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So let me get this straight
Your idea of an example is a scientific impossibility?
100% does not exist in real life due to the inductive way we gather information.
And yes ELO works iterative… No shit.
Nobody said it is perfect. That wasn’t my claim. My claim was that your method does not work, because it doesn’t do what it sets out to do (what elo does).
But lets go into more detail why your method is BS.
You take an average (you don’t even explain what you mean by average, that isn’t a mathematical term, my guess is you mean the mean) of the players in a team.
Buth this is the elo distribution on a chess website:
https://images.app.goo.gl/rRNpGjyftqSSV1z37
WT would be similar and as you can see it is very much skewed. Since you use the arithmetic mean on a skewed distribution, there is already a red flag, since the arithmetic mean only works as an estimator for symmetric distributions.
Here the measures are explained and why the arithmetic mean doesn’t fit skewed distributions.
It’s even more skewed if you take into account that ELO does not work linear. So you would treat a 2 player per team game as follows:
Team A consists of a 1000 and a 2000 elo player.
Team B consists of both players being 1500 rating.
Your system would only work if the chance of a 2000 pts player beating a 1000pts player with a chance of 2:1. It’s actually a 99.96% chance for the 2000 point player to win.
1500 beats 1000 with a chance of 94.61%
2000 beats 1500 with a chance of 93.97%
So lets look at it as a Deathmatch dogfight where the opponents split up (which is a fair assumption since leaving on opponent alone is an almost instant fail and has no longer to do with the skill levels also the competent pilot would always engage the split, it’s fighter doctrine anyway) and look at the likelyhood of outcomes.
1000(A) beats 1500(B) and 2000(A) beats 1500(B) - A wins = 5.06%
1500(B) beats 1000(A) and 1500(A) beats 2000(A) - B wins = 5.70%
1000(A) beats 1500(B) and 1500(A) beats 2000(A) after that 1000(A) beats the remaining 1500(B) - A wins = 0.02%
1000(A) beats 1500(B) and 1500(A) beats 2000(A) after that 1500(B) beats the remaining 1000(A) - B WINS = 0.31%
1500(B) beats 1000(A) and 2000(A) beats 1500(B) after that 2000(A) beats the remaining 1500(B) - A wins = 83.54%
1500(B) beats 1000(A) and 2000(A) beats 1500(B) after that 1500(B) beats the remaining 2000(A) - B wins = 5.36%
So we get a chance of A winning to be roughly 88.62%
While your system thinks it’s 50%. It’s not as easy as just taking the arithmetic mean.
To give you an idea, the team rating for Team A in that composition should be around 1900.
So your system fails at what is supposed to do. It does not calculate the win chances properly.
As i said mathematically it doesn’t work, if it did it would be used in football, as it would be more precise than the team ratings.
