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Ok I might be asking something stupid, please bear with me, math is not my strong suit. This has to be a duplicate but I dont know how to search them. I dont even know how to word the title, kindly help me there too.

Question

I was trying to find the average win-loss record of a player over a period of 4 years. His win-loss record looks like:

19/23, 13/16, 21/24, 19/22

which means in the first year he won 19 of his 23 matches, in the second year he won 13 matches and lost 3, in the third year he played 24 times and won 21 times, and so on.

I want to find his consistency and hence I am looking for his average winning rate over the 4 years.

I thought it was quite straightforward but quite surprisingly it occurred to me that I can go about it in two ways when I found some discrepancies in the way I calculated it:

  1. Add up all the wins during the 4 year stint and divide by the number of total matches played during that time.

    So my player's consistency is measured like

    (19 + 13 + 21 + 19) / (23 + 16 + 24 + 22) 
    

    which is 84.7%

  2. Finding the average of win percentage of the 4 samples.

    So my player's consistency is measured like

    (19/23 + 13/16 + 21/24 + 19/22) / 4
    

    which is 84.43%

Well at first being so naive I thought either way I should get the exact same result, but no! I then thought there must be some rounding issues in the second case which is why I get a different but close number. I tried to form a theoretical basis like shown below.


Let's say a player won "a" matches out of "b" matches played in first year, and in second year he won "c" matches out of "d" matches played. His consistency can be measured by the above two models like:

  1. (a + c) / (b + d)
    

    or

  2. (a/b + c/d) / 2    =>    (ad + bc) / 2bd
    

Clearly the two aren't the same. So my questions are:

a. What is the practical difference between the two ways to calculate consistency? I'm not asking for a strict mathematical definitions, but trying to know in what real world scenarios the two different approaches make sense? The nature of my question is similar to this question. I want to know how to calculate a player's consistency over a period of time and which model fits where.

b. Are there any technical terms in math world to denote the two ways of assessing average?

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The difference between the two results is a consequence of the fact that he played different number of games in different years.

If every year he would play the same number of games, the two results would be the same.

Now to understand the issue, think about a simplified model. Lets say that the player played 100 games the first year, and 200 games the second year. When you calculate his consistency. how do you take into account that the second year the number of games doubled?

The first way of calculating is saying that since the number of matches is doubled, then his consistency for this year is twice more important than in the first year.

The second way is saying that since you only care about the consistency from year to year, the number of matches is irrelevant, and the consistencies in the first and second year are both as important as the other.

As for notation I think that the first method is called an average, while the second is average of average.

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  • $\begingroup$ Oh man, you really opened my eyes. Simple explanation. Now I feel stupid. It is indeed the first model I'm after! Thanks :) $\endgroup$ – nawfal Jul 26 '15 at 20:50
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You might be interested in reading about Simpson's paradox (easily found online). It can happen, for example, that A has a better average every single year than player B does, yet player B has the better combined average. For example:

$$A:\;\;\;\left[\frac23,\frac{5}{10},\frac{5}{10}\right]$$ $$B:\;\;\;\left[\frac{12}{20},\frac13,\frac25\right]$$

Then, as promised, A has a better average in each of the three years. Combined, however, A won 12 out of 23 for a 52.17% average while B won 15 out of 28 earning a 53.57% average.

The point, of course, is that B's combined average was dominated by a great performance in the first year...a year in which A got a better average in a very small number of games.

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  • $\begingroup$ Oh thanks for that. Statistics is not an easy game I realise! $\endgroup$ – nawfal Jul 26 '15 at 20:56

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