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I am building a photo challenge website. To get the winning ratio, I am using the following equation

ratio = number of win (number of win / number of lose) * 100

the problem is, I get the same winning ratio for the following photos

photo a: 1 win 1 lose photo b: 2 win 2 lose

so now I changed my equation to

number of played + number of win.

photo a gets 3 and photo b gets 6

am I going to have a problem with my new equation? I am trying to confirm it.

p.s : someone please add tags for me...I can't create new tags

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Relevant: "How Not To Sort By Average Rating" by Evan Miller. Reddit uses this to sort comments by "best" ranking. – Rahul Mar 23 '11 at 20:56
@Rahul Narain // I think that is exactly what I was looking for. Sigh...the equation looks hard...thanks!! – Moon Mar 25 '11 at 5:31
up vote 1 down vote accepted

Your new system is going to give the photos which have been compared often high scores. A photo which has lost 100 times gets a higher score than one which won 10/10. Is that what you want?

Ranking after a sequence of noisy pairwise comparisons is a common issue, and since winning may not be transitive, there may be no perfect solution. However, a step forward may be to use something like the Elo rating used in chess.

When chess players play, the difference in their ratings suggests the chance that player A beats player B. Rating points are wagered (so the winner's gain equals the loser's loss) as though that suggestion is a fair gamble. A player rated 200 points higher might be expected to win 75% of the time, and would gain 1/3 as much by winning as he or she would lose.

New players are given a provisional rating which may start at 1500, and which is allowed to move more rapidly for the first few matches.

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If you wanted a user's percentage of won games, you should try (# of win/# of total) which is (# of wins/(# of wins + # of losses)) assuming winning and losing are the only two possibilities.

If your problem is that the percentage itself treats brand new users the same as long-time users that just happen to win at the same rate, then you can't solve this just by the percentage alone. You need another measurement that tells you the confidence in the percentage.

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