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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。9 v9 V# Q0 j* W! l8 `
- }# x) b% O/ Z$ B' z- R2 L(1) 首先写「 1 2 3 」的数列。6 u" M9 p7 g; Z$ y" y" c
1 {; l7 P) u- }, b4 n) {5 c1 J* U+ _* O) e(2) 赌本就是左端的数字 (1) 加右端的数字 (3) 即是 (4) 。, J% g. `& M4 i# g9 H* ^* a
# n: w. P2 g% S" J/ v0 _(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) 。7 h, J; N& j7 g& B# w% D; ^
: S: z1 ]$ B) r(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。2 d) f/ W. G/ x. S: z/ [$ U
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(6) 想继续的话,回到 (1) 。7 Q# F9 b1 g F7 c" X' q
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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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0 W$ o5 n, J5 z& ^3 n' M二合一战术8 P% D: A) h' t1 a1 X* R4 `( y
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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 。* @7 G: z! J0 s4 R2 E7 L2 X6 o
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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?3 }, N* A: s, A+ }# D9 v9 i
2 ?+ n0 D' t$ Q6 {, F/ Z) `3 q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?5 _ s2 ` z! G' i% {
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