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)	\[\frac{1}{3}\]	\[\frac{2}{9}\]	\[\frac{4}{27}\]	\[\frac{8}{27}\]

 

\[E (X) = \left( 5 \times \frac{1}{3} \right) + 4 \left( \frac{2}{9} \right) + 3\left( \frac{4}{27} \right) - 3 \left( \frac{8}{27} \right)\]
\[ = \frac{1}{27}\left( 45 + 24 + 12 - 24 \right)\]
\[ = \frac{57}{27}\]

Expected value of the amount he wins/loses is  \[\frac{57}{27}\]