Question

In the game of roulette, a player can place a $10 bet on the number 15 and have a38probability of winning. If the metal ball landson 15, the player gets to keep the $10 paid to play the game and the player is awarded an additional $350. Otherwise, the player isawarded nothing and the casino takes the player's $10. What is the expected value of the game to the player? If you played thegame 1000 times, how much would you expect to lose?The expected value is $0(Round to the nearest cent as needed.)

Answer 1

Answer: The expected value of the game to the player is $ 0.53( It means losing), and the amount expected to lose when the game is played 1000 times is $ 530.

Explanation:

Given:

The winning amount of the player = $350

The probability of winning the game = 1/38

The amount that the player gets to keep to play the game=$10

Since the probability of winning the game is 1/38, the probability of losing the game is 1-1/38.

To find the expected value of the game to the player, we use:

$\begin{gathered} \text{Expected Value=(350)(}\frac{1}{38})\text{ + (-10)(1-}\frac{1}{38}) \\ Calculate\text{ } \\ =-0.53 \\ \end{gathered}$

Expected value= - $ 0.53( It means losing)

The amount expected to lose when the game is played 1000 times is:

1000($ 0.53) =$ 530

Therefore, the expected value of the game to the player is $ 0.53( It means losing), and the amount expected to lose when the game is played 1000 times is $ 530.