A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8,

Q: Wait, why do we use an infinite series? Can't the game end in a finite number of turns?

That is a brilliant observation! While the game *could* end in a finite number of turns, we don't know *which* turn it will end on. Because the dice rolls are independent and the game continues until someone wins, there is a non-zero probability that the game continues for 10, 100, or even 1000 turns. To account for all these possibilities, we must sum the probabilities of all winning scenarios, which forms an infinite series. It is the only way to capture the 'total' probability space.

Q: Why is the common ratio $P(A^c) \times P(B^c)$? Shouldn't it be just $P(A^c)$?

Think about the structure of the game. For A to get a second chance to win (on the 3rd turn), two specific things must happen in sequence: A must fail their first throw, AND B must fail their first throw. If B wins on their first throw, the game is over, and A never gets that second chance. Therefore, the 'cycle' of the game repeating requires both players to fail. Thus, the common ratio must be the product of their failure probabilities.

Q: Is there a faster way to solve this without writing out the whole series?

Yes! You can use a recursive approach. Let $P$ be the probability that A wins. A wins if they roll a 5 immediately (probability $P(A)$), OR if they fail, B fails, and then we are back to the start of the game (probability $P(A^c)P(B^c)P$). So, $P = P(A) + P(A^c)P(B^c)P$. Solving this linear equation for $P$ gives you the exact same result as the infinite series formula. It is a powerful shortcut!

Question

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

Options:

(A) $\frac{8}{17}$

(B) $\frac{9}{19}$

(C) $\frac{9}{17}$

(D) $\frac{8}{19}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are not just solving a probability problem; we are stepping into the arena of an infinite game. This problem is a classic in the JEE Advanced curriculum because it tests your ability to bridge the gap between discrete events and infinite series.

Imagine you and a friend are sitting across from each other, a pair of dice in hand. You are Player A, hunting for a sum of $5$. Your friend is Player B, hunting for a sum of $8$.

The rules are simple: you go first. If you don't win, the dice pass to your friend. If they don't win, the dice come back to you. This cycle repeats until someone claims victory.

### The Sample Space and the Targets

Before we dive into the infinite, let's ground ourselves in the finite. When you roll two dice, the total number of outcomes is $6 	imes 6 = 36$. This is our universe.

For Player A, the winning outcomes are $(1,4), (2,3), (3,2), (4,1)$. That is $4$ favorable outcomes. So, the probability of A winning on any single turn is:

$$P(A) = \frac{4}{36} = \frac{1}{9}$$

Now, look at Player B. Their target is $8$. The combinations are $(2,6), (3,5), (4,4), (5,3), (6,2)$. That is $5$ favorable outcomes. Thus:

$$P(B) = \frac{5}{36}$$

Notice the subtle asymmetry here? Player B actually has a higher probability of winning on any single turn than Player A. However, Player A has the **first-mover advantage**, which is the core tension of the problem.

### The Infinite Timeline

To find the total probability of A winning, we must map out every possible way A can win. A can win on the 1st turn, the 3rd turn (after A fails, then B fails), the 5th turn, and so on.

Let $P(A^c)$ be the probability of A failing, which is $1 - \frac{1}{9} = \frac{8}{9}$. Let $P(B^c)$ be the probability of B failing, which is $1 - \frac{5}{36} = \frac{31}{36}$.

The probability of A winning is given by the infinite series:

$$P(	ext{A wins}) = P(A) + P(A^c)P(B^c)P(A) + (P(A^c)P(B^c))^2 P(A) + \dots$$

This is a **Geometric Progression (GP)**. The first term $a$ is $P(A) = \frac{1}{9}$. The common ratio $r$ is the probability that the game 'resets' to A's turn:

$$r = P(A^c) 	imes P(B^c) = \frac{8}{9} 	imes \frac{31}{36} = \frac{248}{324}$$

Simplifying the ratio by dividing both numerator and denominator by $4$, we obtain:

$$r = \frac{62}{81}$$

### The Elegant Conclusion

Now, we apply the sum formula for an infinite GP, $S = \frac{a}{1-r}$. Substituting our values:

$$S = \frac{\frac{1}{9}}{1 - \frac{62}{81}}$$

The denominator becomes $1 - \frac{62}{81} = \frac{81 - 62}{81} = \frac{19}{81}$. Therefore, our probability is:

$$P(	ext{A wins}) = \frac{\frac{1}{9}}{\frac{19}{81}} = \frac{1}{9} 	imes \frac{81}{19}$$

The $9$ and $81$ cancel out beautifully. The final probability that A wins is:

$$\mathbf{P(	ext{A wins}) = \frac{9}{19}}$$

It is a result that feels inevitable once you see the structure. Remember, in JEE Advanced, the math is rarely the hardest part—it is the ability to visualize the process. You have just mastered the infinite game.

Animated Solution for Mathematics - Probability: A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

Select Answer:

Visualized Solution

The Game Setup

Probability of A Winning a Turn

Probability of B Winning a Turn

Complementary Probabilities (Failing)

The Winning Timeline for A

Forming the Infinite Series

Identifying GP Parameters

Applying the Sum Formula

Final Calculation

Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Sample Space and the Targets

The Infinite Timeline

The Elegant Conclusion

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