What changes could you make to values assigned to outcomes to make the game fair? Prove that the game would be fair using expected values. Thank you!

For a definite answer, let us take a look at the given circle graph. You are given that landing on a blue sector will give 3 points, landing on a yellow sector will give 1 point, purple sector will give 0 points and red sector will give -1 point. You are asked to find the probability of landing -1, 0, 1 and 3 points. There are a total of 7 pie graphs in the circle.

For -1 point, you know that only a red sector will give you a negative one point. In the circle graph, there are two red portions. So you will have a probability of 2/7.

For the 0 point, you know that only a purple sector will give you zero point. In the circle graph, there are two purple portions. So you will have a probability of 2/7.

For the 1 point, you know that only a yellow sector will give you one point. In the circle graph, there are two yellow portions. So you will have a probability of 2/7.

For the 3 points, you know that only a blue sector will give you three points. In the circle graph, only one blue portion is shown. So you will have a probability of 1/7.

Аноним Аноним · Answer 2 · 2019-08-20T09:49:07+02:00

Answer with explanation:

⇒A game will be fair , if there is equal chance of winning and losing.

→The game would be fair , if

Landing on Blue sector gives 3.5 points.

Then the, Expected value will be

[tex]E(x_{i})=x_{i} \times P(x_{i})\\\\E(x)=x_{1} \times P(x_{1})+x_{2} \times P(x_{2})+x_{3} \times P(x_{3})+x_{4} \times P(x_{4})\\\\E(x)=-1 *\frac{2}{7}+(0) *\frac{2}{7}+(1) *\frac{2}{7}+(3.5) *\frac{1}{7}\\\\E(x)=(3.5) *\frac{1}{7}\\\\E(x)=0.50[/tex]

→→By Assigning, Blue sector =3.5 points, the game will become fair, Gives, E(x)=0.50.