Making Fair Decisions

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. 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 $1$ to $8,$ and then a fair octahedron die is rolled. All possible outcomes of rolling a fair octahedron die can be listed as follows. 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 $\{\text{H},\text{H},\text{T}\}$ and $\{\text{H},\text{T},\text{H}\}$ are two different outcomes. 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.

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. 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. 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 $2$ 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. 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 $\frac{1}{7}.$ By multiplying these values, the probability of each of the two cards being drawn can be calculated. 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. 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. Note that there is a total of $12$ possible outcomes. The probability of each outcome is $\frac{1}{12}.$ Since each possible outcome is equally likely, this method results in a fair decision.

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 $\boxed{\text{MATH}}$ 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 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. 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	$2$	$1\text{ and }2$
Maya	$3$	$3,4,\text{ and }5$
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. 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. 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	$2$	$1\text{ and }2$	$\frac{2}{6}\approx 0.33$
Maya	$3$	$3,4,\text{ and }5$	$\frac{3}{6}=0.5$
Diego	$1$	$6$	$\frac{1}{6}\approx 0.17$
Dylan	$0$	$\text{None}$	$\frac{0}{6}=0$

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.