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Request for you
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09-20-2009, 05:22 PM
(This post was last modified: 09-20-2009 05:25 PM by mario.)
Post: #11
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RE: Request for you
Quote: If you are able to have a pairs for strictly 4 drawselections it's not "wonderful pairs with one good number in 4 sessions!"- it's a miracle-but I have my experience and know that probably it das not happen too often or maybe not any (100%) of the pairs is sucsesfull in 4 session. Why this is a miracle, a pair with a good one number in 4 session? I calculated in this way with BB: -use =HYPGEOMDIST(0,2,7,49) in Excel, the probability to hit 0 numbers from 2; 0.732142857, so 73% for a bad results! - use =BINOMDIST(4,4,0.73,FALSE) for pobablility to repeat for 4 consecutive sessions this bad results; 0.287331552, so 28% to wait 4 sessions and don't obtain any good number for that pair! - so 100-28=71% to have a GOOD RESULT! It's not a miracle I think. Any pair has a 71% chance to give us a good number in 4 sessions! And the chances increase: for 8 sessions 91%, for 16 sessions 99%; so any pair in 16 sessions can give us a good numbers; even so exist pair that grow up 20 draws, but its fall in 1%, very, very bad luck! What do you think? Then exist some strategy, we don't need to play any pair! So the chances increase! Perhaps Ido must begin another tread called 49s? I want to post some pairs with good results for that game and to explain how I obtained its. |
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09-21-2009, 01:15 AM
Post: #12
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RE: Request for you
[quote='mario' pid='2297' dateline='1253431371']
[quote] Any pair has a 71% chance to give us a good number in 4 sessions! Then exist some strategy, we don't need to play any pair! So the chances increase! [/quote] Hi,Mario, Don't you understand whay I wrote that having a pairs for strictly 4 drawselections is a miracle? A 71% chance of having a good number in 4 sessions is not much sufficient! Because even a 95% chance of having a good number in 4 sessions is not sufficient! Do you immagine if you realy want 1000 pounds in 2 days risking 3350 pounds but having only 71% chance of having a win?In case of negative result it's logical to suppose you'l would retent again -now,how ready you'l be to risk 3350 pounds for second time if you know to have 71% of chance ??? ![]() ![]() I thing that it's justifyed to invest a time &to do efforts only if such an research activity seriosly increases the probability of win ,the margin 95%-98%(as the 100% of chance simply dasn't exist!)is the very profitable one second me,wathever strategy or shoice the player would have to do or intraprend.
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09-21-2009, 01:37 AM
Post: #13
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RE: Request for you
(09-21-2009 01:15 AM)ido Wrote: [quote='mario' pid='2297' dateline='1253431371'] We have obtain 99% if we wait 16 consecutive draws, but the sum for risk is very big even so the result is almost sure. 71% is a good chance, I obtained a sufficient chain of wins to put a wonderful summ in my pocket. But, indeed this 71% chance must increase with a strategy. If I understand you use that balls with the same colour and if its have 10-12 overdue you can play 8 consecutive draws because 20 overdue is very rare. If I substract the both numbers for a pair obtain value=7. Only a simulation in history of 49s prove that this strategy works and I think to make it. Have you a such simulation that prove the results? |
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09-21-2009, 02:21 AM
Post: #14
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RE: Request for you
(09-21-2009 01:37 AM)mario Wrote: 71% is a good chance, I obtained a sufficient chain of wins to put a wonderful summ in my pocket.Hi,Mario, The only way you can manage sucsesfully a 71% is to have a good lostspread strategy(than das it not the same be ready to play for 8 consecutive selections!?!) or to acsept that 3,5 wins serve to cover 1 lost.And this was the very firs question I put to you -how many times monthly do you have your pairs.... As regarded a simulation in history of 49s for proving that this strategy works it's more simple way to use the 49's statistic -this Lottery from 01.01.2001 has 6280 Draws -think that the UKLottery players will have such a drawdata for analyse 47 years later-in year 2056 I'l make the statistic of my method and when ready will post but wath is realy important to me is the fact of the existance of lot of supplemental "help" in matter of colors.
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09-21-2009, 05:49 AM
Post: #15
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RE: Request for you
Hi Ido,
I'm very speedy and modest So I atach a wonderful file with wonderful simulation!! This file show all pairs like X, X+7 and calculate from a draw overdue, and where no1, no2 appeared to see if that pair was ok for play. Now we see that pair with overdue>=10 and where one number has appeared. What contain his file: - in sheet1 49s draws: - I2 - draws from calculate begin - RUN start calculate - in Sheet2 put rezults - colA - draw -col B,C, pair -col C - overdue - col D, E where no1, no2 appear So for the last draw I see 22-29 with overdue 10 and 35-42 with overdue 12 for play! AM I RIGHT? And I see that existed some pairs with one good no appeared to 19 draws! A long waiting? But indeed are rare. |
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09-21-2009, 09:39 AM
(This post was last modified: 09-21-2009 09:46 AM by ido.)
Post: #16
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RE: Request for you
(09-17-2009 02:36 AM)ido Wrote: This is the today's predict set,the newest one : Hi,Mario, I updated this set of pairs for having the possibility to explane my point. First ,compliments,you did a realy speedy job,but its a way to have the cronological overdue of the pairs,the analytic table is simply another one(see the immage) When i use "method overdue" it means the overdue of a pair present in a set of 15 ,so the pair 42-35 blu i have whith 9 draws overdue against maximum of 20 draws(logicaly I don't consider such a 20 draws a rigid parameter as the statistic of the different pairs distance 7 is a very important factor too) The same overdue has the pair 3-45 orange .And now is a time to decide are those two pairs good much enouf for start a play. I categoricaly would say no!!!Because: 1/ the storical max.overdue of 35-42 is 24 draws ,the cronological overdue now is 12, last 2 draws there was no a main blu ball drawn ,but was a treble digit ending 2 and its possible for number ending 2 not to appear for next 1-2 draws 2/the storical max.overdue of 3-45 is only 18 draws ,the cronological overdue now is 9 exactly as the method's one ,but the number 45 have the 3rd big overdue at the moment,there was not a numbers digit ending 3 drawn last two(todays) draws and this is a litle bit encorraging but considering the risk of 9 stake steps plan it's preferable to be prudent and waitfor tommorows draw not beginning a play and hopefull if do not appear any number to have even more information about. This is how I do in general terms whith the colored coples.
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09-22-2009, 12:45 AM
Post: #17
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RE: Request for you
(09-21-2009 09:39 AM)ido Wrote:(09-17-2009 02:36 AM)ido Wrote: This is the today's predict set,the newest one : Hi Ido, My macro put the pairs with the informations: overdue and where no1 and no2 appeared. You say that is chronological overdue and 35-42 had 12 chronological overdue and 9 overdue in set of 15 pairs; how do you choose this set of 15 pairs and what it means 9 overdue? I know only 12 overdue for 35-42. Then your analyses use storical max overdue and appeared same color and same ending digits. How do you use your analytic table? With white are the numbers for that set of 15 pairs? In this table I must count overdue and perhaps should use a macro to write this informations in groups, somehow clearer. But which is the seeking principles for analytic table? I try to post another strategy with graphic skips and the pairs which result from it. |
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09-22-2009, 09:16 AM
(This post was last modified: 09-22-2009 09:19 AM by ido.)
Post: #18
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RE: Request for you
Hi,Mario,
I kindly hope for you that you don't believe learning fishing is an easy thing. ........................ "...Then your analyses use storical max overdue and appeared same color and same ending digits...." My analyses use a thousands of single(small&greate) parameters which I like to call "factors to respect" ,the bigest part consists of statistical data,but wath exactly the intuition is second you,just for mention one non statistical?? ......................... ".....You say that is chronological overdue and 35-42 had 12 chronological overdue and 9 overdue in set of 15 pairs; how do you choose this set of 15 pairs and what it means 9 overdue? I know only 12 overdue for 35-42......" I calculate the sets after any draw so they have the same number ,todays two have numbers 41&42. Chronological overdue is the real,matematical overdue and 35-42 has already 14 now - method overdue is an abstraction ,it's a virtual cathegory -immagine yor last win was on 21.08.,than your next win have one month of overdue ,as it regards a time than any moment can be acsepted virtualy as initial,startpoint for an event. In conclusion one set's overdue begins from the mother draw used for its creation ,ergo 35-42 has an overdue of 11 against 20.This pair is good ,if consider num.49 too,one good must be in strictly 4 draws. "....How do you use your analytic table? But which is the seeking principles for analytic table?....." Before reading the post I link to,look at the immages,the analytic table is simply one of numerous ways to expres the draw data ,gives the possibility to have lot of information at a glance ,the unique thing to do is reading it. The best way of reading is to have a serios statistic knoleage for the Lottery in question . http://lottopost.co.uk/forum/showthread.php?tid=445 .................................. ![]() ![]() ![]() ![]()
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09-22-2009, 08:33 PM
Post: #19
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RE: Request for you
(09-22-2009 09:16 AM)ido Wrote: Hi,Mario, Hi Ido, since I locked in here I was not very sucessful with prediction or deleting numbers for the next draw. Therefore I didn't post anything. With interest I followed the postings of your method to predict pairs of numbers for the next drawing. Because you prefer the distance of seven for your pairs, I counted the availability of all possible 48 distances within the 14.983.816 combinations and found that the most available distance for pairs is one and not seven. Therefore I think you would have higher chances with pairs with the distance one than with pairs with the distance seven. Here is my counting list with the availability of all 48 possible distances. E.g. 1-2 has the distance 1 ( the lowest possible distance, mostly available). 1-49 has the distance 48 (the highest possible distance, least available). The distance 7 is 1.070.190 times less available than the distance 1 and therefore it will be less times drawn. Also the distances 2 and 3 have a high availability. 649lotto player Distance, Availability (within 14.983.816 lines) 1 8561520 2 8383155 3 8204790 4 8026425 5 7848060 6 7669695 7 7491330 8 7312965 9 7134600 10 6956235 11 6777870 12 6599505 13 6421140 14 6242775 15 6064410 16 5886045 17 5707680 18 5529315 19 5350950 20 5172585 21 4994220 22 4815855 23 4637490 24 4459125 25 4280760 26 4102395 27 3924030 28 3745665 29 3567300 30 3388935 31 3210570 32 3032205 33 2853840 34 2675475 35 2497110 36 2318745 37 2140380 38 1962015 39 1783650 40 1605285 41 1426920 42 1248555 43 1070190 44 891825 45 713460 46 535095 47 356730 48 178365 All 48 possible number combinations with the distance 1. 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 46 46 47 47 48 48 49 |
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09-23-2009, 12:05 AM
Post: #20
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RE: Request for you
(09-22-2009 08:33 PM)649lotto player Wrote: since I locked in here I was not very sucessful with prediction or Hi 649lotto player, I would permit myself to initiate my replay whith an advice: Stop,please , trying to predict numbers for the next draw- it's sucessful for lets say not more than 10% and begin to work on predicting for a number(a second of your bet banc or your caracter) of several next draws ........................... ".......Because you prefer the distance of seven for your pairs, I counted the availability of all possible 48 distances within the 14.983.816 combinations and found that the most available distance for pairs is one and not seven Therefore I think you would have higher chances with pairs with the distance one than with pairs with the distance seven......" You obviosly have a time a disposition for doing such a researches, how exactly the availability of distance 1 can increase my chances ?Do you have some pratical advice? In this exact moment there is the "distance1" pair 34-35 in overdue from 17 draws.Second your list its availability must be not less than 174 724 ,89 within 14.983.816 lines , but how it helpes me if I decide to punt on it ,for how many selections do I have to be prepared financialy to resist, how to know when to stop for limit losing money?????? ............................ "........1-49 has the distance 48 (the highest possible distance, least available)...." In matter of this I'm convicted believer in the circonferential consecutivity so 49-1 is a consecutive pair ,all distances have 49 pairs possible,any color has 7 pairs possible for any of the distances possible and this is whay I use the word distance but not matematical difference. I'm curios about the nature of your methods,how do you make your choises - using as Mario the skip strategy?
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