[轉貼]MARK SIX之法門
>六合彩必中之法 $z4V9W5\E2m F E>沒有錯六合彩必中之法!
>有甚麼方法可以必中頭奘呢?方法只有一個---複式全餐(即所有號碼的配搭) ? lGX)V4Hw
>那你需要投多少注呢? 4^%?3r+Kks9D.U
>N=號碼數目
>N(N-1)(N-2)(N-3)(N-4)(N-5)/6X5X4X3X2X1~~~~~自己萛吧 BJ[Y-j:a4Q5N
>那是天文數字,能中到頭奘,也要破產 &`M }8O/`Ch
>但我想和大家分享的是如何能中安慰奘. G3O A h k+c+r+I
>假設攪珠號碼有50個 9@*UJo,wE.^4A? d
>安慰奘最少要中三個字,
>那你要如果以複式方法投注,
>你要投注47個號碼
>47X46X45X44X43X42/720=10737573注
>當然它中頭奘的機會相對地大了.但不中的機會依然存在
>其實要中安慰奘不一定要47個字全投注
>把50個號碼分成兩半,即25個號碼一組 s9W$d5ct-NE
>將其中一組25個號碼裡任何三個號碼的配搭,以不重複方法配入六個空格中. jj.Jc9h,rX uCp!lU
>即(25X24X23)/(6X5X4)=115注 j sXN$T[
>再將另外一組25個號碼裡任何三個號碼的配搭,以不重複方法配入六個空格中. 即(25X24X23)/(6X5X4)=115注 hIEN q%x
>總共230注 j.wVK F t+bel)a
>那樣就能最少中一注安慰奘
>因為中奘號六個號碼中,不論甚麼號碼,最少有三個號碼出現在其中一組.比如:
>我們把50分成單數和雙數兩組
>中奘號碼最少有三個單數或三個雙數
>如1,2,3,4,5,6
>那我們便中了兩注安慰奘 UuIF*x
>如其中一組中了4個字,那便可中最少三注了.如此類推. 3xAOW]:l,n)n3g-?,Pn
>大家會問,中安慰奘有甚麼用?
>但大家要知道,如果在一組中能中4個字以上,我們就很有機會中安慰奘以上的奘額了. 7@3h{6F}
>當然中頭奘的機會是不會變的,你多投一注,機會就大分. $n(H8Dp HL/\
>以50個號碼為例.中頭奘的機會率是: `F5n9e;F
>(50X49X48X47X46X45)/6X5X4X3X2X1=15890700分之一(也等於複式投注總注額)
>你投注230注,即15890700/230=69090分之一
>而$5注每次投注$1150.安慰奘$20.最少中兩注.即實投注為最多$1110. g'h9N:V"W"yD yY8K
>如你只用$1110去投注,即只可下222注.機會率是71578.7分之一. "a)NXfq]U
>少了2489分之一的機會
>這是以最少投資,買最大機會的方法.
>但如何把25個號碼裡任何三個號碼的配搭,以不重複方法配入六個空格中?? 睇完都唔係好明噏咩 ofcoz no way....M4Q RPq
they are big company....
they had been calculated.... 用千幾蚊投資博20蚊?;s/y @{Vn
重係指派彩20蚊,個1100蚊本無得收番個隻喎。
[[i] Last edited by 金魚佬 on 2005-10-10 at 01:58 PM [/i]] no, the theory is correct. it just find out the "x" point of the 2 lines --- "invest" and "reward". no one say that the "x" point is a must win point, its only a point where the investment is minimal with the biggest chance of winning. the point, if you buy mark six in old fashion, you have actually no chance of winning and 100% chance of total loss. with this method, you have a higher chance of break even, higher chance of partial loss, and somewhat chance of winning. [quote]Originally posted by [i]金魚佬[/i] at 2005-10-10 01:53 PM:
用千幾蚊投資博20蚊?