Answer to Question #97577 in Statistics and Probability for Frances

Question

You pay $10 to play the following game of chance. There is a bag containing 12 balls, five are red, three are green and the rest are yellow. You are to draw one ball from the bag. You will win $14 if you draw a red ball and you win $12 if you draw a yellow ball. How much do you expect to win or lose if you play this game 100 times?

Answer 1

Because the events (draw a red/green/yellow ball) form the full group and do not intersect, we can calculate the expected win/lose for 1 game as

$$\mathbb{E}X = \sum\limits_k {{w_k}P(k)}$$

Where ${P(k)}$ is a probability of the certain event, that can be calculated as $P(k) = \frac{{{N_k}}}{N}$ where ${N_k}$ is the number of k-type balls and $N = \sum\limits_k {{N_k}}$ is the amount of balls, ${{w_k}}$ is the win/lose if we get the k-th type ball (it can be calculated as ${w_k} = {x_k} - y$ where ${x_k}$ is the win for the k-th type ball and $y$ is the price for the 1 game). We get

$$\mathbb{E}X = \sum\limits_k {{w_k}P(k)} = \$4 \cdot \frac{5}{{12}} +\$ 2 \cdot \frac{4}{{12}} + \$( - 10) \cdot \frac{3}{{12}} = -\$ \frac{1}{6}$$

Then because the games are independent, the total result after 100 times is

$$\mathbb{E}Y = n \cdot \mathbb{E}X = - \$\frac{1}{6} \cdot 100 = - \$\frac{{50}}{3} \approx - \$16.667$$