Question

In a game, a man wins a rupee for a six and losses a rupee for any other number when a fair die is thrown. The man decide 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.

Answer 1

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 $\frac{5}{6}.$

Three cases can occur.

Case I If he gets a six in the first throw, then the required probability is $\frac{1}{6}$.

Amount he will receive = Rs 1

Case 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 =-Rs 1+ Rs 1=0

Case III If he does not get a six in the first throw 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 =-Rs 1-Rs 1+Rs +=-1

Expected value he can win

$=\frac{1}{3}\left(1\right)+(\frac{5}{6}\times \frac{1}{6})\left(0\right)+[{\left(\frac{5}{6}\right)}^2\times \frac{1}{6}](-1)=\frac{1}{6}-\frac{25}{216}=\frac{36-25}{216}=\frac{11}{216}$