If an unbiased dice is rolled thrice, then the probability of getting a greater number in the i-th r

Q: Wait, why can't I just calculate the probability of $x_2 > x_1$ and $x_3 > x_2$ separately and multiply them?

That is a classic trap! These events are not independent. The outcome of $x_2$ is constrained by $x_1$, and $x_3$ is constrained by $x_2$. You cannot treat them as isolated dice rolls. The entire sequence is a single event defined by the global constraint $x_1 < x_2 < x_3$.

Q: Why do we use combinations $\binom{6}{3}$ instead of permutations $P(6,3)$?

Great question! Permutations count the number of ways to arrange items. But here, the inequality $x_1 < x_2 < x_3$ forces a specific, unique order. Once you pick any three distinct numbers, there is only *one* way to arrange them in increasing order. So, we only care about the *selection* of the numbers, not the arrangement.

Q: What if the question said $x_1 \le x_2 \le x_3$ instead of $x_1 < x_2 < x_3$?

That would change the problem entirely! That would be a 'multiset' problem because repetition is allowed. For strictly increasing, we need distinct numbers. For non-decreasing, we would need to account for repetitions, which requires a different combinatorial technique like 'stars and bars'.

Question

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

Options:

(A) $\frac{3}{54}$

(B) $\frac{2}{54}$

(C) $\frac{1}{54}$

(D) $\frac{5}{54}$

Sigma Percentile · Accepted Answer

### The Dance of the Dice: A Probability Masterclass

Imagine you are standing in a quiet room, holding an unbiased six-sided die. You are about to roll it three times.

Each roll is a moment of pure chance, yet when we look at the sequence of outcomes $x_1, x_2, x_3$, we are looking at a mathematical structure. Today, we are going to unravel the beauty behind the probability of a strictly increasing sequence: $x_1 < x_2 < x_3$.

### Defining the Universe

Before we dive into the constraints, we must understand the total possibilities. When you roll a die once, you have $6$ outcomes.

When you roll it three times, the total number of outcomes is the product of the possibilities for each roll. Mathematically, this is:

$$n(S) = 6 	imes 6 	imes 6 = 6^3 = 216$$

This is our sample space—the entire universe of outcomes. Every single one of these $216$ possibilities is equally likely, provided the die is fair.

### The Heart of the Constraint

The question asks for the probability that the number obtained in the $i^{th}$ roll is strictly greater than the number obtained in the $(i-1)^{th}$ roll. This translates to the condition $x_1 < x_2 < x_3$.

This is not just a condition; it is a filter. It discards any sequence where a number repeats or decreases.

For example, $(1, 2, 2)$ is out because $2$ is not strictly greater than $2$. Similarly, $(3, 2, 1)$ is out because it is decreasing. We are looking for the rare, elegant sequences that climb steadily upward.

### The Selection vs. Arrangement Revelation

Here is where most students stumble. They try to count the sequences one by one, which is a recipe for disaster.

Instead, let's use the power of combinatorics. Notice that for any set of three **distinct** numbers chosen from $\{1, 2, 3, 4, 5, 6\}$, there is exactly one way to arrange them in increasing order.

If you pick the set $\{1, 4, 6\}$, the only valid sequence is $(1, 4, 6)$. If you pick $\{2, 3, 5\}$, the only valid sequence is $(2, 3, 5)$.

This is the "Aha!" moment. We don't need to worry about the order of the rolls because the inequality $x_1 < x_2 < x_3$ forces the order for us.

Therefore, the number of favorable outcomes is simply the number of ways to **select** three distinct numbers from the six available. This is a classic combination problem: $\binom{6}{3}$.

### The Final Calculation

Let's compute the favorable outcomes. The formula for combinations is $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

Plugging in our values, we get:

$$\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$$

There are exactly $20$ such sequences that satisfy our condition.

Now, we bring it all together. The probability $P$ is the ratio of favorable outcomes to total outcomes:

$$P = \frac{20}{216}$$

To simplify this, we divide both the numerator and the denominator by $4$. Thus, the final probability is:

$$\mathbf{P = \frac{5}{54}}$$

### Conclusion: The Elegance of Logic

We didn't brute-force the problem. We looked at the structure of the inequality, realized that selection was equivalent to arrangement due to the strict ordering, and reduced a complex counting problem to a simple combination.

This is the essence of JEE Advanced mathematics: finding the underlying symmetry that makes the impossible, possible. Keep practicing this mindset, and you will find that even the most daunting problems have a beautiful, logical core.

Animated Solution for Mathematics - Probability: 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

Select Answer:

Visualized Solution

Defining the Variables

Calculating Total Outcomes

Understanding the Constraint

Selection vs Arrangement

Raw Setup: Favorable Outcomes

Atomic Compute: Favorable Outcomes

Raw Setup: Probability

Atomic Compute: Simplification

Key Takeaway & Conclusion

The Sigma Insight: Classical Definition of Probability

The Dance of the Dice

Defining the Universe

The Heart of the Constraint

The Selection vs

The Final Calculation

Conclusion

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