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?

  • $\begingroup$ What have you tried? $\endgroup$ – David G. Stork May 25 '20 at 17:31
  • 1
    $\begingroup$ I know I should have included what I've tried in my question but it was quite basic. On second thought I think there will be some use of binomial distribution but due to the infinite number of rolls possible I'm facing an issue which led me to think about something on the lines of infinite geometric progression. $\endgroup$ – Math Comorbidity May 25 '20 at 17:34
  • $\begingroup$ Expected number of rolls appears to be $6$ $\endgroup$ – Alexey Burdin May 25 '20 at 17:50
  • $\begingroup$ @AlexeyBurdin How did you calculate it? $\endgroup$ – Math Comorbidity May 25 '20 at 17:53

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


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.


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


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.


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