HSS.MD.B.6Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.MD.B.6: Using Probability to Make Fair Decisions
In plain English: HSS.MD.B.6 is an advanced (+) Common Core statistics and probability standard that asks students to use probabilities to make fair decisions, for example by drawing lots or using a random number generator. Students check that a method gives every person the same chance, or an agreed share, and design methods that guarantee it. It is usually taught in a statistics course or Precalculus.
(+) Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator).
Common Core State Standards for Mathematics · Domain: Using Probability to Make Decisions (MD) · Cluster: Use probability to evaluate outcomes of decisions Also written as HSS-MD.B.6 or S-MD.6 · Official standard
Students learn that a decision is fair, in the probability sense, when the method gives every person the same chance of being chosen, or a chance that matches a share everyone agreed to in advance. They test common methods, such as flipping coins, rolling dice, drawing lots and using a random number generator, by listing equally likely outcomes and computing each person's probability.
When a method is unfair, students repair it: they assign the same number of equally likely outcomes to each person and repeat the process for leftover outcomes. The lesson also shows that drawing lots in turn is fair for every position, and how to build a weighted but fair draw.
Learning Objectives
By the end of this lesson, students will be able to:
Explain what makes a decision method fair in terms of probability
Decide whether a method that uses coins, dice, cards, lots or a random number generator is fair by computing each person's probability
Design a fair method for any number of people, using a re-roll or redraw for leftover outcomes when needed
Show that drawing lots without replacement gives every position the same probability
Design a fair weighted draw in which each person's chance matches an agreed share
Prior Knowledge Required
Students should already be comfortable with:
Uniform probability models with equally likely outcomes 7.SP.C.7
Sample spaces for compound events from lists, tables and tree diagrams 7.SP.C.8
Independent events and the product rule HSS.CP.A.2
"Our class has one free ticket to a concert. List three ways we could decide who gets it. Which of your methods are fair, and what do you mean by fair?"
Collect methods on the board, such as "draw a name from a hat", "the teacher picks", "first to answer a question", "a random number on a calculator". Push students to say what fair means. Settle on a working definition: a method is fair if every person has the same probability of being chosen. Note that some situations call for agreed, unequal shares (for example, more raffle entries for more volunteer hours), which the lesson returns to later.
Direct Instruction20 minutes
Give the steps for checking or designing a fair method:
Agree on what fair means here: equal chances, or chances that match an agreed share.
Use a device with equally likely outcomes: a fair coin, a fair die, well-mixed slips of the same size, or a random number generator.
List the equally likely outcomes and assign them to people so that each person gets the same number (or the agreed share).
Handle leftover outcomes: if the outcomes do not divide evenly, assign the extras to "repeat the process".
Compute each person's probability to confirm, then announce the method before running it once.
Testing a method: two coins for three people
Ana, Ben and Cal flip two coins: two heads picks Ana, two tails picks Ben, and one of each picks Cal. Is this fair? Repair it.
Equation: P(Ana) = 1/4, P(Ben) = 1/4, P(Cal) = 2/4 = 1/2: unfair. Fair rule: HH Ana, HT Ben, TH Cal, TT flip again, so each has 1/3
Using a re-roll: one die for four people
Four people number themselves 1-4 and roll one die, rolling again on a 5 or 6.
Equation: P(person 1) = (1/6)/(1 - 2/6) = (1/6)/(4/6) = 1/4 for each person
Drawing lots in turn
Five people draw straws, one at a time without replacement, and one straw is short. Is drawing first better or worse?
Equation: P(3rd person) = (4/5)(3/4)(1/3) = 1/5, and every position has probability 1/5
A fair weighted draw with a random number generator
Three volunteers worked 6, 3 and 1 hours and agree that one prize should be drawn in proportion to hours. Use a generator of whole numbers from 1 to 10.
Equation: 1-6: first volunteer (0.6), 7-9: second (0.3), 10: third (0.1)
A method that looks fair but is not
Two players take turns rolling one die, and the first to roll a 6 goes first in the game. Player A rolls first.
Equation: P(A) = (1/6)/(1 - 25/36) = 6/11, about 0.55, so the method favors A
Use Diagram 1 with Example 1: the tree shows four equally likely outcomes, and the unfair rule gives Cal two of them. Use Diagram 2 with Example 3: the person who draws third can only get the short straw if the first two draws were long, and the product of the branch probabilities is still 1/5. For Example 5, ask for a repair: for instance, each player rolls once, the higher roll goes first, and ties roll again.
Guided Practice15 minutes
Pairs judge four methods and repair any that are unfair. After each, one pair explains its probabilities to the class.
Choose one of three people with one die: 1-2, 3-4 or 5-6. (Fair: each 2/6 = 1/3.)
Choose one of three people with the sum of two dice: 2-4, 5-7 or 8-12. (Unfair: 6/36, 15/36 and 15/36.)
Choose one of five people with a calculator command that returns a random whole number from 1 to 5. (Fair: each 1/5.)
Choose one of three people by drawing a card: hearts, diamonds or a black card. (Unfair: 1/4, 1/4 and 1/2; repair by redrawing on spades.)
Listen for pairs that treat the sums 2 through 12 as equally likely, and for pairs that count outcomes that are not equally likely.
Independent Practice10-15 minutes
Students work alone on three problems:
Design a fair way to choose one of 9 people with two dice and no re-rolls. (36 ordered outcomes, 4 for each person, so each has 4/36 = 1/9.)
A teacher picks a student with a random number generator set to 1-30, but the class now has 32 students. What is unfair, and how should the teacher fix it? (Students 31 and 32 can never be chosen; set the range to 1-32.)
Three team members sold 12, 8 and 5 tickets and agree that the prize draw should be proportional to tickets sold. Give each person's probability and a random number range for each. (0.48, 0.32 and 0.20; 1-12, 13-20, 21-25.)
Closure5 minutes
Exit ticket: To choose one of three people, a student flips a coin until the first head. One flip picks Ari, two flips picks Bea, and three or more flips picks Cy. Is this fair? Give each probability. (No: 1/2, 1/4 and 1/4.)
Differentiation Strategies
For Struggling Students
Have students list every equally likely outcome in a table or tree before assigning them to people
Use physical devices first (coins, dice, slips) and record 30 trials, then compare the results with the computed probabilities
Give a checklist: Are the outcomes equally likely? Does each person get the same number? What happens to leftovers?
For Advanced Students
Ask students to find the probability that a re-roll method needs more than two rolls, and the expected number of rolls
Have students prove that drawing lots in turn gives every position probability 1/n for any n
Ask how to make a fair choice between two people with a coin that lands heads 60% of the time
Assessment Guidance
What to Look For
Check that students base every probability on outcomes that are equally likely, not on labels such as "sum of 7" or "one head". A fair method gives every person the same number of equally likely outcomes, and students should say what happens to leftover outcomes. For drawing lots, look for the product of branch probabilities for later positions, not the answer 1/(number of slips left). For weighted draws, check that the shares add to 1 and that the random number ranges have the right sizes.
02
Classroom Activities
3 Activities
1
Is It Fair? Station Rotation
25 minGroups of 3-4
Six stations each describe a method for making a choice, with the device needed to run it. Groups predict whether the method is fair, compute each person's probability, run it 30 times and compare the results with their computation.
The Six Stations
Station 1: Choose one of 3 people with the sum of two dice: 2-5, 6-8 or 9-12 (10/36, 16/36, 10/36: unfair)
Station 2: Choose one of 4 people by the suit of a card drawn from a shuffled deck (each 13/52 = 1/4: fair)
Station 3: Choose one of 2 people: one calls heads or tails while a fair coin is in the air (each 1/2: fair)
Station 4: Choose one of 6 people with a spinner of 6 equal sections (each 1/6: fair)
Station 5: Choose one of 2 people with one die: 1-4 or 5-6 (4/6 and 2/6: unfair)
Station 6: Three people each draw one of 3 folded slips from a bag in turn; the marked slip wins (each 1/3: fair)
Procedure
At each station, write a prediction (fair or unfair) before computing
Compute each person's probability from equally likely outcomes
Run the method 30 times and record how often each person is chosen
For each unfair method, write a repaired version using the same device
Modification for Distance Learning
Students use free online dice, coin, card and spinner simulators, run each station 30 times at home and enter their counts in a shared class table.
2
Fair Assignments with a Random Number Generator
20 minPairs, then whole class
The class uses a random number generator to assign 24 students to 4 project topics so that each topic gets 6 students and every student has the same chance of each topic. Pairs then critique two shortcut methods.
Procedure
Number the students 1-24 from the class list
Use a random number generator for whole numbers from 1 to 24. Assign the first 6 new numbers to Topic A, the next 6 to Topic B, and so on, skipping any number already used
Pairs explain why skipping repeats keeps the method fair
Pairs judge two shortcuts: (1) the first 6 students to raise a hand get Topic A; (2) each student rolls a die and rolls again on a 5 or 6, with 1-4 naming the topic
Discussion Questions
Shortcut 2 gives each student probability 1/4 for each topic. Why might the topics still not end up with 6 students each?
Why is it important to announce the method before generating any numbers?
Is it fair to run the generator again if the result "looks unfair", such as three friends on the same topic?
Extension Variation
Connect to experiments: random assignment of subjects to treatments uses the same process. Pairs describe how they would randomly assign 20 plants to two fertilizers, 10 plants each.
3
A Fair Choice from an Unfair Device
20 minPairs
A thumbtack tossed on a desk lands point up or point down, and the two outcomes are not equally likely. Pairs estimate the chance of point up, then test a method for making a fair choice between two people with the thumbtack: toss twice; up-down picks Person A, down-up picks Person B, and a matching pair means toss twice again.
Procedure
Toss the thumbtack 40 times and estimate p = P(point up)
Run the two-toss method until 20 decisions have been made, recording who is chosen each time and how many pairs of tosses were needed
Using your estimate of p, write the probability of up-down, of down-up and of a matching pair
Teacher check with p = 0.6: up-down and down-up each have probability 0.24, and a matching pair has probability 0.52
Discussion Questions
Why are up-down and down-up equally likely even though up and down are not?
Why must a matching pair be thrown out rather than given to one person?
Why does the method need more tosses when p is far from 0.5?
Modification for Distance Learning
Students use a bottle cap, or a spreadsheet that returns "up" when a random decimal is below 0.6, as the unequal device.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Testing and Repairing a Coin Method
Two fair coins give four equally likely outcomes, each with probability 1/4. The unfair rule gives Cal two outcomes, so Cal has probability 1/2. The fair rule gives each friend one outcome and repeats on TT, so each friend has probability 1/3.
Diagram 2: Drawing Lots Is Fair in Every Position
For a later person to draw the short straw, every earlier draw must be long. Multiplying along the path gives 1/5 for each of the five positions, so the order of drawing does not matter.
04
Homework Assignment
~30 min
HSS.MD.B.6 Homework: Making Fair Decisions
Directions: For each method, list the equally likely outcomes (or draw a tree), give each person's probability, and state whether the method is fair. When a method is unfair, describe a repaired method and show that it is fair.
Part 1: Is the Method Fair? (Problems 1-3)
Four roommates roll two dice to decide who gets the largest bedroom: a sum of 2-5 picks Dev, 6-7 picks Eli, 8-9 picks Fay and 10-12 picks Gus. (a) Find each roommate's probability. (b) Is the method fair? (c) Assign the 36 outcomes of two dice so that the method becomes fair.
Eight friends draw folded slips from a hat one at a time without replacement. Two slips are marked, and the two friends who draw them wash the dishes. (a) Find the probability that the first person to draw is chosen. (b) Find the probability that the second person is chosen. (c) Does drawing order matter?
Two teams decide who kicks off by taking turns flipping a fair coin; the first team to flip heads kicks off, and Team A flips first. (a) Find the probability that Team A kicks off. (b) Is the method fair? (c) Describe a fair method that uses the same coin.
Part 2: Designing Fair Methods (Problems 4-6)
Design a fair method that uses two dice to choose one of 7 club members. Explain what happens to leftover outcomes and show that each member has probability 1/7.
Three volunteers worked 10, 6 and 4 hours. They agree that one gift card should be drawn in proportion to hours worked. (a) Give each volunteer's probability. (b) Describe a draw that uses a random number generator for whole numbers from 1 to 20.
A coach uses a random number generator that returns a decimal from 0 up to 1 to choose one of five players for a penalty kick: a number below 0.1 picks Ana, from 0.1 up to 0.3 picks Bo, from 0.3 up to 0.5 picks Cy, from 0.5 up to 0.8 picks Dee, and from 0.8 up to 1 picks Eli. (a) Find each player's probability. (b) Is the method fair? (c) Change the cutoffs to make it fair.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Equally Likely Outcomes
All outcomes listed or shown in a tree, and they are equally likely
Outcomes listed but some not equally likely
No sample space
Probabilities
Each person's probability correct
Method correct with one error
Probabilities missing or guessed
Fairness Judgment
Correct verdict justified by the probabilities
Correct verdict without justification
Incorrect verdict
Repaired or Designed Method
Method is fair, handles leftovers and is shown to be fair
Method is fair but not justified
Method missing or unfair
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again. Coins and dice are fair unless a question says otherwise.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
A class of 20 students draws one name from a hat, but Jess's name was put in twice by mistake, so the hat holds 21 slips. What is P(Jess), and is the draw fair?
Answer: A
Jess has 2 of the 21 equally likely slips, so P(Jess) = 2/21, while every other student has 1/21. The draw is not fair. Choice B ignores the extra slip. Choice C divides by the number of students instead of the number of slips.
Question 2 of 20 · Multiple Choice
Two friends decide who drives by rolling one fair die: 1, 2 or 3 picks Jo, and 4, 5 or 6 picks Kim. What is P(Jo)?
Answer: C
Jo gets 3 of the 6 equally likely outcomes, so P(Jo) = 3/6 = 1/2, and the method is fair. Choice A counts only one outcome. Choice B confuses the 3 outcomes with a probability of 1/3.
Question 3 of 20 · Multiple Choice
To choose one of three people, a card is drawn from a shuffled standard deck: a jack, queen or king picks Pat, an ace picks Quinn, and any other card picks Ray. What is P(Pat)?
Answer: B
There are 12 face cards (jacks, queens and kings) among 52, so P(Pat) = 12/52 = 3/13. Choice A assumes the three people have equal chances because there are three people. Choice D is P(Ray) = 36/52. The method is unfair.
Question 4 of 20 · Multiple Choice
Five people number themselves 1-5 and roll one die, rolling again whenever a 6 comes up. What is the probability that person 1 is chosen?
Answer: B
A roll of 6 is thrown out, so the process ends on one of 5 equally likely numbers: P(person 1) = (1/6)/(5/6) = 1/5. Choice A ignores the re-rolls. Choice D would be right for four people.
Question 5 of 20 · Multiple Choice
Six people draw folded slips one at a time without replacement, and one slip is marked. What is the probability that the fourth person draws the marked slip?
Answer: D
The first three slips must be unmarked: (5/6)(4/5)(3/4)(1/3) = 1/6, the same as for every position. Choice A is 1/(slips left) = 1/3, which forgets that the marked slip may already be gone.
Question 6 of 20 · Multiple Choice
A teacher chooses one of 25 students with a random number generator for whole numbers from 1 to 25. What is the probability that a given student is chosen?
Answer: A
Each of the 25 numbers is equally likely, so P = 1/25 = 0.04. Choice B treats 25 as a percent. Choice D leaves out the student's own number.
Question 7 of 20 · Multiple Choice
Which method is NOT a fair way to choose one of four people?
Answer: D
With three coins, 1 head and 2 heads each happen in 3 of the 8 outcomes, while 0 and 3 heads each happen in only 1, so D gives probabilities 1/8, 3/8, 3/8 and 1/8. Choices A, B and C each give four equally likely outcomes.
Question 8 of 20 · Multiple Choice
Three friends sold 15, 10 and 5 fundraiser tickets. They agree to draw one prize in proportion to tickets sold. What probability should the friend who sold 15 tickets have?
Answer: C
The total is 30 tickets, so the agreed share is 15/30 = 1/2. Choice A gives equal chances, which is not what the friends agreed to. Choice D divides 15 by 25.
Question 9 of 20 · Multiple Choice
Two players take turns rolling one die, and the first to roll a 5 or a 6 wins. Player A rolls first. What is P(A wins)?
Answer: B
Each roll succeeds with probability 1/3. A wins on the first roll, or after both miss a round, and so on: P(A) = (1/3)/(1 - (2/3)(2/3)) = (1/3)/(5/9) = 3/5. Choice A assumes turn-taking is fair. Choice C is only the chance that A wins on the first roll.
Question 10 of 20 · Multiple Choice
A random number generator gives a decimal from 0 up to 1. Below 0.5 picks Ana, from 0.5 up to 0.75 picks Ben, and 0.75 or more picks Cal. Which change makes the method fair?
Answer: A
Now the chances are 0.5, 0.25 and 0.25. Cutoffs at 1/3 and 2/3 split the interval into three equal parts, so each person has probability 1/3. Choice B still gives someone 0.5. Choice C changes the distribution of the numbers and does not fix the unequal intervals.
Question 11 of 20 · Multiple Choice
A coin lands heads with probability 0.7. To choose between two people, you flip it twice: HT picks Lee, TH picks Max, and HH or TT means flip twice again. Why is this method fair?
Answer: D
The flips are independent, so P(HT) = (0.7)(0.3) = 0.21 = P(TH). The process ends only on HT or TH, so each person has probability 1/2. Choice B is false: HH has probability 0.49 and TT has 0.09.
Question 12 of 20 · Multiple Choice
Seven people draw straws in turn, and one straw is short. Before anyone draws, what is the probability that the last person gets the short straw?
Answer: C
The last person gets the short straw only if the first six draws are all long: (6/7)(5/6)(4/5)(3/4)(2/3)(1/2) = 1/7, the same as for the first person. Choice B is the chance for the last person after six long straws have been drawn. Choice D is the chance that the first person draws a long straw.
Question 13 of 20 · Multiple Choice
Ella and Finn roll two dice. Ella wins if the sum is 2, 3, 4, 10, 11 or 12, and Finn wins if it is 5, 6, 7, 8 or 9. What is P(Ella wins)?
Answer: A
Ella's sums come from 1 + 2 + 3 + 3 + 2 + 1 = 12 of the 36 outcomes, so P(Ella) = 12/36 = 1/3. Choice C treats the 11 sums as equally likely. The method is unfair.
Question 14 of 20 · Multiple Choice
A teacher has 30 students but a random number app that gives whole numbers from 1 to 32. She numbers the students 1-30 and generates again whenever 31 or 32 comes up. What is the probability that a given student is chosen?
Answer: D
Discarding 31 and 32 leaves 30 equally likely numbers, so P = (1/32)/(30/32) = 1/30. Choice A ignores the re-draws.
Question 15 of 20 · Short Answer
Using two fair dice, one red and one blue, design a fair method to choose one of 12 people, and another to choose one of 8 people. Give each person's probability.
List the 36 equally likely ordered outcomes (red, blue). Twelve people: give each person 3 outcomes, for example person 1 gets (1, 1), (1, 2) and (1, 3), so each has 3/36 = 1/12. Eight people: give each person 4 outcomes (32 in all) and roll again on the 4 leftover outcomes; each has (4/36)/(32/36) = 1/8. Other fair answers assign the same number of outcomes to each person and repeat on leftovers.
Question 16 of 20 · Short Answer
A spinner has three sections with angles 90°, 90° and 180°. Tom, Uma and Val are each given one section, and Val gets the 180° section. Find each probability, then repair the method with the same spinner.
P(Tom) = 90/360 = 1/4, P(Uma) = 1/4, P(Val) = 180/360 = 1/2: unfair. Repair: mark a line that divides the 180° section into two 90° sections. Give each person one 90° section and spin again on the fourth. Then each has (1/4)/(3/4) = 1/3.
Question 17 of 20 · Short Answer
Ten students draw slips one at a time without replacement. Three slips are marked "presenter". Show that the second student to draw has probability 3/10 of being a presenter.
Two cases: the first slip is marked and the second is too, (3/10)(2/9) = 6/90; or the first is unmarked and the second is marked, (7/10)(3/9) = 21/90. Total: 27/90 = 3/10, the same as for the first student.
Question 18 of 20 · Short Answer
Explain why a group should agree on its random method before running it, and why running it again until someone likes the result is not fair.
Sample answer: the probabilities are only fair if the method is fixed in advance and run once. If people can reject results, whoever controls the re-runs can keep going until the result they want appears, so their real chances are no longer equal. Agreeing first also prevents choosing a method after knowing who it favors.
Question 19 of 20 · Short Answer
Rosa worked 9 hours, Sam 6 hours and Tia 5 hours. Design a proportional draw with a random number generator for whole numbers from 1 to 20, and give each probability.
Total hours = 20. 1-9: Rosa (9/20 = 0.45); 10-15: Sam (6/20 = 0.30); 16-20: Tia (5/20 = 0.25). The shares add to 1, and each number from 1 to 20 is equally likely.
Question 20 of 20 · Short Answer
Two players each roll one die; the higher roll goes first, and ties roll again. Show that this method is fair, and find the probability that a round ends in a tie.
There are 36 equally likely outcomes. A tie happens in 6 of them, so P(tie) = 1/6. Of the other 30, player A is higher in 15 and player B in 15, by symmetry. Because ties are re-rolled, each player goes first with probability 15/30 = 1/2, so the method is fair.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.MD.B.6 mean?
HSS.MD.B.6 means students use probability to make decisions fairly, for example by drawing lots or with a random number generator. Students must show that a method gives each person an equal chance, or the share that was agreed, and design such methods themselves. It is a (+) standard for advanced courses.
What makes a decision fair in probability?
A decision method is fair when it gives each person the probability they are entitled to before it is run, usually equal chances. It does not mean everyone likes the result. To check, list equally likely outcomes and count how many belong to each person.
Is drawing names from a hat really fair?
Yes, if the slips are the same size and shape, the hat is mixed well and nobody looks. Those conditions make every slip equally likely. Folded slips of different sizes, or names added at the top after mixing, can make the draw unfair.
Does it matter whether you draw first or last when drawing lots?
No, every position has the same chance before the drawing starts. A later person can only win if the earlier draws missed, and that reduced chance exactly balances the better odds among the remaining slips. Students can verify this with a tree diagram.
How do you use a random number generator to make a fair choice?
Number the people from 1 to n, set the generator to produce whole numbers from 1 to n, and run it once. To choose several people, keep generating and skip numbers already used. If the generator's range is larger than n, throw out the extra numbers and generate again.
How can one die choose fairly among a number of people other than 2, 3 or 6?
Assign the same number of faces to each person and roll again on the leftover faces. For example, for four people use 1-4 and re-roll 5 and 6. For larger groups, use two dice (36 outcomes) or a random number generator in the same way.
Can a fair decision give people different chances?
Yes, when everyone agrees in advance that chances should match a share, such as raffle entries for hours volunteered or tickets sold. The method is fair if each person's probability equals the agreed share, for example 6/10 for someone who did 6 of the 10 hours.
Is HSS.MD.B.6 taught in Algebra 2 or in statistics?
It is usually taught in a statistics course or in Precalculus, and sometimes in an advanced Algebra II course. It is a (+) standard, so it goes beyond the core that all students take. It fits naturally alongside expected value (HSS.MD.B.5) and random assignment in experiments.
What mistakes do students make when judging fairness?
A common mistake is treating outcomes as equally likely when they are not, such as the sums of two dice or the number of heads in several coin flips. Others are ignoring leftover outcomes, judging a method by one run, and thinking that later positions in a draw are better or worse.
How does HSS.MD.B.6 connect to experiments and surveys?
Random assignment and random sampling are fair decisions in exactly this sense. In an experiment, each subject should have the same chance of each treatment, and in a simple random sample each group of the chosen size should be equally likely. Students meet these ideas in HSS.IC.B.3.
07
Related Standards
6 standards
These standards connect to HSS.MD.B.6: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.SP.C.7Prerequisite
Develop a probability model, uniform or based on observed frequencies, and use it
