Tournament scoring for variable nbr of players a the table
Posted: 28 January 2021, 15:13
In a swiss or round robin system tournament, it seems normal to give 1 or 2 points per player you defeated at the table.
This gives:
So for instance, in a 5 players games, if 2 players are ex-aequo for the second position, they both get (3+2)/2 = 2.5 points
Or you can, of course, multiply everything above by 2.
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One problem comes if you can't fill all the tables with the same number of players because some player left or because you did not enforce that the number of participants be a nice number.
As it stands above, 3p tables give an average of 1 point per player whereas 4 players tables give an average of 1.5 points per player.
So if you want it to be fair you would need to somehow scale the scoring.
One intuitive way of scaling would be to give in a 3p table: 3 points for the winner and 1,5 point to the second (everything multiplied by 1.5)
That is not however the most generic, nor in my opinion the fairest.
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My suggested scoring:
* score 1 point for each player that is ex-aequo with you (yourself included) and 2 point for each player you defeat, divide the total by the number of players at the table.
This is simple, generic (no need to have a different code for each table size) and fair:
* no matter how many players there are at each table, the average points per player is always 1.
* Winning in a table with many players is better than winning in a table with fewer players.
* Being last in a table with many players is worse than being last in a table with fewer players
So for example, in the absence of ex-aequos:
Tell me what you think...
This gives:
- 2 players table: 1 point to the winner, 0 point to the second
- 3 players table: 2 points to the winner, 1 point to the second, 0 point to the 3rd
- 4 players table: 3 points to the winner, 2 points to the second, 1 point to the 3rd, 0 to the 4th
- 5 players table: 4 ,3 , 2, 1 & 0 points
So for instance, in a 5 players games, if 2 players are ex-aequo for the second position, they both get (3+2)/2 = 2.5 points
Or you can, of course, multiply everything above by 2.
---------------------------
One problem comes if you can't fill all the tables with the same number of players because some player left or because you did not enforce that the number of participants be a nice number.
As it stands above, 3p tables give an average of 1 point per player whereas 4 players tables give an average of 1.5 points per player.
So if you want it to be fair you would need to somehow scale the scoring.
One intuitive way of scaling would be to give in a 3p table: 3 points for the winner and 1,5 point to the second (everything multiplied by 1.5)
That is not however the most generic, nor in my opinion the fairest.
--------------------------------------------------
My suggested scoring:
* score 1 point for each player that is ex-aequo with you (yourself included) and 2 point for each player you defeat, divide the total by the number of players at the table.
This is simple, generic (no need to have a different code for each table size) and fair:
* no matter how many players there are at each table, the average points per player is always 1.
* Winning in a table with many players is better than winning in a table with fewer players.
* Being last in a table with many players is worse than being last in a table with fewer players
So for example, in the absence of ex-aequos:
- 4p table: 1st gets 7/4 = 1.75; 2nd gets 5/4=1.25 ; 3rd get 0.75 ; 4th gets 0.25 ; average = 1.0
- 5p table: 1st gets 9/5= 1.8 ; 2nd gets 7/5=1.4; 3rd gets 1.0 ; 4th gets 0.6; 5th gets 0.2 ; average = 1.0
Tell me what you think...