In a game two players A and B take turns in throwing a pair of fair dice starting with player A and

Q: Wait, why does the game go on forever? Can't we just calculate the first few turns and stop?

That is a fantastic intuition! In reality, the probability of the game lasting 100 turns is astronomically small, almost zero. However, in mathematics, 'almost zero' is not zero. To find the *exact* probability required by JEE Advanced, we must account for the entire infinite tail of the distribution. Summing the infinite series is the only way to capture that mathematical precision.

Q: Why is the common ratio $r = q_A q_B$ and not just $q_A$?

Think about the timeline. For A to get a second chance (turn 3), A must fail on turn 1, AND B must fail on turn 2. Since these are independent events, we multiply their probabilities. The game only 'resets' to a state where A can win again after both players have taken their turns and failed. Thus, the common ratio must account for both failures.

Q: Is there a faster way to solve this using recursive equations?

Yes! You can define $P$ as the probability that A wins. Then, $P = p_A + q_A q_B P$. This equation says: 'A wins either on the first turn ($p_A$) OR A fails, B fails, and then we are back to the start ($q_A q_B P$).' Solving this linear equation $P(1 - q_A q_B) = p_A$ gives the exact same result and is often faster. Both methods are elegant!

Question

In a game two players $A$ and $B$ take turns in throwing a pair of fair dice starting with player $A$ and total of scores on the two dice, in each throw is noted. $A$ wins the game if he throws a total of 6 before $B$ throws a total of 7 and $B$ wins the game if he throws a total of 7 before $A$ throws a total of six. The game stops as soon as either of the players wins. The probability of $A$ winning the game is:

Options:

(A) 5/31

(B) 61/5

(C) 5/6

(D) 30/61

Sigma Percentile · Accepted Answer

### The Infinite Dance of Dice

Imagine you are standing at a table, two fair dice in your hand. Your opponent, B, stands opposite you. The air is thick with anticipation.

This isn't just a game of chance; it is a beautiful, unfolding mathematical narrative. We are looking for the probability that you, player A, win.

The rules are simple: you win if you roll a sum of $6$ before B rolls a $7$. Because the game can theoretically last forever, we must utilize the magic of infinite series.

### Phase 1: The Anatomy of a Single Throw

Before we can conquer the infinite, we must master the finite. When you roll two fair dice, there are $6 	imes 6 = 36$ possible outcomes.

To win, you need a sum of $6$. The favorable outcomes are $(1,5), (2,4), (3,3), (4,2), (5,1)$, totaling $5$ outcomes.

Thus, the probability of you winning on any single turn is $p_A = \frac{5}{36}$. Conversely, the probability of you failing is $q_A = 1 - \frac{5}{36} = \frac{31}{36}$.

Now, consider player B. B needs a sum of $7$. The outcomes are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$, totaling $6$ outcomes.

Therefore, $p_B = \frac{6}{36} = \frac{1}{6}$, and B's failure probability is $q_B = 1 - \frac{6}{36} = \frac{30}{36} = \frac{5}{6}$.

### Phase 2: The Timeline of Success

Now, let us map out your path to victory. You start the game, so you could win on turn $1$ with probability $p_A$.

If you don't, the game continues. For you to win on turn $3$, you must fail on turn $1$, B must fail on turn $2$, and then you must succeed on turn $3$. The probability is $q_A 	imes q_B 	imes p_A$.

If you still haven't won, you could win on turn $5$. This requires you to fail on turn $1$, B to fail on turn $2$, you to fail on turn $3$, B to fail on turn $4$, and finally, you to succeed on turn $5$. The probability is $(q_A q_B)^2 	imes p_A$.

A clear pattern emerges: this is a geometric progression.

### Phase 3: The Infinite Sum

The total probability of you winning is the sum of these mutually exclusive scenarios:

$$P(A) = p_A + (q_A q_B)p_A + (q_A q_B)^2 p_A + \dots$$

This is an infinite geometric series with the first term $a = p_A = \frac{5}{36}$ and a common ratio $r = q_A q_B$.

Calculating the ratio $r$:

$$r = \left( \frac{31}{36} ight) 	imes \left( \frac{30}{36} ight) = \frac{930}{1296}$$

### Phase 4: The Final Calculation

The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$. Substituting our values:

$$P(A) = \frac{\frac{5}{36}}{1 - \frac{930}{1296}}$$

To simplify the denominator:

$$1 - \frac{930}{1296} = \frac{1296 - 930}{1296} = \frac{366}{1296}$$

Now, our expression becomes:

$$P(A) = \frac{5}{36} 	imes \frac{1296}{366}$$

Since $1296 = 36^2$, we simplify this to:

$$P(A) = \frac{5 	imes 36}{366} = \frac{180}{366}$$

Dividing both the numerator and the denominator by $6$, we arrive at the elegant result:

**$$P(A) = \frac{30}{61}$$**

You have successfully navigated the infinite and emerged victorious. This is the power of mathematical modeling—turning a game of chance into a deterministic certainty.

Select Answer:

Visualized Solution

Understanding the Game Dynamics

Probability of A Winning in One Throw

Probability of B Winning in One Throw

Calculating Failure Probabilities

Scenario 1: A Wins on the First Turn

Scenario 2: A Wins on the Third Turn

Scenario 3: A Wins on the Fifth Turn

Formulating the Infinite Series

Identifying GP Parameters

Applying the Sum Formula

Simplifying the Denominator

Final Calculation

The Sigma Insight: Addition and Multiplication Theorems

The Infinite Dance of Dice

Phase 1

Phase 2

Phase 3

Phase 4

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