In many everyday situations, the expected value is\u00a0equal to the mean,\u00a0aka the average.\u00a0 The difference is more in definition than in\u00a0practice.\u00a0 Expected value\u00a0involves probability, whereas mean involves fact.<\/p>\n
Formally, the expected value is defined as follows:<\/p>\n
Expected Value=Σoutcome_value*probability<\/p>\n
The \u03a3 symbol means “sum”.\u00a0 In other words, the expected value takes each possible outcome and multiplies it by its probability.\u00a0 Then, it adds all those products together.<\/p>\n
Example 1: Give the expected value of rolling a fair six-sided die.<\/strong><\/p>\n Solution:\u00a0 We know that the probability of getting each result is 1\/6.<\/p>\n The expected value, E, of the die roll\u00a0is<\/p>\n E=(1\/6)1+(1\/6)2+(1\/6)3+(1\/6)4+(1\/6)5+(1\/6)6=(1\/6)(1+2+…+6)=21\/6=3.5<\/p>\n Notice that the expected value is not necessarily a possible value. Not surprising, really: if you take the average height of ten people, you’ll likely arrive at a height that none of them is.<\/p>\n Example 2: Find your expected payoff in the following situation: 1000 people pay $1 each for a ticket. Then a number is drawn. The holder of the winning ticket gets $500.<\/strong><\/p>\n Solution: The probability of winning is 1\/1000, in which case you get $500. Really though, the payoff is only $499, since you had to spend $1 to buy the ticket. The probability of not winning is 999\/1000, in which case you get nothing. Once again, you still had to pay $1 to lose, so the payoff is -$1. Applying our definition of expected value we get<\/p>\n E=(999\/1000)(-1) +(1\/1000)(499)=-0.999+0.499=-0.50<\/p>\n So, your expected payoff is -$0.50. It had to be a loss, since not all the money paid in was awarded.<\/p>\n