Not So Great Expectations!Which game do you think has the highest return per dollar?

For every dollar spent, you would expect to LOSE… BlackjackBlackjack - 1.5 cents

CrapsCraps - 2.8 cents

Crossword Crossword -50 cents

Heart and StrokeHeart and Stroke - 65 cents

Expected Value E(X)

Chapter 4 – Supplemental (also Chapter 5)

Mathematics of Data Management (Nelson)

MDM 4U

Expected Value A property of an object or game of chance The value to which the average tends after a

large number of trials If all outcomes are the same, it is the average of

all possible values

Expected Value Average of all sides is

3.5 E(X) = 3.5

Average of all sections is 2595

E(X) = 2 595

What if the outcomes are not equally likely? You must factor in the probability as well as

the value. For a game with outcomes x1, x2, …, xn

That have probabilities p1, p2, …, pn

E(X) = p1x1 + p2x2 + … + pnxn

In English…multiply the value of every outcome by its probability and add them all up.

Expected / Average Value The probability-weighted sum of the possible values The long-run average value of an experiment over

many repetitions Does NOT mean the "most likely value“ Is often not a value that the random variable can

take on May be unlikely or even impossible The Expected Value of

One die is 3.5 (impossible) Two dice is 7 (possible) Three dice is 10.5 (not possible)

Example 1 – “Nevada” tickets 2700 tickets per box - sold for $0.50 each 276 prizes as follows:

n($100) = 4 n($50) = 2 n($25) = 6 n($5) = 4 n($1) = 260

What is the probability of buying a winning ticket? P(win) = 276 / 2 700 = 10.22% Would you be equally happy to win $1 as a $100?

How do these factor into the winning scheme?

Example 1 – Probabilities

First, find the probability of each outcome. n($100) = 4 P($100) = 4/2700 n($50) = 2 P($50) = 2/2700 n($25) = 6 P(25) = 6/2700 n($5) = 4 P($5) = 4/2700 n($1) = 260 P($1) = 260/2700

Example 1 – Expected Value

To find the Expected Value, multiply every putcome by its probability and add them together:

= 0.1481 + 0.0370 + 0.0556 + 0.0074 + 0.0962

= 0.3443 or $0.34 The Expected Value of a ticket is $0.34. However, the

ticket costs $0.50. So you would expect to lose $0.16 on every ticket you buy.

Example 2 – Rolling a Die

Suppose you play a game with a friend where you roll a fair die. If the roll is odd, your friend pays you the amount shown on the die. If the roll is even, you pay your friend the amount shown on the die. How much money would you expect to win or lose on each roll? Is this a fair game?

Example 2 – Rolling a die (cont’d) Odd numbers represent wins and have a positive

value Even numbers represent losses and have a

negative value  E(X) = (1)(1/6) + (-2)(1/6) + (3)(1/6) + (-4)(1/6) + (5)(1/6) + (-6)(1/6) = 1/6 – 2/6 + 3/6 – 4/6 + 5/6 – 6/6 = -3/6 = -1/2 or -$0.50

 So you would expect to lose $0.50 on every play.

So looking at Expected Value, the games look like this:BlackjackBlackjack E(X) = 0.985

CrapsCraps E(X) = 0.972

Crossword Crossword E(X) = 0.5

Heart and StrokeHeart and Stroke E(X) = 0.35