Question

A game is played with a single fair die. A player wins $20 if a 2 turns up, $40 if a 4 turns up, and loses $30 if a 6 turns up. If any other face turns up, there is no winning. Find the expected sum of money the player can win.$3$4$5$6.50

Answer 1

Given that a game is played with a single fair die. A player wins $20 if a 2 turns up, $40 if a 4 turns up, and loses $30 if a 6 turns up. If any other face turns up, there is no winning

To Determine: The expected sum of money the player can win

Solution:

Let X be the the random variable giving the amount of money won on any toss of the die

The possible amounts won when the die turns up 1,2, 3, 4, 5, and 6 are:

X1, X2, X3, X4, X5, and X6 respectively.

While the probabilities of these are P(X1), P(X2), P(X3), P(X4), P(X5), and P(X6).

Note that the probability of any face turning up in a fair die is

$P(a\text{ face turn up in a fair die)=}\frac{1}{6}$

So,

$\begin{gathered} P(X_1)=\frac{1}{6};P(X_2)=\frac{1}{6};P(X_3)=\frac{1}{6} \\ P(X_4)=\frac{1}{6};P(X_5)=\frac{1}{6};P(X_6)=\frac{1}{6} \end{gathered}$

The expected sum of money the player can win is

$E(X)=n_1P(X_1)+n_2P(X_2)+n_3P(X_3)+n_4P(X_4)+n_5P(X_5)+n_6P(X_6)$$\begin{gathered} n_1=n_3=n_5=0(\text{Given)} \\ n_2=\text{ \ 20;} \\ n_4=\text{ \ 40} \\ n_6=-\text{ \ 30} \end{gathered}$$E(X)=0(\frac{1}{6})+20(\frac{1}{6})+0(\frac{1}{6})+40(\frac{1}{6})+0(\frac{1}{6})+(-30)(\frac{1}{6})$$\begin{gathered} E(X)=0+\frac{20}{6}+0+\frac{40}{6}+0-\frac{30}{6} \\ E(X)=\frac{20+40-30}{6} \\ E(X)=\frac{60-30}{6} \\ E(X)=\frac{30}{6} \\ E(X)=5 \end{gathered}$

Hence, the expected sum of money the player can win is $5