Question

In the game of​ roulette, a player can place a ​$4 bet on the number 29 and have a 1/38 probability of winning. If the metal ball lands on 29​, 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​ player's ​$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?

Answer 1

Answer:

The expected value of the game to the player is -$0.2105 and the expected loss if played the game 1000 times is -$210.5.

Step-by-step explanation:

Consider the provided information.

It is given that if ball lands on 29 players will get $140 otherwise casino will takes $4.

The probability of winning is 1/38. So, the probability of loss is 37/38.

Now, find the expected value of the game to the player as shown:

$E(x)=(140)\times \frac{1}{38}+(-4)\times \frac{37}{38}$

$E(x)=\frac{140-148}{38}$

$E(x)=\frac{-8}{38}=- 0.2105$

Hence, the expected value of the game to the player is -$0.2105.

Now find the expect to loss if played the game 1000 times.

1000×(-$0.2105)=-$210.5

Therefore, the expected loss if played the game 1000 times is -$210.5.