Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One player throws dice twice. If he has 2 x 6 on the dice he is receving 8*a. If he has one 6 he will collect 4*a. Otherwise (when he has no 6 at all) he is paying a.

For which value of a game is fair?

I was trying to bit that issue by using expected value (should be 0?) But only resonable outcome seems to be 0...

share|improve this question

2 Answers 2

As written, the value is proportional to $a$, so $a=0$ is a solution. I don't find another. The value is $\frac {8a}{36}+4a\cdot \frac {10}{36} -a\cdot \frac{25}{36}=\frac {23}{36}a$

share|improve this answer

The game is fair if $8a\times\frac{1}{36}+4a\times\frac{10}{36}=a\times\frac{25}{36}$ which is only the case if $a=0$.

Here $\frac{1}{36}$ is the probability of throwing twice a six, $\frac{10}{36}$ is the probability of throwing exactly one six and $\frac{25}{36}$ is the probability of throwing no six.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.