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[lastpost] => 2761363 【星宝】8月31日投注流水76832元 1756693648 舞出精彩
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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。
- v0 y" [ J" }) R7 J! J. u, _+ P& O. V0 k: k0 @! Q3 v4 k
假设大路开这样的路:0 s8 c; Q# |) [8 c2 f
$ m! x- B( y2 I, a6 X. r5 ?$ A. R 12121212121212121212121212121212。. A8 ~9 P1 w! [% \8 j6 I
7 ^6 F5 Y0 D6 m; t8 M6 z4 e6 D# B
按2珠路,就是BPBPBPBPBP。9 ~- S2 p; x% [( L4 b
1 c1 f, b# Y0 k# ] 如果我们去掉第一口,就会出现完全相反的结果:' g0 O0 {# ^7 H _3 r
5 D* v; G& s( A% b( e; u" H0 s 21212121212121212121212121212121。
- G5 T2 H) n3 N! L6 }8 d) R+ m8 H! f* Y9 S. o9 D |+ b* b
变成了PBPBPBPBPB。8 X1 C6 W6 Q3 h5 A( [8 h
/ f; @$ n3 O* M# B3 r/ {+ j+ k d 如果我们再去掉一口,又返回第一种情况了。
; f+ P4 @3 H" w3 N# m9 f4 a4 ^( r: n
所以每一条大路,按2珠路排列,有2种不同的路数。) P \/ E( B q3 b
0 ^% @5 b" B% X( v3 o; G; z
再举一个列子:" J) x/ Y+ @% I! @; I8 L2 U
& ]8 U4 K( d9 S* C: \$ V
大路:122122122122122122。
) f' H9 _; {( H# x$ }- e1 k
8 e, ~: l7 p+ F 按三珠路排列:
2 I& w0 h! |% L; O% Y: `0 ^( p
. ^6 ]" v' o. i7 ~. H8 a! N 122,122,122,122,122,122。# k* @. M9 w" K4 J9 V' B. b. G
8 @6 ~. B) X" t3 v6 z9 V: t
去掉第一口,变成:
! I+ h# j: W0 R* t K8 j* K* k- ^% Y/ J
221,221,221,221,221,去掉前2口,变成:
) k. F2 u4 M' H# D- s9 w! c, L# M- d% D. ^
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。
( H5 S: h! T2 Q( |6 h
0 l2 e/ \, m5 A( |% a/ } 同理:按N珠路排列,有N种不同的路数。
- e8 }6 ~! A3 [' w. I1 T
* ~! g$ K1 j* f8 b 我提出这个的意义在于:" l+ J4 K* x; k" y
5 b! X% s* U1 {5 \7 L1 Q' T 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。5 W2 T4 p1 E0 Q' G6 P( w
% _) o6 _. V; q) M& V' u* k 2、为三多理论提供了下注的多面性奠定基础。
: ~ M$ z7 ~# Q8 \- P* A' O; l' M) y# s+ R0 b1 z
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。
& R5 \5 N2 f: X& s5 ~0 ^8 O* l% y6 C, s
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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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