A and B Throw Dice Alternately, A Wins on Sum 5 Before B’s Sum 8

JEE Main 2025 — 24 January, Morning Shift. Previous Year Question.

Problem

Two persons $A$ and $B$ throw a pair of dice alternately. $A$ wins if he throws a sum of $5$ before $B$ throws a sum of $8$. If $A$ throws first, the probability that $A$ wins is:

Key insight. “A wins” doesn’t mean “A succeeds on his first throw” — it means A succeeds on some throw before B ever succeeds. Since the game can go on forever, the winning outcomes form an infinite sequence of failures followed by a success, which is exactly the structure of an infinite geometric series.

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Approach

First find the individual probabilities of success for A (sum $5$) and B (sum $8$) on a single throw, and their complements. Then list every way A can win: on the 1st round, or the 3rd round (both fail once each first), or the 5th round, and so on. Since these rounds are mutually exclusive and each round’s probability shares a common ratio with the last, the total probability is the sum of an infinite GP.

Solution

Step 1 — Probability A gets a sum of 5 on one throw

Sum $5$ arises from $(1,4),(4,1),(2,3),(3,2)$ — four outcomes out of $36$:

$$P(A) = \frac{4}{36} = \frac{1}{9}, \qquad P(A’) = 1-\frac{1}{9} = \frac{8}{9}$$

Step 2 — Probability B gets a sum of 8 on one throw

Sum $8$ arises from $(2,6),(6,2),(5,3),(3,5),(4,4)$ — five outcomes out of $36$:

$$P(B) = \frac{5}{36}, \qquad P(B’) = 1-\frac{5}{36} = \frac{31}{36}$$

Step 3 — Write out every way A can win

A throws first, so A wins if: A succeeds on throw 1, or both fail once and A succeeds on the next round, or both fail twice and A succeeds on the round after that, and so on:

$$P(A \text{ wins}) = P(A) + P(A’)P(B’)P(A) + \big[P(A’)P(B’)\big]^2 P(A) + \cdots$$

Step 4 — Recognise the infinite GP

This is a GP with first term $a = P(A) = \dfrac{1}{9}$ and common ratio $r = P(A’)P(B’) = \dfrac{8}{9}\cdot\dfrac{31}{36} = \dfrac{62}{81}$.

$$P(A\text{ wins}) = \frac{a}{1-r} = \frac{\frac{1}{9}}{1 – \frac{62}{81}}$$

Step 5 — Simplify

$$1 – \frac{62}{81} = \frac{81-62}{81} = \frac{19}{81}$$

$$P(A\text{ wins}) = \frac{\frac{1}{9}}{\frac{19}{81}} = \frac{1}{9}\times\frac{81}{19} = \frac{9}{19}$$

Answer

Common mistakes

• Only computing $P(A)$ on the first throw and stopping there. A can still win on the third throw, fifth throw, and so on — every one of these has to be added as a separate term of the GP, not just the first.
• Mixing up which sum belongs to which player. A throws for a sum of $5$ (four favourable outcomes), B throws for a sum of $8$ (five favourable outcomes) — swapping these gives a completely different, wrong ratio.

Practise next

• Two players alternately roll a single die, and the first to roll a $6$ wins. If player 1 rolls first, find the probability that player 1 wins, using the same infinite-GP method.