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以下是输给那破游戏不甘心后的乱想..如有错误请提出或无视.. 首先是豹子问题/ S8 Z% b0 ]2 o3 i+ K2 r$ p
$ O9 R$ j) f, E" ]9 e* s; M; j8 x因为太郁闷..所以从排列组合开始... f3 c0 T M! P
% U2 Z: `: b: d* e假设直线A上有N个点,取任意点都有相同程度的概率.那么有2个取点者同时随即取点E,2者同时取得点E的概率是E/N*N$ N/ J1 P8 B6 R7 S! F# M) ~! K/ I
- O h R+ Q* h$ `3 p
那么豹子的出现概率是2/216=0.00926=0.926%
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豹子的赔率是30那么收益期望为0.2778风险为99.074%5 Y6 B: M2 ~( q3 Y# Y1 J$ q" {5 z
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连续2次出现率为42.67%的负4次方(大概是大型超市抽奖活动的特等奖的概率..有信心的可以玩下)
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在是压大小问题.3 ~1 q$ g: C% _" z9 i) j( r# e3 i
" I7 \: \6 e6 a根据以上得出每组大小的概率是49%..- P; G+ i( }% G0 C) C- v" B
* m4 b6 d6 ?, H* p+ w* n2 r! ?* @0 k由此可见..压大小就总合来说是必败的. t/ [, D4 a0 \3 \$ d
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心理学中有个叫赌徒效应还是啥啥的..是说随即序列发生一个事件的概率与之前发生的事件无关。即发生的概率并没有因先前发生的事件而增加。
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5 [" R' ]/ v, K/ @也就是说,压大小实际上永远就是49%概率压对。6 @1 N4 P3 n& `: R5 S
+ ?% g# D) k n5 N) o) j) ]0 L并非这次压错了.下次就是50%了6 s5 i: t( F- c5 U9 G" K5 ?
, y2 Q$ y+ @- u, r' F由此可见压大小以综合来说必输.压豹子其收益价与风险值不成比列。0 X6 ]8 a2 u9 @5 i+ O
4 g% k/ I9 {+ F: Y7 D* z根据以上分析.还是奉劝大家!远离DB!
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还有..以上是有实际验证的~昨天赌了上百回资金流动大概有成千上万..损失了进两百
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今天中午又赌了N次..资金流动上千..损失了30左右..均与以上概率成比例~ 嘛- -偶赌的还算比较厉害了..输的惨就不去说了~~那位上千的全没了
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" W a" y" ^1 p- }/ T+ N: ~ K$ l偶决定在也不赌了~真的...( y" t& u0 j/ I8 ?) j- J% w
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