Question

in the game of roulette, a player can place a $9 bet on the number 27 and have a \LXI0 probability of winning. if the metal ball lands on 27, the player gets to keep the $9 paid to play the game and the player is awarded an additional $315. otherwise, the player is awarded nothing and the casino takes the players $9. what is the expected value of the game to the player? if you played the game 1000 times, how much would you expect to lose? note that the expected value is the amount, on average, one would expect to gain or lose each game. the expected value is $ - 0.47. (round to the nearest cent as needed.) the player would expect to lose about $ \square. (round to the nearest cent as needed.)

Explanation:

Step1: Recall the formula for expected value

The expected value \(E(X)\) of a game is calculated as \(E(X)=\sum_{i}(x_{i}P(x_{i}))\), where \(x_{i}\) are the possible outcomes and \(P(x_{i})\) are their corresponding probabilities. In this case, we already know the expected value per game is \(- 0.47\) dollars.

Step2: Calculate the expected loss for 1000 games

If the expected value per game is \(E=-0.47\) dollars, and the number of games \(n = 1000\). Then the total expected value \(T\) for \(n\) games is \(T=n\times E\).
Substitute \(n = 1000\) and \(E=-0.47\) into the formula: \(T=1000\times(- 0.47)\)

Answer:

The player would expect to lose about \(\$470.00\)