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I have two data points

X: Number of Games Played (lowest: 1 highest: 1000)

Y: Average of all game scores[sum of all game scores / number of games played] (lowest: 4 highest: 6.5)

I don't have record of the individual game scores, I only have the average.

How can i generate a score based on these two variables, to make a good comparison mechanisms between multiple entities that have different x,y values? in a sense that when [x=1,y=6.5] is worse[has a lower score] than [x=5, y=6.1] or something similar

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  • $\begingroup$ First, you have to decide what you want your score to do. Do you want [1, 6.5] to be better, or worse, than [2, 4]? Where do you want [1, 6.5] to fit in among [2, 4], [2, 5], [2, 6]? Where do you want [1000, 4] to fit in among [500, 5], [500, 6], [500, 6.5]? There are infinitely many ways to generate a score, but you have to decide what you want the score to do first, before you can decide what formula to use. $\endgroup$ Commented Apr 27, 2018 at 7:06

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One way to correct the average is to add some "dummy" low scores (e.g. $10$), so

$$ A' = {X \cdot Y + 4 \cdot 10 \over X + 10 }. $$

People with a large number of answers see their modified percentage alter very little from their real percentage, but people with relatively few answers will see their modified percentage move considerably toward low values.

This is known as "Bayesian averaging". In effect, the people with many answers will rank higher than people with the same percentage but fewer answers.

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  • $\begingroup$ To be clear, both the 10s are dummy? $\endgroup$
    – DarkMental
    Commented Apr 27, 2018 at 8:53
  • $\begingroup$ Just tested it, and for games with same average, it doesn't produce accurate results. If x is equal then the score is bigger if y is bigger. $\endgroup$
    – DarkMental
    Commented Apr 27, 2018 at 9:25
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    $\begingroup$ Yes, it's right. If you have two 'players' with the same average score, the best is the one with more matches played. If they played same number of matches, the best is the one with higher average. It's accurate indeed! :) And yes, the $10$ is the same (it's just a weighted average). $\endgroup$ Commented Apr 28, 2018 at 9:35

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