My wife was the Board Game / CCG Coordinator for Trinoc-Con 2004. She was planning to organize several tournaments as part of the gaming, mostly by company: there would be a Rio Grande tourney, a Days of Wonder tourney etc. So I was thinking about how to determine the winner in a simple but sort-of-accurate way. My initial idea was to give points based on hours of play (pre-determined for each game) and rank; the winner of a 5-player game would get 5 points, then 4, 3 etc while for 2-player game we'd award 2 and 1 points. In case of ties, the tied players would average the relevant points. These would be multiplied by the number of hours. The idea was to reward players for playing, even if they didn't necessarily win.
Well, I posted
that idea to spielfrieks May 7. Jonathan Degann pointed
out that there was little reason to play 2-player games with this
schedule, and that a better scoring system would result in all players
getting, on average, the same number of points per hour. I agreed, and suggested
modifying the points with a multiplier to achieve this goal. Hence, the formula
was born. Well, as it turns out, it was actually born again: Peter
Clinch posted
the next day that he had previously developed the PLOPS
("Pete's Ludicrously Overcomplicated Points System"). PLOPS was a
subset of my formula: it sets the expected average points per game at
50, and ignores game length. So, I decided to call mine RPLOPS :) .
Even better would be RPLOPSa(x) where a is 0 if ignoring
game time; or 1 otherwise, and x is the average points per game (per
hour) awarded to all players. So PLOPS is actually RPLOPS0(50).
As Peter Clinch notes on his site, "PLOPS is designed to give a weighted score from a game based on the number of players, since assuming roughly equal parity in the players it is relatively more difficult to win a game with more players". I agree with this concept, but there was considerable debate about it on spielfrieks. An alternative formula (version 2), in which all winners are treated the same, is also on this page.
RPLOPSa(x) = (n + 1 - r) * (h ^ a) * 2 * x / (n + 1)
LO = ludicrously overcomplicated ; )
This simplifies to ...
This formula results in the following sample point allocation, with x arbitrarily set to 50 and ignoring game length:
| RPLOPS0(50) | Number of players (n) |
||||
| Rank (r) | 6 |
5 |
4 |
3 |
2 |
| 1 |
85.71 |
83.33 |
80.00 |
75.00 |
66.67 |
| 2 |
71.43 |
66.67 |
60.00 |
50.00 |
33.33 |
| 3 |
57.14 |
50 |
40.00 |
25.00 |
|
| 4 |
42.86 |
33.33 |
20.00 |
||
| 5 |
28.58 |
16.67 |
|||
| 6 |
14.29 |
||||
If players tie, then the RPLOPS for the tied ranks should be averaged together. For example, if a 5 player game resulted in a tie for 2nd place (ranks 1, 2, 2, 4, and 5) players 1, 4 and 5 get points as usual. Rank 2 players each get the average for ranks 2 and 3 on the chart above (66.67 and 50 ... 58.33 points each).
This is the version of the formula for those who believe that all winners should get the same number of points, regardless of the number of players in the game.
n = number of playersRPLOPS2a(x) = (n - r) / (n - 1) * (h ^ a) * x * 2
LO = ludicrously overcomplicated ; )
This simplifies to ...
This formula results in the following sample point allocation, with x arbitrarily set to 50 and ignoring game length:
| RPLOPS20(50) | Number of players (n) |
||||
| Rank (r) | 6 |
5 |
4 |
3 |
2 |
| 1 |
100 |
100 |
100 |
100 |
100 |
| 2 |
80 |
75 |
66.7 |
50 |
0 |
| 3 |
60 |
50 |
33.3 |
0 |
|
| 4 |
40 |
25 |
0 |
||
| 5 |
20 |
0 |
|||
| 6 |
0 |
||||
If players tie, then the RPLOPS points for the tied ranks should be averaged together. For example, if a 5 player game resulted in a tie for 2nd place (ranks 1, 2, 2, 4, and 5) players 1, 4 and 5 get points as usual. Rank 2 players each get the average for ranks 2 and 3 on the chart above (75 and 50 ... 62.5 points each).