[轉貼]MARK SIX之法門
>六合彩必中之法 xdN4Lv3vf>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)
>那你需要投多少注呢?
>N=號碼數目
>N(N-1)(N-2)(N-3)(N-4)(N-5)/6X5X4X3X2X1~~~~~自己萛吧 *W~.B}qk
>那是天文數字,能中到頭奘,也要破產
>但我想和大家分享的是如何能中安慰奘.
>假設攪珠號碼有50個
>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 EpdlqRH|
>當然它中頭奘的機會相對地大了.但不中的機會依然存在 XW?F6Z
>其實要中安慰奘不一定要47個字全投注 'F~ Uhk$D;YI\2D.]
>把50個號碼分成兩半,即25個號碼一組
>將其中一組25個號碼裡任何三個號碼的配搭,以不重複方法配入六個空格中. &H1B2A(FP]
>即(25X24X23)/(6X5X4)=115注 ?1u {*zvzz-P"|c
>再將另外一組25個號碼裡任何三個號碼的配搭,以不重複方法配入六個空格中. 即(25X24X23)/(6X5X4)=115注 '_c5b \l']
>總共230注 Ai d G8y6{|
>那樣就能最少中一注安慰奘
>因為中奘號六個號碼中,不論甚麼號碼,最少有三個號碼出現在其中一組.比如: 4pk-N&XP#V|E
>我們把50分成單數和雙數兩組 /F g#o9`.T
>中奘號碼最少有三個單數或三個雙數 X+Y*wh8[3z
>如1,2,3,4,5,6
>那我們便中了兩注安慰奘
>如其中一組中了4個字,那便可中最少三注了.如此類推. 1rwA$a0MX
>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.
>當然中頭奘的機會是不會變的,你多投一注,機會就大分.
>以50個號碼為例.中頭奘的機會率是: ;Fw}/b/d~#ZY o
>(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 XX p"h(qT$J?;t
>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. *d6mvsK]eR,c l
>如你只用$1110去投注,即只可下222注.機會率是71578.7分之一. 2u2Ybsf/e-Zo
>少了2489分之一的機會 #X4AQ4U1O&K
>這是以最少投資,買最大機會的方法. 7}8P Vyqa ^
>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....
they are big company....
they had been calculated.... 用千幾蚊投資博20蚊?2}\[3a.xh"{
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重係指派彩20蚊,個1100蚊本無得收番個隻喎。
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[[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? e,Zab?
重係指派彩20蚊,個1100蚊本無得收番個隻喎。
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[[i] Last edited by 金魚佬 on 2005-10-10 at 01:58 PM [/i]] [/quote]D'[{,S_KIlb
that's the bottom line. among 230 bets there is 1 must win 20$ back. for the rest 229 bets, you will still have chance to win special, 1st, 2nd, 3rd, and the $20 prize.
tho I know about the theory, but I did not buying any mark six for 20 years. THE THEORY IS WRONG!!!I2{r}V+D+JJrv
THE CHANCE IS THE SAME NO MATTER HOW WE DEVIDE AND CHOOSING THE NUMBER. 大哥,用1150蚊,只保證收回20蚊呀。超過1000蚊要望天打卦呀?
還有我以前見過有人指出吾係115注。作者只先將25c3產生既2300個組合,放係6個空位,以為6個空位裡就有6c3=20組三個號碼的組合,佢就咁用簡單25c3,÷20,就以為係115注。
但實際裡,當用matrix,頭兩次找6個完全無重覆既6個號碼無問題,之後就再找不出純6個號碼裡有三個號碼曾經重覆!即是無20組咁多,也即係用115注排列吾到無重覆既組合。0P(@D-nk\;l
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重有這個theory吾會中三獎的,佢因為保證只中20蚊,假定中了4個號碼啦。因為作者咁買法,每一種3既組合只能出現過一次,無重覆的,所以就算中四個號碼〔假設abcd〕,就只會有4注20蚊,abc,abd, bcd, acd。而這四注中獎彩票,當中其餘的號碼,必定會係新組合又不重覆的!即係無機會這四個號碼出現一起啦,點中四個號碼先?當然好彩的話abcd也可走在一齊,但只有一個機會〔同亂買一樣〕,但就無哂其餘20蚊的中獎啦。(_E@;kx"VO E:Hu
同理就係無得中三獎。%z6CBq0K{I@
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以上原因,所謂其餘229注有頭獎機會,根本同亂買無分別,中大獎機會係只得一個。9g\!D&A6yZT
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[[i] Last edited by 金魚佬 on 2005-10-10 at 03:56 PM [/i]] so complicated ... I will just buy my $5 and treat it as donation.
對唔住,條式係錯0既
對唔住,條式係錯0既(E"F-zG"rRy|?$@6k9OM^}
計俾你睇:
如果你一定要實中起碼2個單數,你要買:
1+3+5+7+(9/11/13/15/17/19/21/23/25/27/29/31/33/35/37/39/41/43/45/47/49)+ (11/13/15/17/19/21/23/25/27/29/31/33/35/37/39/41/43/45/47/49)
共 (21x20) /2=210注,即=HK$1,050Gbiv9Q0r t7A{,V
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若開 11+13+15+17+19+21 特別號碼=23, 0甘你就中一個11同埋特別號碼23a2?+M-xnR jM2it
若開2+4+6+8+10+12特別號碼=14,0甘你就乜都無S;AU5s"h8fI I
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馬會每次開6個號碼+1個特別號碼,