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What Do You Expect D. Brooke Hill & Rose Sinicrope East Carolina University 1530-1645, Thursday, 31 October 2013 NCCTM 43 rd Annual Conference Auditorium III, Joseph S. Koury Convention Center Greensboro, NC

What Do You Expect

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What Do You Expect. D. Brooke Hill & Rose Sinicrope East Carolina University 1530-1645, Thursday, 31 October 2013 NCCTM 43 rd Annual Conference Auditorium III, Joseph S. Koury Convention Center Greensboro, NC. Roll two regular hexahedral dice. What is the sum?. - PowerPoint PPT Presentation

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Page 1: What Do You Expect

What Do You Expect

D. Brooke Hill & Rose SinicropeEast Carolina University

1530-1645, Thursday, 31 October 2013 NCCTM 43rd Annual Conference

Auditorium III, Joseph S. Koury Convention CenterGreensboro, NC

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Roll two regular hexahedral dice.What is the sum?

What do you expect will happen?On the average, what percent of the time, will you get 2?3?:

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What sum can you expect?PROBABILITY HISTOGRAM

2 3 4 5 6 7 8 9 10 11 12

Sum of 2 Regular Hexahedral Dice

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What is the expected value?Outcome Probability Product

2 (1/36) 2(1/36)= 2/36

3 (2/36) 3(2/36)=6/36

4 (3/36) 2(3/36)=6/36

5 (4/36) 2(4/36)

6 (5/36) 2(5/36)

7 (6/36) 2(6/36)

8 (5/36) 2(5/36)

9 (4/36) 2(4/36)

10 (3/36) 2(3/36)

11 (2/36) 2(2/36)

12 (1/36) 12(1/36)

Expected value = sum of the products = (2+3+3+4+4+5+5+5+5+ . .. +12)/36 = 7

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Common Core State Standards

S.MD.2 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.

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Fred’s Fun FactoryA famous arcade in a seaside resort town consists of many different games of skill and chance. In order to play a popular “spinning wheel” game at Fred’s Fun Factory Arcade, a player is required to pay a small, fixed amount of 25 cents each time he/she wants to make the wheel spin. When the wheel stops, the player is awarded tickets based on where the wheel stops and these tickets are then redeemable for prizes at a redemption center within the arcade.

Number of Tickets Probability

1 ticket 35%

2 tickets 20%

3 tickets 20%

5 tickets 10%

10 tickets 10%

25 tickets 4%

100 tickets 1%

The wheel awards the tickets with the following probabilities:

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a) If a player were to play this game many, many times, what is the expected number of tickets that the player would win from each spin?

b) The arcade often provides quarters to its customers in $5.00 rolls. Every day over the summer, Jack obtains one of these quarter rolls and uses all of the quarters for the spinning wheel game. In the long run, what is the average number of tickets that Jack can expect to win each day using this strategy?

c) One of the redemption center prizes that Jack is playing for costs 300 tickets. It is also available at a store for $4.99. Without factoring in any enjoyment gained from playing the game or from visiting the arcade, would you advise Jack to try and obtain this item based on arcade ticket winnings or to buy the item from the store? Explain.

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Expected ValueCommon Core State Standards

Use probability to evaluate outcomes of decisions.Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.

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The Amore Ristorante Valentine Specialfrom Bock, D. E., Vellemna, P.F., DeVeaux, R. D. (2010). Stats: Modeling the world, Third edition. Boston: Addison Wesley

From a deck of 4 aces, pick the Ace of Hearts, win a $20 discount.If the card picked is the Ace of Diamonds, you get a 2nd chance; pick again for Ace of Hearts and a $10 discount.

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• P($20)= P(Ace of Hearts)= One fourth of the time, the Valentine couple gets $20.

• P($10)=P(Ace of Diamonds followed by Ace of Hearts)= One twelfth of the time, the Valentine couple gets $10.

• P($0) = P(Black ace on 1st or 2nd draw) = Two-thirds of the time, the Valentine couple gets $0!

=

Winnings={$20, $10, $0}

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…so how much?

For 120 couples, the restaurant will on average pay out $20 to 30 couples and $10 to 10 couples. That is $600 + $100 or $700 for 120 couples.

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Expected Value of a Random Variable (X)

The expected value is the mean of E(X)= 20(.25)+10(.25)(.3333…)+0(.25)(.6666….) + 0(.5) 5. 83

μ = E(X)=

Outcome Ace Hearts

Ace Diamonds then Ace Hearts

Ace Diamonds then Black Ace Black Ace

X 20 10 0 0

P(X) .25 (.25)(.3333…) (.25)(.666666…) 0.5

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Box Model for Expected ValueLet’s say that the Restaurant has made reservations for 100 couples. They can expect to give 100(5.83) or $583 in discounts! The problem can be modeled as the sum of 100 draws from a box of tickets: 100 |0 0 0 0 0 0 0 0 10 20 20 20|

With replacement

The expected value is the number of draws times the average of the box.

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Expected Value

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Common Core State Standards

(S.MD.3) Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value. For example, find the theoretical probability distribution for the number of correct answers obtained by guessing on all five questions of a multiple-choice test where each question has four choices, and find the expected grade under various grading schemes.

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What do you expect?

A student guesses on all 5 questions of a multiple 4 choice test. 5 correct?

4 correct?

3 correct?

2 correct?

1 correct?

The big zero?

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What do you expect? Does the probability help you decide?

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Binomial Formula If trials are independent, the probability is

same for each trial, and the number of trials is decided in advance then the probability of exactly k occurrences in n trials is

nCk•pk(1-p)n-k

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Probability Histogram(from TI 84 Calculator)

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Expected ValueGrade Frequency for 1000

0 0.237(1000)=237

1 0.396(1000)=396

2 0.264(1000)=264

3 0.088(1000)= 88

4 0.015(1000)= 15

5 0.001(1000)= 1

How would you calculate the average grade?(0(237)+1(396)+2(264)+ …)/1000Expected Value = 0(0.237) + 1(0.396) + 2(0.264)+… ≈1.25

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A game consists of throwing a dart at a target. Assume the dart must hit the target and land inside one of the squares. The player pays $24 to play the game. If the dart hits the shaded region, the player wins $44, otherwise the player receives nothing. What is the expected value of the game?

Probability of shaded region:

Probability of un-shaded region:Profit: $44 - $24 = $20

Solution: (20) - (24) = .75 cents

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The odds of winning a raffle are 1:3. If the winning prize is $12, then how much should a ticket cost if the raffle is fair?12 – c) +

124−14𝑐+−34𝑐=−3

−14𝑐+−34𝑐=−3

−1𝑐=−3

The ticket should cost $3