In American roulette, the wheel has 38 number: 00, 0, 1, 2, 3, ..., 35, 36 marked on equally spaced slots. If a player bets $10 on a single number and wins, then the player gains $350 otherwise the player loses the $10.This means, that X is either $350 or -$10.What is the expected value of this game?Group of answer choices$0.53$18.95-$18.95-$0.53
Question
In American roulette, the wheel has 38 number: 00, 0, 1, 2, 3, ..., 35, 36 marked on equally spaced slots. If a player bets 350 otherwise the player loses the 350 or -0.5318.95-$0.53
Solution
The expected value of a game is calculated by multiplying each possible outcome by the probability of that outcome, and then summing these values.
In American roulette, the probability of winning (i.e., the ball landing on the number you bet on) is 1/38, because there are 38 equally likely outcomes. The probability of losing is therefore 37/38.
If you win, your net gain is 360 but you also lose the 10.
So, the expected value E[X] of the game is:
E[X] = (1/38 * 10)
E[X] = 9.74
E[X] = -$0.53
So, the expected value of this game is -$0.53. This means that, on average, you would expect to lose about 53 cents each time you play this game.
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