THINKING MATHEMATICS

THINKING

MATHEMATICS

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Series for Teachers and Students

and all those seeking

True and Joyous Understanding!

Volume 8

PROBABILITY AND STATISTICS

LEVEL:

TABLE OF CONTENTS

PART 1: BASIC PROBABILITY THEORY

Simplistic Overview ……… 2

Naïve Probability Theory ……… 7

OR ……… 15

AND ……… 17

Sequence Principle ……… 23

The Empirical Model ……… 34

Law of Large Numbers ……… 34

Monte Carlo ……… 35 Expected Value ……… 38 Conditional Probability ……… 45 Baye’s Theorem ……… 49 PROBLEM SET I ……… 55

PART 2: COUNTING

COUNTING PRINCIPLES The Multiplication Principle ……… 2

Factorials ……… 4

The Labeling Principle ……… 9

Multi-stage Labeling ……… 13

Fun with Poker ……… 16

PASCAL’S TRIANGLE A Grid of Numbers ……… 20

The Binomial Theorem ……… 27

PROBLEM SET II ……… 31

PART 3: STATISTICS

Displaying and Summarising Data……… 2

Measures of Central Tendency ……… 5

Measures of Dispersion ……… 9

Scatter Plots ……… 16

Correlation Coefficient ……… 23

Null Hypothesis ……… 27

Distributions ……… 31

Central Limit Theorem ……… 37

Normal Distribution ……… 41 68-95-99.7 Rule ……… 43 z-scores ……… 45 Roulette ……… 51 Confidence Intervals ……… 54 P-values ……… 57 Gallup Poles ……… 60 Sampling ……… 62 Chi-Squared test ……… 66 Quality Control ……… 71 Run Tests ……… 74 Rank Correlation ……… 82

PROBLEM SET III ……… 85

PART 4: ADVANCED TOPICS

Random Variables ……… 2

Sum, differences, multiples ……… 4

Connection to Central Limit Theorem ……… 9

Cereal Box Problem ……… 10

Geometric Distribution ……… 12

Binomial Distribution ……… 14

Proportions ……… 19

Student’s t-distribution ……… 20

Chi Squared distribution ……… 23

Chebyshev’s Inequality ……… 23

PROBABILITY AND STATISTICS

Informal Course Notes

PART I of IV

James Tanton

© 2007 James Tanton

CONTENTS:

Simplistic Overview ……… 2

Naïve Probability Theory ……… 7

OR ……… 15

AND ……… 17

Sequence Principle ……… 23

The Empirical Model ……… 34

Law of Large Numbers ……… 34

Monte Carlo ……… 35

Expected Value ……… 38

Conditional Probability ……… 45

Baye’s Theorem ……… 49

REFERENCES:

SOLVE THIS: Mathematical Activities for Students and Clubs, J. Tanton, Mathematical Association of America, Washington D.C., 2001

SIMPLISTIC OVERVIEW

Probability and Statistics represent two sides of the same coin.

PROBABILITY: Explores what can be said about an unknown sample from a known collection of objects.

e.g. We know all possible combinations from rolling a pair of dice. What is the most likely outcome?

STATISTICS: Explores what can be said about an unknown collection from a known sample.

e.g. We surveyed 100 people and found that 37 chewed gum. What does this say about the gum-chewing habits of the entire nation?

BASIC IDEAS:

Probability: If a situation can be described in terms of possible outcomes that are deemed equally likely, then the probability of any one particular outcome occurring is defined to be:

1 Prob

Total number of outcomes =

e.g. The possible outcomes from rolling a dice are: 1, 2, 3, 4, 5, 6. Each is usually deemed equally likely. Then:

1 Prob(3) 6 1 Prob(5) 6 = = etc.

Probability relies on the ability to COUNT things.

e.g. Four cards dealt from a deck. What’s the probability of getting four aces? This problem relies on the being able to count all 4-card hands. (A bit tricky.)

STATISTICS: There are two branches:

Descriptive Statistics is concerned with methods of collecting, tabulating and summarizing data.

Inferential Statistics is concerned with making inferences and predictions based on collected data.

e.g. A medical study records the heights of 100 eight-year-olds. • average height = a statistic

• tallest height = a statistic

• third to shortest height = a statistic THESE ARE ALL DESCRIPTIVE

• Making a judgment about whether a particular child’s height is abnormal is an INFERENTIAL JUDGMENT

COMMENT: The word “statistik” was coined by German political scientist Gottfried Achenwall (1719-1772) to mean “a summary of how things stand.” It is based on the Latin word

stare

meaning “to stand.”

HISTORY PROBABILITY

The start of probability theory can, essentially, be pinpointed to a single moment in time. In 1654 French nobleman Chevalier de Méré wrote to prominent

mathematician Blaise Pascal asking for advice on the following problem:

Two friends each lay down $100 in a friendly “best of seven” tennis game.

But rain interrupts play after just four matches –with one person having won

three games and the other just one. How should the $200 be divvied up

between the two players so as to properly reflect the likelihood of each

winning?

Pascal shared this problem Pierre de Fermat. Both solved it independently using different techniques. Through this problem, probability theory was born.

Comment: Italian mathematician Girolamo Cardano (1501-1576) actually worked with ideas akin to probability theory before this but did not publish his work. And of course gambling games have been in existence for centuries and scholars have wondered about their results. But the first definitive analysis of “chance” began with the work of Pascal and Fermat.

STATISTICS

The study of statistics – descriptive statistics, at least – is ancient. 3050 B.C.E. Egyptians collated data on population wealth 2300 B.C.E. Ancient Chinese did the same

594 B. C. E. Greeks took a census for tax collection 309 B. C. E. Greeks took a census for population figures

Later Romans kept census records, birth and death records, conducted geographic surveys, etc.

Middle ages: Very little done

The start of inferential statistics can be pinpointed to:

1662 John Grant analysed birth and death records to create “life tables” which were used to predict life expectancies of different social groups.

GETTING OUR FEET WET:

For fun let’s go back and analyse de Méré’s problem. Let’s imagine, like Pascal and Fermat, we are seeing it for the first time. How would you like to approach it? Recall:

Best of 7 games but only 4 games played. Player A has won 3 games

Player B has won 1 game.

How best divvy up a $200 pot? Here’s some space for writing notes!

COMMENT: There are a number of interesting concepts at play in this example. If you have some familiarity with these terms, you may be able to identify in your work … The Law of Large Numbers, the definition of the probability of an event, the notion of expected value. We’ll, of course, talk about these concepts in detail later in these notes.

NAÏVE PROBABILITY THEORY

Recall the basic principle of probability:

If a situation can be described in terms of possible outcomes that are deemed equally likely, then the probability of one particular outcome occurring defined to be:

1 Prob

Total number of outcomes =

Example: In rolling a die there are six possible outcomes: 1, 2, 3, 4, 5, 6, each deemed equally likely. Then:

1 Prob(4) 6 1 Prob(6) 6 = = etc. Definition:

The set of all possible outcomes of an experiment is called the sample space. Example: In tossing a coin … Sample Space = { , }H T

In rolling a die … Sample Space = {1, 2, 3, 4, 5, 6} Ascertaining someone’s age (in years) …

Sample Space = {0, 1, 2 ,.., 120(?) } Definition: An event is a set of outcomes (or just a single outcome).

Example: In rolling a die: Sample Space = {1, 2, 3, 4, 5, 6} An event could be: {2, 4, 6} (rolling an even number)

or : {3} (rolling a three)

Definition: Given a sample space S for an experiment and an event E, the probability of E occurring is defined to be:

# elements of E ( )

# elements of S P E =

(This is, of course, assuming that the sample space has just a finite number of elements, and that every single outcome is “equally likely.”)

Example: In rolling a die … S = { 1, 2, 3, 4, 5, 6}

The probability of rolling an even number, E = {2. 4 6} is:

P(even) = # elements of E 3 1 # elements of S =6 = . 2 Note, also:

(

)

(

)

1 ({3}) 6 4 2 {1, 2, 4,5} 6 3 0 rolling a 7 0 6 6 (rolling any number) 1

6 P P P P = = = = = = =

WARNING: This naïve approach to probability assumes each individual outcome is “equally likely.” THIS IS NOT AN EASY CONCEPT!

For example: In rolling a pair of dice and computing their sum, the set of all possibly outcomes is:

S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

But these individual events ,”2”, “3”, …,”12”, are not equally likely.

Somehow we are meant to know that the underlying “equally likely” quantity here is not the sums 2, 3, …, 12, but the pairs of numbers behind each sum with order considered important!

There are 36 possible ordered pairs: 1-1 1-2 1-3 1-4 1-5 1-6 2-1 2-2 2-3 2-4 2-5 2-6 3-1 3-2 3-3 3-4 3-5 3-6 4-1 4-2 4-3 4-4 4-5 4-6 5-1 5-2 5-3 5-4 5-5 5-6 6-1 6-2 6-3 6-4 6-5 6-6 These are the entities deemed “equally likely.” Now, only one of these pairs gives a sum of “2”, so:

1 (2) 36 P = Also, we see: 2 1 (3) 36 18 3 1 (4) 36 12 P P = = = =

EXERCISE: Write down P(5), P(6), P(7), P(8), P(9), P(10), P(11) and P(12). [QUESTION: Is this correct? Is the order of the pair indeed important? Or should 6-3, say, be deemed equivalent 3-6? How is one meant to know?]

This might not seem too much of an issue here, but in more complicated examples one might be given a sample space, such as S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, and one is meant to somehow “know” whether or not these outcomes are “equally likely,” or whether or not this sample space is a result of a more fundamentally “equally likely” set.

THIS IS VERY PERTURBING TO A MATHEMATICIAN! This is equally perturbing to students – and it should be!

Example: Consider the command:

Pick a whole number at random

. What does this mean? Is each number “equally likely”? Does the term “equally likely” even apply? Example: THE WALLET GAME

Two people decide to play the following game:

Each pulls out her wallet.

Whoever possesses the least amount of money in her wallet wins.

Her prize? The contents of the other player’s wallet.

Each person can reason: “I stand to win more than I lose. Thus the game is in my favour!”

A game can’t be favourable simultaneously to both players! Something is very strange here. What are the odds of winning? Is everything really balanced and “equally likely”?

ADDITIONAL COMMENT:

The very act of defining probability in terms of “equally probable” events is

circular: using the term “equally likely” assumes you already know what probability means! The very basis of naïve probability is philosophically flawed.

AND YET ANOTHER COMMMENT:

We defined, for a sample space S and an event E, the probability of E occurring as: # elements of E

( )

# elements of S P E =

If we change our wording here and say:

size of E ( )

size of S P E =

then we can extend our notion of probability to geometric settings where “size” of a set is taken to be its “area.” For example, a circle sits inside a circle of side length 2 inches. We can ask:

If a point inside the square is chosen at random, what

at the chances of it landing outside the circle?

The area of the shaded region, the region of interest (the event E), is

2 2

2 − ⋅π 1 = − and the area of the entire square (the sample space S) is 4 π 22 = . 4 Then the probability we seek is:

size of E 4 ( )

size of S 4

P E = = −π

QUESTION: What does “equally likely” mean in this setting? Is the probability of picking any specific point zero? If so, is the probability of picking any point from a collection of points also zero?

And while we are mired in philosophical woes, consider the following disturbing problem:

EXERCISE: BERTRAND’S PARADOX

A chord is chosen at random in a circle. What is the probability that the chord is longer than the side-length of an inscribed equilateral triangle?

Answer 1: By rotating the circle we may as well assume that one end of the chosen chord is positioned at the left end of the circle. Then we can see that the chord will be longer than the side of an inscribed equilateral triangle if its second end lies in the shaded portion shown.

This represents 1

3 of the circumference of the circle. Thus the probability we seek is:

1 3 P=

Answer 2: By rotating the circle we may as well assume that the chosen chord is horizontal. Then the chosen chord will be longer than the side-length of an

inscribed equilateral triangle if its mid-point lies on the shaded portion of the diameter shown:

An exercise in geometry shows that this represents 1

2 of the diameter. Thus the probability we seek is:

1 2 P=

This argument is mathematically sound and the result is absolutely correct!!! □

The problem here is that the term “at random” is absolutely vague! The first answer defines “at random” to mean:

select a point on the circumference of the

circle and connect it with a given previously chosen point

. The second solution assumes “at random” means:

draw a circle on the floor and roll a broom handle from

one side of the room across the circle.

It is possible to define “at random” by many different means for this problem and arrive at different, but absolutely valid, answers. (If one draws a circle on a piece of paper and drops straws from above onto the figure, one finds that about 1

4 of them give chords that of the length we seek!)

Examples like these paradoxes alerted mathematicians to the problems with beginning approaches to probability theory. Terms such as “equally likely” and “at random” and even “probability” itself, are fundamentally vague notions. No wonder one’s intuition is so challenged by this subject!

For fun …

SICHERMAN DICE:

Most people believe that ordered pairs of values of the fundamental “equally likely” entities to be considered when rolling a pair of dice and computing their sum. In this case, the results of rolling two dice can be nicely displayed in a table:

We see now that (3) 2 36

P = , (7) 6 36

P = and so forth.

Suppose instead we roll two dice, one numbered 1-2-2-3-3-4 and the other 1-3-4-5-6-8.

Complete the following addition table and verify that these dice give exactly the same probabilities for any given sum as ordinary dice.

CHALLENGE: Is there a way to renumber a pair of tetrahedral dice so that the probability of any given sum is the same as from a pair of “ordinary” tetrahedral dice (numbered 1-2-3-4 and 1-2-3-4)?

NAÏVE PROBABILITY THEORY CONTINUED … Putting philosophical woes on hold for now …

Given a (finite) sample space S and an event A, we have defined the probability of event A occurring as:

size of A ( )

size of S P A =

Next, we need to explore the possibilities of combining actions.

“Definition:” Two actions are said to be independent if the outcomes of one action in no way affect the outcomes of the other.

Example: Tossing a coin and rolling a die are independent events. Here the sample space is the set of twelve pairs:

(H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)

Example: Picking a card from a deck of cards, destroying it, and then picking a second card from the deck, and NOT independent events: The result of the first action affects possible outcomes for the second. For instance, picking the ace of spades first no longer allows the ace of spades to be chosen second.

Example: Deciding what to wear and the weather forecast are not independent events.

COMMENT: We generally rely on our intuitive understanding of the world to

conclude whether or not two events are independent. (Again, this can be a difficult issue.) This notion, as it stands, is just as vague and problematic as the term

THE USE OF THE WORD “OR”

Let’s examine a basic example of two independent actions: ACTION 1: Toss a coin

ACTION 2: Roll a die

The set of all possible outcomes can be displayed in a tree diagram:

The twelve different outcomes are now explicit. It is easy to compute probabilities. For example:

P ( {H, even} ) = 3 12 P( {T, 5 or 6} ) = 2 12 P ({H, even} OR {T, 5 or 6} ) = 3 2 3 2 12 12 12 + = + = P({H, even}) + P({T, 5 or 6})

This illustrates a general principle:

If A = one set of desired outcomes B = a second set of desired outcomes and these outcomes share no common events Then:

P(A or B) = P(A) + P(B)

Example: Let A = “head and any number” B = “tail and 3”

Then ( ) ( ) ( ) 6 1 7

12 12 12 P A∪B = P A +P B = + =

Comment: The union symbol ∪ is interpreted as “or.”

EXERCISE: What if A and B do share events in common? Use the following Venn diagram to explain the formula: (P A∪B)=P A( )+P B( )−P A( ∩B).

EXERCISE: Find P( {H, odd} or {H, 1, 2, or 3}).

THE USE OF THE WORD “AND”

There is a second model called the “square model” that is useful for analyzing probabilities.

Example: Suppose 100 people walk down a garden path that leads to a fork. A left turn leads to house A, a right turn to house B.

Assume that there is a 50% chance that a person will turn one way over another. In this set-up we’d expect, basically 50 people to end up at house A and 50 people at house B. The following diagram of one-hundred dots (for 100 people) depicts this outcome:

The number “100” here is immaterial. The point is that if a square is used to denote the entire population of people walking down the path, then half the area of the square (half the people) end up with results “A”, and the second half the square result “B.”

EXERCISE: Folk walk down the following system of paths. Use the square model to compute the fraction of people that end up at house A, at house B, and at house C. [Assume that each choice encountered at a fork in the path is equally likely.]

EXERCISE: Folk walk down the following system of paths. Use the square model to compute the fraction of people that end up at each house. [Again assume that each choice encountered at a fork in the path is equally likely.]

EXAMPLE: I roll a die and then toss a coin. What are the chances of getting an even number followed by a head?

Answer: Think of this as a path-walking problem with two houses labeled “WANT” and “DON’T WANT.” The forks in the road represent the options that can occur (each, with 50% chance of occurring):

This leads to the square model diagram:

We see that the desired outcome represents one quarter (half of a half) of the square:

P(even AND head) = 1 4

EXAMPLE: I toss a quarter, then I toss a dime, and then I roll a die. What are the chances of receiving HEAD, HEAD, and “5 or 6”?

Answer: Here’s the garden path:

This gives the square model:

We have: P( H and H and {5,6} ) = 2 6 of 1 2 of 1 2 of the square = 2 1 1 1 6× × =2 2 12 □

We see the “multiplication principle” at hand:

If A represents the set of desired outcomes for one action, and B the set of desired outcomes for a second action, and these actions are independent then:

P(A and B) = P(A) × P(B)

In summary:

For independent actions: “OR” means “addition”

“AND” means “multiplication”

ASIDE: Consider a garden path that leads to two possible houses, A and B. Suppose there are an infinite numbers of three-way forks, each with left turn leading to A, right turn leading to B, and straight path leading to the next fork. Create a beautiful depiction of the square model for this situation that makes it visually clear that this formula: 1 1 1 1 1

3+ +9 27+81+ = 2

ANOTHER CONFUSING MATTER …

DOES THE ORDER IN WHICH ONE COMPUTES TASKS MATTER? For example, in tossing a coin and rolling a die there are three possibilities:

1. Toss the coin and then roll the die 2. Roll the die and then toss the coin 3. Roll and toss the coin simultaneously.

Standing back and thinking about this one would likely say that all three scenarios are philosophically equivalent, even though the tree diagrams and square diagrams for possibilities 1. and 2. are different. (Draw them! Is it possible to draw a tree diagram for simultaneous actions?)

It’s good to spell things out and make explicit the following …

SEQUENCE PRINCIPLE:

If two actions are independent, then performing the two

actions simultaneously is philosophically equivalent to performing them one at a

time (and it does not matter in which order one opts to do them).

The following example is typical of the difficulties that can arise:

EXAMPLE: I roll a pair of dice. What are the chances of getting an “6” and a “2”? Answer: It is best to avoid thinking of events that occur simultaneously and

instead tease them apart into a sequence of actions. Here we can imagine rolling one die and then the other. Now it is clear that there are two desirable

possibilities:

Roll a 6 and then roll a 2 OR

Roll a 2 and then roll a 6

Using “+” or OR and “x” for AND, the probability we seek is: 1 1 1 1 1

(6) (2) (2) (6)

6 6 6 6 18

P= P ×P +P ×P = ⋅ + ⋅ = □

EXERCISE: I roll three die simultaneously. What at the chances of seeing two 6s and one 5?

This is essentially all there is to naïve probability theory. Of course, there are subtle issues to explore - and we’ll come to those in applications – but for now, let’s practice the basic ideas.

EXERCISE: Suppose A is an event for some sample space S. Explain the following formula:

P(not A) = 1 – P(A)

EXERCISE:

The probability that any one person will be bitten at least once in life by a dog is 1

20. The probability of being bitten by a cat is 1 50. Find:

a) The probability that a person will be bitten by both a cat and a dog some time in life

b) The probability that a person will never be bitten by a dog.

c) The probability that someone will be bitten by a cat or a dog but not both.

EXERCISE: Three dice are tossed simultaneously. What are the chances of: a) Receiving three 1s?

b) Receiving no 1s?

c) Receiving two 1s and one 2?

Three dice and two coins are tossed simultaneously. What are the chances of: d) Receiving three 1s and two Hs?

EXERCISE: A couple has two children.

a) Draw a tree diagram illustrating all possibilities re gender. b) What is the probability that the couple has two boys?

c) What is the probability that the couple has a child of each gender? [Assume that the chances of having a boy match those of having a girl.] EXERCISE:

a) You know that Jenny has two children and that her first child is a boy. What is the probability that her other child is a girl?

b) You know that Mike has two children and that one is a boy. What are the chances that the other child is a girl?

EXERCISE:

Lulu has four children and you are told that at least one of the four is a boy. What is the probability that …

a) Exactly two of her children are boys? b) At least two are boys?

EXERCISE: Two dice are rolled.

a) What is the probability that their sum is odd? Even? b) What is the probability that their product is odd? Even?

c) When rolling a pair of dice, the most likely sum is “7.” What is the most likely product?

EXERCISE: A pop-quiz has 10 multiple choice questions, each with choices: A, B, C or D. I didn’t study for this quiz and decided circle the answers at random.

a) What are the chances that I will get all 10 questions right? b) What are the chances that I will get 9 out of 10 correct? c) What are the chances that I will get 8 out of 10 correct? d) What are the chances that I will get at least three right?

EXAMPLE: A CLIFF HANGER

Dorothy is not feeling good. She stands at the edge of a cliff, conveniently labeled position 1, with an infinite expanse of land at her back (conveniently labeled in steps 2, 3, 4, …). Regard location 0 as “off the cliff.”

Dorothy lays her fate in the toss(es) of a coin. She pulls out a quarter to toss and decides:

If it lands HEADS, I will step forward (to my doom).

If it lands TAILS, I will step backwards one place and toss again.

She does this repeatedly – stepping forward with each land of HEADS, backwards with each land of TAILS – until she either meets her doom or ends up wandering forever in the infinite expanse behind her.

What are the chances that Dorothy will walk over the cliff?

Answer: We’ll answer this is in a series of steps.

Let p= p(1→0) be the probability, when standing at position 1, of Dorothy

eventually reaching position zero. (This may either be by stepping one pace forward right away, or taking a step back and then two paces forward, and so forth.) It is this value

p

that we seek.

Define (2p → , (31) p →2), and so on the same way.

STEP 1: Can you see that (1p →0) and (2p → and (31) p →2) and so on, are each, philosophically, the same problem and so have the same value

p

?

Notice that:

(

)

(

)

(

)

(

)

(

)

1 0 stepping forward right away OR stepping back to position 2 and reaching 0 sometime later stepping forward right away stepping back AND moving 2 0

1 1

2 0

2 2 1 1

moving from 2 to 1 AND mo 2 2 p p p p p p → = = + → = + → = +

(

)

(

) (

)

ving from 1 to 0 1 1 2 1 1 0 2 2 p p = + → ⋅ →

This leads to the quadratic equation:

2

1 1 2 2 p= + p

STEP 2: Solve for

p.

Is Dorothy’s doom certain?

□

EXERCISE: Explain why

( 0) ( 1) (2 1) (1 0) 1 1 1 1

p N → = p N → N− × ×⋯ p → × p → = ×⋯× × = . What is this saying?

EXERCISE: GAMBLER’S RUIN

A gambler repeatedly plays a simple game: a 50% chance of winning a dollar, a 50% chance of losing a dollar. If she starts with $N, what are the chances of her losing all her money?

COMMENT: We are flirting with the notion of a “random walk” and have essentially proven that a one-dimensional walk, the walker will visit each and every cell of the line an infinite number of times. Feel free to conduct some internet research on this topic.

EXAMPLE: A SMARTER GAMBLER

Another gambler repeatedly plays the same simple game with a 50% chance of winning a dollar, 50% chance of losing a dollar. She starts with $4.

She decides she will stop playing when she either reaches $0 (loses all her money) or gets to $10.

What are her chances of reaching $10?

Answer - almost: Let ( )p N = probability of reaching $10 starting with $N in hand. We certainly have (0) 0p = (we’ve already lost our money and have no chance of reaching $10) and (10) 1p = (with $10 in hand we are guaranteed having $10!). Now, suppose N is between 1 and 9 inclusive. Then:

(

)

(

)

(

)

( ) lose a dollar AND play with $N-1 OR win a dollar AND play with $N+1

1 1 1 1 2 2 p N p p N p N = = − + +

Thus each number ( )p N is the average of its two neighboring values.

CHALLENGE: If (1)p = , show that this means that (2) 2x p = x, and that (3) 3p = x, and so forth. What must be the value of

x

?

And, so, what is the value of (4)p ?

□ EXERCISE: Back to Dorothy …

Suppose she uses a weighted coin that has only a 1

3 chance of landing HEADS and 2

3 chance of landing TAILS. Show that her chance of survival after a possibly infinite number of tosses is now 50%!

A not-so-exciting example: EXERCISE: A bag contains: Two red balls Three white balls Six blue balls

A ball is chosen at random. What is the probability that the ball is: a) Blue?

b) Either red or blue? c) Neither red nor blue? A not-unexciting example:

EXERCISE: A bag contains one Red, one Blue and one White ball. John picks out a ball at random.

If it is Red, he wins. If it is blue, he loses.

If it is white, he puts the ball back, and adds to the bag another red ball and another blue ball. (So the bag now contains 2Rs, 2Bs and one W.)

John then chooses a ball at random. If it is red, he wins now.

If it is blue, he loses.

If it is white, he puts the ball back and adds another red ball and another blue ball and picks again.

John keeps doing this until he either wins or loses. Use this problem to establish the bizarre formula:

1 1 2 1 1 3 1 1 1 4 1

3+ ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅ +3 5 3 5 7 3 5 7 9 = 2 ⋯

EXAMPLE: NON-TRANSITIVE DICE

Here are designs for four dice: A, B, C and D.

You and a friend decide to play the following game.

Your friend chooses a die and then you choose a die.

You each roll your chosen die. Whoever receives the largest number wins.

Show that if you choose the die to the left of your friend’s choice (or choose die D if your friend chooses die A) you will win this game two-thirds of the time. That is, show that:

die A beats die B two-thirds of the time die B beats die C two-thirds of the time die C beats die D two-thirds of the time die D beats die A two-thirds of the time

HINT: The following table shows all possible wins if dice A and B are rolled:

EXERCISE: ANOTHER MAGICAL PROPERTY OF A MAGIC SQUARE Here’s the standard 3x3 magic square.

(Each row, column and diagonal has the same sum of 15).

Player A picks a number at random from row A, player B a number at random from row B, and player C a number at random from row C.

Let’s say that “A beats B” if A’s number is larger than B’s. a) Show that chances are A will beat B.

b) Show that chances are B will beat C. c) Show that chances are C will beat A!

ANOTHER CLASSIC … THE BIRTHDAY PROBLEM

Two random people are kidnapped. What are the chances that their birthdays land on the same day of the year? Give your answer as a percentage to one decimal place.

Answer:

Three random people are kidnapped. What are the chances that at least two of them have the same birthday?

Answer:

Four people are kidnapped. What are the chances that at least two of them have the same birthday?

THE EMPIRICAL MODEL

One way to attempt computing probability values is to repeat an experiment a large number of times and see how often the desired outcome occurs. This is usually the only option available if a particular experiment is extremely complicated or

difficult to compute. For example:

What is the probability that 5 letters chosen at random spell an English word?

The best thing to do would probably be … Have a computer develop 10,000

examples of five letters chosen at random and match each of these with its built-in dictionary to see what proportion of them are English words. This won’t be an exact answer to our questions, but we suspect it would be a very close

approximation.

We are using a principle here called the “Law of Large Numbers.” This law makes intuitive sense and is often assumed without explicit mention. In the nineteenth and twentieth centuries, as mathematicians attempted to put probability theory on a sound, rigorous footing, one of the significant “check-points” of their work was the ability to prove this Law of Large numbers as true according to the axioms of their theory. Here’s the principle:

LAW OF LARGE NUMBERS

The more times a random phenomenon is performed the closer the proportion of

trials in which a particular desired outcome occurs approximates the true

probability of that outcome occurring.

Example: If you toss a coin some number of times, you would expect approximately half of the tosses to be HEADS and half TAILS.

With 10 tosses, it is unlikely you will receive exactly half of each. (Try it!) With 100 tosses, the proportion of heads would be closer to 50%

With 1000 tosses, significantly closer to 50%. Even better with 100000000000000000 tosses!

Comment: Many gamblers incorrectly interpret this result as follows:

If a run of trials did not produce the desired outcome, then the chances of

that outcome occurring on the very next trial is increased.

Example: You toss a coin nine times and got four HEADS and five TAILS. That the next toss will be HEADS is thus almost certain – NOT TRUE!

Example: You toss a coin 999 times and got TAILS every time. The chances that the next toss will be heads, alas, is still only 50%.

Gamblers often feel that a string of losses must produce a win on the next turn.

MONTE CARLO METHOD:

The act of repeating an experiment multiple times to determine (an approximate) value for a probability value is often called the “Monte Carlo Method.” Many

casinos determined odds for their games simply by playing them multiple times and observing frequency of outcomes. (How do you determine the chances of winning a hand of blackjack? An easy method is to observe play of a large number of games.) Aside: Read

Bringing Down the House

to learn how MIT students tipped the odds of blackjack in their favour by certain tactical plays.

One can use the Monte Carlo Method to work out areas of complicated regions. Example: Here is an aerial photograph of an oil spill.

It is known that the area of the rectangular region photographed is 40 square kilometers.

One can compute (a fairly accurate approximation to) the area by digitizing the photograph and have a compute select 10,000 points at random in the photograph. If, say, 6,473 of those points land in the shaded region, then we can say that the area of the spill is ……?

ACTIVITY:

In a group of three call one person player 0, one person player 2 and the remaining person player 3. Each person places his right hand behind his back and secretly holds up one, two or three fingers. The players then show their hands.

If all three numbers match, player 3 receives a point.

If two of the three numbers match, then player 2 receives a point. If there are no matches, player zero receives a point.

a) Play this game a large number of times and tally points in a table. From your data, who seems to have the largest chance of receiving a point in any single game? Estimate that probability of winning. Estimate the chances of a win for each of the remaining two players.

b) Use theory to determine the actual probability of a win for each of the three players.

c) Suppose players 3, 2, and 0 are assigned, instead of one point for each win a, b and

c points, respectively, for each win. Choose values for a, b and c that makes this game “fair.”

EXERCISE: a) A bag contains four balls each colored either red or blue. Jenny pulls out two balls at random and gets a pair of blue balls. She returns the balls to the bag, gives it a shake, and pulls out another pair of balls. She does this 100 times recording the results along the way:

BB = 52 times BR = 48 times RR = 0 times

Most likely, how many blue balls and how many red balls are in that bag? b) Suppose, instead, Jenny obtained the result:

BB = 16 times BR = 70 times RR = 14 times

OPTIONAL EXERCISE: BUFFON NEEDLE PROBLEM

Do some internet research on how one can use probability theory to approximate

π

.

EXAMPLE: KRUSKAL COUNT

In the early 1980s, Princeton physicist Martin Kruskal discovered a remarkable mathematical property all passages of written text seem to possess. This phenomenon is now referred to as Kruskal’s count. To illustrate, consider the familiar nursery rhyme:

Twinkle twinkle little star, How I wonder what you are, Up above the world so high, Like a diamond in the sky. Twinkle twinkle little star, How I wonder what you are. Perform the following steps:

1. Select any word from the first or second line and count the number of letters it contains.

2. Count that many words forward through the passage to land on a new word. (For example, choosing the word

star

, with four letters, will transport you to the word

what

.)

3. Count the number of letters in the new word, and move forward again that many places.

4. Repeat this procedure until you can go no further (that is, counting forward will take you off the nursery rhyme.)

5. Observe the final word on which you have landed.

Surprisingly, no matter on which word you start this counting task, the procedure always takes you to the same word in the final line, namely, the word

you

.

Kruskal observed that this same phenomenon seems to occur with

any

sufficiently large piece of text - counting forward in this way from any choice of beginning word lands you at the same place at the end of the page. This provides an amusing activity for several people to perform simultaneously, all working with the same text, but starting with different choices of initial word.

EXPECTED VALUE

Suppose you a play a game with some monetary values associated with it. For example: Flip a coin.

If it comes up HEADS, you win $2. If it comes up TAILS, you lose $1.

Definition: The expected value of a game is the average profit (or loss) one would expect if the game were played a large number of times.

For example, suppose we played the above game 200 times. Then, on average, we would expect to win $2 one hundred times and lose $1 one hundred times.

Average win = 2 2 2 ( 1) ( 1) ( 1) 200 + +⋯+ + − + − +⋯+ − = 2 100 ( 1) 100 200 × + − × = 2 1 ( 1) 1 2 2 × + − × = 1 2 = 50 cents Thus, we’d expect to win 50 cents per game.

COMMENT: Many text-book questions phrase a game like the one described above as follows:

You pay $1 to play the following game:

Flip a coin. If it is HEADS, win $3. If it is TAILS, you win nothing.

Do you see that it is exactly the same game as before? The idea of “having to pay first” often offers a point of confusion for students. It is good to tease such questions apart and list the end outcomes explicitly:

If HEADS – I’m up $2 overall If TAILS - I am down $1 overall Now it is clear how to handle analysis of the game.

EXAMPLE: Imagine the following simple dice game: Roll 1: You win $10

Roll 2: You win $5 Roll 3,4,5,6: You lose $3 Is this game worth playing?

Answer: Imagine playing 600 rounds. (Why did I choose the number 600?) On average, we’d expect:

100 times a win of $10 100 times a win of $5 400 times a loss of $3 Average profit = 100 10 100 5 400 ( 3) 1 10 1 5 4 ( 3) 0.50 600 6 6 6 × + × + × − = × + × + × − =

This game is in your favour. It is worth playing. You can expect to win, on average,

50 cents per game. □

Note: In this calculation we see the appearance of the probabilities of each outcome multiplied by their respective values of outcomes.

In general …

If a game offers values x x1, 2,…,xn with probabilities p p1, 2,…,pn, then the

expected value of the game is:

1 1 2 2 n n

x p +x p +⋯+x p This number is often denoted µ or E.

Exercise: Find the expected value of tossing a pair of dice. Answer:

□ EXAMPLE: A coin is tossed.

If H comes up on the first toss, you win $1.

If H first appears in the second toss, you lose $1. If H first appears in the third toss, you win $1. and so on.

(Basically: You win $1 if H first appears on an odd toss, lose $1 if H first appears on even toss.)

What is the expected value of this game? Answer:

The probability of getting heads on the first toss is 1 2

The probability of getting heads on the second toss is: 1 1 1

2× = . (Why?) 2 4 The probability of getting heads on the third toss is: 1 1 1 1

2× × = . 2 2 8 And so on Thus: 1 1 1 1 1 1 1 1 1 ( 1) 1 ( 1) 2 4 8 16 2 4 8 16 µ = × + − × + × + − × +⋯ = − + − +⋯

One can use the “geometric series” formula to evaluate this infinite sum. Another approach (a neat trick) is to multiply this sum by two:

1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 2 4 8 16 32 2 4 8 16 2 4 8 16 µ = _ − + − + − _= − + − + − = −_ − + − + _= −µ  ⋯ ⋯  ⋯ So 2µ = − giving 1 µ 1 3 µ = . Done! □

EXERCISE:

a) Consider the example we first studied in this section, phrased as the textbooks tend to phrase it:

You pay $1 to play the following game:

Flip a coin. If it is HEADS, win $3. If it is TAILS, you win nothing.

What is the expected profit in playing this game?

A student decides to answer the problem as follows:

Well …

1 3 1 0 $1: 50

2 2

µ = ⋅ + ⋅ =

. But since we paid a dollar, we must subtract $1

from this amount. The expected profit is therefore 50 cents.

This agrees with our previous answer. Coincidence?

COMMENT: Some students prefer to follow this approach to questions like these. b) A casino offers a game in which one can win x₁, x₂, x₃ or x₄ dollars with

probabilities p p1, 2,p p3, 4 respectively. Suppose the expected value of this game is

1 1 2 2 3 3 4 4

x p x p x p x p

µ

= + + + . As a promotion, the casino decides to offer “bonus night” during which all payouts increase by $2. (Thus the payouts are now

1 2, 2 2, 3 2, 4 2

x + x + x + x + dollars.) Prove, mathematically, that

µ

increases by 2. Does the mathematics used here also explain the result of part a)?

TWO TIDBITS

Here’s a seemingly paradoxical exercise: EXERCISE:

In Tiny-Town, 90% of the city cabs are purple and the remaining 10% are blue. A crime was committed and an eye-witness claims she saw a blue cab at the scene. Subsequent tests showed that this witness is correct in her observations four times out of five, that is, 80% of the time.

What are the chances that the cab at the scene really was blue?

EXAMPLE: MONTY HALL PROBLEM

Named after the host of a popular American TV game show

Let’s Make a Deal!

, the Monty Hall problem is a classic puzzler often used to test initiates in the field of probability theory. It goes as follows:

On a game show three closed doors stand before you. The host informs you that a cash prize lies behind one of the doors, and nothing behind the other two. You select a door, but before you open it, the host quickly opens one of the remaining two doors to show you that the prize is not there. He now gives the chance to change your mind and open instead the third remaining door. The question is: What should you do? Should you stay with your original choice of door, or switch to the other option? Is there any advantage to switching?

One’s typical first reaction to this puzzle is that there is no advantage at all to switching – since two doors remain with only one containing a prize, the chance of selecting the correct door, either by staying with the chosen door or switching, is always 50 percent. Surprisingly, this reasoning is not correct for it makes no use of the subtle information the host presents to you, which you can actually use to your advantage.

a) Play the game with a partner using playing cards as “doors” – one black and two red. Take turns being host and being contestant. What do you notice about your choices as host?

b) Explain why your odds of winning double if you choose to always “switch” rather than “stick.”

c) Suppose the host presents you with 100 doors with only one containing a prize. You reach for a door but just before you open it, the host reveals to you the empty contents of 98 other doors. There are now two closed doors, one with your hand on it. The host then offers you the chance to change your mind and open instead the remaining closed door. Should you “stick” with your original choice or “switch”?

For the bold …

Consider the following variation of the game:

A game contains four doors, one with a fabulous cash prize behind it, the remaining

three empty. The host invites you to select a door and you place your hand on its

knob.

At this stage, Monty opens one of the remaining three doors and shows its empty

contents. He offers you the chance to “stick” with your current door choice or

“switch” to one of the remaining two closed doors. You make your choice.

Next, Monty opens a second door to reveal its empty contents. Two closed doors

remain, one with your hand on its knob. He again offers you the chance to “stick” or

“switch.”

At this stage the game ends and you accept the consequences.

There are four strategies to this two-stage game: Stick-Stick; Stick-Switch; Switch- Stick; Switch-Switch.

FOR THE BOLDER …

In our analysis of the original (and subsequent) Monty Hall problems we made two assumptions:

i) Monty does know behind which door the prize lies

ii) Monty has no preference as to which non-prize door he opens when faced with a choice.

Let’s consider some alternative scenarios:

Assume that Monty knows where the prize lies but has a preference as to which non-prize door he opens when faced with a choice.

(For an example of preference, suppose the doors are numbered 1, 2, and 3 and Monty will always open the lowest numbered door he can.)

In this scenario, switching is not always better! Sometimes a stick is just as good! Matters depend on individual plays.

For example, suppose the contestant reaches for door number 1. If Monty opens door number 3 to reveal a non-prize, then the contestant should switch for a certain win. (What stopped Monty from opening door 2 if that was his preference?) If, on the other hand, Monty did open door 2 to reveal a non-proze, then there is no advantage to sticking or switching. (Monty’s action here reveals no information about the possible location of the prize.)

EXPERIMENT: Conduct a card experiment with a friend mimicking this scenario. How often can your friend deduce for certain the location of the black card? In general, what are the odds of your friend winning this version of the game?

Assume that Monty has no knowledge of the location of the prize and, by luck, opened a non-prize door.

There is no advantage to sticking or to switching: each produces a 50% chance of winning. To see why, imagine playing the game 30 times. On average, the contestant will have his hand on the correct door for 10 of those games, and on in incorrect door for 20 of those games. In those 20 games, Monty will accidentally reveal the prize half the time, and so we must reject those occurrences. (We are told that this did not happen.) So we are left with 20 games to ponder, 10 of each type. Sticking leads to a win for 10 of those 20. Switching leads to a win for the remaining 10 of the 20.

CHALLENGE: Consider the final scenario in which Monty does not know the location of the prize but will open the lowest number door available to him. It turns out to a non-prize. Should the contestant stick or switch, or does it depend?

CONDITIONAL PROBABILITY

Analysis of the Monty Hall problem flirts with difficulty of what to do when partial information is revealed in a situation. This leads us to the notion of “conditional probability,”

Definition: The probability of an event occurring given knowledge that another event has already occurred is called a conditional probability.

Example: Two cards are drawn at random from a deck. Knowledge of the colour of the first card will affect the likelihood that the second card is red. Specifically: If we are told that the first card was black, then:

P( second card red) = 26 51

If we are told that the first card was red, then:

P( second card red) = 25 51

If we are told nothing about the colour of the first card, then:

P( second card red) = 26 1 52 = 2

[The first card might just as well still be in the deck.]

Notation: The probability that event A will occur given knowledge that event B has already occurred is denoted:

P(A|B)

e.g. We have: P(second card red | first card black) = 26 51

We can conduct a “thought activity” to determine a formula for P(A|B).

Imagine running the experiment a large number of times and observing the number of times B occurs.

We want P(A|B), the proportion of times A occurs among those times B has already happened. That is, we want the number of times both A and B occurred compared to the number of times just B occurred. This suggests:

P(A|B) = ( ) ( ) P A B

P B ∩

Example: Draw a card from a deck. A friend tells you that the card is red. What is the probability that it is an ace?

Answer: P( Ace | red ) = (Ace and Red) (Red) P P = (red ace) P(red) P = 2 52 1 2             = 1 13

[And this makes sense since among the 26 red cards, two are aces.] □

Exercise: A die is rolled. Someone yells out that the answer is odd.

Given this information, what is the probability that the roll was a “3”? a “4”? Answer these questions by practicing the formula for conditional probability (and then check that the answers make sense!)

CONDITIONAL PROBABILITY AND INDEPENDENT EVENTS

Recall that two actions are independent if the outcomes of one in no way affect the outcomes of the other. So, if A and B are independent events, we’d expect then:

P(A|B) = P(A)

[Knowledge of B occurring in no way affects the likelihood of A occurring.] We can prove this mathematically.

Recall, for independent events, we have: ( and )P A B =P A( )×P B( ). Thus:

P(A|B) = ( and ) ( ) ( ) ( ) ( ) ( ) P A B P A P B P A P B P B ⋅ = =

Example: A coin is tossed and a die is rolled. What is the probability of getting a HEAD given that the die rolled a “6”?

Answer: P(HEAD | Six) = (HEAD and SIX) (SIX) P P = 1 1 2 6 1 6 × = 1 2 □

COMMENT: In a sophisticated theory of probability, one defines a “measure” onto a set of objects that defines what “at random” means for the problem at hand. This obviates the issue of “equally likely” and begins to put the theory on sound logical footing. Mathematicians then take the relation ( | )P A B = p A( ) as the

definition

of what it means for two events A and B to be independent, that is, A and B are said to be independent if the “measure” P(A|B) equals the “measure” P(A).

BAYE’S THEOREM (1763)

What is the relationship between P(A|B) and P(B|A) ? e.g. Draw a card from a deck. Then:

P(ace|red) = 1 13 P(red|ace) = 1

2

What is the connection between these two numbers? RECALL: P(A|B) = ( ) ( ) P A B P B ∩ and P(B|A) = ( ) ( ) P B A P A ∩ We have: ( ) ( ) ( ) ( ) ( ) ( | ) ( | ) ( ) ( ) ( ) ( ) ( ) P B A P A B P A B P B P B P B A P A B P A P A P B P A P A ∩ ∩ ∩ = = = ⋅ = ⋅ That is:

( )

( | )

( | )

( )

P B

P B A

P A B

P A

=

⋅

Example: P(red|ace) = P(ace|red). ( ) ( ) P red P ace =

1 1/ 2 1

More generally …

Suppose B1 and B2 are two non-overlapping events that cover the whole sample

space.

e.g. B1 = getting a red card 2

B = getting a back card Suppose A is another event.

e.g. A = getting an ace Then: BAYES THEOREM: 1 1 1 1 1 2 2

( |

)

(

)

(

| )

( |

)

(

)

( |

)

(

)

P A B

P B

P B

A

P A B

P B

P A B

P B

⋅

=

⋅

+

⋅

This looks worse than it is. It is also fairly straightforward to prove. Here’s some blank space to write out the proof:

Proof:

Let’s do an example: EXAMPLE:

Bag 1 contains 5 red and 2 white balls. Bag 2 contains 7 red and 4 white balls.

A bag is selected at random. A ball is selected at random from that bag. You are told the ball is red.

What is the probability that that ball came from bag 1? Answer: Let:

1

B = ball comes from bag 1

2

B = ball comes from bag 2 A = ball is red

We want P(B1|A), the probability that the ball came from bag 1 given that it is red.

According to Baye’s theorem:

1 1 1 1 1 2 2 ( | ) ( ) ( | ) ( | ) ( ) ( | ) ( ) 5 1 7 2 5 1 7 1 7 2 11 2 55 104 P A B P B P B A P A B P B P A B P B = + ⋅ = ⋅ + ⋅ = □ Would you have guessed this answer?

The theorem looks complicated, but it allows you to compute some nasty problems with relative ease.

Here’s a simple, but philosophically confusing, example: EXERCISE: TWO CARD PARADOX

One card is red on one side and black on the other. A second card is red on both sides.

Both cards are put in a bag and one is pulled out at random. You see that one side of the chosen card is red. What is the probability that the other side of this chosen card is also red?

a) Answer this question first by using logical reasoning b) Answer this question a second time using Baye’s theorem.

Answers:

EXERCISE: Yale psychologists have coined the term “cognitive dissonance” for the act of devaluing an object after being told it is not available, and conducted an experiment with monkeys to show that they might too engage in cognitive dissonance. (See

http://www.nytimes.com/2008/04/08/science/08tier.html?_r=1&8dpc&oref=slogin ). Scientists had discovered that monkeys prefer red, green and blue M&M’s over all other colors. Monkeys were then given two M&M’s of different colors – say one red and one green. The monkeys would grab one candy, say the red one, and then have the other one taken away. Next the monkey would be offered another two M&M’s but of the colors it had not eaten, in our example, blue and green. The monkey had already experienced the green M&M being taken away, and the scientists found that about two thirds of the

monkeys opted to take the blue one instead. Had the majority of monkeys indeed devalued the green M&M given the previous loss?

Show that this is a mathematical result and not a psychological result. That is, show that, of all the monkeys that prefer red M&M’s over green M&M’s, two thirds of them also prefer blue M&M’s over green M&M’s, irrespective of whatever experiment is to be conducted!

PROBLEM SET I

Question 1: Analyse and answer the following variation of de Mere’s problem:

Two players play a series of games for the “best out of five.” The winner is to

receive a prize of $1000. After three games of play, in which the first player had

won one game and the other two games, the match was interrupted by an

earthquake. How should the $1000 be divvied up between the two players so as to

properly reflect their likelihoods of having won the series? (Assume the each

player has a 50% chance of winning any particular game.)

Question 2: Repeat question 1 but this time assume that the first player has only a 10% chance of winning any individual game.

Question 3: 8640 people walk down the following garden path. At each fork, equal numbers of people take each option. Find the number of people that end up in each of the houses A, B, C, and D.

Question 4: Assume that exactly 50% of children born are boys and 50% are girls. A couple has three children.

a) Draw a tree diagram displaying the possible genders of their three children, b) What are the chances that the couple has three boys?

c) What are the chances that the couple has at least one boy? d) What are the chances that the couple has exactly two boys? Suppose that we are now told that their first child was a girl.

e) What are the chances that the other two children are also girls?

f) What are the chances that at least one of their three children is a girl? Question 5: Billy’s girlfriend has a dimple on her left cheek (there is 1/100 chance that this occurs), blue eyes (there is a 1/100 chance this occurs), and likes math (there is a 1/100 chance that this occurs). He says that his girlfriend is “one in a million.” Is he correct?

Question 6: A card is drawn at random from a deck of 52 playing cards. a) Describe the sample space if suits are not considered relevant

b) Describe the sample space is suits and numerical value are considered relevant

c) Describe the sample space if the value of the card is considered irrelevant Question 7: A card is drawn at random from a deck of 52 cards. What is the probability of:

a) Drawing an ace? b) Drawing a club?

c) Drawing the ace of clubs? d) Drawing an ace or a club?

e) Drawing neither an ace nor a club? f) Drawing any suit except clubs?

Question 8: An urn contains 5 red balls, 8 blue balls, and 7 white balls. A ball is selected at random. What are the chances of:

a) selecting a red ball? b) not selecting a blue ball?

c) selecting a ball that is white or blue?

After the ball is selected, it is returned to the urn, and the experiment is repeated.

What are the chances, in the run of these two experiments, of d) selecting a red ball followed by another red ball?

e) selecting two balls of the same colour? Question 9:

The chances that someone gets bitten by a dog at least once in life is 0.02 . The chances of being hit by a meteorite at least once in life is 0.001 .

The chances of stepping in gum at least once in life is 0.99 . What is the probability …

a) Of being hit by a meteorite and being bitten by a dog in your life? b) Of never being hit by a meteorite?

c) Of never stepping in gum and never being hit by a meteorite? d) Of all three events happening in your life.

e) None of these events happening in your life.

Question 10: M&M’s come in six colors. Here’s a table showing the probability that a randomly chosen M&M has a particular colour:

Colour Brown Red Yellow Green Orange Blue Probability 30% 20% 20% 10% 10%

a) Fill in the missing number for blue.

b) What are the chances that an M&M chosen at random is either brown or red?

c) What are the chances that two M&M’s chosen at random from an extremely large bag are both blue?

d) What are the chances that three M&M’s chosen at random from an extremely large bag are all blue?

I’ve been told that the colour distribution for Peanut M&Ms is different. Obtain a bag of peanut M&M’s and use your sample to make estimates for the entries in the following table:

Colour Brown Red Yellow Green Orange Blue Probability

Question 11 (ANNOYING – BUT INTERESTING):

It is said that a Friday falls on the 13th day of the month 48 times every 28 years. a) Verify this calculation.

b) What is the probability that a randomly chosen Friday is a “Friday the 13th”? Question 12: (HARD-ISH, BUT REALLY INTERESTING!)

There are eight possible outcomes in tossing a coin three times: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT.

Two players decide to play the following game. Player A chooses the sequence HHH and player B the sequence THH. A coin is tossed repeatedly until one of these sequences appears. For example, the coin might produce T, T, H, T, H, H and player B wins. If the coin produces the sequence H, T, T, H, H, H then player A wins.

a) Play the game 10 times. Does player B seem to win the majority of times? b) Explain why player B has the advantage.

c) Suppose instead player A chooses the sequence HHT and player B the

sequence THH. Play the game 10 times. Does player B again win the majority of times? Can you explain why?

d) Here’s a table showing all the options A could choose and what B chooses in response.

If A chooses this … Then B chooses this …

HHH THH HHT THH HTH HHT THH TTH TTH HTT THT TTH HTT HHT TTT HTT

Play the game 10 times for each of the eight rows in the table. Verify that B wins the majority of times in each case. Show me the results you obtained. Question 13: A bag contains a red ball and a white ball. Jodie takes out a ball at random. If it is red she wins. If it is white, she then moves to a bag that contains two red balls, and a single white ball, and pulls out a ball. If it is red, she wins. If it is white, she then moves to a bag that contains three red balls and a single white ball, and pulls out a ball. If it is red, she wins. If it is white, she then moves on to … There are an infinite number of bags available to her, and she keeps playing this game until she eventually wins.

Explain how this hypothetical scenario “proves” the following equation:

1 1 2 1 1 3 1 1 1 4

1 2 + 2× + × × + × × × +3 2 3 4 2 3 4 5 =

⋯

(That is, in math notation, we’ve established:

1 1 ( 1)! n n n ∞ = = +

∑

.)

Question 14: A bag contains a red ball, a blue ball, and a white ball. Schuyler pulls a ball out at random. If it is red, he wins. If it is blue, he loses. If it is white, then he moves on to a bag that contains two red, two blue and one white ball. He pulls one out at random. If it is red, he wins. If it is blue, he loses. If it is white he moves on to a bag that contains four red, four blue, and one white ball. And so on, with double the number of red balls and double the number of blue balls from bag to bag.

a) Explain why Schuyler’s chances of winning this game are ½.

b) Write an interesting infinite sum based on this hypothetical scenario whose value is ½.

Question 15: Three dice are tossed simultaneously. What are the chances of rolling …

a) three sixes?

b) two sixes and a one? c) no sixes?

d) at least one six?

Question 16: Three dice and two coins are tossed simultaneously. What are the chances of receiving …

a) three sixes and two heads? b) no sixes and two heads? c) no sixes and no heads?

Question 17: Five cards are drawn from a deck of 52 cards:

a) Explain why the chances of pulling out five cards that are all hearts is: 1 12 11 10 9

0.05% 4 51 50× × ×49×48≈ .

b) Find the probability of pulling out five black cards.

c) Show that the probability of pulling out four Kings among the five cards is close to 0.002%.

d) Show that the probability of pulling out three Kings and two Queens is close to 0.001%.

Question 18: Consider the following magic square:

Player A chooses a number at random from the first row; player B chooses a number at random from the second row, and player C chooses a number at random from the third row.

a) What are the chances that player Bs number is higher than player As? b) What are the chances that player Cs number is higher than player Bs? c) What are the chances that player As number is higher than player Cs? [In this game of chance, B has the advantage over A, C has the advantage over B, and A has the advantage over C!]

Question 19:

a) Eleven numbers are arranged in a line. The first number is 0, the last number is 0, and every number in between is the average of its two neighbors. What are the 11 numbers and why?

b) Eleven numbers are arranged in a line. The first number is 0, the last number is 1, and every number in between is the average of its two neighbors. What are the 11 numbers and why?

Question 20: You are a game show contestant and the game show host presents to you 100 boxes. She tells you that inside one box lies a fabulous prize and all the remaining boxes are empty. You select a box at random and are about to open it when the host interrupts you and opens 98 boxes to reveal to you their emptiness. This leaves two boxes: the one you selected and one other.

You are now given the chance to “stick” with the box you first chose, or to “switch” and open instead the second box.

a) If you decide to stick, what are your chances of winning the prize? b) If you decide to switch, what are your chances of winning the game?

Suppose the game show host opens only 97 boxes. This leaves three boxes: the one you first selected and two others.

The host now gives you the choice to either “stick” with your original box or to switch to either one of the remaining boxes.

c) If you decide to stick, what are your chanced of winning the prize? d) If you switch to a different box, what now are your chances of winning? Question 21: In a game, if

a

outcomes are deemed “favorable” and the remaining

b

possible outcomes “unfavorable,” then folk may say – in horse racing circles in particular – that the odds in favor of winning are “

a

to

b

”, or alternatively that the odds against are “

b

to

a

.” For example, in rolling a die the odds in favor of rolling a 6 are 1:5. The odds against rolling a 5 or a 6 are 4:2 (which could be reduced to 2:1). In a horse race if the odds against a horse are 7:2, this means that bookies believe that the horse has only a 2

9 chance of winning. CORRECT or INCORRECT?

a) A bookie at a horse race says that the odds against a particular horse are 5:8. This means that the probability the horse will win the race is 8

13.

b) A game yields a 30% chance of a win. The odds against winning the game are thus 7:3.

c) In casting a die, the odds in favor of rolling a number smaller than 5 are 2:1. d) In tossing a coin twice, the odds against receiving two heads is 3:1.

Question 22:

Bag 1 contains 13 red balls and 14 blue balls. Bag 2 contains 12 red balls and 7 blue balls.

A bag is selected at random and a ball is pulled out of that bag at random. We are told that the ball is red.

Question 23: A die is tossed. What is the probability that the result is a number less than 4 if …

a) We are told no other information?

b) We are told that the result was an odd number? c) We are told that the result wasn’t 5?

d) We are told that the result wasn’t 1? Question 24:

a) Two ordinary dice are tossed. What is the probability of NOT getting a total of 7 or 11?

b) Two Sicherman dice are tossed. What is the probability of NOT getting a total of 7 or 11?

Question 25: One bag contains 4 red and 5 white balls. A second bag contains 3 red and 6 white balls. A ball is drawn from each bag. What is the probability that …

a) Both balls are white b) Both balls are red

c) One is white and the other is red

Question 26 One bag contains 2 red and 3 white balls. A second bag contains 3 red and 1 white balls. A ball is drawn from each bag. Suppose we are told that one ball chosen was red. What are the chances that the second ball is also red?

Question 27: A bag contains 5 red and 4 white balls. A ball is selected and then, without replacing the first ball, a second ball is selected. I tell you that the second ball is white. What is the probability that the first ball was white?

Question 28: You play a simple coin-tossing game. If the coin lands heads, you win $3. If lands tails, you must pay $1.

a) If you play this game 100 times, how much money do you expect to have? b) What is the expected value of this simple game?