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?
Best Answer
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 $\frac{10+2}{6}=2$