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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
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百家乐蒙特卡洛战术，需要纸和笔。但是不很难。( ]* D" ]$ {- c) d
4 `- Q. H0 j) ^. w0 C(1) 首先写「 1 2 3 」的数列。( k/ ^* H( J4 f. H5 {; e; I! T
0 k$ ^8 U* A$ Z+ |1 M8 k(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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% c, R% \# [7 p" V! i. H5 m7 D(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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(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。
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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) 。% E9 C* {& K$ ]6 D
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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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二合一战术" g- T7 X( x6 F9 E9 H
3 V' `$ v+ V1 a4 W2 w% I ]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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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 。 u' h3 ]. Y9 \. H
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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？
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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。" p, v6 A3 ?/ T3 P
5 o6 E' k/ p5 {! z, H这方法我试了不咋地，你们可能可以?/ w D& T" ?$ I' q
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