Honduras Member #20982 August 29, 2005 4715 Posts Offline

Posted: May 24, 2007, 5:03 am - IP Logged

This post is titled: "How to Beat Powerball: And drastically lower the odds (theory) Another method..."

Guys, I think i found (theoretically) the way to beat Powerball...Is a theory...

Let me start buy saying this...What will you do if you knew that tomorrow Powerball winning draw (the 5 numbers not including the bonus ball) was going to have a Sum total of 119...How about what if you could know what Sum total each winning combination of Powerball was going to be in the future...You probably would feel happy but at the same time worried that you have to still figure out which one of the probably many thousands of combinations that also end in 119 is going to be...But what if you knew that tomorrow Powerball winning combination Sum total was going to be 119 and that the winning combination was going to have this appearance: OEEOO...By this i mean that the 1st number was going to be odd, the 2nd number was going to be even, the 3rd number was going to be even, the 4th number was going to be odd, and the 5th number was going to be odd thus OEEOO...(i got the idea from TnTea's pick5/39 Schooling's e/o types, except that this doesn't have a number just the letters)...If you knew this before hand you probably will feel more relieved and more happy...Well maybe i couldn't tell you what is going to be the ONE sum total that is going to play tomorrow, but i can select like 6 sum totals and be correct..And i perhaps may not get it right the 1st, 2nd, or 3rd try but i think i can do it in less than 5 tries...How, Because i found a big pattern in the sum total...I have some papers with Powerball draws starting from the first Powerball draw of 2006 'til April, 21st, 20007 (i haven't updated it yet and is a total of 136 draws)...The lowest Sum total from the first PB draw of 2006 'til 4/21/07 is 58...Two more things..From those 136 Powerball draws 17 (12.5%) had sum total below 100, and 7 (5.14%) had sum totals above 200...That leaves you with sum totals that go from 101 to 200 or in other words 100 sum totals left to play (remember those 100 sum totals because i am going to mention them later)...So we can eliminate draws that are below 100 and above 200 because they rarely happen...Here comes the amazing part: Starting from the first PB draw of 2006 "GOING FORWARD" the sum totals almost never repeat (not saying that they can't)...I couldn't track how many repetitions they were because it will take a long time...I look at it and can only see 2 repetitions..But i believe there will be like 8 of them...Here is the amazing part..Going FORWARD you can go almost 95 draws and see no sum total repetion and i believe it will continue to go like that indefinately (You see that's an extraordinary pattern that ALL lotteries that have long sum total ranges DISPLAY), and even if they were 1 or 2 repetitions, the following draw will DEFINATELY/GUARANTEED do not repeat...What does this mean?If starting from the first draw of 2006 and going forward you can go 95 draws, each with different sum total without sum total repetition then this mean that the 100 sum totals left to play you can subtract from it 95 sum totals that have played already (from the 95 draws) and that leaves you with 5 sum totals left that theoretically should be the next sum total to play..You may get closer to the 100 sum total (you may have 98 which you will eliminate from 100 sum totals left to play and the result should be less sum totals which should be the next sum total to play..You may also get farther from the 100 sum total which in this case you will eliminate from the 100 sum totals left to play (for example you may get 90 which will give you 100-90= 10 sum totals that will play for the next draw) and the sum totals left back should be the next sum totals to play...You can go as much as you can but remember to leave a room of 90 to 95 draws (that will be different sum totals that you can eliminate from the 100 sum totals left to play...So here comes the amazing part..If you was left with 5 sum totals that should be the next sum totals to play this is how you odds are drastically reduced according to my calculations....A Pick5/55 has 3,478,761combinations..If you select the lowest sum total from the first PB draw of 2006 'til 4/21/07 and you select the highest sum total of that period (which were 58 & 212) and you take the difference between the two you will get 154 (Even though those are not the lowest or highest sum total of a pick5/55; the highest sum total is 265 [you add 51+ 52 + 53 +....55] and the lowest sum total is 15 [you add 1+2+3+....5]...) Now hypothetically/theoretically the 3,478,761 combinations should have evenly distributed combinations with sum totals equivalent to those 154 meaning that if you divide 3,478,761 between 154 you get 22,589.35..This means that each of the 154 sum totals have 22,589.35 combinations evenly..However this will not be entirely so because some sum totals will have more combinations than others, but we will leave it at 22,589.35 to be on the safe side..And what does 22,589.35 means? i means that that number is what 1 sum total of the 154 sum totals will have..But we are likely to be left with 5 sum totals or somewhere around that number...I mean if you can narrow it down to 1 sum total your odds for winning powerball (the 5/5 match) will be 1 in 22,589.35 with 1 ticket...But because we have 5 sum totals we have to multiply 22,589.35 x 5 = and this will give you the odds of 1 in 112,947...But it gets better..I can also notice that the E/O types, they are 31 E/O types which are:

2even/3odds 3even/2odds

1. EEOOO 11. EEEOO

2. OEEOO 12. OEEEO

3. OEOEO 13. EOEEO

4. EOOOE 14. EEOOE

5. OOEOE 15. OEEEO

6. EOEOO 16. OOEEE

7. OEOOE 17. OEOEE

8. EOOEO 18. OEEOE

9. OOOEE 19. EOEOE

10.OOEEO 20. EEOEO

21. EOOEE

1even/4odds 4even/1odd

22. EOOOO 27. EEEEO

23. OEOOO 28. EEEOE

24. OOEOO 29. EEOEE

25. OOOEO 30. EOEEE

26. OOOOE 31. OEEEE

Now theoretically each sum total should have an equal amount of combinations with each of the 31 E/O types..What this means, it means that if you divide 112,947 by 31 it will give you the theoretical odds...However this may not be exactly like that and the only way to find out is by using a lotto software, specifically lotto expert, you can find out exactly how many combinations each E/O type would have...But to be on the safe side i am going to divide 112,947 by 10 ONLY...This will give me 11,294.7..And this is your odd if you could what was going to be the ONE E/O type...You may ask why did i divide 112,947 by 31 or by 10..Here is why...

Here is the amazing part: The same scenario occurs with the 31 E/O types...Also of those 31 E/O types, 10 are the most rare ones so if you like you can eliminate the 10 that rest and that leaves you with 21...Now If you go FORWARD so many draws from the first PB draw 'til 4/21/07 (i'll have to check out how many draws can you go) you will notice that the E/O types do not repeat that often, very seldom (Even if you were to keep the 31 E/O types)You can go like 12 draws each contain a different E/O type (from the 21 E/O types left you can subtract 12 draws each containing a different E/O type)..This will leave you with 10 E/O types..And the most amazing thing is: "i think i can eliminate most of E/O types and leave you with only 10 (at least) E/O types that i think are going to be the next draw...And because you can't get ONE E/O type and be sure is going to be the next draw but instead can get 10 E/O types, you multiply 11,294.7 x 10 E/O types..Remember 11,294.8 is the amount of combinations that will have the one same E/O type...11,294.7 x 10 = again 112,947 combinations and that's our odd: 1 in 112,947...No you may ask well that's only the 5/55..What about the Bonus ball that has a staggering odd of 1 in 150 million roughly...But THAT'S NOT THE POINT...The point is that if you take 1 ticket from those 112,947 combinations and multiply it by 23 (or by it with the 23 even or odd bonus balls; it would have cost you 23 dollars); the number of odd or even balls (one or the other), your overall odds including with the bonus ball would not be 1 in 150 million but 1 in 112,947 divided by 2 = 56,473.5 or 1 in 56,473.5...Why? Because you are playing half of the bonus balls (in this case the odds only or the evens only) and you will have a 50/50 chance of getting the bonus ball right and if you look at the history of Powerball you will notice that is one draw with even bonus ball and the next draw is with odd bonus ball, all somewhat...This will give you a 50/50 chance which you will proceed to divide 112,947 by 2 (or 50%)...And it would have given you the theoretical odds of 1 in 56,473.5 with 1 ticket and not 1 in 150 million...

Imagine if we had divided 112,947 by 31 we would have had 3,643.4 x 10 = 36,430 combinations and that divided by 2 (or 50% of the bonus ball in this case even or odd) our theoretical odds would had been 1 in 18,215 with 1 ticket and not 1 in 150 million...Again this is theoretically...Not everything goes as planned though...

Guys but the most amazing thing was that look what i was able to accomplish, now image what a supercomputer with 35+ Trillion Calculations per second like MACH5 can do (And the MACH5 at a price of 6 million dollars)....

question: Could you have bout 56,473.5 combinations or the 18,215 combinations and win Powerball...Theoretically speaking you might have...

suggestion: You don't necessarily have to go forward, you could go backwards when checking the draws..

Kentucky United States Member #32652 February 14, 2006 6145 Posts Offline

Posted: May 24, 2007, 2:22 pm - IP Logged

"I mean if you can narrow it down to 1 sum total your odds for winning powerball (the 5/5 match) will be 1 in 22,589.35 with 1 ticket...But because we have 5 sum totals we have to multiply 22,589.35 x 5 = and this will give you the odds of 1 in 112,947...But it gets better..I can also notice that the E/O types, they are 31 E/O types which are:"

An odd sum has either 5 odd numbers, 1 odd number and 4 even numbers, or 3 odd numbers and 2 even numbers. In a 5/55 matrix there are 33,312 combinations with a sum of 119, 1911 combinations have 5 odd numbers, 9535 combinations have 1 odd and 4 even numbers, and 21,866 with 3 odd and 2 even numbers.

"Now theoretically each sum total should have an equal amount of combinations with each of the 31 E/O types."

There are 16 E/O types with odd sums and 6 of them begin with an odd number so you might consider using high/low filtering too.

"question: Could you have bout 56,473.5 combinations or the 18,215 combinations and win Powerball...Theoretically speaking you might have..."

If you "knew" the sum was going to 119 and in fact the sum is 119, you have 1 second prize winner by playing 33,312 combinations. But to win Powerball you would also have to match the PB number so you have to play more combinations unless you "knew" the PB number too.

It's a fact you can greatly reduce the amount of combinations "if" you know what the sum will be but since there are sums of 15 - 265, that's a mighty big "IF". How long would it take to accurately fill out and play 6662 play slips?

Honduras Member #20982 August 29, 2005 4715 Posts Offline

Posted: May 24, 2007, 2:26 pm - IP Logged

i said: "question: Could you have bout 56,473.5 combinations or the 18,215 combinations and win Powerball"

correction:question: "Could you have "bought" the 56,473.5 combinations or the 18,215 combinations and win Powerball".."

also:

suggestion: You don't need a supercomputer though to win Powerball using this method, all you need is a software like lotto expert...If you have a smoking hot custom lotto software is even better..All i was saying is that with a supercomputer you could escudrinar/dissect (escudrinar is a Spanish word, i don't remeber & can't find the translation in English) Powerball deeper...

suggestion: There might be 1 or 2 more E/O types missing..

"Laura Simpson from Great Lakes, Illinois deserves to be rich..." "She is so rare..."

Honduras Member #20982 August 29, 2005 4715 Posts Offline

Posted: May 24, 2007, 3:24 pm - IP Logged

There are 2 or 3 more further patterns i am going to post later on today or tomorrow that you can use to further reduce the odds...(One of those patterns is an unheard of pattern)...

Trying to win Powerball is like going to the center of the Earth and coming out in the movie "The Core"....

"Laura Simpson from Great Lakes, Illinois deserves to be rich..." "She is so rare..."

Kentucky United States Member #32652 February 14, 2006 6145 Posts Offline

Posted: May 24, 2007, 3:59 pm - IP Logged

Quote: Originally posted by lottolaughs on May 24, 2007

How long would it take to accurately fill out and play 6662 play slips?

It took me half an hour to run 1,000 QP's for someone once. Better get to the store early!

It took me almost a half hour to get 5 Mega Millions QPs one night.

First he ran 5 Classic Lotto QPs then had to cancel them when I told him I wanted Mega Millions. Then he hands me 5 Rolling Cash 5 QPs and had to cancel those too. When he finally got it right, he asked for $6 because he checked "yes" to the Kicker after I told him "no".

I gave him the 6 bucks because by that time there were 22 people in the check-out line.

Wow.. an hour.. were they single tickets or were there five per line...

When I run 100 it only takes a few minutes...

Tntea,it was only half an hour but again,this was a dozen years ago when the machines seemed slower. I was running a betslip through over and over, so five at a time. Now I think you can play up to 10 on a betslip,is that correct?

Wandering Aimlessly United States Member #25360 November 5, 2005 4461 Posts Offline

Posted: May 24, 2007, 4:30 pm - IP Logged

Quote: Originally posted by Stack47 on May 24, 2007

It took me almost a half hour to get 5 Mega Millions QPs one night.

First he ran 5 Classic Lotto QPs then had to cancel them when I told him I wanted Mega Millions. Then he hands me 5 Rolling Cash 5 QPs and had to cancel those too. When he finally got it right, he asked for $6 because he checked "yes" to the Kicker after I told him "no".

I gave him the 6 bucks because by that time there were 22 people in the check-out line.

Not to pass judgment, but that's why I think every player should probably fill out a play slip. Then the chance of a mistake is almost eliminated. I've gotten a wrong ticket even with a play slip, but the chances of getting one from another lottery game are slim to none if you fill out a card. I understand that you feel you should be able to just walk in and buy your tickets, but think about how the people in back of you felt if you were frustrated!

Sunny California United States Member #40295 May 31, 2006 7711 Posts Online

Posted: May 24, 2007, 4:33 pm - IP Logged

Quote: Originally posted by Stack47 on May 24, 2007

It took me almost a half hour to get 5 Mega Millions QPs one night.

First he ran 5 Classic Lotto QPs then had to cancel them when I told him I wanted Mega Millions. Then he hands me 5 Rolling Cash 5 QPs and had to cancel those too. When he finally got it right, he asked for $6 because he checked "yes" to the Kicker after I told him "no".

I gave him the 6 bucks because by that time there were 22 people in the check-out line.

Kentucky United States Member #32652 February 14, 2006 6145 Posts Offline

Posted: May 24, 2007, 4:45 pm - IP Logged

Quote: Originally posted by justxploring on May 24, 2007

Not to pass judgment, but that's why I think every player should probably fill out a play slip. Then the chance of a mistake is almost eliminated. I've gotten a wrong ticket even with a play slip, but the chances of getting one from another lottery game are slim to none if you fill out a card. I understand that you feel you should be able to just walk in and buy your tickets, but think about how the people in back of you felt if you were frustrated!

I usually do use play slips but sometimes they tell me the reader isn't working or the play slips I used in the last two draws are filled out incorrectly so I told him "5 auto Mega Millions, no Kicker".

If I would handed him a couple of play slips, the line behind me would have been here to Florida.