Aman and Bhuvan Throw a Pair of Dice Alternately. In Order to Win They Have to Get a Sum of 8. Find Their Respective Probabilities of Winning if Aman Starts the Game

A standard probability problem, not tied to a specific exam paper.

Problem

Aman and Bhuvan throw a pair of dice alternately. In order to win, they have to get a sum of 8. Find their respective probabilities of winning if Aman starts the game.

Key insight. “Aman wins” means the sequence of throws is a failure, a failure, …, then finally a success on an odd-numbered throw — this is an infinite geometric series in the probability of failure, and it sums cleanly using the standard formula for an infinite GP.

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Approach

First find the probability $p$ of rolling a sum of 8 on a single throw of two dice, and let $q=1-p$ be the probability of failing. Since Aman throws first, he wins on throw 1, or throw 3 (if both he and Bhuvan fail once first), or throw 5, and so on — an infinite series of alternating failures followed by his eventual success. Summing this geometric series gives his winning probability directly; Bhuvan’s probability is whatever remains.

Solution

Step 1 — Find the probability of rolling a sum of 8

Out of the 36 equally likely outcomes of two dice, the pairs summing to 8 are $(2,6),(6,2),(3,5),(5,3),(4,4)$ — 5 outcomes:

$$p = \frac5{36}, \qquad q = 1-p = \frac{31}{36}$$

Step 2 — Set up the series for Aman’s winning probability

Aman throws first. He wins if:

• he succeeds on throw 1 (probability $p$), or
• he fails, Bhuvan fails, then Aman succeeds on throw 3 (probability $q\cdot q\cdot p = q^2p$), or
• both fail twice more before Aman succeeds on throw 5 (probability $q^4p$), and so on.

$$P(\text{Aman wins}) = p+q^2p+q^4p+q^6p+\cdots = p\left(1+q^2+q^4+\cdots\right)$$

Step 3 — Sum the infinite geometric series

The series $1+q^2+q^4+\cdots$ is an infinite GP with first term $1$ and common ratio $q^2$ (valid since $0<q^2<1$):

$$1+q^2+q^4+\cdots = \frac1{1-q^2} = \frac1{(1-q)(1+q)} = \frac1{p(1+q)}$$

(using $1-q=p$).

Step 4 — Simplify Aman’s winning probability

$$P(\text{Aman wins}) = p\cdot\frac1{p(1+q)} = \frac1{1+q}$$

Substituting $q=\dfrac{31}{36}$:

$$P(\text{Aman wins}) = \frac1{1+\frac{31}{36}} = \frac1{\frac{67}{36}} = \frac{36}{67}$$

Step 5 — Find Bhuvan’s winning probability

Since exactly one of them eventually wins (the game continues until someone rolls an 8), Bhuvan’s probability is the complement:

$$P(\text{Bhuvan wins}) = 1-\frac{36}{67} = \frac{31}{67}$$

Answer

Common mistakes

• Forgetting that the first player has an advantage. Since Aman throws first, his winning probability ($\frac{36}{67}\approx0.537$) is naturally larger than Bhuvan’s ($\frac{31}{67}\approx0.463$) — a probability of exactly $\frac12$ each would be a sign of a setup error, since the turn order genuinely matters here.
• Using the wrong power of q in the series. Aman’s chances of winning on throw $2k+1$ require both players to have failed $k$ times each before that — that’s $q^{2k}$, not $q^k$. Using $q^k$ instead of $q^{2k}$ badly overstates how quickly the series decays.
• Trying to sum the series term by term instead of using the infinite-GP formula. Since $|q^2|<1$, the closed form $\dfrac1{1-q^2}$ is exact and immediate — there’s no need to approximate by adding several terms.

Practise next

• Aman and Bhuvan throw a pair of dice alternately, but now they need a sum of 9 to win, and Aman still starts. Find their respective probabilities of winning.