Question

In American roulette, the wheel has the 38 numbers, 00,0,1,2,…,34,35, and 36 , marked on equally spaced slots. If a player bets $9 on a number and wins, then the playerkeeps $9 and receives an additional $315. Otherwise, the player is awarded nothing, and the casino takes the player's $9. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose. The expected value is $ (Round to the nearest cent as needed.)

Answer 1

The expected value of playing a single game of American roulette for a player who bets $9 is approximately -$0.47, signifying that the player should expect an average loss of 47 cents per game.

Step-by-step explanation:

To calculate the expected value (E(X)) for a play in American roulette, we consider two outcomes: winning and losing. If the player bets $9 on a number and wins, the gain is $315. The probability of this happening is 1/38 since there are 38 numbers on the wheel. However, if the player loses, the loss is $9, and this will happen with a probability of 37/38.

The expected value can be calculated using the formula: E(X) = (gain or loss) × (probability of the gain or loss). Applying this formula, we get:

• E(X) for winning: $315 × (1/38)
• E(X) for losing: -$9 × (37/38)

Adding these two expected values gives us the overall expected value for the game:

E(X) = ($315 × (1/38)) + (-$9 × (37/38)) = -$0.474.

Rounded to the nearest cent, the expected value is -$0.47. This means on average, a player can expect to lose about 47 cents per game. This suggests that in the long term, it is not profitable for players to play this game as they would lose money.