In a game, a man wins Rs. 5 for getting a number greater than 4 and loses Rs. 1 otherwise, when a fair dice is thrown. The man decided to throw a die thrice but to quit as and when he gets a number greater than 4. The expected value of the amount (in Rs.) he wins or loses is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 19/9
A number greater than 4 (5 or 6) has probability 2/6 = 1/3, and any other number 2/3. He stops at the first win, so let X be his total amount; the possible outcomes are:
\( \displaystyle \text{Win on 1st throw: } X = 5,\quad P = \frac{1}{3} = \frac{9}{27} \)
\( \displaystyle \text{Lose, then win: } X = -1 + 5 = 4,\quad P = \frac{2}{3}\cdot\frac{1}{3} = \frac{6}{27} \)
\( \displaystyle \text{Lose, lose, win: } X = -1 - 1 + 5 = 3,\quad P = \frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{4}{27} \)
\( \displaystyle \text{Lose all three: } X = -3,\quad P = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \)
The probabilities add up to (9 + 6 + 4 + 8)/27 = 1. The expected value is
\( \displaystyle E(X) = 5 \cdot \frac{9}{27} + 4 \cdot \frac{6}{27} + 3 \cdot \frac{4}{27} - 3 \cdot \frac{8}{27} \)
\( \displaystyle E(X) = \frac{45 + 24 + 12 - 24}{27} = \frac{57}{27} = \frac{19}{9} \)
So on average he wins Rs. 19/9.