Question

In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses.

Answer 1

The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equatl to 4 in the first throw and in the second throw he may get the number greater than 4 and quits the game.
In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.
Let X be the amount he wins/looses.
Then, X can take values -3, 3, 4, 5 such that
P (X = 5) = P(Getting number greater than 4 in first throw) = $\frac{1}{3}$
​P (X = 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) = $\frac{4}{6}\times \frac{2}{6}=\frac{2}{9}$
​P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) = $\frac{4}{6}\times \frac{4}{6}\times \frac{2}{6}=\frac{4}{27}$
​P (X = -3) = P(Getting number less than equal to 4 in all three throws) = $\frac{4}{6}\times \frac{4}{6}\times \frac{4}{6}=\frac{8}{27}$

X	5	4	3	-3
P(X)	$\frac{1}{3}$	$\frac{2}{9}$	$\frac{4}{27}$	$\frac{8}{27}$

$$\begin{gathered} E\left(X\right)=\left(5\times \frac{1}{3}\right)+4\left(\frac{2}{9}\right)+3\left(\frac{4}{27}\right)-3\left(\frac{8}{27}\right) \\ =\frac{1}{27}\left(45+24+12-24\right) \\ =\frac{57}{27} \end{gathered}$$
Expected value of the amount he wins/loses is $\frac{57}{27}$.