11. Making Fair Decisions
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Chapter 6
11.
Making Fair Decisions
This lesson delves into the concept of making fair decisions using probability and various methods. It discusses scenarios like choosing a team in a sports game, deciding who gets a prize in a company, or even settling disputes among friends. Different methods are explored, such as drawing numbers out of a hat or using dice and coins. The key takeaway is that for a decision to be considered fair, each possible outcome should have an equal chance of occurring. For instance, if you're a team captain deciding who bats first, you could use a fair die roll to make the choice. Similarly, if a company wants to randomly award a prize to its employees, various fair methods are suggested to ensure that everyone has an equal chance of winning.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 7 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Making Fair Decisions
Slide of 10
Making decisions can be a difficult thing to do amidst a sea of information and variables. The goal of this lesson is to understand the use of probabilities in helping to sort through the given information and make fair decisions. Different scenarios such as sports and chess, a flip of a coin, a game of cards, and a prize giveaway will be analyzed in order to calculate fair decisions.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Making a Fair Decision
A local 3-on-3 beach volleyball tournament was hosted in South Beach. Heichi, Izabella, and Kevin pulled off a huge upset and won the tournament. The reward is a bicycle that flies as wind gusts propel the rider anywhere they so please to go. Everyone wants it, but, unfortunately, there is only one. Therefore, each player suggests a method to decide who gets to keep the highly coveted bicycle.
Considering the three suggestions, which method is fair?
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Flipping a Coin to Make a Fair Decision
Dominika and Emily's mom made a delicious red velvet cake the other day, but, unfortunately, only one piece is left in the fridge. Both girls are craving to eat it. Brazenly, they agreed that only one of them would get the whole piece. Therefore, they decided to design a coin flip game to determine who will be the lucky person to eat the last piece!
They came up with three game types and want to decide which one is fair.
- Game A: They flip a coin. If they get heads, then Dominika wins. If they get tails, then Emily wins.
- Game B: They flip a coin two times. If two heads appear, then Dominika wins. If two tails appear, then Emily wins. If one heads and one tails appear, then they play again.
- Game C: They flip a coin three times. If one heads and two tails appear, then Dominika wins. If three heads or three tails appear, then Emily wins. In other cases, they play again.
Try to determine which game leads to a fair decision. In order to do this, complete the following steps.
- List all possible outcomes for each game.
- Calculate the probability of Dominika winning and the probability of Emily winning. Are they the same or different?
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Fair Decision
A fair decision is a decision for which all events are equally likely to happen, meaning that the probabilities of the events are the same. Fair decisions can be made by using several methods, such as selecting names from a hat, flipping a coin, rolling a die, drawing a card from a standard deck of cards, using a random number generator, and so on.
Example
Suppose that the captains of two baseball teams A and B, have to decide which team bats first. To do so, the captains will roll one die each, simultaneously.
One example of a fair method and one example of an unfair method will be presented for this situation.
- Fair Method: If the sum of the numbers is less than 7, then Team A bats first. If the sum of the numbers is greater than 7, then Team B bats first. If it is 7, they play again.
- Unfair Method: If the product of the numbers is even, then Team A bats first. If the product of the numbers is odd, then Team B bats first.
Why
The fairness of the mentioned methods will be explained one at a time.
Fair Method
This method suggests rolling two dice and calculating the sum of the numbers obtained. Since rolling each die has 6 possible outcomes, the number of all possible combinations is 36.
Recall that if the sum of the numbers is less than 7, then Team A bats first. For a clearer understanding, highlight all the outcomes which satisfy this event.
P(TeamAbats) = Favorable outcomes/Total outcomes
▼
Substitute values and simplify
SubstituteII
Favorable outcomes= 15, Total outcomes= 36
P(TeamAbats) = 15/36
ReduceFrac
a/b=.a /3./.b /3.
P(TeamAbats) = 5/12
CalcQuot
Calculate quotient
P(TeamAbats) = 0.416666...
RoundDec
Round to 2 decimal place(s)
P(TeamAbats) ≈ 0.42
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.
P(TeamAbats) = Favorable outcomes/Total outcomes
▼
Substitute values and simplify
SubstituteII
Favorable outcomes= 27, Total outcomes= 36
P(TeamAbats) = 27/36
ReduceFrac
a/b=.a /9./.b /9.
P(TeamAbats) = 3/4
CalcQuot
Calculate quotient
P(TeamAbats) = 0.75
P(TeamBbats) = Favorable outcomes/Total outcomes
▼
Substitute values and simplify
SubstituteII
Favorable outcomes= 9, Total outcomes= 36
P(TeamBbats) = 9/36
ReduceFrac
a/b=.a /9./.b /9.
P(TeamBbats) = 1/4
CalcQuot
Calculate quotient
P(TeamBbats) = 0.25
Example
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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.
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b Calculate the probability of Dominika winning and round it to three decimal places.
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c Is the card game they are playing fair?
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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.
P(Two♡)=P(♡first)* P(♡second | ♡first)
SubstituteValues
Substitute values
P(Two♡)= 13/26* 12/25
▼
Evaluate
ReduceFrac
a/b=.a /13./.b /13.
P(Two♡)=1/2* 12/25
MultFrac
Multiply fractions
P(Two♡)=12/50
ReduceFrac
a/b=.a /2./.b /2.
P(Two♡)=6/25
P(Emily wins)=P(Two♡)+P(Two♠)
SubstituteValues
Substitute values
P(Emily wins)=6/25+6/25
AddFrac
Add fractions
P(Emily wins)=12/25
CalcQuot
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.
P(Dominika wins)= P(First face)* P(Second face | First face)
SubstituteValues
Substitute values
P(Dominika wins)= 6/26* 5/25
▼
Evaluate right-hand side
ReduceFrac
a/b=.a /2./.b /2.
P(Dominika wins)=3/13* 5/25
ReduceFrac
a/b=.a /5./.b /5.
P(Dominika wins)=3/13* 1/5
MultFrac
Multiply fractions
P(Dominika wins)=3/65
UseCalc
Use a calculator
P(Dominika wins)=0.046153...
RoundDec
Round to 3 decimal place(s)
P(Dominika wins)≈ 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)&=0.48 P(Dominika wins)&≈ 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.
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Using Simulations in Decision-Making
There are opportunities to make fair decisions in different contextual situations by the use of simulations.
Concept
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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)=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:& 1 Missed Free Throw:& 0 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.
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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.
a m=8, n=1
b m=8, n=2
c m=12, n=1
d 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 to 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 18. 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 to 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._nC_r=n!/r!(n-r)!
▼
Substitute values and evaluate
SubstituteII
n= 8, r= 2
_8C_2 = 8!/2!( 8- 2)!
SubTerm
Subtract term
_8C_2 = 8!/2!* 6!
Write as a product
_8C_2 = 8* 7* 6!/2* 1 * 6!
SplitIntoFactors
Split into factors
_8C_2 = 4* 2* 7* 6!/2* 1 * 6!
CrossCommonFac
Cross out common factors
_8C_2 = 4* 2* 7* 6!/2* 1* 6!
CancelCommonFac
Cancel out common factors
_8C_2 = 4* 7/1
Multiply
Multiply
_8C_2 = 28/1
DivByOne
a/1=a
_8C_2 = 28
Method II: Drawing Two Cards Without Replacement
Take eight spades ranging from 2 to 9 from a deck of 52 cards and assign numbers from 2 to 9 to the employees. After shuffling those cards, draw two cards in succession without replacing them.
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 112. 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.
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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 |
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 to 6 can be assigned to the members according to the number of games they won.
| Members | Games Won | Numbers Assigned |
|---|---|---|
| Ali | 2 | 1and2 |
| Maya | 3 | 3, 4,and5 |
| 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)=Number of Favorable Outcomes/Number of Possible Outcomes [1em] ⇓ P(A)=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 | 1and2 | 2/6≈ 0.33 |
| Maya | 3 | 3, 4,and5 | 3/6=0.5 |
| Diego | 1 | 6 | 1/6≈ 0.17 |
| Dylan | 0 | None | 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.
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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.
- P(X≤ 0.5)
- P(X≥0.75)
- P(0.2
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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.
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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.
P(Izabella wins) = Favorable Outcomes/Total Outcomes
SubstituteII
Favorable Outcomes= 9, Total Outcomes= 36
P(Izabella wins) = 9/36
ReduceFrac
a/b=.a /9./.b /9.
P(Izabella wins) = 1/4
CalcQuot
Calculate quotient
P(Izabella wins) = 0.25
P(Kevin wins) = Favorable Outcomes/Total Outcomes
▼
Substitute values and simplify
SubstituteII
Favorable Outcomes= 10, Total Outcomes= 36
P(Kevin wins) = 10/36
ReduceFrac
a/b=.a /2./.b /2.
P(Kevin wins) = 5/18
CalcQuot
Calculate quotient
P(Kevin wins) = 0.277777...
RoundDec
Round to 2 decimal place(s)
P(Kevin wins) ≈ 0.28
P(Heichi wins) = Favorable Outcomes/Total Outcomes
▼
Substitute values and simplify
SubstituteII
Favorable Outcomes= 17, Total Outcomes= 36
P(Heichi wins) = 17/36
CalcQuot
Calculate quotient
P(Heichi wins) = 0.472222...
RoundDec
Round to 2 decimal place(s)
P(Heichi wins) ≈ 0.47
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 | 5/15≈ 0.33 |
| Izabella wins | 5 | 15 | 5/15≈ 0.33 |
| Kevin wins | 5 | 15 | 5/15≈ 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.
P(Heichi wins) = Favorable Outcomes/Total Outcomes
SubstituteII
Favorable Outcomes= 1, Total Outcomes= 4
P(Heichi wins) = 1/4
CalcQuot
Calculate quotient
P(Heichi wins) = 0.25
P(Kevin wins) = Favorable Outcomes/Total Outcomes
SubstituteII
Favorable Outcomes= 2, Total Outcomes= 4
P(Kevin wins) = 2/4
ReduceFrac
a/b=.a /2./.b /2.
P(Kevin wins) = 1/2
CalcQuot
Calculate quotient
P(Kevin wins) = 0.5
Making Fair Decisions
Exercise 3.1
Davontay, Diego, and Dominika are siblings. Their parents want them to divide the household chores which includes doing the dishes, cleaning the bathroom, and vacuuming. Nobody ever wants to clean the bathroom. To come up with a way to decide which one must do it, they agree to use a standard deck of cards. They all propose different methods of selecting who is on bathroom duty.
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For a decision making processes to be fair, the probability for each sibling to get tasked with cleaning the bathroom has to be the same. Let's analyze each proposed method, one at a time.
Davontay's Method
In a standard deck of cards, there are 13 cards for each suit: 13 hearts, 13 spades, 13 diamonds, and 13 clubs. Removing the clubs leaves a total of 13+13+13= 39 possible outcomes when choosing a card randomly. Note that, there are the same number of outcomes for which either sibling would need to clean, 13. Thus, the probability for each of them cleaning the bathroom are the same. P = 13/39 = 1/3
Therefore, this method leads to a fair decision. Note that removing the clubs has no impact on whether or not the decision is fair, since the clubs represent no favorable outcome for any of the siblings. It just ensures that there is no tie.
Diego's Method
In a standard deck of cards there are four suits but only two colors — red and black — both used in exactly 26 cards out of the total 52. When drawing two cards, we have four possible outcomes. Let's show these outcomes with their probabilities by using a tree diagram. Recall that we can calculate the probability of an event by multiplying the probabilities along the corresponding branches.
Notice that whenever a card is drawn, then possible outcomes for the next card decrease by one. As we can see, if this method is used, the probabilities that Dominika has to do the cleaning are much higher than the probabilities of Davontay and the probabilities of Diego cleaning. Therefore, this is not a fair decision making process.
Dominika's Method
Remember that a standard deck of cards has 52 cards, out of which 13 are clubs. With this information we can calculate the probability that the first card is a club. Dominika cleans [1em] P(♣)=13/52=0.25 On the other hand, the probability of not drawing a club can also be calculated in a similar way. Since there are three other suits, there are 13 * 3 = 39 cards that are not clubs. With this in mind, we can calculate the probability of the first card not being a club and the second card being a club by using the Multiplication Rule of Probability. Diego cleans [1em] P(not ♣, ♣)=39/52 * 13/51≈ 0.191 [1em] Notice that each time a card is drawn the possible outcomes decrease by one. Finally, let's calculate the probability that three cards that are not clubs are drawn consecutively. Davontay cleans [1em] P(not ♣,not ♣,not ♣) = 39/52* 38/51* 37/50≈ 0.414 We can see that the probability for each sibling doing the bathroom cleaning under this set up are very different. Therefore, this is not a fair decision making process.
Conclusion
From our discussion, we know that only Davontay's method results in a fair decision making process.
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Making Fair Decisions
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