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| 10 Theory slides |
| 7 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
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?
Try to determine which game leads to a fair decision. In order to do this, complete the following steps.
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.
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.
The fairness of the mentioned methods will be explained one at a time.
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.
Favorable outcomes=15, Total outcomes=36
ba=b/3a/3
Calculate quotient
Round to 2 decimal place(s)
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.
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.
Favorable outcomes=27, Total outcomes=36
ba=b/9a/9
Calculate quotient
Favorable outcomes=9, Total outcomes=36
ba=b/9a/9
Calculate quotient
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.
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.
Substitute values
ba=b/13a/13
Multiply fractions
ba=b/2a/2
Substitute values
Add fractions
Calculate quotient
Substitute values
ba=b/2a/2
ba=b/5a/5
Multiply fractions
Use a calculator
Round to 3 decimal place(s)
There are opportunities to make fair decisions in different contextual situations by the use of simulations.
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.
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.
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.
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.
n=8, r=2
Subtract term
Write as a product
Split into factors
Cross out common factors
Cancel out common factors
Multiply
1a=a
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.
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 |
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.
Find a method of assigning outcomes so that the chance of winning is proportional to the number of games won.
Members | Games Won | Numbers Assigned |
---|---|---|
Ali | 2 | 1 and 2 |
Maya | 3 | 3,4, and 5 |
Diego | 1 | 6 |
Dylan | 0 | None |
Members | Games Won | Numbers Assigned | Probability of Winning the Chess Set |
---|---|---|---|
Ali | 2 | 1 and 2 | 62≈0.33 |
Maya | 3 | 3,4, and 5 | 63=0.5 |
Diego | 1 | 6 | 61≈0.17 |
Dylan | 0 | None | 60=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.
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.
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.
Begin by identifying all possible outcomes of the experiments in each method. Then, calculate the probabilities to determine the fair method(s).
Each method will be examined one at a time.
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.
Favorable Outcomes=9, Total Outcomes=36
ba=b/9a/9
Calculate quotient
Favorable Outcomes=10, Total Outcomes=36
ba=b/2a/2
Calculate quotient
Round to 2 decimal place(s)
Favorable Outcomes=17, Total Outcomes=36
Calculate quotient
Round to 2 decimal place(s)
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 | 155≈0.33 |
Izabella wins | 5 | 15 | 155≈0.33 |
Kevin wins | 5 | 15 | 155≈0.33 |
It was obtained that each player is equally likely to get the bicycle. Therefore, Izabella's method is fair.
Kevin suggested flipping a coin two times. To analyze his method, make a table of all possible outcomes.
Favorable Outcomes=1, Total Outcomes=4
Calculate quotient
Favorable Outcomes=2, Total Outcomes=4
ba=b/2a/2
Calculate quotient
A fair 20-sided die has been labeled using dots. Each side can have any number ranging from 1 to 6 dots. Below we see the result of a large number of throws using this die.
Outcome (number of dots) | Frequency |
---|---|
1 | 990 |
2 | 2580 |
3 | 3430 |
4 | 1050 |
5 | 490 |
6 | 1460 |
Let's estimate the experimental probability of landing on each of the dot amounts. To do so, we will calculate how many times the die was thrown by adding up all the frequencies. 990+2580+...+1460=10 000 Now we can calculate the experimental probability of getting every number of dots by dividing the frequency of each outcome by the total number of throws.
Outcome | Frequency | Frequency/10 000 | Probability |
---|---|---|---|
1 | 990 | 990/10 000 | 0.099 |
2 | 2580 | 2580/10 000 | 0.258 |
3 | 3430 | 3430/10 000 | 0.343 |
4 | 1050 | 1050/10 000 | 0.105 |
5 | 490 | 490/10 000 | 0.049 |
6 | 1460 | 1460/10 000 | 0.146 |
Note that none of the probabilities are around or greater than 0.5 — the probability if the specific number of dots appeared on 10 sides, which is half of the amount of sides. In that case, let's calculate the theoretical probability that a certain number of dots is present on 1 through 9 sides of the die.
Favorable sides | Total number of sides | Probability | = |
---|---|---|---|
1 | 20 | 1/20 | 0.05 |
2 | 20 | 2/20 | 0.10 |
3 | 20 | 3/20 | 0.15 |
4 | 20 | 4/20 | 0.20 |
5 | 20 | 5/20 | 0.25 |
6 | 20 | 6/20 | 0.30 |
7 | 20 | 7/20 | 0.35 |
8 | 20 | 8/20 | 0.40 |
9 | 20 | 9/20 | 0.45 |
If we compare the experimental probabilities to the theoretical probabilities, we can conclude the following. Since0.10 ≈ 0.099, & 2sides have 1dot Since0.25 ≈ 0.258, & 5sides have2dots Since0.35 ≈ 0.343, & 7sides have3dots Since0.10 ≈ 0.105, & 2sides have 4dots Since0.05 ≈ 0.049, & 1sides have5dots Since0.15 ≈ 0.146, & 3sides have6dots As we can see, two sides have 1 dot and two sides have 4 dots. Therefore, these outcomes are equally likely in theory.
Paulina and 19 classmates sing in their school's choir. They are very united and, for the benefit of the school's choir, their members have volunteered a total of 150 hours doing after-school activities. Paulina has contributed 25 hours of the total amount.
Their next performance is coming soon, and it requires one member to perform a solo. The choir's conductor wants to make a fair decision when deciding who will perform the solo.
For this case, we consider a decision to be fair if each member of the choir has the same probability of being selected. Therefore, we are considering a uniform probability model with 20 possible outcomes, which is the total number of choir members, including Paulina. &P=Number of favorable outcomes/Number of possible outcomes [1em] &P = 1/20 Under this set up, the probability of Paulina getting chosen is the same as every other member, 1 20. Note that the hours spent volunteering did not have any effect in the decision's method.
In this case, we are considering the decision to be fair it proportional to the number of hours each member volunteered. With that in mind, the probability for any member chosen can be determined as follows. P=Hours volunteered by the member/Total number of volunteered hours We know that the total number of volunteered hours is 150 from which Paulina contributed 25. This can be substituted into the equation. P = 25/150 = 1/6 Taking into account the definition of fair by these standards, we have calculated the probability of Paulina being selected to be 16.
Consider the four following dice nets — each diagram represents the faces of a die. They have been labeled D1, D2, D3, and D4.
Two friends are using these dice to play a game. They agree upon a winning event and then each choose one die to roll. Which of the dice would produce a fair game for the given event?
A game is fair if the probabilities of winning are the same for every player. To calculate probability of winning, we can divide the number of favorable outcomes by the total number of possible outcomes. P=Number of favorable outcomes/Number of possible outcomes Notice that all dice have 6 sides. Therefore, the number of possible outcomes is 6 for all four dice. To determine the favorable outcomes, we must count which sides are favorable for the given event, in this case odd numbers. Let's mark all of the sides having odd numbers.
As we can see, both D1 and D4, have 4 sides with odd numbers. Since both have the same number of favorable outcomes, the probability of winning is the same for both players P = 4 6 = 23. Therefore, if the friends play using these dice, the game will be fair and the winner can happily enjoy their steak.
In this case, the favorable event is defined to be getting a number greater than 2. Let's mark all of the sides that have numbers greater than 2 on the different dice.
As we can see, D2, D3, and D4 all have 3 sides that are greater than 2. Since these three dice have the same number of favorable outcomes, if the friends play using any pair from these dice, the game will be fair and the winner can happily enjoy their ice cream!
In this case, the favorable event is defined to be rolling a number less than 3. Let's mark all of the sides that have numbers less than 3 on the different dice.
Again, in D2, D3, and D4, we have the same number of favorable outcomes. Therefore, the friends can pick any pair from these dice and enjoy the winnings of their game!
We will analyze each of these statements, one at a time.
When tossing a coin, there are only two possible outcomes — heads or tails. Both outcomes have the same probability P= 12. The brothers are equally likely to do the cleaning if they decide by tossing a coin. Therefore, this method is fair.
There are several factors that can affect the probabilities of one of the brothers winning the game of Tic-Tac-Toe. The person who plays first has advantage. Additionally, one of the brothers may have more experience playing Tic-Tac-Toe. Therefore, this is not a fair decision making process.
Initially, one may think that this is an example of a fair decision making process. However, studies show that rock is the preferred choice when playing the game. About 35 % of the time, rock is chosen. Therefore, this is not a fair decision making process as one brother might be aware that rock is a likely choice of his opponent.
This is not a fair decision making process. Consider that if the friend has a preference or is friendlier with one of the brothers, the decision made is more likely to be bias.
Base on the points in our discussion, of the presented situations, we can conclude that only i represents a fair decision making process.