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Side question about error total for cube actions

Posted By: Timothy Chow
Date: Monday, 15 February 2010, at 5:11 p.m.

In Response To: Side question about error total for cube actions (Gregg Cattanach)

Gregg Cattanch wrote:

Shouldn't the error for dropping the cube be 1.000-0.187 = 0.813?

That would be the magnitude of the error by the opponent, not the magnitude of your error. It makes sense to me that my errors ought to be separated from my opponent's errors.

When totting up the opponent's errors, then 0.813 is the correct figure. I assume that XG calculates this correctly when you ask for the opponent's errors.

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