Pigeonhole principle 3

Question

I need help on this question, I'm lost and really don't know how to proceed:

Use the pigeonhole principle to prove that in a round-robin chess tournament (with 18 participants) there will be at least two players with the same number of draws?

Answer 1

My original answer was wrong. Please see the other answers.

Answer 2

Each player has 17 games which means they can have anywhere from 0 to 17 draws. But note that if a game ends in a draw, both players gain a draw. This means that if there is one player with 0 draws, there cannot be a player with 17 draws and vice versa.

So let your pigeon holes be [0 or 17], [1], [2], ... ,[16] draws. That's 17 possibilities with 18 contestants.

Answer 3

The argument below is a parity argument.

Let us suppose that each player makes a tick mark in her notebook for every drawn game she is involved in. Note that a drawn game results in $2$ tick marks, so the total number of tick marks is even.

Each player makes $0$ to $17$ tick marks. If all the players have different numbers of draws, then the total number of tick marks is $0+1+2+\cdots+17$, which is $153$. This is impossible, since $153$ is odd.