# 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

## 3 Answers

As I understand the problem you do need to consider the case in which he does not through a six three times in a row since this is when he loses money. The easiest way to approach this kind of problem is to draw some kind of tree in which the end nodes are the possible outcomes. I got that the expected value is about -1.685

(-3 * (125/216)) + (1 * (1/6)) + (0 * (5/36)) + (-1 * (25/216)) = -1.685185

It's still possible for the fourth case (he doesn't get a six on any of his rolls) to happen, correct? So to answer your question, yes, you need to consider the case where he doesn't get any sixes. I think if you work through everything, he should end up expecting to lose about 1.69 rupees.

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

He loses 1.68518 rupees. This agrees with a simulation.