<s>[INST] Pierre runs a game at a fair, where each player is guaranteed to win $10 - Die rolling question Pierre runs a game at a fair, where each player is guaranteed to win $10. 
Players pay a certain amount each time they roll an unbiased die, and must keep rolling until a ‘6’ occurs. 
When a ‘6’ occurs, Pierre gives the player $10 and the game concludes. 
On average, Pierre wishes to make a profit of $2 per game. How much does he need to charge for each roll of the die?
 [/INST] The probability of not rolling $6$ for $k$ times is $\left(\frac{5}{6}\right)^k$.
The probability of not rolling $6$ for $k-1$ first times and then rolling $6$ on the $k$th roll is $\frac16\left(\frac{5}{6}\right)^{k-1}$.
So to find the expected number of rolls we are just to find the sum 
$\sum\limits_{k=1}^\infty k\frac16\left(\frac{5}{6}\right)^{k-1}$.
Let $S_n=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k}$,
$$\begin{align*}
\frac{5}{6}S_n
&=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k+1}\\
%&=\frac56+\sum\limits_{k=1}^{n} (k+1)\left(\frac{5}{6}\right)^{k+1}\\
&=\sum\limits_{k=1}^{n+1} k\left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} k\left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} (k+1)\left(\frac{5}{6}\right)^{k}-
\sum\limits_{k=0}^{n+1} \left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} (k+1)\left(\frac{5}{6}\right)^{k}-
\frac{(5/6)^{n+2} - 1}{5/6-1}\\
&=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k}
+(n+2)\left(\frac{5}{6}\right)^{n+1}
-\frac{(5/6)^n - 1}{5/6-1}\\
&=S_n
+(n+2)\left(\frac{5}{6}\right)^{n+1}
+6\left((5/6)^n - 1\right),
\end{align*}$$
$$\begin{align*}
S_n&=-6\left((n+2)\left(\frac{5}{6}\right)^{n+1}
+6\left(\left(\frac{5}{6}\right)^{n} - 1\right)\right)
\end{align*}$$
$$\begin{align*}
\lim\limits_{n\to\infty}S_n&=-6\left(0
+6\left(0 - 1\right)\right)=36
\end{align*}$$
So the expected number of rolls is $\frac{1}{6}\cdot 36=6$ and a roll cost to have income $2$ per game on average is $\frac{10+2}{6}=2$
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