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Sum Totals Subtle Patterns for Pick4...


Well i found this pattern and another one too..And the other one has so much reduction properties and is one of the most important patterns for pick4 and is a huge subtle pattern...I'll post it soon...But before i post it i have a question that i am going to post after this blog that i need answered...That question is important for my analyzation of Pick4...And it could be important in helping win/crack Powerball...That question is very important so try and help me please...I'll post that question on the following/subsequent blog...Reason: Powerball if you look at the 5 ending digits and you eliminate the last ending digit and just go by the 4 of them you would notice (so far that's what i notice though i didn't do a 1 year draw history of Powerball) like how i was saying you will notice that 43% to 45% of the time Powerball ending digits repeat one time....

Now going back to the Sum Totals Subtle Patterns for Pick4 (subtle pattern # 2) here is a second pattern i found...I am not too proud of this pattern maybe because i was hoping you could reduce it further...I mean it is still a subtle pattern and a great pattern just that it wasn't why i was expecting to see for PIck4...

Subtle Pattern # 2 for Pick4: If you look at the draws (assuming you look at the midday and evening together) you MIGHT find that the sum totals range for PIck4 on average (85% to 90% of the time) ranges from 12 to 26...You MIGHT stretch it and say from 9 to 26 (depending on you and your state but i have hopes it will be from 12 to 26 most of the time)....Well if you take a history of the draws and you beging to eliminate Sums that have already played you might notice that you can do this 8 draws back, to be specific 8 different Previous Sums you can eliminate...Some of you depending on yourself and of your state you can reduce 10 previous Sums but if you do try not to go beyond it and remember that is your aim and win that's going to suffer...You might see 1 sum repeat and if it does, don't take it into account, still eliminate those 8 previous Sums that subsequently previously played...I was hoping to be able to reduce 15 previous sum totals but that will hurt the win and the aim...You have to understand Pick4 is a little different than Pick3...But as a good consolation remember that the subtle patterns for Pick3 (if not all) you can still apply them to Pick4..(Hurrah)...

Now if there is a range of Sums on average for Pick4 from 12 to 26 that means that there are 14 Sum Totals possible most of the time...And if you eliminate 8 different prevsious sums Totals that leaves you with 6 different sum totals...And if you knew that 6 sum totals are among the winning sum total for the next draw, wouldn't that be wonderful...Sounds great, but here is the problem and the source of me not been all that proud of this pattern..Proud but all that proud not...You see you can't paint everything roses you must show the bad part if everything comes out wrong...If some of you stretch it and say that the average sum total spans from 9 to 26 then that means that there are 17 Sum Totals and if you subtract the 8 sum totals from the reduction of the 8 previous draw that will leave you with 9 Sum totals for the next draw...And even if it had left you with 6 sum totals is not enough because on Pick4 numbers repeat half the time (50%)...And not only you have the wheel of 10 numbers for pick4 with 1 ticket having 4 of 4 that amounts to 200 or 210 that you can reduced the sums from but you have another wheel that has repeats on it and even though i don't know how much this wheel would span of repeaters i am guessing is going to be another 200/210 more numbers...

So basically what i am saying is even though you might be left with 6 sum totals for the next draw the possibility exist for it to be 9 and both numbers are not enough...Is not enough reduction plus you have the repeaters that you must think about..But you see you can't reduce more because then your aim will suffer..As i am writing this i am thinking and heading in the direction of you winning every day with minimal loses, i am talking like 5 loses total in an entire month...

Looking at the bright side of this subtle pattern2, even though it might leave you with not enough reduction for pIck4, you can look through it through the bright side...Remember when i said that if you look at Powerball like an illusion and look at it like a Pick4 through the ending digits, the odds can be 1 in 10,010 and not 1 in 150 million? The bright side is that even if this method doesn't leave you with enough reduction (6 to 9) you can use it for Powerball and MegaMillions...How will you feel if you knew that in the 6 sum totals (even though i can be 9) there is going to be the Pick4 ending digits combination for Powerball/MegaMillions and thus get 5/5 + 0...But you still would have to tackle the straight option that you must do but i have hopes you can do it...Remember it will not be and it should not be 24 ways for a number but instead 12 if you choose the repeaters and 12 anyways (this one is a hypothesis)...

Now that i remember there is a third subtle pattern that it might do you little help (or maybe more help than thought), you can still use it to try and crack Powerball...I shall post it here too...This pattern is not a Sum Total Subtle Pattern but is a subtle pattern only...

Entry #85


four4meComment by four4me - December 20, 2007, 1:56 am
Ok i tried to put the pick 4 sum chart on here but it wouldn't take it oh well.
here goes that are lots of pick 4 numbers between the sum of 12 and 26
sum 12 = 415 numbers
sum 13 = 480
sum 14 = 540
sum 15 = 592
sum 16 = 633
sum 17 = 660
sum 18 = 670
sum 19 = 660
sum 20 = 633
sum 21 = 592
sum 22 = 540
sum 23 = 480
sum 24 = 415
sum 25 = 348
sum 26 = 282

7940 numbers total
Comment by pumpi76 - December 20, 2007, 2:40 am
That's because is counting the 24 ways and 12 ways a number can be played...If you divide each sum by 12 (which is not completely accurate because there are numbers that are all 4 completely singles and thus will have 24 ways they can be played) so if you divide each sum by 12 that will give you a rough draft of how many numbers there are with that sum...You can also divide each Sum by 24 and take the number that you get with 24 and the number that you get with 12 and take the average and that will roughly give you the amount of how much numbers each sum has (boxed)...
Sum12=17 +35 divided by 2 equal= 26
Sum 13=20 + 40 divided by 2 equal = 30
sum 14 = 23 + 45 divided by 2 equal= 34
sum 15 = 25 + 49 divided by 2 equal= 37
sum 16 = 26 + 53 divided by 2 equal = 40
sum 17 = 28 + 55 divided by 2 equal = 42
sum 18 = 28 + 56 divided by 2 equal = 42
sum 19 = 28 + 55 divided by 2 equal = 42
sum 20 = 26 + 53 divided by 2 equal = 40
sum 21 = 25 + 50 divided by 2 equal = 38
sum 22 = 23 + 45 divided by 2 equal = 34
sum 23 = 20 + 40 divided by 2 equal = 30
sum 24 = 17 + 35 divided by 2 equal = 26
sum 25 = 15 + 29 divided by 2 equal = 22
sum 26 = 12 + 24 divided by 2 equal = 18

All this sum should equal 420 and i am guessing because i don't know how much does a wheel with repeaters covering all 10 numbers come out to...I know that non repeaters covering all 10 numbers boxed a wheel comes out to be 210...
The sums of the numbers come out to be 501 which is grossly overboard of the 420...And want to know something...420 is really not the limit, it should be less than that but i don't know how much..Is not counting the tripplets but triplets rarely play...
Comment by pumpi76 - December 20, 2007, 2:54 am
Another approximation you can do to try and figure out how many numbers does each sum has is to divide 420 by 14 (26-12 = 14) and that should give you 30 as a rough draft and this will mean that each sum total from 12 to 26 has 30 numbers each boxed...
Comment by pumpi76 - December 20, 2007, 7:29 pm
8 X 30 = 240
420 - 240= 180 combinations left and the payout is 200...Only 20 dollars to the imagination..

Still more to go...I want to be able to predict pick4 with 20 tickets...

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