Question

A modified roulette wheel contains 44 ​numbers, of which 21 are​ red, 21 are​ black, and 2 are green. When the roulette wheel is​ spun, the ball is equally likely to land on any of the 44 numbers. For a bet on black​, the house pays 1 to 2 odds. 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 black​.)

Answer 1

Answer:

To make the bet​ fair, the odds paid by the house should be 1 to 1.91.

Step-by-step explanation:

The roulette wheel contains 44 numbers, of which:

21 are red.

21 are black.

2 are green.

There are 44 numbers, of which 21 are black, and 44-21 = 23 are not black.

To find the odds against the ball landing on black, we have to solve the following direct(cross multiplication) rule of three:

1 is to x what 23 is to 44. So:

1 - x

23 - 44

$23x = 44$

$23x = 44$

$x = \frac{44}{23}$

$x = 1.91$

So:

To make the bet​ fair, the odds paid by the house should be 1 to 1.91.