An interesting question on combinations.

Since "each contestant must play every other contestant exactly once", it is essentially the same problem to find the number of ways to choose 2 players from 18, which is $\displaystyle {18\choose2} =153$ .

This problem is solved by starting with the first person in which we will have (n-1 = 17) and so on with each person until reaching 1 since each iteration of each person is increased by playing with the next person as -1. The formula is a summation (17 * 18) / 2

The number of games will be simply $17 + 16 + 15 + ... + 2 + 1$ , which is equal to $18 \times 8 + 9 = 153$

Player one will play 17 matches because he can't play against himself. Player two only needs to play 16 matches because he already had one with player one. This continues all the way to player 17, who'll only need to play player 18 because all of the others played him. And player 18 doesn't need to play any matches because he already played with all of the other contestants. The equation comes out to 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 153 .