Three friends A, B and C are playing with a pair of dice. They throw two dice alternately. Coming of a doublet on two dice leads to a success and the game stops. If A starts the game, then the probability of his winning, is :

$\frac{36}{91}$

The correct answer is Option 2: $\frac{36}{91}$

A "success" is rolling a doublet (e.g., 1-1, 2-2). There are 6 doublets out of 36 possible outcomes when throwing two dice.

• Probability of success ($p$): $6/36 = 1/6$

• Probability of failure ($q$): $1 - 1/6 = 5/6$ 

Since A starts, they can only win on the 1st turn, the 4th turn (after B and C both fail once), the 7th turn (after two full rounds of failure), and so on. 

• A wins on 1st turn: $1/6$

• A wins on 4th turn: $(5/6)^3 \times 1/6$

• A wins on 7th turn: $(5/6)^6 \times 1/6$

This creates an infinite geometric series where the first term ($a$) is $1/6$ and the common ratio ($r$) is $(5/6)^3$, which is $125/216$. Using the sum of an infinite series formula, $S = \frac{a}{1 - r}$: