Question

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 win / 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 be obtain.
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 $=\left(\frac{5}{6}\times \frac{1}{6}\right)=\frac{5}{36}$
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 $=\left(\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\right)=\frac{25}{216}$
Amount he will receive = -Re 1 - Re 1 + Re 1 = -1
Expected value he can win $=\frac{1}{6}\left(1\right)+\left(\frac{5}{6}\times \frac{1}{6}\right)\left(0\right)+[{\left(\frac{5}{6}\right)}^2\times \frac{1}{6}](-1)$
$=\frac{36-25}{216}=\frac{11}{216}=0.05$