4 ms·
> I computed a simple “error score” that includes mistakes while giving blunders more weight: [number of mistakes] × 2*[number of blunders] This doesn't give
by upwardbound 3y ago
> I computed a simple “error score” that includes mistakes while giving blunders more weight: [number of mistakes] × 2*[number of blunders]
This doesn't give blunders more weight..
Multiplication is commutative and associative, so the author's formula is also the same as this one: 2*[number of mistakes] × [number of blunders]
To give more weight to blunders, you could use an exponent, which is a common trick in baseball statistics (SABRmetrics), like this: [number of mistakes] × [number of blunders ^ 1.1]
See e.g. some of the formula concepts invented by Bill James https://en.wikipedia.org/wiki/Bill_James#Innovations https://en.wikipedia.org/wiki/Bill_James#Innovations
- rKarpinski 3y agoThink the star represents exponent, otherwise they are using two different symbols for multiplication.
- jprete 3y agoIt'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.
- smartbear 3y ago(Author of article) Sorry, that was supposed to be +, not * ! You're right. Just a typo! Now fixed.