question archive In the game of? roulette, a player can place a ?$4 bet on the number 18 and have a Start Fraction 1 Over 38 End Fraction probability of winning
In the game of? roulette, a player can place a ?$4 bet on the number 18 and have a Start Fraction 1 Over 38 End Fraction probability of winning
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In the game of? roulette, a player can place a ?$4 bet on the number 18 and have a Start Fraction 1 Over 38 End Fraction
probability of winning. If the metal ball lands on 18?, the player gets to keep the ?$4 paid to play the game and the player is awarded an additional ?$140. ?Otherwise, the player is awarded nothing and the casino takes the? player's ?$4. Find the expected value? E(x) to the player for one play of the game. If x is the gain to a player in a game of? chance, then? E(x) is usually negative. This value gives the average amount per game the player can expect to lose.
The expected value is ?$
nothing.
?(Round to the nearest cent as? needed.)
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Expected value = - $ 0.21
Step-by-step explanation
probability of winning = 1/38
when you place a bet and win, total income
=-4 (cost) + 4 (recover bet) + 140 (prize)
= 140
when you place a bet and lose:
total cost = -4
so
Event.........................Probability.........................Profit/loss
Winning.......................1/38,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,140
Losing.........................37/38.....................................-4
Multiply the probability by the profit loss
winning = 1 / 38 * 140 = $3.68421
losing = 37/38 * -4 = - $3.89474
Add those previous values:
$3.68421 - $3.89474 = - $ 0.21053
nearest cent
Expected value = - $ 0.21
