A modified roulette wheel contains 20 numbers, of which 9 are red, 9 are black,

Question

A modified roulette wheel contains 20 numbers, of which 9 are red, 9 are black, and 2 are green. When the roulette wheel is spun, the ball is equally likely to land on any of the 20 numbers. For a bet on red, the house pays even odds (that is, 1 to 1). What should the odds actually be to make the bet fair? (Hint: To make the bet fair, the odds paid by the house should be the odds against the ball landing on red.) Click the icon to view an explanation of odds. The odds, in simplified form, should actually be to [ Type whole numbers.) ] to make the bet fair. Explanation of Odds Odds are closely related to probabilities, and used much more often than probabilities in gambling contexts. If the probability that an event occurs is p, the odds that the event occurs are p to 1-p. This fact is also expressed by saying that the odds are p to 1-p in favor of the event or that the odds are 1 -p to p against the event. Conversely, if the odds in favor of an event are a to b (or, equivalently, the odds against it are b to a), the probability the event occurs is a+b Enter your answer in each of the answer boxes

Explanation / Answer

The odds in simplified form should be 11 to 9 to make the bet fair.

According to the question and the explanation succeeding it, the odds paid by the house should be the odds against the ball landing on red.

The probability that the ball lands on red is simply 9/20

The probability that the ball does not land on red is 11/20

Hence the odds actually to make the bet fair is 11 to 9.

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