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Need a formula to compare percentages based on different amounts (weighted?)

Posted on 2008-10-27
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Last Modified: 2012-05-05
Okay, I probably knew how to do this at one point, but I haven't cracked a math book open in quite some time. How can I fairly compare averages based on different amounts?

Say a runner runs 100 races and wins 25 of them. He has a winning percent of 25%. But if another runner runs just 10 races and wins 5, he has a winning percent of 50%. Now, the second racer has a higher winning percentage, but the first runner has won more races and could be construed as a "better runner".

I know I could extrapolate and assume that if the second racer runs 100 races, he will win 50 of them, but I'm looking for a more accurate comparison of what they've actually done and not what they may do.

Is there some formula or theory I can use?

Thanks in advance

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Question by:StarbuckLives
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7 Comments
 
LVL 18

Expert Comment

by:deighton
ID: 22817066
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Author Comment

by:StarbuckLives
ID: 22817987
This looks like it's about figuring out the reliability of an estimated number, not the comparison of two accurate percentages.
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LVL 27

Expert Comment

by:aburr
ID: 22819166
"looking for a more accurate comparison of what they've actually done and not what they may do."
-
There is no better way. It depends on what you are looking for. If you want a runner who was won the most races, take the first one.
If you want the runner who wins the highest proportion of his (or her) races , take the second one.
You are asking for two different things.
Consider baseball players and their batting averages. some players have a large number of hits but still have a (slightly) smaller batting average. Which one do you trade for?
If a player has only a few at bats, people tend to discount a large batting average. After a certain number of at bats, the precise number of at bats tends to be disregarded.
In the case of your runners, what or the quality of the races. The 10 race guy might have enter only races in Podunk. The 100 race guy may have run only in the olympics and world championships.
I know a golfer who has won several tournements. He could not get into the PGA if he tried.
You are asking for two different things (albeit related). There is no more accurate comparison
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LVL 27

Expert Comment

by:aburr
ID: 22819178
The confidence limit will help you out a bit. If you  put a statistically significant + or - on the percentage you can see if they overlap. If they do there is little difference between them.
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LVL 18

Expert Comment

by:deighton
ID: 22820413
there isn't really any reasonable extrapolation, A may improve and win 75 more races, but there's no evidence to support this any more than him losing all future races, so you can only really calculate the probability that B's .5 rate was a blip, and his underlying win rate is lower.  whether or not this is true binomial isn't clear though, because strength of field is critical.

I would recommend looking at personal best times, strength of field etc, but not forgeting that fast and slow races occur for diferenet reasons.


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LVL 3

Accepted Solution

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jtm111 earned 500 total points
ID: 22835657
But if you're talking about aggregating the result of multiple runners, then you certainly would use the weighting scheme you mentioned.

For example, if you had 5 runners:

Runner    Wins    Number of Races
1               10               50
2                5                10
3                5                20
4                2                 3
5                1                 1

If I wanted to express the aggregate performance of the team, I'd weight the percentages by the number of races:

10/50 * 50 +
5/10 * 10 +
5/20 * 20 +
2/3 * 3 +
1/1 * 1

Then divide the whole thing by the sum of the weights (50 + 10 + 20 + 3 + 1)

That way you get the answer: 23/84 = .274 or 27.4%

This reflects the proper weighting of each member of the group.

If not the answer you're looking for, maybe enough to get you thinking in the right direction.

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Author Closing Comment

by:StarbuckLives
ID: 31510523
This is more what I was looking for, thanks.
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