In a game, a man wins a rupee for a six and looses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/looses.

Probability of a man winning in first change = 1/6
Amount won/loose in this case = 1

Probability of a man winning in second change = 5/6 * 1/6
Amount won/loose in this case = -1 + 1 = 0

Probability of a man winning in third change = 5/6 * 5/6 * 1/6
Amount won/loose in this case = -1 -1 + 1 = -1

Probability of not winining = (5/6)^3
Amount won/loose in this case = -1 -1 -1 = -3

Expected amound = Probability * amound
= (1/6)*1 + (5/6 * 1/6)*0 + (5/6 * 5/6 * 1/6)* (-1) + (5/6 * 5/6 * 5/6) * (-3)

simplify above to get your answer.

Best Of luck!