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		jprete on April 3, 2024 \| parent \| context \| favorite \| on: Avoid blundering: 80% of a winning strategy 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 on April 3, 2024 [–]

> 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.

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