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Clive Lovett
Canada Kamloops British Columbia

Hi
We have five in our regular weekly gaming group and less when we have impromptu gatherings. We have started keeping track of who wins by game and by number of players.
The results recorded are:
1. Percentage of total wins by total games played (total 100% with not everyone having played in every session)
2. Percentage of wins per games played for each individual
3. Percentage of wins for 2p, 3p, 4p and 5p sessions for each individual
We are going to have year end awards but I would like help figuring out the best way to judge who deserves the OVERALL CHAMPION award. The second stat seems the fairest but if I have played 51 games and have won 17 games (33.3%) of games I have played and another player has played 20 and won 10 (50%) of games they have played then the lack of weighting seems unfair to me. Any solutions?


Mike Lee
Canada Vancouver British Columbia

We do something similar. We have a yearly gaming season and keep track of all the games we play.
We use a weighted ELO system. You can look it up.


Clive Lovett
Canada Kamloops British Columbia

Thanks for the info and I have always loved me some Electric Light Orchestra
(seriously, thanks!)
MikeyMike79 wrote: We do something similar. We have a yearly gaming season and keep track of all the games we play.
We use a weighted ELO system. You can look it up.


Jason Hinchliffe
Canada Mississauga Ontario

We used to keep track via a championship belt. I swear. We went out and bought a replica WWE Heavyweight Championsip belt, and the rules were as follows: To win the belt, the player had to win two games, before the current champion won one. The champ had possesion of the belt and was required to bring it to game night, particularly if a contender was to be present.
Believe me, it made for some epic metagaming.


Clive Lovett
Canada Kamloops British Columbia

clockwerk76 wrote: We used to keep track via a championship belt. I swear. We went out and bought a replica WWE Heavyweight Championsip belt, and the rules were as follows: To win the belt, the player had to win two games, before the current champion won one. The champ had possesion of the belt and was required to bring it to game night, particularly if a contender was to be present.
Believe me, it made for some epic metagaming.
Love this idea! Looking for a fake championship belt now!


Jason Hinchliffe
Canada Mississauga Ontario

Clive65 wrote: clockwerk76 wrote: We used to keep track via a championship belt. I swear. We went out and bought a replica WWE Heavyweight Championsip belt, and the rules were as follows: To win the belt, the player had to win two games, before the current champion won one. The champ had possesion of the belt and was required to bring it to game night, particularly if a contender was to be present.
Believe me, it made for some epic metagaming. Love this idea! Looking for a fake championship belt now!
Awesome. I'm glad you like it. We had guys show up with ghetto blasters and play theme music as they walked in with the belt. Pure comedy. It made for some serious laughs.


Ron
Austria Vienna
“It's all in the mind.” ― George Harrison
Devoted Follower of the Most Holy Church of the Evil Bob. Possessed and down the road to become chaotic, evil & naughty. All hail the Evil Bob and his Stargate.

I log my games here at the Geek.
Would be nice if after the redesign we could use such a feature.


Manchuwok
Canada Mission BC

What would really be the most representative of overall skill would be if you recorded complete placement for every game. I'm assuming you haven't done that? If you have, you could use a version of the Racing Games Payout System I devised that takes into account number of players in each game. You would either need to ensure every player had played the same number of games or use a minimum divisor (you could set it to the average games played by all players).


Clive Lovett
Canada Kamloops British Columbia

manchuwok wrote: What would really be the most representative of overall skill would be if you recorded complete placement for every game. I'm assuming you haven't done that? If you have, you could use a version of the Racing Games Payout System I devised that takes into account number of players in each game. You would either need to ensure every player had played the same number of games or use a minimum divisor (you could set it to the average games played by all players).
I wish we had kept track of placing. However, some games do not require completion after a player has won (though I guess they could). I may devise some questionable system that weights results on numbers of games played and number of opponents per game That brings me to another question that I believe has no correct answer. What is harder to win, a two player game or a five player game? If all players are equal, on one hand I would argue that the two player game is harder because one mistake made could be fatal. BUT on the other hand, in a five player game you have four players working against you.


Aaron Isley
United States Tulsa Oklahoma

I think you really need to factor in the "third stat". Imagine a hypothetical scenario in which one player plays a lot of 2p games and another player plays a lot of 4p games. The first player plays slightly below average (45% win rate) while the second player plays well above average (40% win rate). Note that the first player underperforms the 50% rate that is expected for a 2player game, but he looks better than the other guy who is killing the 25% win rate that is expected in a 4player game.
Here is your solution: 1. Multiply winning percent (as a fraction, i.e. 50%=0.5) for each of your "third stats" by the number of players. Note that a value of 1 is meeting expectation, less than 1 is worse, and greater than 1 is better. 2. Multiply those numbers by the number of times a player has played games with that amount of people. 3. Add up all the numbers and divide them by the TOTAL number of ALL games played.
Here is an example for clarity: Player A has gone 1/2 (50%) in 2player games, 1/3 (33%) in 3player games, 4/9 (44%) in 4player games, and 2/5 (40%) in 5player games. For the first of my steps, you would multiply 1/2*2=1, 1/3*3=1, 4/9*3=1.7, and 2/5*5=2. Note that this is meeting expectations in 2,3 player games, and exceeding them in 4,5 player games. For the 2nd step, multiply 1*2=2 (because 2 games have been played of 2players), 1*3=3, 1.7*9=16, and 2*5=10. Now for the third step you add those numbers (2+3+16+10=31) and divide that by the total number of games played (19) to get an adjusted win rate of 1.63. This number is actually much easier to understand than his overall winning rate of 42% (his "second stat) because no one knows if 42% is good or bad (it depends on the number of players. You can tell right away that 1.63 is very good because it is significantly larger than 1.


Clive Lovett
Canada Kamloops British Columbia

Aaron I wrote: I think you really need to factor in the "third stat". Imagine a hypothetical scenario in which one player plays a lot of 2p games and another player plays a lot of 4p games. The first player plays slightly below average (45% win rate) while the second player plays well above average (40% win rate). Note that the first player underperforms the 50% rate that is expected for a 2player game, but he looks better than the other guy who is killing the 25% win rate that is expected in a 4player game.
Here is your solution: 1. Multiply winning percent (as a fraction, i.e. 50%=0.5) for each of your "third stats" by the number of players. Note that a value of 1 is meeting expectation, less than 1 is worse, and greater than 1 is better. 2. Multiply those numbers by the number of times a player has played games with that amount of people. 3. Add up all the numbers and divide them by the TOTAL number of ALL games played.
Here is an example for clarity: Player A has gone 1/2 (50%) in 2player games, 1/3 (33%) in 3player games, 4/9 (44%) in 4player games, and 2/5 (40%) in 5player games. For the first of my steps, you would multiply 1/2*2=1, 1/3*3=1, 4/9*3=1.7, and 2/5*5=2. Note that this is meeting expectations in 2,3 player games, and exceeding them in 4,5 player games. For the 2nd step, multiply 1*2=2 (because 2 games have been played of 2players), 1*3=3, 1.7*9=16, and 2*5=10. Now for the third step you add those numbers (2+3+16+10=31) and divide that by the total number of games played (19) to get an adjusted win rate of 1.63. This number is actually much easier to understand than his overall winning rate of 42% (his "second stat) because no one knows if 42% is good or bad (it depends on the number of players. You can tell right away that 1.63 is very good because it is significantly larger than 1.
Thanks for this. I have this information so this might work for my group! Cheers!


Jason Hinchliffe
Canada Mississauga Ontario

Aaron I wrote: I think you really need to factor in the "third stat". Imagine a hypothetical scenario in which one player plays a lot of 2p games and another player plays a lot of 4p games. The first player plays slightly below average (45% win rate) while the second player plays well above average (40% win rate). Note that the first player underperforms the 50% rate that is expected for a 2player game, but he looks better than the other guy who is killing the 25% win rate that is expected in a 4player game.
Here is your solution: 1. Multiply winning percent (as a fraction, i.e. 50%=0.5) for each of your "third stats" by the number of players. Note that a value of 1 is meeting expectation, less than 1 is worse, and greater than 1 is better. 2. Multiply those numbers by the number of times a player has played games with that amount of people. 3. Add up all the numbers and divide them by the TOTAL number of ALL games played.
Here is an example for clarity: Player A has gone 1/2 (50%) in 2player games, 1/3 (33%) in 3player games, 4/9 (44%) in 4player games, and 2/5 (40%) in 5player games. For the first of my steps, you would multiply 1/2*2=1, 1/3*3=1, 4/9*3=1.7, and 2/5*5=2. Note that this is meeting expectations in 2,3 player games, and exceeding them in 4,5 player games. For the 2nd step, multiply 1*2=2 (because 2 games have been played of 2players), 1*3=3, 1.7*9=16, and 2*5=10. Now for the third step you add those numbers (2+3+16+10=31) and divide that by the total number of games played (19) to get an adjusted win rate of 1.63. This number is actually much easier to understand than his overall winning rate of 42% (his "second stat) because no one knows if 42% is good or bad (it depends on the number of players. You can tell right away that 1.63 is very good because it is significantly larger than 1.
Yeah, but you can't walk out to your theme song brandishing your algorithm.



