digit

J

jack

Member
Hello, here are two ram statistics for the whole of the six numbers and also by suit
Example one of your results =
05 16 24 34 32 48 49
This statistic sums the pairs together the last first
0,6,2,4,2,8,4 = 7 pairs = pairs (pairs have repeated (2, 4) the valley is number of times left pair
Odd = 5 1 3 3 9 = 5 odd (odd is repeated (3)
This just another statistic of suits
example
05 16 24
Three pairs pairs = 0,2,4
= 5.1 odd two odd
Make the other suits also
In both statistics, the whole of 6 or suits when added to stop the amount of odd, is the first and last digit together to be more exact, the suits will be fine
 
J

jack

Member
Hello, ram, good breakdown of the 20 positions in the suit,
It is good that you can create filters, this filter looks the last digit of each of the 20 suits
The suit lasts digits, when created can not equal up to a limit of draws, past
Viewing in each vertical position, eg after the study was that the last digit
Position of the 1 st 2 nd and 3 rd = 429, then the last digit terminations or 429, this training can not be equal to 200 results, then looking in the vertical position until 200 draws ago
429 have an equal education, not play, the limit of the draws have to see, tell me
If you get it?
 
J

jack

Member
1 - raises the frequency of each of the numbers in the game.
2 - It is classified by the numbers in descending order of their frequencies.
From this point, what becomes interesting is the ranking of each number or each number is associated with your order according to their frequency in the file.
3 - You take every combination and it is the sum of the qualifying orders of their numbers
4 - Rate this file are in ascending order of amounts. Thus, the
combinations that have the numbers of mixtures with higher frequencies are at the beginning of the file with numbers and combinations of lower frequencies are at the end of the file.

If the file has a reasonable number of combinations (say above 1000 for the Mega) combinations are most likely between 30% and 70% of file size. Of course, the assembly method makes the game more combinations likely to move over to the beginning or towards the end of the file.

One can therefore conclude that using the same methodology and it is efficient, substantial, consistent, and sets tivem the same size, the most likely combinations are in very close.
This can be tested. Simply attach to different games competitions using the same methodology and force such that there are secondary prizes in the game (without abandoning the methodology). If the method is good, the winning combinations of the games will be in similar regions.

So it is possible to mount volumes and games with similar methodology and pick only the region of the file where the prizes are located.
ram, study well and also put your ideas, make at
all (number 6 in the case of 49/6) is a tender (20 suits)
 
J

jack

Member
is the point of the fine adjustment. By drawing the distribution curve it can be seen the placement of fashion( moda). She is the reference for identifying the micro region with greater possibilities of the existence of prizes. For example, a method of assembling games put the likely winners in the transition from 2nd to 3rd track from the fashion toward the beginning of the curve

.
The variable used to differentiate potential is the frequency of numbers, Logo criteria mounting where the frequencies of all numbers, involved are very similar, as is the case of developments and closings
 
J

jack

Member
Hello, here is a way to look like a pick3, ie is seen every suit,
Only the last digit vertically from each of 20 positions, so ram is analyzed and filtered
Each of the 20 suits, the last digit like a pick3 in each suit, =
Start to record keeping midday everyday and evening seperate from eachother. have a space available for these categories next to the drawing listed.DATE-MID - EVE - OE / HL SUM / ROOTSUM / PRIOR / HOT-COLD-DUE
E.g. - 267 EEO / LHH 15 6 - 2 6 7
EEA 042/6 6 LLL - 0/2 4 -
EOO 639 / HLH 18 9 6 6 3 9
EOO 497 / HLH 20 2 4 4 / 9-7
prediction is comprised of these categories
history
Analysis
Probability
Outcome
On average, HOT NUMBERS = APPROX 59% PRIOR NUMBERS = 23% = 12% COLD NUMBERS, NUMBERS DUE = 6%
80% OF THE TIME "SUMS" = BETWEEN 10-20, USUALLY DOUBLES FALL
 
J

jack

Member
Hello Ram, another filter is the sum of both even and odd at all all 6 numbers
And by separate suits As you know the results are composed of even and odd numbers, the largest percentage, others do not, or are only Odd or Even.

- A filter that we could choose how we wanted to select the amount of odd / even, and apply them according ace their sums.

Taking as an example-

Figures --- 19-25-28-33-44 49 P / I = 2/4

Pairs: 28 +44 = soma72
Odd: 19 +25 +33 +49 = soma126

-The idea was this: suppose I wanted to find capostas possible, the same figure for the P / I - 2/4, between the sums (60-86) for the couple, and (51-87) for Odd ..!

Do not give optimal filter ..!

and see what sums out more in terms of percentages of exits, but I think many Pairs or Trios are within the same sums, which could greatly reduce the bets
Within your idea of the sum, but in odd and even separate
Attention, but only the last digit (0-9)
 
J

jack

Member
Hello, ram, plus a filter, spectacular, good tone when the mirror
So grab the last two pairings
example =
05 16 24 32 48 49 = last digit = 5,6,4 ,2, 8, 9
03 17 18 22 46 54 = last digit = 3,7,8, 2, 6, 4
Difference = 2,1,4 , 0, 1, 5
Ram you have to reduce the last digit of the last two contests,
Then he = 2,1,4,0,1,5, not play these next digit of their coluns with mirror and adjacent work when the proper filter of the mirror, but it has to be in its upright position
You do cadaum 20 suits, then is the difference between the two last sweepstakes, but only the last digit, tell me if you know? please
 
J

jack

Member
Hello Ram, here is a concept deuma lottery 46/5, maspode use on 49/6 or 60/6
The goal is to see the output spacing suits (20 suits each sweepstakes) who have left
What can repeat any suits, (here the first and last digit are together), goal
You see the back of the suit, good after mastering can do well if only the last digit,
That is when you will return the suit after leaving last digit =
Consider the "mega" number of the Mega Millions lottery. There are 46 "mega" numbers. The probability of the same "mega" number k appearing later drawings (and not before) is Given by the binomial distribution. Specifically: (1/46) * (1 - 1/46) ^ (k-1), where the operator "^" means "to the power of". (For example, 4 ^ 3 = 4 * 4 * 4 = 64.)
For example, suppose the most-recently drawn "mega" number is 30 (21 Aug 2012). The probability That will be the 30 "mega" number in the next drawing is 1/46. The probability That will be the 30 "mega" number in 2 drawings later (and not before) is (1/46) * (1-1/46). In 3 drawings later (and not before): (1/46) * (1 - 1/46) ^ 2.
If we compute That probability for k = 1,2,3, ..., we Find That 50% of the time, 30 (really any particular "mega" number) will be Appear again 13 to 63 drawings later (that is, the middle two quartiles: 25% to 75% cumulative probability).
I have the 742 Mega Million Analyzed drawings from 24 June 2005 through 31 July 2012 (I'm a little behind). And in fact, the same "mega" number Appears again 11 to 58 drawings later. Pretty close to expectations based on statistical theory.
The statistical theory is similar for the "distance" between drawing a particular number among the five "regular" numbers (non-megs). There are 56 "regular" numbers. So the probability of including a particular "regular" number in the next drawing is COMBIN (55.4) / COMBIN (56.5), or about 8.93%.
So I would expect a particular number from the last drawing to Appear in the next drawing with a probability of 8.93%. In 2 drawings later (and not before) with probability 8.93% * (1 - 8.93%). In 3 drawings later (and not before) with probability 8.93% * (1 - 8.93%) ^ 2.
So 50% of the time, I would expect a particular number from the last drawing to Appear again 4 to 8 drawings later (again, the middle two quartiles: 25% to 75% probability). That I have not vetted against the current historical drawings.
But that is just one particular number. I find it more useful to ask: given a selection of 5 "regular" numbers, how should we expect to recently have seen any__ __ 1, 2 or 3 of our selected numbers in past drawings.
Based on statistical theory, I would expect any 1 of 5 selected "regular" numbers to Appear 2 to 4 drawings earlier 50% of the time. Any two numbers to Appear 6 to 13 drawings earlier. Any three numbers to Appear 88 to 212 drawings earlier.
Analyzing the historical 742 Mega Millions drawings, I Find That any 1 of 5 selected "regular" numbers 1 to 4 Appears drawings earlier 50% of the time. Any two numbers Appear 6 to 25 drawings earlier. Any Appear 3 numbers 46 to 225 drawings earlier. It is very rare for any 4 or 5 numbers to Appear in a previous drawing. Again, pretty close to expectations based on statistical theory.
So in choosing five numbers to play in the next drawing, we might reject any combination That matches 4 or 5 numbers of any previous drawings, or That matches 3 Numbers Within the last 45 drawings recently or in earlier than 226 drawings. Et cetera.
Of course, such a choice does not alter the probabililty of matching some or all numbers in the next drawing.
It is just a rationalization for choosing ("filtering") among the 3.8 million 5-of-56 combinations That we might play. It just makes us feel better about our choices.
 
J

jack

Member
However it must be said that the terminations of the filter functions as a filter applied within the diagonals 1-48 because each diagonal 1-48 has only two terminations.

Let's face it: In this picture we have 13 diagonals that make up the diagonals of 1-48, 1-48 baptized because of it starts at 1 and ends at 48.
If fixed, these each diagonal contains only two terminations.
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






Regarding Diagonals of 6-49 have 14 diagonals that comprise it, it also baptized because of 6-49 starts at 6 and ends at 49. If we analyze these endings also diagonal, we can notice that in each one there is a repeat termination.

And each diagonal consists of 6-49:
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, given that only come out in EuroMillions 5 balls in the framework above, it suffices to choose only 5 diagonals also playing in both the 1-48 and 6-49, and 5 endings, not obvious? When doing this application will almost always the numbers reduced to a amount of 5 to 11 numbers endings. But also, if the choices are "poorly done" we have a reduction of less than 5 numbers, which will allow for the addition of a few more choices.

However, the problem lies in that diagonal endings and choose! On this issue, you must be a sensitivity and a little luck (just a little!) ...

For example ... and even just an example!
If we choose Diag.1-48 in 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 in 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 the filters together, we were only a total of 12 numbers, which are the ones that put in bold (and 03.08.10.13.15.20.26.31.33.36.38 41)

Now, by applying the filter of endings, if we choose the endings 0,1,3,6 and 8 are left with a final result of 11 numbers:
 
J

jack

Member
Hello, ram, always remembering that lasts the digits (0-9 endings)
Each of the terms 20 may be the three terminations all odd or all pairs
Below the two lists of odd and even endings, ram
I need to do in each of the two lists some statistics of hot, medium and cold
Etc. ..
odd
1 1 1
1 1 3
1 1 5
1 7 1
1 1 9
1 1 3
1 3 3
1 3 5
1 3 7
1 3 9
1 5 1
1 5 3
5 1 5
1 5 7
1 5 9
1 7 1
1 3 7
1 7 5
1 7 7
1 7 9
1 1 9
1 9 3
1 9 5
1 9 7
1 9 9
3 1 3
3 3 3
3 5 3
3 3 7
3 3 9
3 5 3
3 5 5
3 5 7
3 5 9
3 7 3
3 7 5
3 7 7
3 7 9
3 9 3
3 9 5
3 9 7
3 9 9
5 1 5
3 5 5
5 5 5
5 5 7
5 9 5
5 7 5
5 7 7
5 7 9
5 9 5
5 7 9
5 9 9
1 7 7
7 3 7
7 5 7
7 7 7
7 7 9
7 9 7
7 9 9
9 1 9
9 3 9
9 5 9
9 7 9
9 9 9
pairs =
0 0 0
0 0 2
0 0 4
0 0 6
8 0 0
0 2 0
0 2 2
0 2 4
2 0 6
0 2 8
4 0 0
0 2 4
0 4 4
0 4 6
0 4 8
6 0 0
0 6 2
0 6 4
6 0 6
8 0 6
8 0 0
0 8 2
0 8 4
8 0 6
0 8 8
2 0 0
2 0 2
0 2 4
2 0 6
2 0 8
2 2 0
2 2 2
2 2 4
2 2 6
2 2 8
4 2 0
2 2 4
2 4 4
2 4 6
2 4 8
2 0 6
2 6 2
2 4 6
2 6 6
2 6 8
8 2 0
2 8 2
4 8 2
2 8 6
2 8 8
4 0 0
4 2 0
4 4 0
4 0 6
4 0 8
4 2 0
4 2 2
4 2 4
2 4 6
4 8 2
4 4 0
4 4 2
4 4 4
4 4 6
4 4 8
4 6 0
2 4 6
4 6 4
4 6 6
4 6 8
4 8 0
4 8 2
4 8 4
4 6 8
4 8 8
6 0 0
2 0 6
6 4 0
6 0 6
6 0 8
6 2 0
6 2 2
2 4 6
6 2 6
6 2 8
6 4 0
2 4 6
6 4 4
6 4 6
6 4 8
6 0 6
6 6 2
6 6 4
6 6 6
6 8 6
6 8 0
6 8 2
6 8 4
6 8 6
6 8 8
8 0 0
8 2 0
8 0 4
8 0 6
8 0 8
8 2 0
8 2 2
4 8 2
8 2 6
8 2 8
8 4 0
4 8 2
8 4 4
8 4 6
8 4 8
8 6 0
8 6 2
8 6 4
8 6 6
8 6 8
8 8 0
8 8 2
8 8 4
8 8 6
8 8 8
 
J

jack

Member
Hello, ram, sums, the last digit, odd and even separated in each suit
To be seen, then a separate list endings (last digit) IN PAIRS AND ODD, already has in previous post and do the sum in each suit
example
Suit odd sum = 1,5,7 01,15,17 = 13
Suit par = 12,42,48 = 2,2,4 adds 8
Ran you have two separate lists the last digit (already has in previous post)
The front digit (first digit) does not enter, you have a good idea of the sums
Just who do have the odd and even separate suits, will be a good filter.
 
J

jack

Member
Hello, ram, also as in endings (last digit) of the even and odd
If you donate can do high and low-
High = 5,6,7,8,9
Low = 0,1,2,3,4
Then you can sort the suits (20 positions) endings in ups and downs as the even and odd =
0 0 0
0 1 0
0 0 2
0 0 3
0 0 4
0 1 0
1 0 1
0 1 2
0 1 3
0 1 4
0 2 0
0 1 2
0 2 2
2 0 3
0 2 4
0 3 0
0 1 3
0 3 2
0 3 3
4 3 0
4 0 0
0 4 1
0 2 4
4 3 0
0 4 4
0 1 0
1 0 1
0 1 2
1 0 3
1 0 4
1 1 0
1 1 1
1 1 2
1 1 3
1 1 4
0 1 2
1 2 1
1 2 2
1 2 3
1 2 4
1 3 0
1 1 3
1 2 3
1 3 3
1 3 4
1 4 0
1 4 1
1 2 4
1 3 4
1 4 4
2 0 0
0 1 2
2 0 2
2 0 3
0 2 4
2 1 0
2 1 1
2 1 2
1 2 3
2 1 4
2 2 0
2 2 1
2 2 2
2 2 3
2 2 4
2 0 3
1 2 3
2 3 2
2 3 3
2 3 4
4 2 0
2 4 1
2 2 4
2 3 4
2 4 4
3 0 0
3 1 0
0 3 2
3 0 3
4 3 0
3 1 0
3 1 1
1 2 3
3 1 3
3 1 4
2 0 3
1 2 3
3 2 2
3 2 3
2 3 4
3 3 0
1 3 3
3 3 2
3 3 3
3 3 4
4 3 0
3 4 1
2 3 4
4 3 3
3 4 4
4 0 0
1 4 0
4 2 0
4 3 0
4 4 0
4 1 0
4 1 1
4 1 2
4 1 3
4 4 1
4 2 0
4 2 1
4 2 2
4 2 3
4 2 4
4 3 0
4 3 1
4 3 2
4 3 3
4 3 4
4 4 0
4 4 1
4 4 2
4 4 3
4 4 4




5 5 5
5 5 6
5 5 7
5 5 8
5 5 9
5 6 5
5 6 6
5 6 7
5 6 8
5 6 9
5 7 5
5 7 6
5 7 7
5 7 8
5 7 9
5 8 5
5 8 6
5 8 7
5 8 8
5 8 9
5 9 5
5 9 6
5 9 7
5 9 8
5 9 9
6 5 5
6 5 6
6 5 7
6 5 8
6 5 9
6 6 5
6 6 6
6 6 7
6 6 8
6 6 9
6 7 5
6 7 6
6 7 7
6 7 8
6 7 9
6 8 5
6 8 6
6 8 7
6 8 8
6 8 9
6 9 5
6 9 6
6 9 7
6 9 8
6 9 9
7 5 5
7 5 6
7 5 7
7 5 8
7 5 9
7 6 5
7 6 6
7 6 7
7 6 8
7 6 9
7 7 5
7 7 6
7 7 7
7 7 8
7 7 9
7 8 5
7 8 6
7 8 7
7 8 8
7 8 9
7 9 5
7 9 6
7 9 7
7 9 8
7 9 9
8 5 5
8 5 6
8 5 7
8 5 8
8 5 9
8 6 5
8 6 6
8 6 7
8 6 8
8 6 9
8 7 5
8 7 6
8 7 7
8 7 8
8 7 9
8 8 5
8 8 6
8 8 7
8 8 8
8 8 9
8 9 5
8 9 6
8 9 7
8 9 8
8 9 9
9 5 5
9 5 6
9 5 7
9 5 8
9 5 9
9 6 5
9 6 6
9 6 7
9 6 8
9 6 9
9 7 5
9 7 6
9 7 7
9 7 8
9 7 9
9 8 5
9 8 6
9 8 7
9 8 8
9 8 9
9 9 5
9 9 6
9 9 7
9 9 8
9 9 9
 
J

jack

Member
Hello, ram, I thought I'd won a big prize! Ahhh just a little humor,
The previous post is easy, in the other post and I sent the odd pair, the latter is high and low
Example = low = 0,1,2,3,4, and high = 5,6,7,8,9, the other was par = 0,2,4,6,8 and odd = 1,3,5, 7.9
We can use both,
Example = 05,16,24,32,48,49 = last digit = 5,6,4,2,8,9
Low = 4.2
High = 5,6,8,9
Par = 6,8,4,2
Odd = 5.9
In this low and high income, high accuracy in suits, 5,8,9, ie always 100% vai
Paegar three, or low or high, very good ram
then the suit = 5,8,9 above example, this list of high
 
J

jack

Member
Hello, these two lists of 125 formations 3 (high and low) will always have a suit in 100% of any lot may confer, as has also the haphazard way the even and odd
These lists, making hot and cold and see average delay is very important as ever between the two lists will have a training last digit of both even and odd as low and high
Semptre 100% will provide training for
 
J

jack

Member
Hello, ramrock, we also use the sums
example
High = 689 = 23 = sum
So after doing the sums of the even and odd and ups and downs
We can apply an algorithm to "calculate" the intersection of the two groups. (3-E 3-odd even) E (3-E 3 low-high) with = the sum of the columns and endings,
 
J

jack

Member
Hello, ramrock, I've been seeing this pattern in each of the 20 positions, to see the pair, pair positional mounting the suit, you can use a digit that repeats, ok goal is to see pairs to assemble after the last digit, suit, any of the 20 tender may be suit pivot, or fixed
The good this is that this one position, lists = odd / even, low / high, sums, frequency groups, pairs, etc., Along with the filters mirror, adjacent, and decrease the vertical
The last digit (0-9) is a powerful filter to eliminate half of the numbers in a lottery
Tends to have patience in the preparation, because the utilization is 80%, already very good,
 
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