1, y_2, \cdots, y_k \in \Bbb{Z_\geq {0}}.$

Clearly, there is a bijection between the set of solutions to the $x_1,\cdots,x_k$ problem and the set of solutions to the $y_1,y_2, \cdots, y_k$ problem. Therefore, the two problems have the same number of solutions.

For what it's worth, suppose that you are given

$~n,m \in \Bbb{Z^+} ~: ~n < m.$
$P$ equals the number of solutions to
$x_1 + x_2 + \cdots + x_k \leq m.$
$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}$.
$Q$ equals the number of solutions to
$x_1 + x_2 + \cdots + x_k \leq (n-1).$
$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}$.

Then, the number of solutions where $n \leq ~\text{the sum} ~\leq m$
will be $P - Q.$

One additional point. Your posting represents a Probability problem, where the answer can be expressed as

$$\frac{N}{D} ~: ~D = 6^3.$$

Since each of the $216$ rolls are equally likely, each Stars and Bars solution is just as likely to occur as any of the other $215$ possible rolls. Therefore, the problem reduces to enumerating $N$.

What is the probability of rolling three dice and obtaining a sum between 5 to 10(inclusively)?

1 Answers1