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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。5 E% U% V6 y1 d3 ]2 j9 f
" P2 p7 v. v: ]/ ]1 t; q(1) 首先写「 1 2 3 」的数列。
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! e. r o4 P* ?3 l; A(2) 赌本就是左端的数字 (1) 加右端的数字 (3) 即是 (4) 。
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(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) 。
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- A9 W$ z( V0 ]2 b( p- \(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。* P9 B& @$ B: [; m4 v
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(5) 如果数列所有的数字都消了就代表结束。
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(6) 想继续的话,回到 (1) 。
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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 ) 。
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+ f2 C( o) Y- h7 S二合一战术
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3 B4 r6 t+ q' i5 N- L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。
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2 e2 ?0 O# ~& l8 A* P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 。) V3 U( b8 o( p
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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?( ]2 w% ]! P0 I( \
; s1 ]. e$ v5 \/ ]+ e& U5 b# O5 Q$ k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?6 c$ \+ G( G% h, P
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