Conquer Club

Win Rate Suggestion

Suggestions that have been archived.

Moderator: Community Team

Win Rate Suggestion

Postby The Neon Peon on Fri Jan 16, 2009 10:59 pm

Concise description:
  • Add a "Win Rate" feature as part of the site, or a user script. Really simple to code, especially since we have code for map rank already.

Specifics:
  • Here is how the percentage system could be adjusted:

    Games won with Y players (2v2 counts as 2 player, etc.) x Y x Relative Rank for those games
    Add all of these up and divide by total # of games that person has played

    This gives us a percentage that shows how far one of us is above or below the normal. For example, a 1.00 would mean that you win as many games as probable. 2.00 would mean that you win twice as many games as probable.
  • NEW FEATURE (edited into the original post): Terminator games will be counted as "number of opponents eliminated / ( total players in the game - 1 ) " This way, if you eliminate all people, you get full credit, but you get credit for kills, not winning. This simply means that you double the equations and run the first ones on standard-assassin, and the next batch on terminator.
  • Imagine a player who does the following:
    Has won 750 out of 1000 two player games against people of half his rank.
    Has won 10 out of 100 six player games against people of his own rank.
    Plays no other games for simplicities sake of this calculation (seeing as I have accidentally deleted this entire post twice already)

    His win ratio is currently 69% (Good, right? If you did not know anything else about him)

    His Win Ratio would be the following:
    ( 750 (2 player games won) x 2 x .500 + 10 x 6 x 1.00 ) / 1100 (Total Games played) = .74

    What does this mean?

    Well, if all games were determined by luck, and the score system was accurate, then everyone's win ratio would be exaclty 1. However, some people are better than others, and so you do not win exactly 50% of your two player games. So what this statistic does is take the normal win percentages, but take into account that you are more likely to win a 2 player game than an 8 man one, that you might play more 2 player than 8 man games, and that you might be farming (which only affects your score at the moment, and helps your win rate).

    My equation would say that the person wins only .74 of the games he should win. Take the next example:
  • A player has the following statistics:
    Has won 25 out of 100 two player games against people double his rank.
    Has won 33 out of 99 three player games against people of his own rank.
    Has won 25 out of 100 eight player games against people of half his rank.

    25 x 2 x 2.00
    33 x 3 x 1.00
    25 x 8 x .500

    Added and Divided by a total of 299 games played = 1.00

    Although he wins only 25% of his games in two player, he plays against people of double his skill, so it does not matter. Although this player farms on 8 player games, his win rate is not positively affected by that 25% win percentage on them. He wins double the games he should against people of half his rank, half the games he should agains people double his rank, and the exact amount of games he should win against people his own rank. Therfore, his win rate is exaclty 1.00
  • Before, the player would have had a win percentage of 28%. This stuff tells you much more that simply 28%, doesn't it? I suggest you run it on yourself, even if it does not get updated to the site. I am a major right now, but my win rate is only 1.08, so not that many more games won than I should win, is it?

This will improve the following aspects of the site:
  • A way more accurate statistic than win percentages
  • Low rank farming will no longer positively affect your win rate (let's face it, a lot of them are really good at it. Possibly square the relative rank?)
  • Another statistic. Always good.
  • I was kind of interested to find out about this myself. Run it on myself every so often, interesting to see how good you actually are rather than how many points you can accumulate.
Last edited by The Neon Peon on Sat Jan 17, 2009 10:59 am, edited 2 times in total.
User avatar
Lieutenant The Neon Peon
 
Posts: 2342
Joined: Sat Jun 14, 2008 12:49 pm

Re: Win Rate Suggestion

Postby Artimis on Sat Jan 17, 2009 3:55 am

The win rate as it stands is misleading, I second this suggestion.
==================================================
This post was sponsored by Far-Q Industries.

Far-Q Industries: Telling you where to go since 2008.
User avatar
Captain Artimis
 
Posts: 810
Joined: Sat Oct 11, 2008 9:09 am
Location: Right behind ya!!! >:D

Re: Win Rate Suggestion

Postby The Neon Peon on Sat Jan 17, 2009 6:16 pm

Anyone have any thoughts on this?
User avatar
Lieutenant The Neon Peon
 
Posts: 2342
Joined: Sat Jun 14, 2008 12:49 pm

Re: Win Rate Suggestion

Postby Jumentum on Sat Jan 17, 2009 7:05 pm

good idea, and well thought out too
although your points sort of also take into account how many more difficult games you won
but the calculations would make it a lot easier to tell, esp if you want to know the difference between a skilled noob and a 3year vet who just played thousands of 2v2s
User avatar
Corporal 1st Class Jumentum
 
Posts: 49
Joined: Thu Jun 12, 2008 8:08 am
Location: Boston, USA

Re: Win Rate Suggestion

Postby iambligh on Sun Jan 18, 2009 12:56 am

I agree -- there are tons of players who boast about their win percentage as though it's a real indicator of how good they are. Meanwhile, they spend most of their time playing the same map heads-up farming noobs!
Sergeant 1st Class iambligh
 
Posts: 144
Joined: Thu Jul 17, 2008 7:15 pm

Re: Win Rate Suggestion

Postby FarangDemon on Tue Jan 20, 2009 12:53 pm

I think it is a great idea but I think you are trying to build too much on top of a foundation that has one big problem:

if you win one 8-player game you get an 8.

if you win three 2-player games you get an 2.

Given that you are as likely to win as any your opponents, you have a (1/2)*(1/2)*(1/2) = 1/8 chance of being able to win three 2-player games. Thus, since the probability of winning three 2-player games is equal to the probability of winning an 8-player game, any measurement that hopes to reconcile the disparities of the current win percentage metric with regard to number of opponents in each game should give equal scores to players accomplishing equally improbable wins.

After this can be reconciled, and I have suggested a way to do this (below), it would be really interesting to take RR into account in the manner that you have suggested.

http://www.conquerclub.com/forum/viewtopic.php?f=4&t=65998&start=45

This is a method for quickly approximating how much better than expectation you were able to win and it works for all numbers of opponents. The output is a Z-score which you can look up in a table to determine the probability that a player (that wins as equally as likely as his opponents) does as good as you did.

Normal Approximation to Binomial Distribution:

n = total games
k = games you have won
p = (product of all 1/x's for every x-player game you have played) ^ (1/n)
q = 1-p


Z = (k - np) / sqrt(npq) - look up in Z table to get percentage
User avatar
Brigadier FarangDemon
 
Posts: 700
Joined: Wed Apr 23, 2008 1:36 am

Re: Win Rate Suggestion

Postby The Neon Peon on Tue Jan 20, 2009 5:32 pm

FarangDemon wrote:I think it is a great idea but I think you are trying to build too much on top of a foundation that has one big problem:

if you win one 8-player game you get an 8.

if you win three 2-player games you get an 2.

Given that you are as likely to win as any your opponents, you have a (1/2)*(1/2)*(1/2) = 1/8 chance of being able to win three 2-player games. Thus, since the probability of winning three 2-player games is equal to the probability of winning an 8-player game, any measurement that hopes to reconcile the disparities of the current win percentage metric with regard to number of opponents in each game should give equal scores to players accomplishing equally improbable wins.

Firstly, if you win three 2 player games, you get a 3 x 2 x RR = 6. Not a 2.

Secondly, the statistic you showed for three two player game victories is off in that your statistic is that of winning three of them in a row, not simply winning them.
User avatar
Lieutenant The Neon Peon
 
Posts: 2342
Joined: Sat Jun 14, 2008 12:49 pm

Re: Win Rate Suggestion

Postby FarangDemon on Tue Jan 20, 2009 7:34 pm

The Neon Peon wrote:
FarangDemon wrote:I think it is a great idea but I think you are trying to build too much on top of a foundation that has one big problem:

if you win one 8-player game you get an 8.

if you win three 2-player games you get an 2.

Given that you are as likely to win as any your opponents, you have a (1/2)*(1/2)*(1/2) = 1/8 chance of being able to win three 2-player games. Thus, since the probability of winning three 2-player games is equal to the probability of winning an 8-player game, any measurement that hopes to reconcile the disparities of the current win percentage metric with regard to number of opponents in each game should give equal scores to players accomplishing equally improbable wins.

Firstly, if you win three 2 player games, you get a 3 x 2 x RR = 6. Not a 2.

Secondly, the statistic you showed for three two player game victories is off in that your statistic is that of winning three of them in a row, not simply winning them.


( 750 (2 player games won) x 2 x .500 + 10 x 6 x 1.00 ) / 1100 (Total Games played) = .74

Aren't you dividing by total games played? So you should have 3 * 2 * 1 / 3 = 2

1/2 * 1/2 * 1/2 is the probability of winning three 2-player games out of three possible 2-player games. They don't need to be in a row. There can be different events in between.
User avatar
Brigadier FarangDemon
 
Posts: 700
Joined: Wed Apr 23, 2008 1:36 am


Return to Archived Suggestions

Who is online

Users browsing this forum: No registered users