# Expected value of a game where one wins by rolling 6 and loses otherwise

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 tried to get the answer, but could not get as I considered the case mentioned above. I tried in this way :

• First case : 6 in first toss
• Second case : Any other no. in first toss, 6 in second toss
• Third case : Any other no. in first two tosses, 6 in third toss.
• Fourth case : 6 does not appear (he can't toss again as mentioned he can toss thrice only)

Do we need to consider this 4th case? Why or why not? I mean to ask is questions somehow denying to consider this 4th case or not?

• please write down your apporach. – Lost1 Jul 8 '14 at 14:42
• I have given my approach now. Answer if you can. – lokesh israni Jul 8 '14 at 14:48
• Yes you need to consider the 4th case. You have written down a lot of cases but what are their probabilities? Your 'approach' is not really maths. I suggest in each of the cases, you write down the probability of each case with the payout/loss... – Lost1 Jul 8 '14 at 15:01
• I thought you can do that job on your own. But i expected wrong. All the infromation you have, can't you calculate the probability of each case? – lokesh israni Jul 8 '14 at 15:05
• ^ yeah I can, but can you? This is not my homework... – Lost1 Jul 8 '14 at 15:11

$$\frac{1}{6}-\frac{5}{6}+\frac{5}{6}\cdot \frac{1}{6}-\left(\frac{5}{6}\right)^2+\frac{1}{6} \left(\frac{5}{6}\right)^2-\left(\frac{5}{6}\right)^3=-1.68518$$