 # Problem of the Week Problem C and Solution Fair Game?

## Problem

For a school mathematics project, Zesiro and Magomu created a game that uses two special decks of six cards each. The cards in one deck are labelled with the even numbers $$2$$, $$4$$, $$6$$, $$8$$, $$10$$, and $$12$$, and the cards in the other deck are labelled with the odd numbers $$1$$, $$3$$, $$5$$, $$7$$, $$9$$, and $$11$$.

A turn consists of Zesiro randomly choosing a card from the deck with even-numbered labels and Magumo randomly choosing a card from the deck with odd-numbered labels. These two cards make a pair of cards. After a pair of cards is chosen, they perform the following steps.

1. They determine the sum, $$S$$, of the numbers on the cards. For example, if Zesiro chooses the card labelled with a $$6$$ and Magumo chooses the card labelled with a $$3$$, then $$S=6+3 = 9$$.

2. Using $$S$$, they determine, $$D$$, the digit sum. If $$S$$ is a single digit number, then $$D$$ is equal to $$S$$. If $$S$$ is a two-digit number, then $$D$$ is the sum of the two digits of $$S$$. For example, if Zesiro chooses the card labelled with a $$6$$ and Magumo chooses the card labelled with a $$3$$, then the sum and the digit sum are both $$9$$. If Zesiro chooses the card labelled with a $$10$$ and Magumo chooses the card labelled with a $$5$$, then the sum is $$S = 10 + 5 = 15$$ and the digit sum is $$D = 1+5=6$$. If Zesiro chooses the card labelled with a $$10$$ and Magumo chooses the card labelled with a $$9$$, then the sum is $$S = 10 + 9 = 19$$ and the digit sum is $$D = 1+9=10$$.

Zesiro gets a point if the digit sum, $$D$$, is a multiple of $$4$$.

Magomu gets a point if the number on one of the cards is a multiple of the number on the other card.

Is this game fair? That is, do Zesiro and Magomu have the same probability of getting a point on any turn? Justify your answer. ## Solution

To solve this problem, we will create a table where the columns show the possible choices for the even-numbered card, the rows show the possible choices for the odd-numbered card, and each cell in the body of the table gives the sum of the corresponding pair of cards.

Even Card $$2$$ $$4$$ $$6$$ $$8$$ $$10$$ $$3$$ $$5$$ $$7$$ $$9$$ $$11$$ $$13$$ $$5$$ $$7$$ $$9$$ $$11$$ $$13$$ $$15$$ $$7$$ $$9$$ $$11$$ $$13$$ $$15$$ $$17$$ $$9$$ $$11$$ $$13$$ $$15$$ $$17$$ $$19$$ $$11$$ $$13$$ $$15$$ $$17$$ $$19$$ $$21$$ $$13$$ $$15$$ $$17$$ $$19$$ $$21$$ $$23$$

From the table, we see that the total number of possible pairs is $$6 \times 6 = 36$$.

We create another table where the columns show the possible choices for the even-numbered card, the rows show the possible choices for the odd-numbered card, and each cell in the body of the table gives the digit sum of the corresponding pair of cards.

Even Card $$2$$ $$4$$ $$6$$ $$8$$ $$10$$ $$3$$ $$5$$ $$7$$ $$9$$ $$1+1=2$$ $$1+3 =4$$ $$5$$ $$7$$ $$9$$ $$1+1=2$$ $$1+3 =4$$ $$1+5=6$$ $$7$$ $$9$$ $$1+1=2$$ $$1+3 =4$$ $$1+5=6$$ $$1+7=8$$ $$9$$ $$1+1=2$$ $$1+3 =4$$ $$1+5=6$$ $$1+7=8$$ $$1+9=10$$ $$1+1=2$$ $$1+3 =4$$ $$1+5=6$$ $$1+7=8$$ $$1+9=10$$ $$2+1=3$$ $$1+3 =4$$ $$1+5=6$$ $$1+7=8$$ $$1+9=10$$ $$2+1=3$$ $$2+3=5$$

If the digit sum is a multiple of $$4$$, then Zesiro gets a point. In the table there are two digit sums, $$4$$ and $$8$$, that are multiples of $$4$$. The digit sum $$4$$ occurs six times in the table and the digit sum $$8$$ occurs four times in the table. This totals ten possible outcomes for Zesiro, and so his probability of scoring a point on any pair is $$\frac{10}{36}$$.

Magomu has far less work to determine when he gets a point. None of the odd numbers are multiples of the even numbers. All multiples of even numbers are even and hence will never be odd.

Whenever a $$1$$ is chosen, Magomu will score a point. That is, each of the six even numbers is a multiple of $$1$$.

When a $$3$$ is chosen, Magomu will score a point if the number on the face of the even-numbered card is a $$6$$ or $$12$$. That is, only two of the even numbers are multiples of $$3$$.

When a $$5$$ is chosen, Magomu will score a point if the number on the face of the even-numbered card is a $$10$$. That is, only one of the even numbers is a multiple
of $$5$$.

None of the numbers in the deck containing only even numbers is a multiple of $$7$$, $$9$$, or $$11$$.

So Magomu will score a point on $$6+2+1=9$$ of the $$36$$ possible pairs. Therefore, Magomu’s probability of scoring a point on any pair is $$\frac{9}{36}$$.

The game is not fair since Zesiro’s probability of scoring a point on any pair is greater than Magomu’s probability of scoring a point on any pair.