Fair die being rolled repeatedly A fair die is rolled repeatedly, and let $X$ record the number of the roll when the 1st  $6$  appears.  A game is played as follows.  A player pays  \$1  to play the game.  If $X\leq 5$ , then he loses the dollar.   If  $6 \le X \le 10$, then he gets his dollar back plus \$1.  And if $X  >  10$, then he gets his dollar back plus \$2 .  Is this a fair game?  If not, whom does it favour?
I think that that it is not a fair game because it solely depends on whether the first number is $6$ which is a $\frac{1}{6}$ chance. But I don't know how to prove this further.

The probability the number is $X$ is $(\dfrac{5}{6})^{X-1}\cdot\dfrac{1}{6}$
Therefore the expected gains of a game is 
$$\sum_{X=6}^{\infty}(\dfrac{5}{6})^{X-1}\cdot\dfrac{1}{6}+2\sum_{X=11}^\infty(\dfrac{5}{6})^{X-1}\cdot\dfrac{1}{6}=\dfrac{1}{6}(\sum_{X=6}^\infty(\dfrac{5}{6})^{X-1}+2\sum_{X=11}^\infty(\dfrac{5}{6})^{X-1})=$$
$$\dfrac{1}{6}(18-(1+\frac{5}{6}+((\frac{5}{6})^2\dots(\frac{5}{6})^4)+2(1+\frac{5}{6}+((\frac{5}{6})^2\dots(\frac{5}{6})^9)=$$
$$\frac{1}{6}(18-\frac{4651}{1296}-2(\frac{50700551}{10077696}))\approx0.72$$ so no, it isn't worth it to play.
