The following information is for questions 32 to 33. In the game of roulette, a

Question

The following information is for questions 32 to 33.

In the game of roulette, a player can place an $8 bet on any number and have a 2.05% probability of winning. If the metal ball lands on that selected number, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. Otherwise, the player is awarded nothing and the casino takes the player's $8.

32. What is the expected value of the game to the player?

33. If the reward is increased to $350 and you played the game 1000 times, how much would you expect to lose (or win)?

Explanation / Answer

Probability of winning = 2.05% or 0.0205

Probility of losing = 97.95% or 0.9795

32. Gain on winning = $280

Gain on losing = -$8 ( or loss of $8)

Expected value of the game = Sum(outcomes * probability) = 280 * 0.0205 + (-8) * 0.9795 = -2.096

Expected value of the game = -$2.096 or a loss of $2.096

b.Expected value of the game when the reward is $350

Expected value = 350 * 0.0205 + (-8) * 0.9795 = -$0.661

If you played 1000 times, Loss = 1000 * expected value = 1000*-0.661 = -$661

I would expect to lose $661 if I play the game 1000 times

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