CHAPTER 16 Expected Value of a Discrete Random Variable A measure of the “middle” of the values of a random variable

Center The mean of the probability distribution is the expected value of X, denoted E(X) E(X) is also denoted by the Greek letter µ (mu)

k = the number of possible values (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) x k ·p(x k ) Weighted mean Mean or Expected Value Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) Lousy P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 )

k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) x k ·p(x k ) Weighted mean Each outcome is weighted by its probability Mean or Expected Value

Other Weighted Means zGPA A=4, B=3, C=2, D=1, F=0 zStock Market: The Dow Jones Industrial Average yThe “Dow” consists of 30 companies (the 30 companies in the “Dow” change periodically) yTo compute the Dow Jones Industrial Average, a weight proportional to the company’s “size” is assigned to each company’s stock price

k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) x k ·p(x k ) EXAMPLE Mean Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) Lousy P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 )

k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) x k ·p(x k ) EXAMPLE µ = 10* * *.25 – 4*.15 = 3.65 ($ mil) Mean Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) Lousy P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 )

k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) x k ·p(x k ) EXAMPLE µ = 10· · · ·.15 = 3.65 ($ mil) Mean µ=3.65

Interpretation zE(x) is not the value of the random variable x that you “expect” to observe if you perform the experiment once

Interpretation zE(x) is a “long run” average; if you perform the experiment many times and observe the random variable x each time, then the average x of these observed x- values will get closer to E(x) as you observe more and more values of the random variable x.

Example: Green Mountain Lottery zState of Vermont zchoose 3 digits from 0 through 9; repeats allowed zwin $500 x$0$500 p(x) E(x)=$0(.999) + $500(.001) = $.50

Green Mountain Lottery (cont.) zE(x)=$.50 zOn average, each ticket wins $.50. zImportant for Vermont to know zE(x) is not necessarily a possible value of the random variable (values of x are $0 and $500)

Example: expected number of heads in 3 tosses of fair coin  Suppose a fair coin is tossed 3 times and we let x=the number of heads. Find  (x). zFirst we must find the probability distribution of x.

Example: expected number of heads in 3 tosses of fair coin (cont.) zPossible values of x: 0, 1, 2, 3. zp(1)? zAn outcome where x = 1: THT zP(THT)? (½)(½)(½)=1/8 zHow many ways can we get 1 head in 3 tosses? 3 C 1 =3

Example: expected number of heads in 3 tosses of fair coin (cont.)

zSo the probability distribution of x is: x0123 p(x)1/83/83/81/8

Example: expected number of heads in 3 tosses of fair coin (cont.) zSo the probability distribution of x is: x0123 p(x)1/83/83/81/8

US Roulette Wheel and Table zThe roulette wheel has alternating black and red slots numbered 1 through 36. zThere are also 2 green slots numbered 0 and 00. zA bet on any one of the 38 numbers (1-36, 0, or 00) pays odds of 35:1; that is... zIf you bet $1 on the winning number, you receive $36, so your winnings are $35 American Roulette (The European version has only one 0.)

US Roulette Wheel: Expected Value of a $1 bet on a single number zLet x be your winnings resulting from a $1 bet on a single number; x has 2 possible values x-135 p(x)37/381/38 zE(x)= -1(37/38)+35(1/38)= -.05 zSo on average the house wins 5 cents on every such bet. A “fair” game would have E(x)=0. zThe roulette wheels are spinning 24/7, winning big $$ for the house, resulting in …

Expected Value, Surprise Onside Kicks zhttp:// The change in expected points for the kicking team: successful 1.9; fail zX=change in expected points for kicking team when attempting surprise onside kick zWhat values of p make surprise onside kicks a good strategy? X p(x)p1-p

Two More Examples 1.X = # of games played in a randomly selected World Series Possible values of X are x=4, 5, 6, 7 2.Y=score on 13 th hole (par 5) at Augusta National golf course for a randomly selected golfer on day 1 of 2011 Masters y=3, 4, 5, 6, 7

Probability Distribution Of Number of Games Played in Randomly Selected World Series zEstimate based on results from 1946 to x4567 p(x)12/65= /65= /65=0.415 Probability Histogram

Probability Distribution Of Score on 13 th hole (par 5) at Augusta National Golf Course on Day 1 of 2011 Masters y34567 p(x) Probability Histogram

E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414) =5.86 games E(Y)= µ=3(.04)+4(0.414)+5(0.465)+6(0.051)+7(0.03) =4.617 strokes Mean or Expected Value x4567 p(x)12/65= /65= /65=0.415 y34567 p(x)

E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414) =5.86 games Mean or Expected Value µ=5.86