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.
 A: Let $Y$ be the net amount the "player"  wins. Then $Y$ is a random variable that takes on the possible values $-1$, $1$, and $2$.
The probability that $Y=2$ is the probability of $10$ non-sixes in a row. For those are precisely the conditions under which the player has a net gain of $2$ dollars. Thus $\Pr(Y=2)=(5/6)^{10}$.
The probability that $Y=1$ is the probability of between $5$ and $9$ non-sixes in a row. This is the probability of $5$ non-sixes minus the probability of $10$ non-sixes. Thus $\Pr(Y=1)=(5/6)^5-(5/6)^{10}$.
The probability that $Y=-1$ is $1$ minus the probability of $5$ non-sixes in a row. Thus $\Pr(Y=-1)=1-(5/6)^5$. 
Now we calculate the expectation $E(Y)$ of $Y$. This is 
$$(-1)\Pr(Y=-1)+(1)\Pr(Y=1)+(2)\Pr(Y=2).\tag{1}$$
If (1) is $\gt 0$, the game favours the player, and if it is $\lt 0$ it favours the opponent. 
Remark: It is a little easier to work with the gross win, which is any of $0$, $1$, or $2$. It turns out that the expectation is remarkably simple. You may glimpse some of that simplicity by doing "algebra" on (1) instead of feeding it immediately into a calculator. 
A: The probability of failures before the first success is modeled by the geometric distribution. So you have to figure out what the expectation of the random variable with payout:
$$
\begin{matrix}
-1 & P(X \leq 5)\\
1 &  P(6 \leq X \leq 10)\\
3 & P(X \geq 11)
\end{matrix}
$$
Where $X$ is geometrically distributed with probability of success is the probability of seeing a $6$ on a six-sided die. If the expectation of the payout is 0, the game is fair; otherwise someone expects to make a profit in the long run.
Update
To answer the comment. Since this is discrete, you'll have to calculate three sets of probabilities. The first is $X \leq 5$ which is made up of $X = 0, X = 1, X = 2, X = 3, X = 4,$ and $X = 5$. For all of these possibilities, the player loses a dollar. Now for $X = 6, X = 7,\ldots,X = 10$, the player gains two dollars. At this point we have the sum of all the probabilities for $X \leq 10$. So the probability that $X \geq 11$ is just 1 (total probability of everything) minus the running sum we needed for the first two sets.
At this point we have three numbers: $p_1$ which is the sum for $X=0$ to $X=5$, $p_2$ which is the middle bucket, and $p_3 = 1 - p_1 - p_2$. Now we know that $p_1$ of the time, the player loses a dollar, $p_2$ of the time the player wins a dollar, and $p_3$ of the time the player wins two dollars. Now, is the expectation = 0 or not? That determines if it is "fair" or not.
A: 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.
