Scoring systems for a tournament with variable player numbers [closed]

Question

I'm running a tournament where I need to score games with anything from 3 to 8 players.

My initial plan was to always have 8 players, and award 10 for 1st, 8 for 2nd, then 6/5/4/3/2/1. However, I need to cope with other numbers of players in a fair way. Obvious thoughts include:

1. Just have 10/8/6/etc as before, and ignore the lower scores (seems unfair that someone winning a 3 player game gets the same points as someone winning an 8 player game)
2. Have the points scaled to player count. So first in a 3 player game would be 10*3/8 = 3.75 points. This seems to take things too far the other way.

Is there a recommended approach for this, or any bright ideas?

Answer 1

You need to ensure that the number of players is always equal otherwise there will be lots of problems with fairness of the scores. Here are a few problems I see with keeping track of how players perform in the same game with different number of players.

• 3rd place in a 3 player game is not the same as 3rd place in a 10 player game.

Something that needs to be kept in mind is that the harder it will be to get those places.

• Larger games can impact outcomes and a player could get a lower ranking due to competition and cooperation.

A critical point that needs to be remembered is in a larger game you might have players working together to help keep the competition lower. What I mean by this is if there are two players that are neck and neck for a spot in the tournament one of them might work with other players to help sabagoute the other to help those rankings. However if one of those players is in a 3 player game with weak opponents they could get an easy win.

In the end it becomes extremely difficult to compare the results of a small game to a large game and it should be avoided if all possible.

Answer 2

You should have the average points per player be constant. If an eight player game has 10/8/6/5/4/3/2/1, that's a total of 39 points, for an average of 4.875 points per player. So the math would be simpler if you gave first place 11 points, giving 5 points per player. Then in a three player game you would have 15 points to distribute. Perhaps 9 for first, 4 for second, and 2 for last. (BTW, if everyone plays the same number of games, then subtracting 1 from all the point awards doesn't change the relative standing. So having eight player games be 10/7/5/4/3/2/1/0 and three player games be 8/3/1 would be mathematically equivalent, though perhaps not psychologically equivalent, to the above.)

Keeping all the average points the same means that players of the same ability level will not have their expected points depend on the size of the game, but it can be still be a factor when one takes into account player ability. For instance, someone who's really good would probably prefer to be in a 11-for-first game rather than a 9-for-first one. That's harder to control for. You could try arranging things to that both the maximum and average points are the same, or you could find some "fair" way of distributing the sizes of games. For instance, if you have a feeder tournament, the players can come in with "seed" values that affect the size of the games they go into. Or players could bid on games; there could be some auction in which players can take a point penalty in exchange for getting to choose what game they play in.

Answer 3

Certainly Joe W is correct, and there's no way to do this in general. There are cases you can solve, though.

If you have a system that is designed for 8-player, and you get 36, you can ensure that the event is 6 brackets of 6, not three 8s and a four. If you do that, then whatever scoring you use is consistent.

This doesn't work for "there were 20 people this week and 48 people last week, and we ran 8s last week, and we need to have "winner today == winner last week", but it solves "4 round swiss, with payouts at end of the night".

Certainly strategy in a 6-player game (or a 3-player game!) is different from an 8-player game, but that, too, can be a skill factor - oh, we're playing small-table today. The winner of the night will be the best "small-table" player on the day. The best will become the best at playing all size games, eventually.

Answer 4

BGA uses ELO rating with a modification that handles multiplayer games.

When you finish a game, the number of points you gain/lose depends on the ELO difference between you and your opponents. The easiest situation is a 2 player game. You will gain a number of ELO points equal to

K * (W - p(D)) 

where: W is the result of the game: 1 for a victory, 0 for a defeat, 0.5 for a tie. D is the difference of ELO between you and your opponent. p(D) is the probability of victory considering the previous difference (see Wikipedia for details). K, the 'elasticity factor,' ... you can use K=20.

modification for multiplayer games:

When you play with more than 1 opponent, we consider that you win against each opponent who scores lower than you during this game, and lose against each opponent who scores higher than you. For example, if you place 2nd in a 4 player game, you gain ELO points as if you won against the 3rd and 4th players, and lost against the 1st.

See:

1. BGA: How is my ELO ranking computed?
2. Medium: Implementing the Elo Rating System