Question

In a game, a man wins rupees five for a six and loses rupee one 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

Hey !!

In a throw of die, the probability of getting 6 is 1/6 and the probability of not getting 6 is 5/6

CASE 1 :- If he gets a six in a first throw, then the required probability is 1/6

         Amount received is Rupees 5

CASE 2 :- If he does not get a six in a first throw and gets a six in second throw, then probability.

                 = (5/6 × 1/6)

                 = 5/36

Amount he will receive = - 1 + 5 = 4 (Rupees)

CASE 3 :- If he does not get a six in the first two throws and gets a six in the third, then probability

                     = (5/6 × 5/6 × 1/6)

                      = 25/216

Amount he will receive =  - 1 - 1 + 5 = 3 (Rupees)

CASE 4 :- If he does not get any six, then probability

                 = (5/6 × 5/6 × 5/6)

                 = 125 / 216

Now, expected value can be written as,

             = $(\frac{1}{6})5 +( \frac{5}{36} )4 + (\frac{25}{216} )3 + \frac{125}{216} (-3) = 0$

Good luck !!