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 wins/loses.

Answer 1

When a die is thrown, then probability of getting a six = 16

then, probability of not getting a six = 1 - 16 = 56

If the man gets a six in the first throw, then

probability of getting a six = 16

If he does not get a six in first throw, but gets a six in second throw, then

probability of getting a six in the second throw = 56×16 = 536

If he does not get a six in the first two throws, but gets in the third throw, then

probability of getting a six in the third throw = 56×56×16 = 25216

probability that he does not get a six in any of the three throws = 56×56×56 = 125216

In the first throw he gets a six, then he will receive Re 1.

If he gets a six in the second throw, then he will receive Re (1 - 1) = 0

If he gets a six in the third throw, then he will receive Rs(-1 - 1 + 1) = Rs (-1), 
that means he will lose Re 1 in this case.

Expected value = 16×1 + 56×16 × 0 + 56×56×16×-1 = 11216

So, he will loose Rs 11216.