wrong! a simple, yet proven piece, of statistical data, from the beggining of lottery; reveals that all numbers should add up to a minimum or max range.
True!
Lets take pick3 for an example. You can take any other game, but for this example's sake it is easier math.
There are 3 X 10 digits, from 0 through 9 for each position.
There are 3 digits drawn each draw.
It is easy math to calculate how many times a digit will show up after 10 draws in each position. The answer is: each digit will appear once, for each position.
This ofcourse is theory. In reality it wont be like that.
At least, not after 10 draws.
I'll use North Carolina Eve for this example, because NC eve has had 992 draws so far.
Lets round it up to a thousand, for the examples sake.
The theoretical statistics tell us that in position 1 each digit should have fallen 100 times.
The reality however shows something else:
Digit 0 : 111 times
Digit 1 : 117 times
Digit 2 : 105 times
Digit 3 : 108 times
Digit 4 : 85 times
Digit 5 : 88 times
Digit 6 : 96 times
Digit 7 : 92 times
Digit 8 : 100 times
Digit 9 : 90 times
Some digits seem to fall more than other digits, but they all float around that magical 100.
If we would review these statistics in another 1000 draws, so when there have been 2000 draws, perhaps digit 1 will no longer be the best performing digit in position 1. Perhaps digit 4, with now "only" 85 appearances, will have the highest hits.
Still, they will all float around the value of 200, give or take 10 to 20 %.
If a digit falls "behind" on hits, it will make it up after another set of draws. (that is when the "hot period" for a digit starts, after a cold period...)
And this example was only for digits in 1 position.
We can review the other position, all on itself, or take all 3 positions into account.
The boxed pairs: A non-match boxed pair (2 different digits) have a mathematical skip of 18.51 draws, meaning that in theory when a pair hits, it will go away and come back after 18.51 draws.
A match pair, e.g 11, 22, 33, ..., has an average skip of 35.71 draws.
Again, that's the theory.
A Pick3 number is always made of 3 pairs, even when it is a double (225, 338, ...)
We put the number in its boxed format, meaning the lowest digit first, then the second highest digit and the highest digit in last position.
ABC : the pairs are AB, AC and BC (or any other order ofcourse)
This is the current data for the first 10 pairs, for NC eve:
| Pairs | Times | Cur. skip | Av. skip | Math Av skip |
| Box pairs | | | | |
| 00 | 32 | 43 | 31, | 35,71 |
| 01 | 48 | 25 | 20,67 | 18,51 |
| 02 | 57 | 2 | 17,4 | 18,51 |
| 03 | 58 | 2 | 17,1 | 18,51 |
| 04 | 63 | 11 | 15,75 | 18,51 |
| 05 | 50 | 12 | 19,84 | 18,51 |
| 06 | 62 | 5 | 16, | 18,51 |
| 07 | 49 | 12 | 20,24 | 18,51 |
| 08 | 54 | 26 | 18,37 | 18,51 |
| 09 | 57 | 29 | 17,4 | 18,51 |
The column "Av. skip" is the real average skip, calculated on real hits, in comparison to the amount of draws.
Some pairs perform better than others, just like the digits.
But again, give them another 1000 draws and the picture will be different for those pairs.
What stays the same is that the real average skip will move towards the Mathematical average skip.
And this was only for digits and boxed pairs.
There are LDR, Root, Sums, straight pairs, straight positional pairs, VTracs, HighLow, OddEven, InOut, OpenClosed, ... name it...
Now, when looking at these pairs in the example we see that the current skip (amount of draws the pairs has not shown in the draws) for pair 09 is at 29 draws.
We then say "it is 1.5 times due" (approx.)
Things go rarely "missing" (times due) for 8 times their average skip.
If a pair has a current skip of 148 draws, then it is 8 times due.
That's like a breaking point in the statistical data i have assemblied.
In Wisconsin for example, the boxed pair 01 is out for 146 draws.
So it is 7.89 times due.
Now, there is no guarantee, but this pair will soon come in!!! (no guarantee because they might have had a testdraw in which that pair showed up, and we know nothing about it --> but it doesn't matter, it doesn't!!! lol)
Why stay with "part" of a number?
Why not use a set of numbers? Lets be honest, waiting for the 01 to come in for 100 draws is a mighty long time lol, and the profit will be low.
Lets take the set of numbers HHH, straight.
This would represent 125 straight numbers.
1000 / 125 = 8
So the average skip is 8 draws.
If in State X the structure HHH has not fallen for 64 draws, then none of those numbers have fallen for 64 draws. (In theory each straight number will fall once every 1000 draws. (in reality that is not so!))
So one could start playing those 125 numbers using that information. And i know, playing 125 numbers is a big risk.! and with a payout of 500 to 1, the risk is just too high. (but not with a payout of 900 to 1!!!)
This was just a small example of how statistics are so usefull in these games.
Ofcourse, if you take a 6/42 game, which has over 5 000 000 combinations, it takes a whole other approach to use the statistics to your advantage.
Still, it can be done. But it takes time, perhaps a lifetime lol
cheers
Ricky
lasas3
An onion a day keeps everyone away!!!
