Pierre runs a game at a fair, where each player is guaranteed to win $10 - Die rolling question

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?

Answer 1

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 $\frac{10+2}{6}=2$

Answer 2

The following question was answered by u/Alkalannar on reddit.

Answer:

Assumption not stated: This is a 6-sided die.

Consider the general n-sided die, and you want to roll max (or 1).

Expected income for the game is [Sum from k = 1 to infinity of xk(1 - 1/n)k-1(1/n)] = xn, where x is the price and n is the number of sides of the die.

So xn - 10 = 2, xn = 12, x = 12/n.

So when n = 6, x = 2.

And checking, [Sum from k = 1 to infinity of 2k(1 - 1/5)k-1(1/5)] = 12, which is what we want for expected income, so that expected profit is 2.

Answer 3

In the long term, one out of six throws gives a six, costing Pierre 10 dollars and ending a game. Pierre wants to make 2 dollars profit per game, so he must ask for 12 dollars for each six thrown, so he charges 2 dollars per throw. No sums needed.