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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。8 ]7 W$ U" B. Q. I* W2 l% {) F
* i! t7 r1 e5 j! @$ ^; @5 u( v* { 假设大路开这样的路:. @% o2 z) }! S* r h
6 v: g- K" b" j2 k4 h+ S
12121212121212121212121212121212。* O' l- A8 d5 P x( `- W$ t
; S0 i% g2 S" n( W/ e% z0 U& h
按2珠路,就是BPBPBPBPBP。# n. T7 K* y' h, M, x0 l [
) c: o" J- W4 z* k, l 如果我们去掉第一口,就会出现完全相反的结果:
8 S9 U1 e2 a8 y7 J' I/ A t
' S% n* \5 \& L# I) y 21212121212121212121212121212121。1 h. d. T3 O4 f, ^6 L+ f
: A9 R# d9 _- z5 H% ?* `2 Q 变成了PBPBPBPBPB。
4 k* n' Z; u, J, `2 }* U3 N* { Q; W3 b! x( q
如果我们再去掉一口,又返回第一种情况了。9 v r& p. K0 n( u' V
! p/ l: _) s4 f# O# c: [6 i
所以每一条大路,按2珠路排列,有2种不同的路数。3 r. K& s) R/ q) ^' F+ R* n
2 s* n/ j1 t6 }9 `
再举一个列子:
/ v$ h) k3 N% M/ k1 @& x0 x' A9 l; k( R/ n8 m
大路:122122122122122122。: H: U, S9 b; F% h% `
+ n6 w7 q# o @: ]5 C" l/ ~2 K 按三珠路排列:" b; N8 }) N1 W: N
+ O+ m2 _/ A1 v4 M( A @5 n
122,122,122,122,122,122。8 y) |* O8 o2 \6 \
& S% M1 F S9 d+ n! W
去掉第一口,变成:+ e3 |: m+ I+ ~( x3 p* r, c$ X
) w/ U' R, Y: Q! Z( S 221,221,221,221,221,去掉前2口,变成:/ y V! N. j+ l% V" P
4 q6 @# Y/ l( z 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。0 N* a# X* _5 w$ h$ K
& G, G0 B" c% r* m' ~
同理:按N珠路排列,有N种不同的路数。
( Q9 R" D& o. H3 h! V6 C/ i
) l0 s, `, Y7 v4 z( }# @' q 我提出这个的意义在于:7 H5 `' S5 ^1 O' G5 a
* }5 B, G8 {$ ?
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。& w/ O# n0 B# B i) b. k- N
/ ?6 }8 Q3 j' w$ L8 @1 S3 K 2、为三多理论提供了下注的多面性奠定基础。
, _& O' q1 i8 P. X
9 y6 v3 j- Q+ X6 n, ^1 u 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。
" E) Q$ ~: k1 [4 {2 E) F: S% d/ h9 A8 k
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不用花时间去研究这个思路,n年前就见过。结果都一样,下三路就是个例子。 |
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稳稳的发现没问题了的啊。
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楼主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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