Probability with Three Dice: 216 Outcomes | Class 10 Maths

Quick answer: When three dice are rolled together, there are $6 \times 6 \times 6 = 216$ equally likely outcomes. The probability that all three show the same number is $\frac{6}{216} = \frac{1}{36}$, that all three are different is $\frac{120}{216} = \frac{5}{9}$, and of getting at least one six is $1 - \frac{125}{216} = \frac{91}{216}$. A sum of $10$ ($27$ outcomes) is slightly more likely than a sum of $9$ ($25$ outcomes).

Rolling one die gives $6$ outcomes, and rolling two dice gives $6 \times 6 = 36$. With three dice, the number of outcomes grows to $6 \times 6 \times 6 = 216$.

Three-dice questions test careful counting: how many outcomes give three equal numbers, three different numbers, at least one six, or a particular sum. A famous example, discussed by Galileo, asks why a sum of $10$ comes up more often than a sum of $9$.

Syllabus note: The CBSE Class 10 NCERT chapter on Probability works with one and two dice. Three-dice problems go beyond the textbook, but they are good practice for higher-order thinking (HOTS) questions and competitive exams.

What You Will Learn

• Why rolling three dice gives $216$ outcomes
• The probability of three equal numbers, exactly two equal numbers and three different numbers
• How to find the probability of at least one six using the complement
• How to count outcomes for a given sum, from $3$ to $18$
• Why a sum of $10$ is more likely than a sum of $9$

1. Total Number of Outcomes

Think of the three dice as different colours, say red, green and blue. Each die can show any of $6$ numbers, so:

$\text{Total outcomes} = 6 \times 6 \times 6 = 6^3 = 216$

Each outcome is an ordered triple $(x, y, z)$, where $x$, $y$ and $z$ are the numbers on the red, green and blue dice.

2. Equal and Different Numbers

All Three Numbers the Same (a Triplet)

The favourable outcomes are $(1,1,1)$, $(2,2,2)$, $(3,3,3)$, $(4,4,4)$, $(5,5,5)$ and $(6,6,6)$, which is $6$ outcomes.

$P(\text{all three the same}) = \frac{6}{216} = \frac{1}{36}$

All Three Numbers Different

The first die can show any of $6$ numbers, the second any of the remaining $5$, and the third any of the remaining $4$:

$\text{Favourable outcomes} = 6 \times 5 \times 4 = 120$

$P(\text{all different}) = \frac{120}{216} = \frac{5}{9}$

Exactly Two Numbers the Same

The remaining outcomes have exactly two equal numbers:

$216 - 6 - 120 = 90$

$P(\text{exactly two the same}) = \frac{90}{216} = \frac{5}{12}$

Check: $\frac{6}{216} + \frac{120}{216} + \frac{90}{216} = \frac{216}{216} = 1$. ✓

3. At Least One Six (Using the Complement)

Instead of counting outcomes with one, two or three sixes, count the outcomes with no six. Each die then has $5$ choices ($1$ to $5$):

$\text{Outcomes with no six} = 5 \times 5 \times 5 = 125$

$P(\text{no six}) = \frac{125}{216}$

$P(\text{at least one six}) = 1 - \frac{125}{216} = \frac{91}{216}$

4. Sums of Three Dice

The smallest possible sum is $1 + 1 + 1 = 3$ and the largest is $6 + 6 + 6 = 18$.

To count the outcomes for a sum, list the groups of three numbers that add up to it, then count how many ways each group can be arranged:

• Three different numbers can be arranged in $6$ ways.
• Two equal numbers and one different number can be arranged in $3$ ways.
• Three equal numbers can be arranged in only $1$ way.

5. Why Is a Sum of 10 More Likely Than a Sum of 9?

In the 17th century, gamblers noticed that with three dice, a sum of $10$ came up more often than a sum of $9$. This was puzzling, because both sums can be made from six groups of numbers.

Galileo explained that what matters is the number of ordered arrangements, not the number of groups:

• Sum 9: three groups of different numbers ($3 \times 6 = 18$), two groups with a repeated number ($2 \times 3 = 6$) and one triplet ($1$), giving $18 + 6 + 1 = 25$ outcomes.
• Sum 10: three groups of different numbers ($3 \times 6 = 18$) and three groups with a repeated number ($3 \times 3 = 9$), giving $18 + 9 = 27$ outcomes.

$P(\text{sum } 9) = \frac{25}{216} \qquad P(\text{sum } 10) = \frac{27}{216} = \frac{1}{8}$

Key point: A sum of $10$ is slightly more likely than a sum of $9$, because $(3,3,3)$ can be arranged in only one way.

6. Summary Table

7. Exam Tips and Common Mistakes

Exam Tip: For "at least one", use the complement: $P(\text{at least one six}) = 1 - P(\text{no six})$. It is much faster than counting directly.

Common Mistake: Using $36$ as the total. That is for two dice. For three dice, the total is $216$.

Common Mistake: Counting groups instead of arrangements. $(1, 2, 6)$, $(2, 1, 6)$ and $(6, 2, 1)$ are different outcomes.

Frequently Asked Questions

How many outcomes are there when three dice are rolled?

There are $6 \times 6 \times 6 = 216$ equally likely outcomes.

What is the probability of getting the same number on all three dice?

There are $6$ such outcomes, so the probability is $\frac{6}{216} = \frac{1}{36}$.

What is the probability of getting at least one six with three dice?

$1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216} = \frac{91}{216}$.

Why is a sum of 10 more likely than 9 with three dice?

A sum of $10$ can occur in $27$ ordered ways and a sum of $9$ in only $25$, because $(3, 3, 3)$ has just one arrangement.

Are three-dice problems in the Class 10 syllabus?

The NCERT textbook works with one and two dice. Three-dice problems are extra practice for HOTS and competitive exams.