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通过我几个月来对珠路理论的研究，5 R+ y* d$ ? k6 I7 G/ O3 r7 g8 H2 [4 u$ b6 I, c1 a
y- d- O; y) Y6 X+ F% a. j* G$ G7 }
发现了一个有趣的现象，
1 y$ g {" ]! _9 |6 k, Q/ f每一条大路，
* F K( n+ R, B& O5 w% f7 h) ~按2珠路排列，有2种不同的路数1 _% l2 j Y8 @" u) {# f
# c/ Q" V2 a: r. `8 c按3主路排列，有3种不同的路数& n( |2 F1 f; ?, q' L3 A: Z4 z
; I) Y$ f7 f" `& A) H$ o/ P4 g按4珠路排列，有4种不同的路数
' C# D& d8 h; O# z! h: a9 A/ o按N珠路排列，有N种不同的路数。: n1 f3 R9 @ p% ]* C5 X* i3 w+ r8 ^6 k, u1 }- [
什么意思呢，可能大家一时不清楚，4 [: }$ Q# v8 }
我举一个例子，假设 B= 1 ， P=2，T暂且忽略吧。7 f# f3 c1 ~2 l U7 b5 y1 I$ E' V, a/ u
假设大路开这样的路：! z ~0 G3 v. K' o5 A4 ]" }5 w# u
# Z8 \8 F% q$ w6 F12121212121212121212121212121212, j0 j; e6 p* q3 Q5 h" u! _8 u$ c. m- ^
够极端了吧
( J3 j! X$ p6 V6 M# ^2 v% _按2珠路，就是BPBPBPBPBP....., m8 f) O! x+ S9 L2 x) F, ~& e0 J, O+ B1 M8 R' x4 F
如果我们去掉第一口，就会出现完全相反的结果：
% y4 w) s( u5 Y; f; {1 x$ u( P3 N212121212121212121212121212121214 m* T( Y& ?4 t/ \* i& x# t
变成了PBPBPBPBPB.......1 u5 g6 c6 F/ L4 g! f7 Z# E+ C
( f/ _9 f4 t7 c如果我们再去掉一口，又返回第一种情况了。! [( g, L- u! O3 @# G4 D) O; U* `" x" z0 B, f7 K( H
所以每一条大路，! R6 V# q$ h. c; z% O9 G3 I! N# q* D1 y+ S+ m# v) F |
按2珠路排列，有2种不同的路数。
; `3 T; M% Q( `3 V' z$ b' y: M j再举一个列子：2 T8 y& `7 i9 Q* F5 v
大路：122122122122122122。。。
M2 R8 g9 y: k4 Q# N$ M按三珠路排列：
' Q3 @* g! k6 X+ K+ z/ K122，122，122，122，122，122。。。
' P' u1 `# p; l/ P, m去掉第一口，变成：6 G2 V! \" L' c" k: d5 B: J& k# _5 P, ~
221，221，221，221，221，+ R$ P2 d/ f7 Z, P5 M
) v2 C; d& \- T. u去掉前2口，变成：3 v1 q) z( ?2 f! }) U/ W) _4 y4 L$ P' y: K0 _( i; @+ ?
212，212，212，212，212，: y& l, B; I6 e- m. J# c) `
* T$ k6 S* X2 S6 n& p去掉3口，又返回122，122，122，了
/ `" R8 Y- i0 r7 w) @' }所以每一条大路，+ y$ A* s% Y+ I# z0 a8 P
按3珠路排列，有3种不同的路数。# N. G& [( a9 J; j2 U
同理：按N珠路排列，有N种不同的路数。
6 T/ I2 r' j+ Q. Q& Z我提出这个的意义在于：" v) X/ h3 }* D1 {5 J! a1 s. O
3 f5 V7 o8 N6 A/ a9 P2 r字串83 L: R) G1 V- s2 Y+ Y" i h: i5 w' q# V2 l
- p, d3 [6 U. s5 J' [0 {1 A" m' k( @
+ E# ^0 K$ S' F. `" Z1、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。2 g: X/ W# h7 N" o7 |9 l* |2 u+ @/ S: s# |* |) K2 P& |
2、为三多理论提供了下注的多面性奠定基础。1 L* c7 Y# C5 p& N8 k. a5 q
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.
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