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通过我几个月来对珠路理论的研究,+ X7 E! N7 A; a
2 l* b8 k# O- t8 Z, N; y' e& p
发现了一个有趣的现象,
( i+ G$ ^1 Y; c3 c9 {6 |- W每一条大路,
( E$ f( P% y {/ n% N: ~+ m按2珠路排列,有2种不同的路数! j$ ~& L6 t6 s ?3 Z
按3主路排列,有3种不同的路数, y3 W# |6 H: X" V3 O! O( y
按4珠路排列,有4种不同的路数
2 U( r" Y0 r8 R, E, Q' p2 b! t" I按N珠路排列,有N种不同的路数。
' }2 Q0 M8 ~' b8 _$ K0 T6 C" n什么意思呢,可能大家一时不清楚,
' _4 A1 t. {. ?* R: ?我举一个例子,假设 B= 1 , P=2,T暂且忽略吧。
s7 A( L* A0 G. w& J$ M5 @假设大路开这样的路:! O0 ~ U% [. [* ~+ `
121212121212121212121212121212129 k' x( B: @# M a! L* k7 F6 S: K
够极端了吧- x2 o; ]% `/ Q) X! K* Y
按2珠路,就是BPBPBPBPBP.....3 y# s- b- k! L, G
如果我们去掉第一口,就会出现完全相反的结果:3 a5 _1 _# S7 D. b2 G
21212121212121212121212121212121" c4 x6 d2 a2 Q3 A1 ` S/ W, U3 Q
变成了PBPBPBPBPB.......- A; A+ ?$ U6 f' D( t; T, Z
如果我们再去掉一口,又返回第一种情况了。
/ Y2 h! _2 Y" ]$ P# @; N6 Y. B所以每一条大路,) t6 l0 c" B4 S1 D
按2珠路排列,有2种不同的路数。* q' V1 w+ g+ m- e
再举一个列子:" v9 _3 c! w9 g7 q! }/ r
大路:122122122122122122。。。
6 r- r5 ?0 l t: \! Z按三珠路排列:
+ n" A. c+ p0 U- d: [$ ]" Z122,122,122,122,122,122。。。
( _6 U) g6 _0 I6 |$ Z+ S, k去掉第一口,变成:
; X _" H1 q3 Q# n1 ]$ o221,221,221,221,221," |6 F7 @8 W+ i; V
去掉前2口,变成:; ]1 [; F1 ^' w
212,212,212,212,212,
/ x) g; l5 n6 O! S, ?7 T% w去掉3口,又返回122,122,122,了+ I5 ]& R9 p9 D8 }; e' J* s$ {
所以每一条大路,2 Q( E# K, T7 |: ]$ h+ t+ Q
按3珠路排列,有3种不同的路数。6 d/ n% N% o$ m1 k" U! x( F; q
同理:按N珠路排列,有N种不同的路数。: p3 l2 s3 e. ?- D
我提出这个的意义在于:
4 A7 F0 [% z1 } ]: |$ j- X# g字串8
8 z1 H* c+ p/ X( N; K$ M3 @, F. b: K' e4 k& X: R1 M
1、每靴的第一口为起点来编排二三珠路,与第2口,第3口开始的排列是不同的结果。5 k J! S# _: [6 P2 L7 {) U' @; q
2、为三多理论提供了下注的多面性奠定基础。
1 @; E' m4 K! H) @0 `, U% S3、某一靴牌,按第一口开始的珠路可能是 烂路,按第2口,第3口开始的珠路可能是 上上路.
1 ?# F& _$ p0 h' t- U# | |
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你的理论打法不错的呀,赢钱了应该是打法对你有用哦~来收藏下了解下了 |
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我现在还是研究一下理论打法,感激楼主的分享,我也来学习 |
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楼主估计也是一个理论很内行的玩家,谢谢老哥的分享。 |
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