3 ms·
It's possible, but that formula is (after a log2 transformation) equivalent in comparison power to log2(mistakes) + blunders. This is almost reducing the mistak
by jprete 3y ago
It's possible, but that formula is (after a log2 transformation) equivalent in comparison power to log2(mistakes) + blunders. This is almost reducing the mistake term to a tiebreaker, because mistakes and blunders are on the same scale (a proportion of the number of actions taken in a game).
- rKarpinski 3y ago> This is almost reducing the mistake term to a tiebreaker Isn't that exactly what it's supposed to be? >> However, in 40% of the games both players had an equal number of blunders. So I also included “mistakes”—the next-worst kind of error.