Probability of getting a 6 when throw a die = 1/6

Probability of not getting a 6 when throw a die = 1- 1/6 = 5/6 

Given that a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown.

Now there are three cases:

1. If man gets a 6 in the first throw, then Probability = 1/6

Amount received = 1

2. If man does not get a 6 in the first throw and gets 6 in the second throw, then Probability = (1 - 1/6)*1/6 = 5/6 * 1/6 = 5/36

Now amount received = 1 - 1 = 0

3. If man does not get a 6 in the first two throws and gets 6 in the third throw, then Probability = (1 - 1/6)*(1 - 1/6)*1/6 =5/6* 5/6 * 1/6 = 25/216

Amount received = -1 -1 + 1 = -1

Now expected value man can win = (1/6)*1 + (5/25)*0 + (25/216)*(-1)

                                                    = 1/6 - 25/216

                                                    = (36-25)/216

                                                    = 11/216  

So the expected value of the amount he wins / loses is 11/216