K In the game of roulette, a player can place a \( \$ 7 \) bet on the number 31 and have a \( \frac{1}{38} \) probability of winning. If the metal ball lands on 31 , the player gets to keep the \( \$ 7 \) paid to play the game and the player is awarded an additional \( \$ 245 \). Otherwise, the player is awarded nothing and the casino takes the player's \( \$ 7 \) 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?

1. Define the outcomes: - A win occurs if the ball lands on 31. In that case, the player recovers the \$7 bet and receives an additional \$245. The net gain is therefore: \[ 245 \] - A loss occurs if the ball does not land on 31. In that case, the player loses the \$7 bet. The net loss is: \[ -7 \] 2. The probabilities for each outcome are: - Probability of win: \[ \frac{1}{38} \] - Probability of loss: \[ \frac{37}{38} \] 3. Calculate the expected value (EV) of a single game: \[ \text{EV} = \left(\frac{1}{38}\right)(245) + \left(\frac{37}{38}\right)(-7) \] Simplify the expression: \[ \text{EV} = \frac{245 - 259}{38} = \frac{-14}{38} = -\frac{7}{19} \] In decimal form, this is approximately: \[ -0.3684 \] So, the expected value of the game to the player is \(-\frac{7}{19}\) dollars, or about \(-\$0.37\) per game. 4. Find the expected loss after 1000 games: \[ \text{Total Expected Loss} = 1000 \times \left(-\frac{7}{19}\right) = -\frac{7000}{19} \] Converting to decimal: \[ -\frac{7000}{19} \approx -368.42 \] Therefore, after 1000 games, you would expect to lose approximately \(-\$368.42\).