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

The game continues until someone rolls a 3 or 6. B throws first, then they alternate.

On any single throw:

Probability of winning (rolling 3 or 6) = $\frac{2}{6} = \frac{1}{3}$

Probability of not winning (rolling 1, 2, 4, or 5) = $\frac{4}{6} = \frac{2}{3}$

Since B goes first, for A to win, B must fail first. A can win in the following ways:

Round 1: B fails, then A wins

Round 2: B fails, A fails, B fails, then A wins

Round 3: B fails, A fails, B fails, A fails, B fails, then A wins

This pattern continues indefinitely.

$P(\text{A wins}) = \frac{2}{3} \cdot \frac{1}{3} + \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{1}{3} + \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{1}{3} + ...$

$P(\text{A wins}) = \frac{2}{3} \cdot \frac{1}{3} \left[1 + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^4 + \left(\frac{2}{3}\right)^6 + ...\right]$

The series in brackets is a geometric series with first term $a = 1$ and common ratio $r = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$

Sum $= \frac{a}{1-r}$

Sum $= \frac{1}{1 - \frac{4}{9}}$

Sum $= \frac{1}{\frac{5}{9}}$

Sum $= \frac{9}{5}$

$P(\text{A wins}) = \frac{2}{3} \cdot \frac{1}{3} \cdot \frac{9}{5}$

$P(\text{A wins}) = \frac{2}{9} \cdot \frac{9}{5}$

$P(\text{A wins}) = \frac{18}{45}$

$P(\text{A wins}) = \frac{2}{5}$

Therefore, the probability of A winning the game is $\frac{2}{5}$.