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[lastpost] => 2776400 【星宝】10月25日投注流水230053元 1761376401 rainwang
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通过我几个月来对珠路理论的研究,发现了一个有趣的现象:每一条大路,按2珠路排列,有2种不同的路数;按3主路排列,有3种不同的路数;按4珠路排列,有4种不同的路数,按N珠路排列,有N种不同的路数何解?我举一个例子,假设 B= 1 , P=2,T暂且忽略吧。' X) [# Z4 Y7 Z+ o
. R5 Y6 U& p$ r0 f+ } T k 假设大路开这样的路:5 w! W9 o* U/ S8 k5 B. W
) D! u2 a. z8 J! t6 {3 \
12121212121212121212121212121212。
1 t/ g- r9 _/ d( o$ f
# ~6 p# p; Q; B" X7 ~4 J$ H& P* r 按2珠路,就是BPBPBPBPBP。/ m6 Z) p2 _! g( i- t+ x. n% Z* ~! O
! k9 n" @8 \& [- `
如果我们去掉第一口,就会出现完全相反的结果:
* p2 u, s j( Q9 L0 M. x+ `( t. D6 P
5 V; r/ v" E0 z+ U. s 21212121212121212121212121212121。
$ o/ x; L: u( a, K/ q3 L V* x
9 z" }% t& J& R: `* i& S6 _ 变成了PBPBPBPBPB。, T$ a, u9 s. K( z" c# U
! V! Q0 a' V: L0 `1 Z& K# D 如果我们再去掉一口,又返回第一种情况了。0 _$ {. H% ^% d# @
% ?; N; E0 D, C" a! c0 F 所以每一条大路,按2珠路排列,有2种不同的路数。
# w8 O0 b- v6 j2 D2 c' X* ^2 T: y
5 E) b! i9 r4 h+ h6 h 再举一个列子:0 b6 M* u4 O- t! ~
6 c i* K# \# Q. y6 f& P! E3 l
大路:122122122122122122。
. v: z$ ]5 V e4 G2 q( U; m' x2 [, M9 ?( m( |* \! t/ H
按三珠路排列:
5 t8 ~, D& }0 t% S3 G- H1 n3 n# K( R8 U, k0 e! t) `5 R
122,122,122,122,122,122。
& M8 u e! g1 D5 X, r0 d, B0 P/ B- W8 i8 t/ ~1 |: u: x
去掉第一口,变成:
$ I9 @7 m8 W+ o: z6 ^6 p' ~2 l0 A
5 Y/ j% }8 z# p, {2 w 221,221,221,221,221,去掉前2口,变成:
' m8 \0 X: q @/ @
6 q" K8 d* z: @. L6 t& Q* H% W 212,212,212,212,212,去掉3口,又返回122,122,122,了所以每一条大路,按3珠路排列,有3种不同的路数。9 _6 t) P+ | _) S; c! j
8 ?- [3 Y- s( F9 W6 j' V 同理:按N珠路排列,有N种不同的路数。2 y, f6 t4 U" n$ Z( A7 s" }9 V
1 I! @; [' [% H$ U* y: f2 x F% A$ B 我提出这个的意义在于:
( O8 D8 b+ S# r( ]4 _1 ]4 m, y+ w# k' l6 Y
1、字串81、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。
' C+ G B% [/ n4 d+ S7 ?! A- B( R5 ?9 v; ?3 S7 M* {8 U( {; K
2、为三多理论提供了下注的多面性奠定基础。
$ k6 G0 b3 e' T2 O5 {1 [0 R9 L1 I" i6 ~( I" [$ p
3、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 G2 e( t: y( U5 U( ]6 y J' f( \9 H! l
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我现在还是研究一下理论打法,感谢你的分享,我也来学习 |
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