Unlocking the Secrets of Fair Decisions through Probability and Outcomes

$Favorable outcomes = 15$ , $Total outcomes = 36$

P (Team A bats) = \frac{1 5}{3 6}

$\frac{a}{b} = \frac{a / 3}{b / 3}$

P (Team A bats) = \frac{5}{1 2}

Calculate quotient

P (Team A bats) = 0.416666 \dots

Round to $2$ decimal place(s)

P (Team A bats) \approx 0.42

The probability of Team

B

batting first can be found in a similar way. First, highlight the outcomes satisfying the event of the sum of the numbers being greater than

7 .

Again, there are $15$ favorable outcomes for the event. Therefore, the probability that Team $B$ bats first is also about $0.42 .$ Since two events, Team $A$ batting first and Team $B$ batting first, have equal probabilities, this method is fair.

Unfair Method

This method suggests rolling two dice and multiplying the numbers obtained. Since rolling each die has $6$ possible outcomes, there are $36$ possible products.

According to this method, if the product of the numbers is even, then Team $A$ bats first. Hence, start by identifying the favorable outcomes which are the even numbers in the table.

There are

27

favorable outcomes for Team

A

out of

36

possible outcomes. With this information, the probability of Team

A

batting first can be calculated.

P (Team A bats) = \frac{Favorable outcomes}{Total outcomes}

▼

Substitute values and simplify

$Favorable outcomes = 27$ , $Total outcomes = 36$

P (Team A bats) = \frac{2 7}{3 6}

$\frac{a}{b} = \frac{a / 9}{b / 9}$

P (Team A bats) = \frac{3}{4}

Calculate quotient

P (Team A bats) = 0.75

Now, the probability of Team

B

batting first can be found in a similar way. To do so, highlight all the odd numbers in the table.

As can be seen, the number of favorable outcomes for Team

B

9

out of

36

possible outcomes. Using this information, the probability of Team

B

batting first can be calculated.

P (Team B bats) = \frac{Favorable outcomes}{Total outcomes}

▼

Substitute values and simplify

$Favorable outcomes = 9$ , $Total outcomes = 36$

P (Team B bats) = \frac{9}{3 6}

$\frac{a}{b} = \frac{a / 9}{b / 9}$

P (Team B bats) = \frac{1}{4}

Calculate quotient

P (Team B bats) = 0.25

Since the probability of Team

A

batting first is three times greater than the probability of Team

B

batting first, this method is not fair.

In this lesson, events and their probabilities will be analyzed to determine whether presented methods are fair or not.

Example

Exploring the Fairness of Drawing a Card

In the last daring coin flip, Emily won and devoured the red velvet cake. The girls enjoyed the game so much that they decided to continue playing these games of probability just for fun. This time they will use a standard deck of cards.

Dominika will first remove all of the clubs and diamonds from the deck. Emily will then draw two cards in succession without replacing them. If she draws two cards with the same suit, then Emily wins. If she draws two face cards, then Dominika wins. For any other outcome, they draw again.

Try to determine if the game leads to a fair decision. To do so, complete the following steps.

a Calculate the probability of Emily winning and round it to two decimal places.

b Calculate the probability of Dominika winning and round it to three decimal places.

c Is the card game they are playing fair?

Hint

a How many cards in the game have the same suit?

b How many face cards are there in the game?

c Compare the probabilities calculated in Part A and Part B.

Solution

a When all the clubs and diamonds are removed from a deck of

52

cards, there are

26

cards remaining —

13

hearts and

13

spades. Next, two cards are drawn in succession without being replaced. If both of those cards are hearts or spades, Emily wins.

Note that the event of drawing two hearts and the event of drawing two spades are mutually exclusive, therefore, the probability on the right-hand side equals the sum of the individual probabilities.

Since there are

26

cards in total, the probability of getting a heart on the first draw is

\frac{1 3}{2 6} .

If the first card is drawn is a heart, then only

12

hearts remain in the deck, and because the cards are not being replaced there are

25

remaining cards in total. Hence, the probability of drawing a second heart is

\frac{1 2}{2 5} .

P (♡ first) P (♡ second ∣ ♡ first) = \frac{1 3}{2 6} = \frac{1 2}{2 5}

By the Multiplication Rule of Probability for dependent events, the probability of drawing two hearts equals the product of the two probabilities written above.

P (Two ♡) = P (♡ first) \cdot P (♡ second ∣ ♡ first)

SubstituteValues

Substitute values

P (Two ♡) = \frac{1 3}{2 6} \cdot \frac{1 2}{2 5}

▼

Evaluate

$\frac{a}{b} = \frac{a / 1 3}{b / 1 3}$

P (Two ♡) = \frac{1}{2} \cdot \frac{1 2}{2 5}

MultFrac

Multiply fractions

P (Two ♡) = \frac{1 2}{5 0}

$\frac{a}{b} = \frac{a / 2}{b / 2}$

P (Two ♡) = \frac{6}{2 5}

Since there are also

13

spades out of

26

cards, the probability of drawing two spades is the same as the probability of drawing two hearts, which is

\frac{6}{2 5} .

Finally, the probability of Emily winning can be computed by adding the probabilities of drawing two hearts and drawing two spades.

P (Emily wins) = P (Two ♡) + P (Two ♠)

SubstituteValues

Substitute values

P (Emily wins) = \frac{6}{2 5} + \frac{6}{2 5}

AddFrac

Add fractions

P (Emily wins) = \frac{1 2}{2 5}

Calculate quotient

P (Emily wins) = 0.48

b Recall that two cards are drawn without replacement out of

26

cards in which

13

are hearts and

13

are spades. Dominika wins if both cards are face cards. Recall that face cards are the jack, queen, and king cards. Therefore, there are

6

face cards in the game —

3

face cards of hearts and

3

face cards of spades.

Since there are

6

face cards out

26

cards, the probability of getting a face card in the first draw is

\frac{6}{2 6} .

If the first card is drawn is a face card, then only

5

face cards remain in the deck, and because the cards are not being replaced there are

25

remaining cards in total. Hence, the probability of drawing a second face card is

\frac{5}{2 5} .

P (First face) P (Second face ∣ First face) = \frac{6}{2 6} = \frac{5}{2 5}

As in Part A, by multiplying these values, the probability of Dominika winning can be calculated.

P (Dominika wins) = P (First face) \cdot P (Second face ∣ First face)

SubstituteValues

Substitute values

P (Dominika wins) = \frac{6}{2 6} \cdot \frac{5}{2 5}

▼

Evaluate right-hand side

$\frac{a}{b} = \frac{a / 2}{b / 2}$

P (Dominika wins) = \frac{3}{1 3} \cdot \frac{5}{2 5}

$\frac{a}{b} = \frac{a / 5}{b / 5}$

P (Dominika wins) = \frac{3}{1 3} \cdot \frac{1}{5}

MultFrac

Multiply fractions

P (Dominika wins) = \frac{3}{6 5}

UseCalc

Use a calculator

P (Dominika wins) = 0.046153 \dots

Round to $3$ decimal place(s)

P (Dominika wins) \approx 0.046

c To decide if the game is fair or not, the probabilities found in Part A and Part B should be compared. If they are equal, then the game is considered fair.

P (Emily wins) P (Dominika wins) = 0.48 \approx 0.046

As can be seen, the probability of Emily winning is greater than the probability of Dominika winning.

0.48 > 0.046

Oh snaps, the game is not fair! Also, note that if two face cards of the same suit are drawn, both Emily and Dominika will win, which is ambiguous. This also says that the game is not well set.

Discussion

Using Simulations in Decision-Making

There are opportunities to make fair decisions in different contextual situations by the use of simulations.

Concept

Simulation

A simulation is a model that imitates a real-life process or situation. Simulations are often used as probability models to make predictions about real-life events. Consequently, the experimental probability of an event occurring can be estimated by simulating the event.

P (event) = \frac{Number of times event occurs}{Number of trials}

In general, simulations are used when actual trials of some experiment are impossible or unreasonable to conduct.

Using a Random Number Generator

For example, Mark and Tadeo want to eat dinner at different restaurants and they are not able to come together on a decision. It so happens that they both really like basketball. They then came up with an idea to make a free throw to determine the person who will choose the restaurant.

External credits: @macrovector

Mark only recently began practicing free throws, so the probability that Mark makes a free throw is

0.5

— this is the success rate. Unfortunately, there is no basketball court nearby, so they decide to simulate the free throws using a random generator.

Made Free Throw : Missed Free Throw : 10

If the random number generator produces

1

— meaning that Mark made a free throw — then they will go to the restaurant that Mark wants. Otherwise, they will go to the restaurant that Tadeo wants.

Example

Awarding a Prize to a Random Winner

A company with $m$ employees wants to award a prize to $n$ employee(s) at random — it has been a good year. Propose a fair way to choose the random winners for the given values of $m$ and $n .$

m = 8,

n = 1

m = 8,

n = 2

m = 12,

n = 1

m = 246,

n = 5

Answer

a There are a few possible methods.

Method I: Assign each person a number, write the numbers on slips of paper, and then draw one of them out of a hat.
Method II: Assign each person a number from $1$ to $8$ and then roll a fair octahedron die.
Method III: Flipp three unbiased coins and assign each employee one of $8$ possible outcomes.

b There are a few possible methods.

Method I: Assign each person a number, write the numbers on pieces of paper, and then draw two pieces out of a hat.
Method II: Eight spades ranging from $2$ to $9$ are taken from a deck. The numbers are assigned to employees and then two cards are drawn without being replaced.

c Sample Method: The employees are assigned numbers

01,

02,

03,

04,

05,

06,

11,

12,

13,

14,

15,

16 .

Then the tens digit is picked by flipping a coin (heads -

0,

tails -

1

) and the second digit is determined by rolling a die.

d Sample Method: Random generator is used to get a number from

1

246

assigned to employees.

Hint

a Assign a number to each employee. Think of how that number can be randomly chosen.

b Think of whether the method of drawing cards from a deck or smaller group of cards can be used.

c The method leads to a fair decision if each possible outcome is equally likely to happen.

d Can a random generator be used?

Solution

a In the first case,

1

person randomly chosen out of

8

employees gets the prize. To pick that employee, various methods can be used.

Method I: Drawing a Number Out of a Hat

One possible method is to assign a number to each employee. Then the numbers should be written on pieces of paper and put in a hat. After shaking the hat, one piece of paper is randomly picked and the employee with the matching number wins the prize. This method has eight possible outcomes.

{1, 2, 3, 4, 5, 6, 7, 8}

Note that each number has an equal chance of being chosen. That means each employee has the same probability of winning the prize, which is

\frac{1}{8} .

Therefore, this method is fair.

Method II: Rolling a Fair Octahedron Die

Another possible method is to use an octahedron die — an eight-sided die. Each employee is assigned a number from

1

8,

and then a fair octahedron die is rolled. All possible outcomes of rolling a fair octahedron die can be listed as follows.

{1, 2, 3, 4, 5, 6, 7, 8}

Similar to the previous method, since the die is fair, each outcome is equally likely to happen. Therefore, this method also results in a fair decision.

Method III: Flipping Three Unbiased Coins

The third method is to use three unbiased coins. By flipping three unbiased coins,

8

possible outcomes can be obtained. Note that the order in which heads or tails appear matters, meaning that

{H, H, T}

and

{H, T, H}

are two different outcomes.

{H, H, H}, {H, H, T}, {H, T, H}, {T, H, H}, {T, T, H}, {T, H, T}, {H, T, T}, {T, T, T}

If each of these outcomes is assigned to one employee, then a fair decision can be made, since each outcome is equally likely. The employee with the matching outcome wins the prize.

b In this case,

2

people will be chosen out of

8

employees to be awarded a prize. To be sure this is the case, many different methods can be proposed. Here, two will be presented.

Method I: Drawing Two Numbers Out of a Hat

Numbers assigned to each employee are written on different pieces of paper and put into a hat. Then, two pieces of paper are randomly picked, determining the winners of the prize. Note that the order of chosen employees is not important. Hence, the total number of possible outcomes can be calculated using the Combination Formula.

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

▼

Substitute values and evaluate

$n = 8$ , $r = 2$

_{8} C_{2} = \frac{8 !}{2 ! ( 8 - 2 ) !}

SubTerm

Subtract term

_{8} C_{2} = \frac{8 !}{2 ! \cdot 6 !}

Write as a product

_{8} C_{2} = \frac{8 \cdot 7 \cdot 6 !}{2 \cdot 1 \cdot 6 !}

SplitIntoFactors

Split into factors

_{8} C_{2} = \frac{4 \cdot 2 \cdot 7 \cdot 6 !}{2 \cdot 1 \cdot 6 !}

CrossCommonFac

Cross out common factors

_{8} C_{2} = \frac{4 \cdot 2 \cdot 7 \cdot 6 !}{2 \cdot 1 \cdot 6 !}

CancelCommonFac

Cancel out common factors

_{8} C_{2} = \frac{4 \cdot 7}{1}

Multiply

_{8} C_{2} = \frac{2 8}{1}

DivByOne

$\frac{a}{1} = a$

_{8} C_{2} = 28

Therefore, there are

28

possible ways to choose

2

employees out of

8 .

Assuming that the slips of paper are chosen randomly, each group of

2

employees has equal chances of being selected.

P (Outcome) = \frac{1}{2 8}

Since each possible outcome is equally likely, this method results in a fair decision.

Method II: Drawing Two Cards Without Replacement

Take eight spades ranging from

2

9

from a deck of

52

cards and assign numbers from

2

9

to the employees. After shuffling those cards, draw two cards in succession without replacing them.

For the first card drawn, each card has a probability of being chosen equally with a probability of

\frac{1}{8} .

Since the card is not being replaced, when drawing the second card, there are only

7

cards left. Therefore, for each card the probability of being drawn on the second draw equals

\frac{1}{7} .

P (being the first drawn card) P (being the second drawn card) = \frac{1}{8} = \frac{1}{7}

By multiplying these values, the probability of each of the two cards being drawn can be calculated.

P (Outcome) = \frac{1}{8} \cdot \frac{1}{7} ⇓ P (Outcome) = \frac{1}{5 6}

Although cards have different probabilities of appearing on the first and second draw, each group of

2

cards has an equal chance of being selected. Therefore, each possible outcome is equally likely. This means that this method leads to a fair decision.

c One possible method to select

1

person out of

12

is to assign to the employees the following numbers.

To determine who will win the prize, first flip a coin and then roll a six-sided die. If the coin lands on heads, the tens digit will be

0 .

If it lands on tails, the tens digit will be

1 .

The second digit will be determined by rolling a die. Here is the list of all possible outcomes of this event.

{01, 02, 03, 04, 05, 06, 11, 12, 13, 14, 15, 16}

Note that there is a total of

12

possible outcomes. The probability of each outcome is

\frac{1}{1 2} .

Since each possible outcome is equally likely, this method results in a fair decision.

d Finally,

5

employees will be selected among

246

employees. Note that for

246

employees it might be difficult to develop a method such as drawing a number out of a hat, rolling dice, flipping coins, or drawing cards. Therefore, in this case, a random number generator can be used.

Sample Method

Start by assigning a number from $1$ to $246$ to each employee and then use a technology-based random number generator to select one number. As an example, a graphing calculator can be used. First, push the $MATH$ button. Then, scroll to the right to the PRB menu and choose the fifth option called randInt(.

The function randInt $(a, b, c)$ outputs $c$ integers in the range from $a$ to $b,$ inclusive. Since $5$ integers from $1$ to $246$ are required in this case, evaluate randInt $(1, 246, 5)$ in the calculator.

The random number generator provides each number from $1$ to $246$ an equal chance of being chosen, so each employee has an equal probability of winning the prize. Therefore, this method is fair.

Example

Solving Real World Problems Fairly

A chess club of $4$ members decided gift a hand-carved wooden chess set to one of its members. The club wants each member to have a chance of winning this gift based on how many games they have won this week.

Members	Number of Games Won
Ali	$2$
Maya	$3$
Diego	$1$
Dylan	$0$

Describe a fair method to decide who wins the chess set. Use probabilities to explain why the proposed method is fair for the members listed.

Answer

Sample Method: Assign integers from $1$ to $6$ to each member according to the number of games they won. Then roll a die to determine the winner.
Why it is Fair: The outcomes of rolling a die are equally likely. Additionally, the probability of winning the gift for each member is proportional to the number of games they have won.

Hint

Find a method of assigning outcomes so that the chance of winning is proportional to the number of games won.

Solution

A method of choosing the winner of the gift will be fair if members who won more games will have a better chance of winning the chess set. First, calculate the total number of games won by all the members.

2 + 3 + 1 + 0 = 6

Since there are

6

games in total, integers from

1

6

can be assigned to the members according to the number of games they won.

Members	Games Won	Numbers Assigned
Ali	$2$	$1 and 2$
Maya	$3$	$3, 4, and 5$
Diego	$1$	$6$
Dylan	$0$	None

Next, a standard six-sided die can be rolled to decide who wins the set. Recall that there are six possible outcomes of rolling a six-sided die and all outcomes are equally likely.

Possible Outcomes : {1, 2, 3, 4, 5, 6}

Therefore, the probability of winning the set for each member equals the quotient of the number of favorable outcomes to the number of possible outcomes. For example, for Ali to win, there are two favorable outcomes,

1

and

2 .

Hence, the probability

P (A)

that Ali wins the set can be calculated as follows.

P (A) = \frac{Number of Favorable Outcomes}{Number of Possible Outcomes} ⇓ P (A) = \frac{2}{6}

Using the same procedure, the probability of winning the set can be calculated for each member.

Members	Games Won	Numbers Assigned	Probability of Winning the Chess Set
Ali	$2$	$1 and 2$	$\frac{2}{6} \approx 0.33$
Maya	$3$	$3, 4, and 5$	$\frac{3}{6} = 0.5$
Diego	$1$	$6$	$\frac{1}{6} \approx 0.17$
Dylan	$0$	$None$	$\frac{0}{6} = 0$

Notice that the probability of winning the gift for each member is proportional to the number of games they have won this week. Therefore, rolling a die can be considered as a fair method to decide who gets the gift.

Explore

Random Number Generator Between $0$ and $1$

Let $X$ be a random number between $0$ and $1$ produced by a random number generator. In the applet, a random number is represented by the point on the number line.

Investigate the distribution of

X

on the interval

[0, 1]

by calculating the following probabilities.

$P (X \leq 0.5)$
$P (X \geq 0.75)$
$P (0.2 < X < 0.8)$

Think of whether it is possible to use this random number generator to make fair decisions.

Closure

Studying the Fairness of Different Methods

At the beginning of the lesson it has been said that the team of Heichi, Izabella, and Kevin won the local $3 -$ on $- 3$ beach volleyball tournament and there is a bicycle in the prize package. All players want it, but there is only one. Therefore, each player proposed a method to decide who gets to keep the new bicycle.

Which player's method is fair?

Hint

Begin by identifying all possible outcomes of the experiments in each method. Then, calculate the probabilities to determine the fair method(s).

Solution

Each method will be examined one at a time.

Heichi's Method

Heichi suggests rolling two dice and multiplying the numbers obtained. Since rolling each die has $6$ possible outcomes, there are $36$ possible products.

According to Heichi's method, if the product of the numbers is odd, then Izabella keeps the bicycle. Hence, the outcomes which are the odd numbers should be identified in the table.

There are

9

such outcomes, which is the number of favorable outcomes for Izabella out of

36

possible outcomes. With this information, the probability of Izabella getting the bicycle can be calculated.

P (Izabella wins) = \frac{Favorable Outcomes}{Total Outcomes}

$Favorable Outcomes = 9$ , $Total Outcomes = 36$

P (Izabella wins) = \frac{9}{3 6}

$\frac{a}{b} = \frac{a / 9}{b / 9}$

P (Izabella wins) = \frac{1}{4}

Calculate quotient

P (Izabella wins) = 0.25

Next, the probability of Kevin winning the bicycle can be found in a similar way. First, highlight all the numbers that are even and greater than

15

in the table.

As can be counted, the number of favorable outcomes for Kevin is

10

out of

36

possible outcomes. Using this information, the probability that Kevin wins the bicycle can be determined.

P (Kevin wins) = \frac{Favorable Outcomes}{Total Outcomes}

▼

Substitute values and simplify

$Favorable Outcomes = 10$ , $Total Outcomes = 36$

P (Kevin wins) = \frac{1 0}{3 6}

$\frac{a}{b} = \frac{a / 2}{b / 2}$

P (Kevin wins) = \frac{5}{1 8}

Calculate quotient

P (Kevin wins) = 0.277777 \dots

Round to $2$ decimal place(s)

P (Kevin wins) \approx 0.28

Finally, the probability of Heichi winning can be found. He will win if the product is even and less than

15 .

Start by identifying those numbers in the table.

There are

17

favorable outcomes for Heichi out of

36

possible outcomes. With this information, the probability that Heichi wins can be found.

P (Heichi wins) = \frac{Favorable Outcomes}{Total Outcomes}

▼

Substitute values and simplify

$Favorable Outcomes = 17$ , $Total Outcomes = 36$

P (Heichi wins) = \frac{1 7}{3 6}

Calculate quotient

P (Heichi wins) = 0.472222 \dots

Round to $2$ decimal place(s)

P (Heichi wins) \approx 0.47

To determine whether the Heichi's method is fair, compare the probabilities of getting the bicycle for each person.

P (Izabella wins) P (Kevin wins) P (Heichi wins) = 0.25 \approx 0.28 \approx 0.47

Since the probabilities are not the same, each player is not equally likely to get the bicycle. Therefore, Heichi's method is not fair.

Izabella's Method

Izabella suggested writing each player's name $5$ times on slips of paper and putting all $15$ pieces in a bowl. Then one slip is randomly picked, which determines who gets the bicycle. To investigate if this method is fair, begin by calculating the probability of keeping the bicycle for each player, then analyze the results.

	Favorable Outcomes	Total Outcomes	Probability
Heichi wins	$5$	$15$	$\frac{5}{1 5} \approx 0.33$
Izabella wins	$5$	$15$	$\frac{5}{1 5} \approx 0.33$
Kevin wins	$5$	$15$	$\frac{5}{1 5} \approx 0.33$

It was obtained that each player is equally likely to get the bicycle. Therefore, Izabella's method is fair.

Kevin's Method

Kevin suggested flipping a coin two times. To analyze his method, make a table of all possible outcomes.

According to Kevin's method, if two heads show, then Heichi gets the bicycle. Note that there is only

1

outcome that two heads show out of

4

possible outcomes. Using this information, the probability that Heichi wins can be calculated.

P (Heichi wins) = \frac{Favorable Outcomes}{Total Outcomes}

$Favorable Outcomes = 1$ , $Total Outcomes = 4$

P (Heichi wins) = \frac{1}{4}

Calculate quotient

P (Heichi wins) = 0.25

Kevin also said that if two tails show, then Izabella gets the bicycle. Similarly, there is only

1

outcome that two tails turn up out of

4

possible outcomes. Therefore, the probability that Izabella keeps the bicycle is also

0.25 .

P (Izabella wins) = 0.25

Finally, Kevin proposed that if they land on one heads and one tails, then he gets the bicycle. Identify those outcomes in the table.

There are

2

favorable outcomes for Kevin out of

4

total outcomes. With this information, the probability that Kevin gets the bicycle can be calculated.

P (Kevin wins) = \frac{Favorable Outcomes}{Total Outcomes}

$Favorable Outcomes = 2$ , $Total Outcomes = 4$

P (Kevin wins) = \frac{2}{4}

$\frac{a}{b} = \frac{a / 2}{b / 2}$

P (Kevin wins) = \frac{1}{2}