Question

In a dice game, a player pays a stake of Rs 1 for each throw of a die. She receives Rs 5, if the die shows a 3, Rs 2, if the die shows a 1 or 6 and nothing otherwise, then what is the player's expected profit per throw over a long series of throws ?

Answer 1

Let X is the random variable of profit per throw.

$\begin{array}{cccc}X& -1& 1& 4\\ P\left(x\right)& \frac{1}{2}& \frac{1}{3}& \frac{1}{6}\end{array}$

Since, she toss Rs 1 on getting any of 2, 4 or 5.

So, at X = - 1, $P\left(X\right)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$

Similarly, at X = 1, $P\left(X\right)=\frac{1}{6}+\frac{1}{6}+\frac{1}{3}$ [If die shows either 1 or 6]

and at X = 4, $P\left(X\right)=\frac{1}{6}$ [if die shows a 3]

$\therefore$ Player's expected profit $=E\left(X\right)=\sum XP\left(X\right)$

$=-1\times \frac{1}{2}+1\times \frac{1}{3}+4\times \frac{1}{6}$

$=\frac{-3+2+4}{6}=\frac{3}{6}=\frac{1}{2}=Rs0.50$