Solve part A , and B In the game of roulette, a player can place a S4 bet on the

Question

Solve part A , and B In the game of roulette, a player can place a S4 bet on the number 26 and have a 38 probability of winning If the metal ball lands on 26, the player gets to keep the $4 paid to play the game and the player is awarded an additional $140. Otherwise, the player is awarded nothing and the casino takes the players $4. 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? The expected value is Round to the nearest cent as needed.)

Explanation / Answer

P(winning) = 1/38
P(losing) = 1-1/38 = 37/38
Expected value = 140*1/38 - 4*37/38

= 140/38 - 148/38

= - 8/38

= -0.21
The player's expected value is $ -0.21

When the game is played 1000 times, the amount expected to lose

= 1000*8/38

= $8000/38

= $210.5 (Rounded to nearest cent)

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