Amodified 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.)

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

So:

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

because 1.2 isnt half of 6 duh

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