Calculating Expectation

Question

You are running a casino with the following game:


The player rolls a fair die until a 4, 5 or 6 is thrown. For every number 1, 2 or 3 that is thrown the player's score increases by 1. If the game stops with a 4 or 5 the player is paid the accumulated score. If the game stops with a 6 the player is paid nothing.


How much do you charge patrons to play this game?

Answer

The expected score when playing this game is

$E[S]=\frac{1}{6}*0+\frac{1}{3}*0+\frac{1}{2}*(\frac{1}{6}*0+\frac{1}{3}*1+\frac{1}{2}*(\frac{1}{6}*0+\frac{1}{3}*2+\frac{1}{2}*(...$

$E[S]=(\frac{1}{2})^{1}*\frac{1}{3}*1+(\frac{1}{2})^{2}*\frac{1}{3}*2+(\frac{1}{2})^{3}*\frac{1}{3}*3+...$

$E[S]=\frac{1}{3}*\sum_{i=0}^{\infty}i*(\frac{1}{2})^{i}$

$E[S]=\frac{2}{3}$

So played over many trials, you can expect to pay out 67% of each unit wagered. Since you are booking the action and anything can happen in the short run, add a bit of risk premium and charge something like 4 to 3 on a minimum bet of $X to make it worth your overhead to play.