Table Tennis

This activity provides visual support for students to predict the number of games in various draws. After completing draw tables for question 1, in which the students work out the number of games for tournaments with 3 and then 4 players, they are able to make predictions for the number of games with 8, 16, and then 100 players.
In the table for 3 players, there are 3 x 3 = 9 spaces to fill (figure 1). The three diagonal spaces can be filled with A v A, B v B, and C v C. This is not sensible, so the spaces on the diagonal are shaded (figure 2). So there are now 3 x 3 – 3 = 6 spaces to fill (figure 3). A v B is the same game as B v A, A v C is the same game as C v A, and B v C is the same game as C v B. So, in practice, only one-half of the
3 x 3 – 3 = 6 spaces need to be filled (figure 4). The number of games in a tournament for 3 players is therefore 1/2 of (3 x 3 – 3), which is 1/2 of 6 = 3.

A tournament with 4 players has 1/2 of (4 x 4 – 4) or 1/2 of 12, which is 6 games. A tournament with 8 players has 1/2 of (8 x 8 – 8), or 1/2 of 56, which is 28 games. Draw tables for these are given in the Answers. The table below shows how the pattern works.

We can use algebraic symbols to show this pattern. So, for x players, there are 1/2 of (x x x – x) games, or

Another way to visualise the pattern is to notice, for example, that in the table for 4 players there are 4 rows of 3 spaces to be filled. The space that lies on the diagonal is excluded, so there are 4 x 3 = 12 spaces to be filled. But half of the games in these spaces are duplicates (A v B = B v A, and so on), so the number of games is 1/2 of (4 x 3) or 1/2 of 12, which is 6 games. The table below shows how the pattern  works:

We can also use algebraic symbols to show this pattern. So, for x players, there are 1/2 of games.

While the two algebraic rules look different, they are equivalent. The second rule,, can be expanded to give , which simplifies to which is the first rule.

Answers to Activity
1. a. i.

ii. If the shaded spaces were filled in, they would be: Alison v Alison, Barry v Barry, and Casey v Casey. This makes no sense, so the spaces are not filled.
iii. Yes. A game between Barry and Alison, B v A, is the same as a game between Alison and Barry, A v B. C v A is the same game as A v C, and C v B is the same game as B v C. So, altogether, there are just 3 possible games, B v A, C v A, and C v B. Each player will play 2 games.
b. i.

ii. 6

2. a.

Casey is not correct. The draw table for 8 players shows that there are 28 games altogether. Casey probably reasoned that doubling the number of
players would double the number of games: so if there were 6 games for 4 players, there would be 2 x 6 = 12 games for 8 players.
b. One way to make predictions is to find a short cut to count the spaces in draw tables that represent the games in the tournament. So, for the tournament with 8 players, there are 8 x 8 – 8 = 56 or 8 x 7 = 56 unshaded spaces to
be filled. But each game shown below the shaded spaces in the diagonal is the same as a game above the shaded spaces, so there are just (8 x 7) ÷ 2 = 28 games.
When there are 16 players, there will be (16 x 15) ÷ 2 = 120 games. Your draw table should show 120 games.
c. (100 x 99) ÷ 2 = 4 950 games