Question

Marlene is creating a game of chance for her family she has five different colored marbles in a bag blue red yellow white and black she decides that blue was the winning color but player chooses any other colored he or she loses two points how many points should the Blue Marble be worth for the game to be fair​

Answer 1

Answer:

The Blue Marble should be worth 8 points for the game to be fair

Step-by-step explanation:

Discrete Distribution

It refers to the situation where a finite and known number of outcomes are equally likely to happen, like the throw of the die where each side has 1/6 probability to be shown.

Marlene has five different colored marbles, each one with the same probability of occurrence of 1/5. Each time a color other than blue is chosen the player loses 2 points. Let's call X the number of points a player receives if a blue marble is chosen.

The expected value of this distribution is

$E=a_1p_1+a_2p_2+a_3p_3+...+a_np_n$

Where $a_1...a_n$ are the earnings of each possible chosen marble and $p_1...p_n$ are the probability to choose each marble, they are all the same. So

$E=(-2)(1/5)+(-2)(1/5)+(-2)(1/5)+(-2)(1/5)+x(1/5)$

$E=-8/5+x/5$

To be fair (no win, no lose), E should be zero

$-8/5+x/5=0$

So

$x=8$

This means that when we choose the blue marble, we should get 8 points to get a fair bet