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Eastern Time (GMT-5:00) | Lotto System Development SimplifiedPennsylvania United States Member #93975 July 10, 2010 1835 Posts Offline | | Posted: January 16, 2013, 7:10 pm - IP Logged | |
The cause generating a Momentary trend could be many factors, but am not concerned. What am concerned is the transitional data(could last a certain time, remember my life span is short with regard to large matrix odds). Like I said, stats and probabilty may seem mutually inclusive , but two completely different concepts. Stats deals with data(physical and dynamic), probability is more of inference , base on a data(Physical and dynamic).Since probability is a conditonal concept, i have the pleasure of elasticity with my data trend. See ,I play the pick 3 mostly with pool size 0-7, but sometimes 0-6 ,depending on the draw digits range(Trend), and make decent sum every month. The clue is when you downsize your pool, your betting strategy has to change. Until you define your terms and get more specific, many people here are going to write your rhetoric off as DOUBLE TALK. | | |
United States Member #116363 September 8, 2011 669 Posts Online | | Posted: January 16, 2013, 7:35 pm - IP Logged | |
Until you define your terms and get more specific, many people here are going to write your rhetoric off as DOUBLE TALK. Is not double talk, the info (rhetoric) given is easily tested and one can verify. Verifying a data trend is just easy. Take a look at Arkansa P3 below: For P3, the Trend of draw digits were between 0-7(in red), same for P4. so I look at my trend ,then adjust my Pool(model is reduced/increased depending). Ark data | Pick 3 | Pick 4 | | Midday | Evening | Midday | Evening | | Wed, Jan 9, 2013 | 850 |
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| Tue, Jan 8, 2013 | 804 | 739 | 5680 | 4435 |
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| Mon, Jan 7, 2013 | 155 | 989 | 1342 | 8955 |
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| Sun, Jan 6, 2013 | 202 |
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| Sat, Jan 5, 2013 | 143 | 542 | 5088 | 8605 |
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| Fri, Jan 4, 2013 | 101 | 571 | 3761 | 1276 |
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| Thu, Jan 3, 2013 | 822 | 134 | 0819 | 8441 |
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| Wed, Jan 2, 2013 | 836 | 614 | 6418 | 4734 |
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| Tue, Jan 1, 2013 | 517 | 008 | 0330 | 4299 |
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| Mon, Dec 31, 2012 | 235 | 775 | 4307 | 2273 |
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Pennsylvania United States Member #93975 July 10, 2010 1835 Posts Offline | | Posted: January 17, 2013, 2:28 am - IP Logged | |
Is not double talk, the info (rhetoric) given is easily tested and one can verify. Verifying a data trend is just easy. Take a look at Arkansa P3 below: For P3, the Trend of draw digits were between 0-7(in red), same for P4. so I look at my trend ,then adjust my Pool(model is reduced/increased depending). Ark data | Pick 3 | Pick 4 | | Midday | Evening | Midday | Evening | | Wed, Jan 9, 2013 | 850 |
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| Tue, Jan 8, 2013 | 804 | 739 | 5680 | 4435 |
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| Mon, Jan 7, 2013 | 155 | 989 | 1342 | 8955 |
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| Sun, Jan 6, 2013 | 202 |
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| Sat, Jan 5, 2013 | 143 | 542 | 5088 | 8605 |
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| Fri, Jan 4, 2013 | 101 | 571 | 3761 | 1276 |
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| Thu, Jan 3, 2013 | 822 | 134 | 0819 | 8441 |
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| Wed, Jan 2, 2013 | 836 | 614 | 6418 | 4734 |
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| Tue, Jan 1, 2013 | 517 | 008 | 0330 | 4299 |
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| Mon, Dec 31, 2012 | 235 | 775 | 4307 | 2273 |
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What I see are 34 numbers from 10 days of lottery results. 19 of them, or 56%, comprised exclusively of the digits 0-7. I'm puzzled that you perceive a pattern in this display indicating a continuation of draws sans 8-9. First of all, 0-7 comprise 80% of all base 10 digits, so 56% seems a bit shy of expected, not an excess. Secondly, the digits 8-9, which you decided to exclude, total 21, whereas 6-7, which you included, total only 17. (Not that it makes any difference!) I think this random distribution of digits over a 10 day period is of no value whatsoever as an aid in selecting tickets for Jan 10,11,,, Besides, even if you buy all 8*8*8=512 tickets [in the Pick-3], you can't stop 999 from popping up! | | |
United States Member #116363 September 8, 2011 669 Posts Online | | Posted: January 17, 2013, 8:49 am - IP Logged | |
What I see are 34 numbers from 10 days of lottery results. 19 of them, or 56%, comprised exclusively of the digits 0-7. I'm puzzled that you perceive a pattern in this display indicating a continuation of draws sans 8-9. First of all, 0-7 comprise 80% of all base 10 digits, so 56% seems a bit shy of expected, not an excess. Secondly, the digits 8-9, which you decided to exclude, total 21, whereas 6-7, which you included, total only 17. (Not that it makes any difference!) I think this random distribution of digits over a 10 day period is of no value whatsoever as an aid in selecting tickets for Jan 10,11,,, Besides, even if you buy all 8*8*8=512 tickets [in the Pick-3], you can't stop 999 from popping up! 'First of all, 0-7 comprise 80% of all base 10 digits, so 56% seems a bit shy of expected'. What's your position on subject of discussion? May be specifics on your part , without quoting odds , but using data to explain your position will make this discussion worthwhile . People perceive data in different way, I may interprete 7/10 ratio in days,for draw digits between pool 0-7 from dates dec 31 to Jan 7( This was trending, so choosing days, weeks, or months data to select my pool size is my perogative so you should not be puzzled. If 0-7 comprises of 80% base digits , then you are saying all combos between[ 0-7 ]012, 013, 014, 015, 016, 017, 023, 024, 025, 026, 027, 034, 035, 036, 037, 045, 046, 047, 056, 057, 067, 123, 124, 125, 126, 127, 134, 135, 136, 137, 145, 146, 147, 156, 157, 167, 234, 235, 236, 237, 245, 246, 247, 256, 257, 267, 345, 346, 347, 356, 357, 367, 456, 457, 467, 567 has 56% chance of occurrance, and bit shy of expected.Taking a digit(s) from the pool changes everything and make a difference, not from odds perspective( which seem to be your position), because probabilty is a conditional concept!. Check the definition of probability(please do not link wiki, am aware) and its interpretations. | | |
Pennsylvania United States Member #93975 July 10, 2010 1835 Posts Offline | | Posted: January 17, 2013, 4:32 pm - IP Logged | |
'First of all, 0-7 comprise 80% of all base 10 digits, so 56% seems a bit shy of expected'. What's your position on subject of discussion? May be specifics on your part , without quoting odds , but using data to explain your position will make this discussion worthwhile . People perceive data in different way, I may interprete 7/10 ratio in days,for draw digits between pool 0-7 from dates dec 31 to Jan 7( This was trending, so choosing days, weeks, or months data to select my pool size is my perogative so you should not be puzzled. If 0-7 comprises of 80% base digits , then you are saying all combos between[ 0-7 ]012, 013, 014, 015, 016, 017, 023, 024, 025, 026, 027, 034, 035, 036, 037, 045, 046, 047, 056, 057, 067, 123, 124, 125, 126, 127, 134, 135, 136, 137, 145, 146, 147, 156, 157, 167, 234, 235, 236, 237, 245, 246, 247, 256, 257, 267, 345, 346, 347, 356, 357, 367, 456, 457, 467, 567 has 56% chance of occurrance, and bit shy of expected.Taking a digit(s) from the pool changes everything and make a difference, not from odds perspective( which seem to be your position), because probabilty is a conditional concept!. Check the definition of probability(please do not link wiki, am aware) and its interpretations. You're right about your prerogatives. If you perceive a trend, who am I to question it? What I believe I am justified in asking, however, is just what sorts of phenomena you would posit as causes of the "trends" that you perceive. | | |
Pennsylvania United States Member #93975 July 10, 2010 1835 Posts Offline | | Posted: January 17, 2013, 10:47 pm - IP Logged | |
Youll are sOme smart guys Not really, Pick4ologist. The really smart people who visit this website, after perusing a few threads like this one, simply shake their heads, and move on. --Jimmy4164 | | |
Pennsylvania United States Member #93975 July 10, 2010 1835 Posts Offline | | Posted: January 22, 2013, 1:14 am - IP Logged | |
Lotto System Development Simplified A lot of energy is expended here trying to design a system that might help a player win a Lotto Jackpot. The problem is that the dimensions of the state games are such that designers must deal with very LARGE numbers of combinations and very SMALL probabilities. Very often, the designer's ideas become obscured by complicated and tedious calculations. This thread is an attempt to provide a Lotto game model of scaled down dimensions to allow people to focus on system theories, rather than boring calculations. So, let's take a look at an arbitrary scaled down version of a Lotto game. Although I know of no Lotto game, state sanctioned or otherwise, that has these dimensions,it should be clear that any system designed for it can be extrapolated to a game with a larger matrix. For this purpose, I've designed a (5,2) Lotto game. It will be much easier to understand a betting system when it's applied to a (5,2) game as opposed to say, a (56,5) game! (1) (2) (3) (4) (5) Remember now, we're playing Lotto, not Pick-2. Initially, there are 5 Balls in the drawing machine, numbered 1 through 5. The first selection is made from these 5, and the second is drawn from the 4 remaining. The order they are drawn is not significant so we will list the results for each draw in ascending order, left to right. (1,2) (2,4) (1,3) (2,5) OK, in a (5,2) Lotto game, (1,4) (3,4) here are the possible outcomes. (1,5) (3,5) (2,3) (4,5) That's a total of Ten (10) possibilities. In a (56,5) game, this number would be 3,819,816. I think our scaled down game will be a little easier to work with! In this simplified model, ODDS and PAYOFFS will be easy to determine by inspection. Look at the list of 10 winning ticket possibilites and convince yourself that the theoretical chance of matching both numbers you select is one in ten, or 1:10, or 1/10, or 0.1, or 10%, however you like to express it. This is our Jackpot. Our 2nd tier prize will be awarded for correctly selecting 1 number on your ticket. By inspection, you should see above that the chance of matching one of the two numbers you select is four in ten, or 4:10, or 2/5, or 0.4, or 40%. Consequently, to ensure the Expected Value of this game is 0.50, the typical state lottery case, our $1 tickets will pay $4.00 for matching both numbers (the Jackpot,) and 25 Cents for a one number match. So, ON AVERAGE, for each set of 10 tickets we purchase, we can expect to match one number four times, and both numbers one time, for a return of $5.00 on our $10 investment, our desired ROI. Since some people in another Forum have been studying subsets of the total number field, I thought I would do some of the setup that will help them use this (5,2) game in their research. They have been looking at a subset of 28 of the Megamillions total of 56. They observe that IF they could select a subset of the 56 number field that contains the 5 numbers that are ultimately drawn, Then they would have better odds of winning a Jackpot. So, let's see how we can partition our (5,2) game in an analogous way. If we eliminate one of the five numbers from play, we would only have to deal with four, so... (1,2,3,4) (1,2,3,5) ...here are the partitions of our (1,2,4,5) 5 numbers taken 4 at a time. (1,3,4,5) Note that each set is missing (2,3,4,5) one of the five numbers. Hint to the subset researchers: Each of the 10 winning (Jackpot) pairs of numbers in our first list above appears in 3 of the 5 possible subsets of 5 numbers taken 4 at a time. I'll leave speculation on the probabilities and other implications of this model for a later post. Perhaps someone has a system they would like to present using this simple model. --Jimmy4164 I was hoping someone else would do this, but... See the above quoted post for reference. This analysis ignores the second tier prize. By inspection, you can see that the probability of selecting the winning pair from the 10 combinations of 5 balls taken 2 at a time is 1/10, or 10%. Note that if you select one of the 5 subsets of 4 above, you will have only 6 pairs to choose from, instead of 10. So,IF you choose one of the 3 winning subsets, your probability of winning will then be 1/6, or about 17%, which is an improvement over 10%. However, this is where our friends in the other thread lose their way. The red IF is crucial here. You only have a 17% chance of picking the winning pair AFTER you choose one of the 3 correct subsets! But the probability of doing this is not 1.0, it is 3/5! Your 17% (1/6) probability must be multiplied by 3/5. The probability of selecting a winning pair using subsets is: (3/5) X (1/6) = (3/30) = (1/10) = 10% Q.E.D. If anyone would like to do the calculations for a (56,5) game, please be my guest! --Jimmy4164 | | |
5+1 Winner Arizona United States Member #116287 September 7, 2011 14810 Posts Offline | | Posted: January 22, 2013, 10:56 am - IP Logged | |
You're right about your prerogatives. If you perceive a trend, who am I to question it? What I believe I am justified in asking, however, is just what sorts of phenomena you would posit as causes of the "trends" that you perceive. The trend is your friend until the end............. | | |
5+1 Winner Arizona United States Member #116287 September 7, 2011 14810 Posts Offline | | Posted: January 22, 2013, 10:57 am - IP Logged | |
Oh, what the hell. I think I will just "move on" | | |
mid-Ohio United States Member #9 March 24, 2001 15962 Posts Offline | | Posted: January 22, 2013, 1:46 pm - IP Logged | |
I was hoping someone else would do this, but... See the above quoted post for reference. This analysis ignores the second tier prize. By inspection, you can see that the probability of selecting the winning pair from the 10 combinations of 5 balls taken 2 at a time is 1/10, or 10%. Note that if you select one of the 5 subsets of 4 above, you will have only 6 pairs to choose from, instead of 10. So,IF you choose one of the 3 winning subsets, your probability of winning will then be 1/6, or about 17%, which is an improvement over 10%. However, this is where our friends in the other thread lose their way. The red IF is crucial here. You only have a 17% chance of picking the winning pair AFTER you choose one of the 3 correct subsets! But the probability of doing this is not 1.0, it is 3/5! Your 17% (1/6) probability must be multiplied by 3/5. The probability of selecting a winning pair using subsets is: (3/5) X (1/6) = (3/30) = (1/10) = 10% Q.E.D. If anyone would like to do the calculations for a (56,5) game, please be my guest! --Jimmy4164 The probability of selecting a winning pair using subsets is: conditional on all the winning numbers being in the subset. --RJOh * The fundamentals of winning a lottery jackpot *
* play a lottery you can win *
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bgonçalves Brasil Member #92592 June 9, 2010 1134 Posts Offline | | Posted: January 22, 2013, 1:57 pm - IP Logged | |
hello, jimmy, my question is this for discussion, if I divide a lottery 40/6
in groups of 5 numbers, the question is why, the Group of 25 to 30 has a frequency of at least a number of 91% and a group of 35 to 40, has a 36% of it puts, since all numbers have the same chance to leave, but the Center's group comes out more at least one group? | | |
bgonçalves Brasil Member #92592 June 9, 2010 1134 Posts Offline | | Posted: January 22, 2013, 2:00 pm - IP Logged | |
application and triangulation of filters, which were more faithful to each other were the diagonal I call 1-48, 6-49 and diagonal endings that then I will explain who is who. However it is important to say that the filter of the terminations, works as a filter applied within the diagonal 1-48 since each diagonal of 1-48 has only two endings. Now let's see: in this image we have 13 diagonals that compose the diagonals of 1-48, I baptized it from 1-48 because starts at 1 and ends at 48. If noticed, each diagonal of these contains only 2 endings. Diag. 1-01 Diag. 2-02.07 Diag. 3-03.08.13 Diag. 4-04.09.14.19 Diag. 5-05.10.15.20.25 Diag. 6-06.11.16.21.26.31 Diag. 7-12.17.22.27.32.37 Diag. 8-18.23.28.33.38.43 Diag. 9-24.29.34.39.44 Diag. 10-30.35.40.45 Diag. 11-36.41.46 Diag. 12-42.47 Diag. 13-48 As for the Diagonals of 6-49 have 14 diagonals that compose it, also I baptized her 6-49 because starts at 6 and ends at 49. If we analyze these diagonal endings, also can fix that in each there is a single ending that repeats itself. And each diagonal of 6-49 is composed of: Diag. 1-06 Diag. 2-05.12 Diag. 3-04.11.18 Diag. 4-03.10.17.24 Diag. 5-02.09.16.29.30 Diag. 6-01.08.15.22.29.36 Diag. 7-07.14.21.28.35.42 Diag. 8-13.20.27.34.41.48 Diag. 9-19.26.33.40.47 Diag. 10-25.32.39.46 Diag. 11-31.38.45 Diag. 12-37.44 Diag. 13-43.50 Diag. 14-49 Now, bearing in mind that in the Euromillions only leave 5 balls, on top, just that choice also only 5 diagonals to play, both in 1-48 as in 6-49, and 5 endings, not obvious? When doing this application will almost always reduced numbers of a quantity of 5 to 11 final numbers. But also, if the choices are "poorly made" we can having a reduction of less than 5 figures, which will make it possible to add some more choices. However, the problem lies in that the diagonals and terminations to choose! As for this problem, by recall and a bit of luck (just a little!)... For example ... and only as an example! If we choose in diag. 1-48 the following 5 diagonals: Diag. 3-03.08.13 Diag. 5-05.10.15.20.25 Diag. 6-06.11.16.21.26.31 Diag. 8-18.23.28.33.38.43 Diag. 11-36.41.46 And diag. 6-49 as follows: Diag. 4-03.10.17.24 Diag. 6-01.08.15.22.29.36 Diag. 8-13.20.27.34.41.48 Diag. 9-19.26.33.40.47 Diag. 11-31.38.45 As you can see, combining filters, we were only with a total of 12 numbers, which are put in bold (03.08.10.13.15.20.26.31.33.36.38 and 41) Now, when we apply the filter of terminations if we choose 0 endings, 1, 3, 6 and 8 we get a final result of 11 numbers: After we get to the final result, we must first of all, look at the | | |
Pennsylvania United States Member #93975 July 10, 2010 1835 Posts Offline | | Posted: January 22, 2013, 2:34 pm - IP Logged | |
hello, jimmy, my question is this for discussion, if I divide a lottery 40/6
in groups of 5 numbers, the question is why, the Group of 25 to 30 has a frequency of at least a number of 91% and a group of 35 to 40, has a 36% of it puts, since all numbers have the same chance to leave, but the Center's group comes out more at least one group? dr san, If you are good with analogies, you will find the answer to your question in this article: Casino City Times If you have a problem associating your question with what's discussed in the Casino City Times article, this might help: http://subs.emis.de/journals/ZDM/zdm056r1.pdf --Jimmy4164 P.S. If anyone feels a desire to jump in here and complain about my posting of links to other people's work, please first Google the idea of "Reinventing The Wheel." | | |
bgonçalves Brasil Member #92592 June 9, 2010 1134 Posts Offline | | Posted: January 22, 2013, 4:30 pm - IP Logged | |
Hello jimmy, perfect, good article from donald, jimmy still the lottery 6/40, but can be another Lottery, see 01 the smallest pair 12 and 28 to 40 top pair getting together to see the pair, which can be random, or two pairs of extremaidades (the pair higher with lower pair) can these two pairs give evidence of the central couple (the couple missing to complete the bet) making the lowest pair of 01 to 12 are 66 pairs but if you increase the limit increase the logicothe 66 smaller pairs you can see frequency statistics terminations, etc Equal to the larger of 28 to 40 pairs, giving a large reduction | | |
bgonçalves Brasil Member #92592 June 9, 2010 1134 Posts Offline | | Posted: January 22, 2013, 4:44 pm - IP Logged | |
Hello, jimmy, I agree with much that the donald writes, but also we have to see we are trying to find a 100% of a lottery scheme, where in Lottery can only come with holding of 3 to 4 numbers (Lotto 6/40 a) or similar The rest is random, this is crucial, we provide two pairs of ends What are 4 numbers, or also dismembered the result of 6 numbers in 15 positions of 4 numbers each, that is a good way to predict 4 numbers and not the 6 numbers, What is missing after is completed, this is our sin, trying to predict the 6 numbers | | |
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