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

32) P(winning) = 0.0205

P(loosing) = 1 - 0.0205 = 0.9795

Expected value of the game to the player = 0.0205 * 288 - 0.9795 * 8

                                                                  = $ -1.932 (ans)

33) Expected value for 1 game = 0.0205 * 350 - 0.9795 * 8

                                                  = -0.661

So, if you played 1000 times , expected money to loose = 0.661 * 1000 = $661 (ans)

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