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[打法交流]
近期在网上研究到了个百家乐著名战术 蒙特卡洛战术
[复制链接]
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百家乐蒙特卡洛战术,需要纸和笔。但是不很难。% Y2 f8 D( p7 Z1 y% ^6 U; m8 C m
& U+ c3 F R3 [" j7 s A(1) 首先写「 1 2 3 」的数列。
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0 h l& ~& V, A* s( Q1 ?! f; u(2) 赌本就是左端的数字 (1) 加右端的数字 (3) 即是 (4) 。' z4 k5 s3 ~) [: t! I
3 n9 T, h/ m# m6 [( @- i(3) 输了的话就在数列右端之外又加上顺序的下一个数,现在这个例子要加的数字 (=4) ->( 「 1 2 34 」 ) 。如此类推,并且重複上面的第 (2) 点 ( 但今次会是 (1)+(4)=5) 。1 b) u" Z6 Q/ [3 c# \
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(4) 胜了的话,倍率 2 倍的游戏就把左端的数字 (1) 和右端的数字 (3) 消去。倍率 3 倍的游戏从数列上去掉左端的二个数字和右端的二个数字。4 t3 o+ g# G; t) B" |
1 D+ X& y* _2 Z |8 A(5) 如果数列所有的数字都消了就代表结束。8 f! L9 M) Y5 M$ j3 m; {5 w0 S8 S
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(6) 想继续的话,回到 (1) 。( j& w5 R. X& y3 R6 E. q
4 H; i8 ^7 \. D+ ]; C" Y蒙特卡洛战术的问题,在于如果没有充分而源源不绝的赌金,就会出现途中被迫停止的可能。而且连败的话,亦有机会超越 Max Bet ( 赌注的上限额 ) 。
* _! y( f' A _1 X: H- x. n8 w5 R* m( g
二合一战术
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3 G) D' j, f! X# u5 X7 F; L4 o这个可说是蒙特卡洛战术的一种,是以 1 局取回 2 局所输的方法。方法简单。写下输了的金额,赌注就那个左端和右端的数字的共计金额。6 l6 x5 k0 x# E0 A: P
$ w2 i' F+ E3 ~- M* b0 I$ f譬如,先写下「 1 」。所以第一局的赌注是 1 。如这局输了,右侧添写 1 ,成为「 1 1 」。那个下一局赌注将会是左端和右端的共和 =2 。* u9 h8 y1 r; p
$ k) |, `/ K8 r6 ^4 R5 x反复这样做,如四连败,数列将成为「 1 1 2 3 」。因此下一赌注会是 4 ,而且胜了的话,消去左端和右端的数字,数列将成「 1 2 」,下一赌注就等于 1 和 2 的总数 3 。再胜的话,会变成?. t: E8 g/ N! i2 {
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儘管结果是 2 胜 4 败,共计收支是一加一减等于零。
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这方法我试了不咋地,你们可能可以?
% n) o9 `6 O6 T* e/ ^* j0 ?) n# f* B" y8 o8 P
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