Что думаешь? Оцени!
Nothing in the argument required the arc to be a semicircle. If the arc has length xxx times the circumference (where x≤1/2x \leq 1/2x≤1/2), each anchor's event has probability xN−1x^{N-1}xN−1 instead of (1/2)N−1(1/2)^{N-1}(1/2)N−1. Mutual exclusivity still holds because the gap is at least 1−x≥1/21 - x \geq 1/21−x≥1/2 of the circumference, too large for any other anchor's arc to bridge:。关于这个话题，快连下载安装提供了深入分析
，更多细节参见夫子
Added symbol-append to (hoot symbols).，这一点在体育直播中也有详细论述
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