A and B throw a die alternatively till one of them gets 3 or 6 and wins the game. If B starts the game, then the probability of winning the game by A is

2/5

The correct answer is Option (1) → 2/5

Let $p$ = probability that A wins when B starts.

Probability of success (getting 3 or 6) in one throw: $1/3$ (since 2 favorable outcomes out of 6)

Probability of failure in one throw: $2/3$

If B starts, two cases for A:

1) B fails (prob $2/3$), then A throws: probability A wins on first try = $2/3 \cdot 1/3 = 2/9$

2) If both fail, game resets with B's turn: probability = $(2/3 \cdot 2/3) = 4/9$, then A's chance = $4/9 \cdot p$

$p = 2/9 + (4/9)p \Rightarrow p - (4/9)p = 2/9 \Rightarrow (5/9)p = 2/9 \Rightarrow p = 2/5$