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百家樂高胜术 2.0 三路齐进时各路取胜的或然率& v8 T2 @5 e0 F* K: U, w! C. F 
6 t# J  S, V) A6 L 
  百家樂高胜术是三路齐进的, 让我从第一路开始分析;此法是以20铺内定输赢, 故先让式我们看看20铺内的各种组合, 以及此种组合的或然率与其回报的关系。 20铺内连负五口的或然率:1/32x20铺=0.625铺, 连负五口的回报 (1,4,8,13,27,0计):-$53, 即此种组合的回报是:-$53x0.625=-$33.125 20铺内连负四口才胜的或然率:1/32x20铺=0.625铺, 连负四口オ胜的回报:+$1, 即此种组合的回报是:+$1x0.625=$0.625 为什么此两种组合的或然率相同?因连负五口, 不需理会第六口是否胜负, 断缆便成立。但连负四口后, 必须要胜, 此种组合オ成立, 故两者的或然率部是1/32。 20铺内连负三口才胜的或然率:1/16x20铺=1.25铺, 此组合之回报是0 20铺内连负二口オ胜的或然率;1/8x20铺=2.5铺, 此负ニ口才胜组合之回报是+$3,此组合之回报是3x2.5铺=+$7.5 20铺内负一口后胜的或然率:1/4x20铺=5铺, 负一口才胜组合之回报是+$3, 此组合之回报是:+$3x5=$15 20铺内胜一口后胜的或然率:1/4x20铺=5铺, 胜一口后胜组合之回报是$1, 此组合之回报是:+$1x5=+$5 此路(20铺)内, 固报是:-$33.125+$0.625+$0+$7.5+$15+$5=-$5 故只行1路, 在数学上所推论是要输的, 但如果三路齐去, 结果便大大不同 ,因断第一路时不能再断ニ,三路, 即二及三路断缆的可能性递减, 以下以数学论之; 在断一路时. 二路只有16铺的空间可供其断缆, 三路只有12铺的空间可供其断缆。 
5 [( S2 N5 \3 Y' V5 J- @  B) l9 J7 \) P% l3 O 
  在有收入的组合的或然率不变而损失的组合的或然率收缩下, 本法的生存空间便存在! 第二路连负五口的或然率:1/32x16铺=0.5铺, 回报:-$53x0.5铺=-$26.5 第二路的回报是:-$26.5+$0.625+$0+$7.5+$15+$5=+$1.625 第三路连负五口的或然率:1/32x12铺=0.375铺, 回报:-$53x0.375铺= -$19.875 第三路的回报是:-$19.875+$0.625+$0+$7.5+$15+$5=+$8.25 结论: 第一,二及三路共回报:-$5+$1.625+$8.25=+$4.875 此基础上再加上趋势分析(庄间系数), 只玩庄多间小, 断1追+10, 等便得到每靴牌平均回报+9的结果& Z6 D+ Y7 a9 i$ E' [) i 
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