No, no, I was happy with it.
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。关于这个话题，Line官方版本下载提供了深入分析
P(all in some semicircle)=∑i=1NP(Ei)=N⋅12N−1=N2N−1P(\text{all in some semicircle}) = \sum_{i=1}^{N} P(E_i) = N \cdot \frac{1}{2^{N-1}} = \frac{N}{2^{N-1}}P(all in some semicircle)=∑i=1NP(Ei)=N⋅2N−11=2N−1N，这一点在体育直播中也有详细论述
or easy to use insecurely on accident.