in the game of roulette a player can place a four dollar bet on the number of 18
ID: 2936538 • Letter: I
Question
in the game of roulette a player can place a four dollar bet on the number of 18 and have a one out of 38 probability of winning if the metal ball lands and 18 the player gets to keep the four dollars paid to play the game and the player is awarded an additional $140 otherwise the player is awarded nothing in the casino takes the players four dollars what is the expected value of the game to the player if you play the game 1000 times how much would you expect to lose in the game of roulette a player can place a four dollar bet on the number of 18 and have a one out of 38 probability of winning if the metal ball lands and 18 the player gets to keep the four dollars paid to play the game and the player is awarded an additional $140 otherwise the player is awarded nothing in the casino takes the players four dollars what is the expected value of the game to the player if you play the game 1000 times how much would you expect to lose in the game of roulette a player can place a four dollar bet on the number of 18 and have a one out of 38 probability of winning if the metal ball lands and 18 the player gets to keep the four dollars paid to play the game and the player is awarded an additional $140 otherwise the player is awarded nothing in the casino takes the players four dollars what is the expected value of the game to the player if you play the game 1000 times how much would you expect to loseExplanation / Answer
Total number of numbers = 18
Probability of winning by betting on 18 = 1 / 38.
=> Expected value = $44 * 1/ 38 - $4 * 37/38
= -2.73684$
On playing 1000 times, the expected loss is 1000 * 2.7368$ = 2736.84$.
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