Question

In the game of roulette, a player can place a $6 bet on the number 25 and have a 1/38 probability of winning. If the metal ball lands on 25, the player gets to keep the $6 paid to play the game and the player is awarded an additional $210. Otherwise, the player is awarded nothing and the casino takes the player's $6. 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

AS THE WINNING PROBABILITY IS 1/38 WHICH MEANS AFTER PLAYING 38 GAMES U WILL WIL ONE GAME

COST OF 38 GAMES = 38*6 = 228

WINNING AMOUNT = 210 +6 = 216

HENCE LOSS IN 38 GAMES = 228 -216= 12

LOSS PER GAME = 12/38 = 0.316

HENCE E(X) = - 0.316

B) IF THE GAME IS PLAYED 1000 TIMES THEN TO TOTAL LOSS

AS LOSS PER GAME 0.316

HENCE LOSS AFTER 1000 TIMES = 0.316 *1000 = 316

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