In the game of roulette, a player can place a $4 bet on the number 22 and have a 1/38 probability of winning. If the metal ball lands on 22, the player gets to keep the $4 paid to play the game and the player is awarded $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?

Question

anonymous · Answer

The answer is actually -0.21.  Not sure why though haha

zarkon · Answer

-.21052631579 ;)

zarkon · Answer

I know ;)

anonymous · Answer

haha nice job

zarkon · Answer

$$\begin{array}{c|c}
  x & 140 & -4\\hline p & \frac{1}{38} &\frac{37}{38}
  \end{array}$$

this is the distribution of the problem

zarkon · Answer

$$E(X)=140\frac{1}{38}-4\frac{37}{38}$$

zarkon · Answer

$$=\frac{-4}{19}$$

anonymous · Answer

ohh I was lost at how to start

zarkon · Answer

you know how to answer the last part of the problem?

anonymous · Answer

Heres another like this: In the game of roulette a player can place a 5 dollar bet on the number 27 and have a 1/38 probability of winning. If the metal ball lands on 27, the player gets to keep the 5 dollars and is awarded $175. Otherwise the player is awarded nothing and the casino takes the 5 dollars. 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?

zarkon · Answer

change 140 to 175 and -4 to -5 from what I did before