## Nov 16 Tournament Maths

The chess tournament that we run at school is underway and I host the results on my website. Here's the state as of the time I write this article:

Each group has everyone play everyone else in the group twice, once as white and once as black. Each player is awarded 3 points for a win, 1 point for a draw and 0 points for a loss.

The top 8 scorers across both groups will advance to the knockout stage. Tiebreak is used to decide between draws and is calculated as the sum of the scores of each player you win against (per win) + 1/2 the sum of the scores of any players you draw against (again, per draw). This should mean that a win against a better player will add more to your tie break than a win against a worse player.

If it is still a draw then I will look at direct games between the two players (if any) and if it is still a draw then we'll play a game of Armageddon to decide. If there are multiple drawn people we will do an Armageddon tournament until someone is victorious.

Now, I was thinking about the top eight scorers rule and how it compares with the alternative rule that you just take the top four scorers from each group. If all games are completed, is it possible to have 5 people from one group and 3 from the other?

For that to happen we want 5 of the players from one group (let's say group A) to get as many points as possible and 1 player to lose every match. So players 1, 2, 3, 4 and 5 win both of their games against player 6 and win one and lose one against each of their other opponents. This gives 6 wins each for a total standing of 18, 18, 18, 18, 18, 0.

For the other group we want 3 good players to sap as many points as possible from the 3 bad players. Let's say players 4, 5 and 6 in group B each draw both of their games against each other and lose all of their games to players 1, 2 and 3. This would give them each 4 draws for a total of only 4 points which isn't nearly enough to qualify over the 5th player in group A.

Because there was so much wiggle room on this we can take this a step further. Can we get 6 players from group A to qualify?

Let's say everyone in group A wins every match as White (and so loses every game as Black). That gives them all 5 wins for 15 points each. Let's also say the literally every game in group B is a draw. That gives them 10 draws each for 10 points. At this point all of the people in group A will qualify.

By changing the problem to have larger group sizes this approach will still solve it. The take away message for this is that draws are bad if you want lots of people in your group to qualify.