In the game of roulette, a player can place a $6 bet on the number 26 and have a

Question

In the game of roulette, a player can place a $6 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 $6 paid to play the game and the player is awarded $210. Otherwise, the player is awarded nothing and the casino takes the player's $6. 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.) The player would expect to lose about $ (Round to the nearest cent as needed.)

Explanation / Answer

expected value =210*(1/38)-6*(37/38)=$-0.32

player would expected to lose =$315.79

a)

b)

=3

=1.22

c)

=np=6*0.5=3

=(np(1-p))1/2 =1.22

x P(x) 0 0.0156 1 0.0938 2 0.2344 3 0.3125 4 0.2344 5 0.0938 6 0.0156

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