Question

in the game of roulette, a player can place a $9 bet on the number 27 and have a $\frac{1}{38}$ 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 $ (round to the nearest cent as needed.)

Explanation:

Step1: Calculate the value when winning

When winning, the net gain is \(315\) dollars (since the player keeps the \(9\) - dollar bet). The probability of winning \(P(W)=\frac{1}{38}\).

Step2: Calculate the value when losing

When losing, the net gain is \(- 9\) dollars (the player loses the \(9\) - dollar bet). The probability of losing \(P(L)=1 - \frac{1}{38}=\frac{37}{38}\).

Step3: Use the expected - value formula

The expected - value formula is \(E(X)=\sum_{i}x_{i}P(x_{i})\). Here, \(x_1 = 315\), \(P(x_1)=\frac{1}{38}\), \(x_2=-9\), and \(P(x_2)=\frac{37}{38}\).

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Answer:

\(-0.47\)