United States Member #41383 June 16, 2006 1969 Posts Offline

Posted: August 10, 2007, 11:53 pm - IP Logged

I have about 12 different 'systems' or 'ways' of 'predicting' numbers, the problem is, at any one time, only two of those 'ways' is valid, so then it becomes a game of guessing which 'way' to go with ? Some get 'hot', some go ice cold, as I've said before, you have precisely two drawings to figure out what is 'hot' before it goes 'cold', because in all my ways of tracking, streaks only last MAYBE 6-7 games at the most.

One of my ways (to pick one number) hit with 90%+ accuracy for 6-8 months in a row, now it's as cold as the north pole, I can't use it.

mid-Ohio United States Member #9 March 24, 2001 19826 Posts Offline

Posted: August 12, 2007, 1:03 pm - IP Logged

I never thought about it but in a 649 game, knowing one number could reduce your odds of winning the jackpot from 1:13,983,816 to 1:1,715,304 or reduce the possible combinations by 800%.

Brazil Member #51781 April 24, 2007 384 Posts Offline

Posted: August 12, 2007, 1:25 pm - IP Logged

It is possible if you find some game like Lotofácil.

This draw use 15 numbers, the matrix is only 25.

If you eliminate the "corners" (four cold numbers) and choose to play some of the others (18 or 19 numbers for the combination) your odds of winning the Jackpot would be better.

If you eliminate five or six cold numbers you will find a tier 1 or tier 2 prize. Even if a tier 2 prize is not much it would be enough to play the game again and if you successful eliminate more six cold numbers you will have a Jackpot.

In this month, I had simulations with 2-4 prizes tier 2. And for full combinations much more prizes and even a Jackpot.

Sure that a person don't need to play all the combinations, only a few for tier 2 and tier 3 prizes. Full combinations are more appropriated for someone that wanted the Jackpot. The cool is that for some games and combinations, all the tickets would be winner tickets!

Ok, another point of view is that this only will work if you start from a winning ticket with 12 or 13 points! If someone could win with just one ticket, why play one hundred combinations???

mid-Ohio United States Member #9 March 24, 2001 19826 Posts Offline

Posted: August 12, 2007, 1:42 pm - IP Logged

Quote: Originally posted by time*treat on August 12, 2007

I wouldn't mind adding guesser's little horse to my own stable of methods, especially if that one number is red.

Guesser stated at the beginning of this thread,

"I've tried collaborating with others that indicate they have as much passion for developing 'systems', and almost NONE of them have offered anything back in the way of 'real' theories."

so if you have something in your stable worth trading you two may be able to do a little horse trading.

* you don't need to buy more tickets, just buy a winning ticket *

NY United States Member #23835 October 16, 2005 3474 Posts Offline

Posted: August 13, 2007, 7:45 pm - IP Logged

Quote: Originally posted by Stack47 on August 10, 2007

"I was thinking of something like starting with multiplying the previous drawing by Pi, etc."

Aw, the old "had I" system. Had I mathematically done this, that, or something else to the previous draw or drawings, I would have the winning numbers in the next draw. These systems are based on knowing the results and could also be called "it worked once so it should always work" strategy.

You could start out by creating an equation; previous draw (pd) plus/minus/multiplied/divided/ by X equal the next draw (nd). After finding "X" the next step is to see if it consistently produces the next winning numbers. Adding and/or subtracting will give you one number each but with multiplying and/or dividing, you'll probably have to round it off.

Pi is an irrational number that is the ratio of the circumference to the diameter of a circle with millions of digits following the decimal point without repetition. I found a site that has the first 10 million digits of the square root of 2 so somebody probably did that to Pi too. If you have the time, you'll probably find the next draw in exact order somewhere in those millions of digits.

"Has anyone had such luck with this approach?"

Luck is when somebody has two cars and the 3 digit plate numbers hit straight two days in a row and they bet all the first hit winnings on the second draw. Isn't the idea of a system to accurately and consistently pick the winning numbers?

"I found a site that has the first 10 million digits of the square root of 2 so somebody probably did that to Pi too. If you have the time, you'll probably find the next draw in exact order somewhere in those millions of digits."

Pi has been calculated to at least a billion digits, but the chance of finding any particular winning combination in there still isn't a sure thing. To win the jackpot you need to match the winning combination. It doesn't matter what order the numbers are drawn in, as long as it's the right numbers. Writing that combination as a 12 digit string you're listing the specific permutation that was actually drawn. To simplify it a bit, let's look at the numbers in numerical order, which gives you the best chance.

PB has fewer numbers to choose from, so let's use that. The first 2 digits will be one of the numbers from 01 to 51, so there are 51 choices. The second number will be one of th enumbers from 02 to 52. Another 51 choices. The third number will be from 03 to 53. Yup, another 51 choices. There seems to be a pattern here. That's 5 consecutive two digit numbers, each with 51 posssible choices. For the sixth number (the power ball) there will be 42 choices. That means there are 51*51*51*51*51*42, or 14,491,060,542 possible 12 digit strings that can be made from the possible winning combinations. If you've got a string of digits 1 billion eleven digits long you can find "only" 1 billion 12 digit strings. That gives you a 1 in 14.5 chance of finding your winning combination.

For a 6 of 49 game you'll do better. With only 6,321,363,049 permutations there's a 1 in 6 chance you'd find the winning combination.

If you look for a 12 digit string that includes the numbers inth eorder they were drawn there will be even more choices, so the chances go down. Since you don't need to guess the drawing order you might as well stick to numerical order.

This lottery thing isn't as easy as it looks, is it?

MD United States Member #1701 June 18, 2003 8361 Posts Offline

Posted: August 13, 2007, 8:13 pm - IP Logged

Quote: Originally posted by guesser on August 7, 2007

I tried that, it's never worked so I dropped it.

I've tried collaborating with others that indicate they have as much passion for developing 'systems', and almost NONE of them have offered anything back in the way of 'real' theories.

So, what I do now is just narrow my 'systems' down to about 3 WB numbers, and for the other 2, I guess/throw darts, mix 'em up, and I seem to do OK.

And for the PB itself, there are things you can do to narrow it down, but even at that, I'm narrowing it down to about 15 numbers, the PB that hits 'usually' comes from one of those 15, but that's still too many.

Maybe you will do better than I.

Wheeling is one of the best methods for your group of 5 numbers and wheeling anything from 10 to 15 numbers is a good idea. Use a modified wheel so you don't have to buy 100 plus tickets. Make your own wheel.

As for the powerball number since they only draw one number out of the whole group what makes you think there is a system for that. They set the game up that way to make it almost impossible to guess the powerball number. Let the lottery machine pick the powerball number.

BigJohn says. You don't hit the number. The number hits you!!!!

I'm not Big John, I'm Four4me, Big John's a friend.

Beautiful Florida United States Member #5709 July 18, 2004 20115 Posts Offline

Posted: August 13, 2007, 8:22 pm - IP Logged

Quote: Originally posted by WIN D on August 8, 2007

Learning the Math behind the game ....... that will get you closer to what you want than anything out there.

In this game ....Math is Sexy!

Absolutely right Win D...!

The trouble with combining systems, is you gain on one end then lose on the other end. Timing is everything trying to corner the next draw, thats why it's so important to learn the math....

" When Injustice Becomes Law, Resistance Becomes Duty "

Kentucky United States Member #32652 February 14, 2006 7310 Posts Offline

Posted: August 14, 2007, 2:17 am - IP Logged

Quote: Originally posted by KY Floyd on August 13, 2007

"I found a site that has the first 10 million digits of the square root of 2 so somebody probably did that to Pi too. If you have the time, you'll probably find the next draw in exact order somewhere in those millions of digits."

Pi has been calculated to at least a billion digits, but the chance of finding any particular winning combination in there still isn't a sure thing. To win the jackpot you need to match the winning combination. It doesn't matter what order the numbers are drawn in, as long as it's the right numbers. Writing that combination as a 12 digit string you're listing the specific permutation that was actually drawn. To simplify it a bit, let's look at the numbers in numerical order, which gives you the best chance.

PB has fewer numbers to choose from, so let's use that. The first 2 digits will be one of the numbers from 01 to 51, so there are 51 choices. The second number will be one of th enumbers from 02 to 52. Another 51 choices. The third number will be from 03 to 53. Yup, another 51 choices. There seems to be a pattern here. That's 5 consecutive two digit numbers, each with 51 posssible choices. For the sixth number (the power ball) there will be 42 choices. That means there are 51*51*51*51*51*42, or 14,491,060,542 possible 12 digit strings that can be made from the possible winning combinations. If you've got a string of digits 1 billion eleven digits long you can find "only" 1 billion 12 digit strings. That gives you a 1 in 14.5 chance of finding your winning combination.

For a 6 of 49 game you'll do better. With only 6,321,363,049 permutations there's a 1 in 6 chance you'd find the winning combination.

If you look for a 12 digit string that includes the numbers inth eorder they were drawn there will be even more choices, so the chances go down. Since you don't need to guess the drawing order you might as well stick to numerical order.

This lottery thing isn't as easy as it looks, is it?

From what I've read they mostly use Pi on Pick-3 and Pick-4 games. I don't know what the odds of finding any 3 or 4 digits together in order, exactly the same spot two draws in a row, but they are probably about the same as buying a QP.

"If you've got a string of digits 1 billion eleven digits long you can find "only" 1 billion 12 digit strings. That gives you a 1 in 14.5 chance of finding your winning combination."

Bet somebody read that and is trying it now. How long do you thing it would take?

"This lottery thing isn't as easy as it looks, is it?"

That Cash25 game in West Virginia sure looks easy and glad I don't live there because I know they would prove wrong.

NY United States Member #23835 October 16, 2005 3474 Posts Offline

Posted: August 14, 2007, 2:45 pm - IP Logged

I don' t know what other people use pi for, as far as the lottery goes. The wheels are round, so maybe it would be useful for roulette.

How long to find the winning combination? I guess it depends on whether you're looking for last week's combination or next week's combination. If you can read 10 digits per second you could go through the list looking for the first digit in100 million seconds. That's 3.17 years. Interestingly enough, thats kind of close to pi.

In days, it's 1157.407407407407407407407407407. I offer that free of charge to those of you who play pick 3 and 4 (and the 407 should clearly be played straight), but I will expect proper credit.

United States Member #13130 March 30, 2005 2171 Posts Offline

Posted: August 24, 2007, 11:56 am - IP Logged

Quote: Originally posted by RJOh on August 12, 2007

Guesser stated at the beginning of this thread,

"I've tried collaborating with others that indicate they have as much passion for developing 'systems', and almost NONE of them have offered anything back in the way of 'real' theories."

so if you have something in your stable worth trading you two may be able to do a little horse trading.

I won't claim to have so much worth trading, as I mostly compile stats and write code to that end. When I see something interesting, I post it somewhere. Even if I ultimately set an idea aside, I like to see when someone has a new spin on things.

When the problem is not too elaborate (or poorly phrased) I write solution code even when the other person has nothing to trade that I would use (like p3 games). I stick to p5 games.

OTOH, I find that when I have a question about something, I get ... the soft chirping of crickets.

I'm starting to think there are not many coders, here.

mid-Ohio United States Member #9 March 24, 2001 19826 Posts Offline

Posted: August 24, 2007, 4:50 pm - IP Logged

Quote: Originally posted by time*treat on August 24, 2007

I won't claim to have so much worth trading, as I mostly compile stats and write code to that end. When I see something interesting, I post it somewhere. Even if I ultimately set an idea aside, I like to see when someone has a new spin on things.

When the problem is not too elaborate (or poorly phrased) I write solution code even when the other person has nothing to trade that I would use (like p3 games). I stick to p5 games.

OTOH, I find that when I have a question about something, I get ... the soft chirping of crickets.

I'm starting to think there are not many coders, here.

I do a little coding myself with GWBasic. I have an analyzing program and a number selection program that's really a RNG with parameters. I only play the pick5 and jackpot games.

Like you I write mostly solution code to check out different theories I've read about on this forum. For example someone mentioned they were winning a lot using a positional theory so I wrote a little program that look as past winning combinations to see how many had the top 7-15 numbers for each position and I found the most any had was three and that only happen 20% of the time and the top 10 or so numbers per position cover a lot of combinations. So far I haven't been able to pick a group or numbers that preformed much better then a group of the same size picked at random.

There are other coders here but they don't spend a time talking about theories, they test them.

* you don't need to buy more tickets, just buy a winning ticket *

United States Member #13130 March 30, 2005 2171 Posts Offline

Posted: August 24, 2007, 8:46 pm - IP Logged

Quote: Originally posted by RJOh on August 24, 2007

I do a little coding myself with GWBasic. I have an analyzing program and a number selection program that's really a RNG with parameters. I only play the pick5 and jackpot games.

Like you I write mostly solution code to check out different theories I've read about on this forum. For example someone mentioned they were winning a lot using a positional theory so I wrote a little program that look as past winning combinations to see how many had the top 7-15 numbers for each position and I found the most any had was three and that only happen 20% of the time and the top 10 or so numbers per position cover a lot of combinations. So far I haven't been able to pick a group or numbers that preformed much better then a group of the same size picked at random.

There are other coders here but they don't spend a time talking about theories, they test them.

Positional theory is what "System X" proposes. You can find it free if you do some digging and, of course, there are plenty of people willing to sell it to you. It's a decent start, if you've got a good-sized bankroll. Otherwise, as you've noticed, it's a bit incomplete by itself especially when you're north of a half-million combos to filter through.

20% doesn't seem like much until you consider the size the OH p5 rolls to sometimes. Even if you were left with 5% hit ratio after all your other filters, that's still 18 Rolling Cash jackpots per year times $100K (minimum) equals +$1.8M ... Or 5 MegaMillions wins, whatever that's worth.