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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...

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

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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.

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