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?

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:

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.

its a little blurry tbh

step-by-step explanation:

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