$Three persons A, B, C throw a die in succession. The one getting \text{'}six\text{'} wins. If A starts, then the probability of B winning is$

$\begin{array}{l}\text{Let }W\text{\hspace{0.17em} be the event of getting six when a die thrown}\\ P\left(W\right)=\frac{1}{6},P\left(L\right)=\frac{5}{6}\\ \text{When the game was started by }A,\text{\hspace{0.17em} the possibility of B winning is as below}\\ \text{Case (i) A\hspace{0.17em}lose, }B\text{\hspace{0.17em}win}\\ \text{Case (ii)\hspace{0.17em}\hspace{0.17em}A lose, B\hspace{0.17em}lose, C\hspace{0.17em}lose, A lose, B win}\\ \text{Case (iii) A lose, B lose, C lose, A lose, B lose, C lose, A lose, B win }\\ \text{and so on}\\ \text{Hence, the probability of winning by }B\text{\hspace{0.17em} is }\\ P\left(LW\right)+P\left(LLLLW\right)+\mathrm{....}=\frac{5}{6}\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^4\frac{1}{6}+\mathrm{....}\\ =\frac{5}{36}\left(1+{\left(\frac{5}{6}\right)}^3+\mathrm{...}\right)\\ =\frac{5}{36}\frac{1}{1-\frac{125}{216}}\\ =\frac{5}{36}\cdot \frac{216}{91}\\ =\frac{30}{91}\end{array}$