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 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.
