In a game, a man wins a rupee for a six and loses 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 / loses.

Explanation

In a throw of a die, the probability of getting a six is frac{1}{6} and the probability of not getting a 6 is frac{5}{6}

So, there will be three caases.

i. If he gets a six in the first throw, then the required probability is` frac{1}{6}`. Amount he will receive =1 {Rs}

ii. If he does not get a six in the first throw and gets a six in the second throw, then probability `=(frac{5}{6} times frac{1}{6})=frac{5}{36}`

Amount he will receive =-1+1=0$ Rs.

iii. If he does not get a six in the first two throws and gets a six in the third throw, then probability `=(frac{5}{6} times frac{5}{6} times frac{1}{6})=frac{25}{216}`

Amount he will receive =-{Re} 1-{Re} 1+{Re} 1=-1

Expected value he can win `=frac{1}{6}(1)+(frac{5}{6} times frac{1}{6})(0)+[(frac{5}{6})^{2} times frac{1}{6}](-1) =frac{36-25}{216}=frac{11}{216}=0.05`