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Openai/69286a1e-2f50-800a-b685-0647e6b12a18
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=== 设每次交易: === * 以概率 ppp 获得倍数 Mwin>1M_{\text{win}}>1Mwin>1(赢时本金乘以这个数); * 以概率 1−p1-p1−p 变为 MlossM_{\text{loss}}Mloss(亏损时本金乘以这个数,通常 0<Mloss<10<M_{\text{loss}}<10<Mloss<1)。 在固定做 nnn 次交易后,若恰好有 www 次是“赢”,最终本金为 Bn=10×Mwin w×Mloss n−w.B_n = 10 \times M_{\text{win}}^{\,w} \times M_{\text{loss}}^{\,n-w}.Bn=10×Mwinw×Mlossn−w. 要求 Bn≥1,000,000B_n \ge 1{,}000{,}000Bn≥1,000,000,等价于 MwinwMloss n−w≥100,000.M_{\text{win}}^{w} M_{\text{loss}}^{\,n-w} \ge 100{,}000.MwinwMlossn−w≥100,000. 对固定 nnn 可以解出需要的最小赢数 wmin(n)w_{\min}(n)wmin(n): wmin=⌈ln(100,000)−nln(Mloss)ln(Mwin)−ln(Mloss)⌉.w_{\min}=\left\lceil \dfrac{\ln(100{,}000)-n\ln(M_{\text{loss}})}{\ln(M_{\text{win}})-\ln(M_{\text{loss}})}\right\rceil .wmin=⌈ln(Mwin)−ln(Mloss)ln(100,000)−nln(Mloss)⌉. 然后成功概率为二项和: Pr(达标 in n)=∑w=wminn(nw)pw(1−p) n−w.\Pr(\text{达标 in }n)=\sum_{w=w_{\min}}^{n}\binom{n}{w} p^{w}(1-p)^{\,n-w}.Pr(达标 in n)=w=wmin∑n(wn)pw(1−p)n−w.
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