Question

In the game of​ roulette, a player can place a ​$10 bet on the number 18 and have a 1/38 probability of winning. If the metal ball lands on 18​, the player gets to keep the ​$10 paid to play the game and the player is awarded an additional ​$350 ​Otherwise, the player is awarded nothing and the casino takes the​ player's $10.

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.

A. The value expected is $__

Answer 1

In the game of roulette, a player can place a 10$ bet on the number 18 and have a 1/38 probability of winning. If the metal ball lands on 18, the player gets to keep the 10$ paid to play the game and the player is awarded 350$ . otherwise the player is awarded nothing and the casino takes the players 10$.

So the expected value of the game to the player is

E(x) = 350(1/38) + (-10)(37/38) = (350-370)/38 = -20/38 = -$0.5263
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So expected value is -$0.5263 that is expected value is a loss .