0.14 More probability  (Page 4/5)

Expected value

An expected gain or loss in a game of chance is called Expected Value . The concept of expected value is closely related to a weighted average . Consider the following situations.

1. Suppose you and your friend play a game that consists of rolling a die. Your friend offers you the following deal: If the die shows any number from 1 to 5, he will pay you the face value of the die in dollars, that is, if the die shows a 4, he will pay you $4. But if the die shows a 6, you will have to pay him $18.

   Before you play the game you decide to find the expected value. You analyze as follows.

   Since a die will show a number from 1 to 6, with an equal probability of $1/6$ , your chance of winning $1 is $1/6$ , winning $2 is $1/6$ , and so on up to the face value of 5. But if the die shows a 6, you will lose $18. You write the expected value.

   $E=\$1\left(1/6\right)+\$2\left(1/6\right)+\$3\left(1/6\right)+\$4\left(1/6\right)+\$5\left(1/6\right)-\$\text{18}\left(1/6\right)=-\$\text{.}\text{50}$

   This means that every time you play this game, you can expect to lose 50 cents. In other words, if you play this game 100 times, theoretically you will lose $50. Obviously, it is not to your interest to play.

2. Suppose of the ten quizzes you took in a course, on eight quizzes you scored 80, and on two you scored 90. You wish to find the average of the ten quizzes. The average is

   $A=\frac{\left(\text{80}\right)\left(8\right)+\left(\text{90}\right)\left(2\right)}{\text{10}}=\left(\text{80}\right)\frac{8}{\text{10}}+\left(\text{90}\right)\frac{2}{\text{10}}=\text{82}$

   It should be observed that it will be incorrect to take the average of 80 and 90 because you scored 80 on eight quizzes, and 90 on only two of them. Therefore, you take a "weighted average" of 80 and 90. That is, the average of 8 parts of 80 and 2 parts of 90, which is 82.

In the first situation, to find the expected value, we multiplied each payoff by the probability of its occurrence, and then added up the amounts calculated for all possible cases. In the second part of [link] , if we consider our test score a payoff, we did the same. This leads us to the following definition.

Expected Value

If an experiment has the following probability distribution,

Payoff	$\begin{array}{ccccc}{x}_1& x_2& x_3& \cdots & x_n\end{array}$
Probability	$\begin{array}{ccccc}p\left(x_1\right)& p\left(x_2\right)& p\left(x_3\right)& \cdots & p\left(x_n\right)\end{array}$

then the expected value of the experiment is

$\text{Expected Value}=x_{1}p\left(x_1\right)+x_{2}p\left(x_2\right)+x_{3}p\left(x_3\right)+\cdots +x_{n}p\left(x_n\right)$