EPTFrom now we will be working under the assumption that the better player has a probability p of winning any single game, and the worse player has probability 1-p. Of course p ≥ 0.5, and the games are assumed to be independent.
(By the way, the answer to the original question is here.)
Thus, the relevant probabilities are:
- 2-0 for better player: p²
- 2-1 for better player: 2p²(1-p)
- 2-1 for worse player: 2p(1-p)²
- 2-0 for worse player: (1-p)²
[p²+(1-p)²] / [2p²(1-p) + 2p(1-p)²]
of matches ending 2-0 to matches ending 2-1. If we write q = 2p(1-p), we find that the above ratio can be expressed as
(1+q)/q = K
So I went through the last five seasons of code A. There are 52 matches per season, but there were four walkovers, so I looked at 256 games in total. Of those, there were 101 2-1's and 155 2-0's. Thus we have our estimate for K:
K = 155/101 = 1.5347.
Solving the equation for q, we find
q = 1/(K+1) = 0.39453.
To find p, write p = 0.5 + r. Then we get q/2 = p(1-p) = 0.25 - r², or in other words
r = sqrt[0.25 - q/2] = 0.22964,
so
p = 0.72964.
What does this mean in practical terms? In an average code A match, the better player is so much better that he has about 73% to win a single game over his unfortunate opponent.
Let's translate this to a rating gap. With a rating gap D, we have
p = 1/(1+10^(-D/400))
or
D = -400 log(1/p-1) = 172.47.
So the average skill gap in code A is a whooping 172 points! That's quite a lot.
Disclaimer: I know this is back-of-the-napkin maths and I've pretty much disregarded all nonlinear effects. "Average" here does not mean "mean."
Edit: Did the same for code S.
K = 124/116 = 1.069
p = 0.591287
D = 64.152
As you can see, it seems a lot closer.
