Three identical dice are rolled. The probability that the same number will appear on each of them is

Q: Wait, the question says the dice are 'identical'. Does that change the sample space?

This is a fantastic question! In probability, we distinguish between physical indistinguishability and probabilistic distinctness. Even if the dice look identical, each die acts as an independent system. To define the sample space correctly, we must treat them as distinct. If we didn't, we would be incorrectly weighting the outcomes. Always treat dice as distinct for the purpose of counting the sample space.

Q: Why is the denominator $6^3$ and not something else?

Think of it this way: the first die has 6 possibilities. For each of those 6, the second die also has 6 possibilities, and for each of those, the third die has 6. By the Fundamental Principle of Counting, we multiply these: $6 \times 6 \times 6 = 6^3 = 216$. This represents every possible combination of faces that can appear.

Q: Can I use the formula for combinations here?

You could, but it is overkill! Combinations are usually for selecting items where order doesn't matter. Here, we are looking for specific outcomes in a sequence. Simple counting is the most robust and elegant way to solve this. Stick to the ratio of favorable outcomes to total outcomes; it is the bedrock of probability.

Question

Three identical dice are rolled. The probability that the same number will appear on each of them is

Options:

(A) $1/6$

(B) $1/36$

(C) $1/18$

(D) $3/28$

Sigma Percentile · Accepted Answer

### The Dance of Chance: Unlocking the Mystery of Identical Dice

Welcome, future engineers and physicists! Today, we are going to peel back the layers of a classic probability problem that often trips up even the brightest minds.

It is a simple scenario: three identical dice are rolled, and we want to know the probability that they all show the same number. It sounds straightforward, but the word 'identical' is a siren song designed to lead you into a trap. Let us embark on this journey together.

### The Illusion of Indistinguishability

When you hear that the dice are 'identical', your intuition might scream, 'Wait, does this mean I need to use Bose-Einstein statistics or some complex counting method?' I want you to take a deep breath and let that thought go.

In the realm of classical probability, we treat these dice as distinct entities that happen to look the same. Imagine you have a red die, a blue die, and a green die; the math remains exactly the same.

By treating them as distinct, we create a clear, logical sample space. This is the first step to mastering any probability problem: define your universe clearly.

### Building the Sample Space

Let us construct our sample space, denoted as $n(S)$. We have three dice, where each die can land on any of the $6$ faces.

According to the Fundamental Principle of Counting, if we have independent events, we multiply their possibilities. The total number of outcomes is:

$$n(S) = 6 	imes 6 	imes 6 = 216$$

This is our denominator. It represents every possible triplet $(x, y, z)$ that can appear on the table.

### Identifying the Favorable Event

Now, let us look for the 'favorable' outcomes, $n(E)$. The condition is that all three dice must show the same number.

The possible outcomes are $(1, 1, 1)$, $(2, 2, 2)$, $(3, 3, 3)$, $(4, 4, 4)$, $(5, 5, 5)$, and $(6, 6, 6)$. There are exactly $6$ such triplets.

Thus, our numerator is $n(E) = 6$.

### The Final Calculation

Now, we bring it all together using the fundamental definition of probability: $P(E) = \frac{n(E)}{n(S)}$. Substituting our values, we get:

$$P(E) = \frac{6}{216}$$

When we simplify this fraction by dividing both the numerator and the denominator by $6$, we arrive at the elegant result:

$$P(E) = \frac{1}{36}$$

### The Pro-Tip: Generalization

Before we wrap up, I want to give you a tool for your JEE arsenal. What if we had $n$ dice?

The total number of outcomes would be $6^n$. The number of favorable outcomes where all dice show the same number is always $6$.

So, the probability for $n$ dice is:

$$P(E) = \frac{6}{6^n} = \frac{1}{6^{n-1}}$$

This is a powerful shortcut. If you are ever asked this for four or five dice, you won't even need to break a sweat. Keep practicing, keep visualizing, and remember: the beauty of physics and math lies in finding the simple, elegant truth hidden beneath the complexity. You have got this!

Animated Solution for Mathematics - Probability: Three identical dice are rolled. The probability that the same number will appear on each of them is

Select Answer:

Visualized Solution

The Experiment

Probability Formula

Outcomes for the First Die

Total Sample Space $n (S)$

Calculating Total Outcomes

The Favorable Event $E$

Listing Favorable Outcomes

The Complete Set $E$

Number of Favorable Outcomes $n (E)$

Substituting into the Formula

Final Calculation

Conclusion & Generalization

The Sigma Insight: Classical Definition of Probability

The Dance of Chance

The Illusion of Indistinguishability

Building the Sample Space

Identifying the Favorable Event

The Final Calculation

The Pro-Tip

Similar Questions

Four fair dice $D_{1}, D_{2}, D_{3}$ and $D_{4}$ ; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that $D_{4}$ shows a number appearing on one of $D_{1}, D_{2}$ and $D_{3}$ is

An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown $n$ times and the list of $n$ numbers showing up is noted. What is the probability that, among the numbers 1, 2, 3, 4, 5, 6, only three numbers appear in this list?

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{t h}$ roll than the number obtained in the $(i - 1)^{t h}$ roll, $i = 2, 3$ , is equal to

Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then $q - p$ is equal to

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

Two dices are rolled. If both dices have six faces numbered 1,2,3,5,7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is