In a game, a man wins a rupee for a six and loses a rupee for any other number, when a fair dice is thrown. He decided to throw a die twice but to quit as and when he gets a six. The expected values of amount he won/lose is :

$-\frac{11}{9}$

The correct answer is Option (4) → $-\frac{11}{9}$

Case 1: Win in 1st throw

$P_I=\frac{1}{6}, \text{Amount}_{I}=1$

Case 1: Win in 2nd throw

$P_{II}=\frac{5}{6}×\frac{1}{6}=\frac{5}{36}, \text{Amount}_{II}=1-1=0$

Case 3: No win at all

$P_{III}=\frac{5}{6}×\frac{5}{6}=\frac{25}{36},\text{Amount}_{III}=-2$

Net expected amount

$≡\frac{1}{6}×1+\frac{5}{36}×0+\frac{25}{36}×(-2)$

$≡\frac{1}{6}-\frac{25}{18}≡\frac{3-25}{18}$

$≡-\frac{11}{9}$