Suppose that in one week at a certain casino there are 25,000 independent plays at roulette. On each play, the gamblers stake $1 on red. Is the chance that the casino will win more than $1000 from these 25,000 plays closest to 2% 50% or 98% explain.
To determine the chance that the casino will win more than $1000 from these 25,000 plays, we need to consider the probability of winning on each play.
In roulette, the probability of winning on a single play by betting on red is approximately 18/38 (assuming a standard American roulette wheel with 18 red slots out of a total of 38 slots).
Let's calculate the expected number of wins for the casino based on these probabilities:
Expected number of wins = Probability of winning * Number of plays
Expected number of wins = (18/38) * 25,000
Expected number of wins ≈ 11,842
Since each play has a stake of $1 on red, if the casino wins, they gain $1, and if they lose, they lose $1.
Now, let's calculate the expected net gain/loss for the casino:
Expected net gain/loss = Expected number of wins * Net gain/loss per win
Expected net gain/loss = 11,842 * $1
Expected net gain/loss = $11,842
Since the expected net gain/loss is $11,842, this means that, on average, the casino is expected to win $11,842 from these 25,000 plays.
To determine if this is more than $1000, we compare it to the given threshold.
$11,842 > $1000
Therefore, the chance that the casino will win more than $1000 from these 25,000 plays is closest to 100% or 98%.
To find the chance that the casino will win more than $1000 from 25,000 plays at roulette, we need to calculate the probability.
First, let's consider the probability of winning on a single play. In American roulette, there are 18 red pockets out of a total of 38 pockets (18 red, 18 black, and 2 green). Therefore, the probability of winning on a single play by betting on red is 18/38 or approximately 0.474.
Next, we need to calculate the expected number of wins for the casino on 25,000 plays. To do this, we multiply the probability of winning on a single play by the total number of plays:
Expected number of wins = Probability of winning * Number of plays
Expected number of wins = 0.474 * 25,000 = 11,850
Since the casino collects $1 from each losing bet, the winnings will exceed $1000 when the expected number of wins multiplied by $1 is greater than $1000:
Expected winnings = Expected number of wins * $1
Expected winnings = 11,850 * $1
So, the chance that the casino will win more than $1000 is determined by analyzing the expected winnings. If the expected winnings are greater than $1000, then the chance is higher than 50%, and if the expected winnings are less than $1000, then the chance is lower than 50%.
Let's calculate the expected winnings:
Expected winnings = 11,850 * $1 = $11,850
As the expected winnings equal $11,850, which is greater than $1000, we can conclude that the chance that the casino will win more than $1000 from these 25,000 plays is higher than 50%. Therefore, the closest option is 98%.
The probability of getting red on a single roulette play is 9/19 = 47.37%.
The expected house winning per spin, with $1 bets, is .5263-.4737 = $0.052
After 25,000 spins, the house wins about $1315, on the average. There is a better tnan 50% chance the house will win $1000. 98% and 2% are too far off.