Question

In the game of​ roulette, a player can place a

​$9

bet on the number

29

and have a

1 Over 38

probability of winning. If the metal ball lands on

29​,

the player gets to keep the

​$9

paid to play the game and the player is awarded an additional

​$315.

​Otherwise, the player is awarded nothing and the casino takes the​ player's

​$9.

Find the expected value​ E(x) to the player for one play of the game. If x is the gain to a player in a game of​ chance, then​ E(x) is usually negative. This value gives the average amount per game the player can expect to lose.

Answer 1

Expected value calculations are very important for the scenarios where we really need to average outcome of an event. We basically need to multiply the each outcome with its probability and sum up the multiplications to find the expected outcome of that probabilistic event.

Probability of winning =1/38

Probability of losing = 1-1/38 =37/38

Profit on winning = $315

Loss on losing = $9

E(x) to the player for one play of the game,

E(x) = 315(1/38) + (-9)37/38)

= (315-333)/38

= -18/38

= -$0.4736

(Amount you can loss on each spin. ie 47 cents)
---------------
If you played the game 1000 times, how much would you expect to lose?
Ans: 1000×0.4736 = $473.68