In a game, a man wins ₹ 1000 if he gets an even number greater than or equal to 4 on a fair dice and loses ₹ 200 for getting any other number on the dice. If he decides to throw the dice until he wins or maximum of three times, then his expected gain/loss in (₹) is --

The outcomes on a fair dice are $\{1, 2, 3, 4, 5, 6\}$.

Winning condition: Even number $\ge 4$, which are $\{4, 6\}$. Probability of winning $p = \dfrac{2}{6} = \dfrac{1}{3}$.

Losing condition: Any other number, which are $\{1, 2, 3, 5\}$. Probability of losing $q = 1 - p = \dfrac{2}{3}$.

The game stops if he wins or after 3 throws. Let $X$ be the gain/loss.

Case 1: Wins on 1st throw. Probability $P_1 = p = \dfrac{1}{3}$. Gain = $1000$.

Case 2: Wins on 2nd throw. Probability $P_2 = qp = \dfrac{2}{3} \times \dfrac{1}{3} = \dfrac{2}{9}$. Gain = $-200 + 1000 = 800$.

Case 3: Wins on 3rd throw. Probability $P_3 = q^2p = \left(\dfrac{2}{3}\right)^2 \times \dfrac{1}{3} = \dfrac{4}{27}$. Gain = $-200 - 200 + 1000 = 600$.

Case 4: Loses all 3 throws. Probability $P_4 = q^3 = \left(\dfrac{2}{3}\right)^3 = \dfrac{8}{27}$. Gain = $-200 - 200 - 200 = -600$.

Expected gain $E = P_1(1000) + P_2(800) + P_3(600) + P_4(-600)$

$E = \dfrac{1}{3}(1000) + \dfrac{2}{9}(800) + \dfrac{4}{27}(600) - \dfrac{8}{27}(600)$

$E = \dfrac{9000}{27} + \dfrac{4800}{27} + \dfrac{2400}{27} - \dfrac{4800}{27} = \dfrac{11400}{27} = \dfrac{3800}{9}$.

Since the result is positive, it is a gain of $\dfrac{3800}{9}$.

Answer: $\dfrac{3800}{9}$ gain