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 wins / loses.
I. Solution
Let X denote the random variable of winning/losing in the game.
P(X = 1)= P(win) = getting a 6 on rolling a fair die.
P(X = 0)= P(loss) = getting a 6 on rolling a fair die.
P(X = 1) = p = = 0.167
P(X = 0) = 1-p = = 0.833
Let Y denote the random variable of winning the game in Nth trial, there can be 4 possible cases with the following probability: (i)Wins in first throw:
P(Y = 1) = p = 0.167 (1)
(ii)Wins in the second throw :
P(Y = 2) = (1 −p) ×p = 0.139 (2)
(ii)Wins in the third throw :
P(Y = 3) = (1 −p) × (1 −p) ×p = 0.107 (3)
(ii)Does not wins in any throw :
P(Y = 3) = (1 −p) × (1 −p) × (1 −p) = 0.596
(4)
Net amount = P(Y=1)×1 + P(Y = 1) × (−1 + 1) +
P(Y = 1)×(−1−1+1)+P(Y = 1)×(−1−1−1) = −1.73
The probabilities were simulated using the python code.
Download python code from here
Figure 1: Simulation for tossing a fair coin
