JEE Advanced 2021 Paper 1 Mathematics: Paragraph 2

In one full round both players fail with probability $\dfrac56\cdot\dfrac12=\dfrac5{12}$. $A$ wins on his own turn in round $n$ with probability $\left(\dfrac5{12}\right)^{n-1}\dfrac16$. Summing the geometric series, $P(A\text{ wins})=\dfrac{1/6}{1-5/12}=\dfrac{1/6}{7/12}=\dfrac27\approx0.29$. For a consistency check, $B$ wins with probability $\dfrac{(5/6)(1/2)}{7/12}=\dfrac57$, and $\dfrac27+\dfrac57=1$, so the game ends with probability one.

Let $N$ be the number of rolls. Use $E[N]=\sum_{n\ge0}P(N>n)$. The game certainly reaches roll 1. It reaches roll 2 if $A$ fails: $\dfrac56$. It reaches roll 3 if both fail: $\dfrac5{12}$. It reaches roll 4 if additionally $A$ fails again: $\dfrac5{12}\cdot\dfrac56$, and so on, so the probabilities come in pairs multiplied by $\dfrac5{12}$ each round. Thus $E[N]=\left(1+\dfrac56\right)\sum_{r\ge0}\left(\dfrac5{12}\right)^r=\dfrac{11/6}{7/12}=\dfrac{22}{7}\approx3.14$. This is the geometric-series analogue of the usual waiting-time argument.