Using 6/49 Vtracs to make playable Combinations

[Icewynd]

Hello, I do not understand why ice pab the left!!

PAB stated his reasons quite clearly. I would advise you to re-read his post and think about what he has said.

About 4 groups is to apply
* The filters, using the standard, and also the terminations filter blocks

4 groups??? Do you mean groups of 4? What is the standard? I am not sure what you mean by "terminations filter blocks" Currently the only filter we have discussed is the restriction of the first digit to 01-08 and the restriction of the last digit to 17-24. This has reduced the total of 134,596 combinations down to 118,608 combinations.

* Central, terminations and filter by groups, ice may not be able to excel

Agreed. Writing VBA code is not my stong point.

* Faer every project, you need a solft?

I do not understand this statement. What is a "solft"???

 

[jack]

Hello, i think that program would be better to see these filters,
* As for the 4 groups of 6 numbers, it's easy, ie we use
* That planilia to see how groups behave,
Ice, have you ever noticed that there are a lot intervals, ie
* Segments of distances, when a sample is divided
* In 4 groups of 6 numbers, is a group with zero or one
* Vtrac, then it is well within the 7.315 blocks, draw, formations
* What is inside the group, as it is not known to have in 4 editions
* But to avoid doing in 4 editions, this in secret endings
That will give evidence that group of 6 will be out in 4 to avoid
* Issues when one does not know which group will be out, or we meet the standards of output sweepstakes

 

[Icewynd]

Hello, i think that program would be better to see these filters,
* As for the 4 groups of 6 numbers, it's easy, ie we use
* That planilia to see how groups behave,

Yes, the DECADES analyis is a good filter.

Ice, have you ever noticed that there are a lot intervals, ie
* Segments of distances, when a sample is divided
* In 4 groups of 6 numbers, is a group with zero or one
* Vtrac, then it is well within the 7.315 blocks, draw, formations

Jack, are you using the "*" symbol to indicate the start of a new idea? It seems as if these lines are all talking about the Decades filter, but your use of the "*" suggests that each line is a new idea.

* What is inside the group, as it is not known to have in 4 editions
* But to avoid doing in 4 editions, this in secret endings
That will give evidence that group of 6 will be out in 4 to avoid
* Issues when one does not know which group will be out, or we meet the standards of output sweepstakes

Sorry, I don't understand this. Are you saying that we do not know which of the 4 numbers in each Decade to choose? If so, we do have some ideas such as SKIP pattern, LAST DIGIT, etc.

Have you downloaded the ON49 spreadsheet created by PAB? If you have it would be helpful if you could use the terms used on each tab of that spreadsheet to describe your ideas. It would certainly help to bridge the language barrier if we are all using terms such as DECADE, SKIP, DELTA, etc. so that we both know what you are referring to.

 

[jack]

Hello, the ice, the forum is for this reason, many debate, ideas and then filter the best ideas, the perfect formula nimguem will find on the 4 groups of 6 numbers
TEMSO create a formula that rotational meet the standard, then my suggestion, moreover
I think it's also only focus on the central 7.315 blocks because a program can simulate situations, then we seek safe filters for central blocks
we need a secure way to use your planilia statistically
calculator to create a filter for the central blocks

 

[jack]

Hello, icew can confirm, please enter the following
Of the 60 pairs of vtrac below, at least one pair in this 100%
Either draw? The next course is also at least 100%
A pair of 60 below this list, this list is good with may decrease by statistics and filters if you do not want to play with 60 pairs, because the couple is beginning to forecast
01 02
01 03
01 07
01 08
01 09
02 03
02 07
02 08
02 09
03 07
03 08
03 09
07 08
07 09
08 09
04 05
04 06
04 10
04 11
04 12
05 06
05 10
05 11
05 12
06 10
06 11
06 12
10 11
10 12
11 12
13 14
13 15
13 19
13 20
13 21
14 15
14 19
14 20
14 21
15 19
15 20
15 21
19 20
19 21
20 21
16 17
16 18
16 22
16 23
16 24
17 18
17 22
17 23
17 24
18 22
18 23
18 24
22 23
22 24
23 24

 

[Icewynd]

Hello, icew can confirm, please enter the following
Of the 60 pairs of vtrac below, at least one pair in this 100%
Either draw? The next course is also at least 100%
A pair of 60 below this list, this list is good with may decrease by statistics and filters if you do not want to play with 60 pairs, because the couple is beginning to forecast

I'm sorry, Jack. I do not understand your request.

Do you want to count the appearances of the 60 pairs?

What is special about these pairs. Why only 60 out of 276 pairs?

 

[jack]

Hello, pairs, you noticed that don't need 276 possible pairs!! With only 60 pairs where a couple have hit 100 a reduction of 216 pairs, of course with the same hit in 100% condition at least a pair, is that with peers is easier to see the hottest, most pairs to return to play, let's see if you can choose the central pair of Centre Court, understand!
Hello ice, I am naotando this double standard polindromo or capicua, example 10,01 or 12 21 or 03.30 etc, hardly a normal numbers draws out the couple of reversed (polindomos) then sertia see if mount vtrac pairs, without being polindromos, with two vrtrac get 4 normal numbers, pick pairs in which the pair have both numbers on a sweepstakesIf the vtrac pair have two numbers in each pair have an accumulation of gain
Needing two more apenhas vtrac to close the bet, and then with vetrac 4 (8 numbers)
We have a jacpot, or see a way to accumulate the two numbers on a couple of vtrac to decrease the vtrac,
Use accumulated a vtrac hitting the two numbers, example you decided to
By vtrac 6 to play, but one of them gave the couple full (both numbers) I have a vtrac for nothing, only 4 vtrac to close complete bet, the idea of 60 pairs is the pivot
The couple of references, or ice do the normal 49 and see the couple that can give and use as pivot
But this couple is taken from the normal combination of 49 numbers, then combine with the other 4 vtrac. Would be a joint conbinaçao, the normal pair and 4 vtrac, that when dismembered par two numbers and more 4 vtrac 8 total = 10 numbers excluding the 2 numbers, since with 6 vtrac need 12 numbers, when we set up the games,
Taking advantage of the normal pair gain out of 49 numbers, elimanos two, vtrac
Being able to combine with blocks, tabem pairs can be end of 01 to 16
And 36 a49 note who now to see the pair of ends, tends to be of normal 49
So hitting the pair of extremes without being vtrac only need 4 vtrac Tony central courts, i.e. mixed combination of normal vtrac pairs, we have two conditions now,
To take advantage of the gain of the normal pairs. Icew, you are very clever, man will notice this!

 

[jack]

Hello, ice, be careful with the translation of google, to make sure at least one pair
In 100% of the sweepstakes, do not need 276, needs 60 pairs, a decrease of 216,
With the same accuracy of 100% of the sweepstakes, but the biggest asset, is to choose a pair
* Without vtrac, and then join with 4 vtrac, will be a combination of normal pair mixed with 4
* Vtrac, this is the idea, the couple may have chosen not to repeat digits this example
* 05xxxx pair 42 has repetition of four digits, the digits initial and final
* Now this example = 05 xxxx45 can not because it has the digit 5 twice, this type
* In normal pair can not be regarded as par pivot,

 

[jack]

Hello, icew, you know that couples normal 01-49 without vtrac = 1176
* Couple on 100% but I managed with the same 100% by a pair in any future
* Giveaway on 276, icew attention are normal peers without vtrac, these 276 pairs at least one
* Pair this with 100% of any future sweepstakes, you want to see the pairs confirm pair?
Within these pairs 276 select or filter without repeating digits pairs
* Example 12.43 can
***************** 12, 42 can not have repeating digit 2

 

[Icewynd]

Hello, ice, be careful with the translation of google, to make sure at least one pair
In 100% of the sweepstakes, do not need 276, needs 60 pairs, a decrease of 216,

OK, so you are saying that AT LEAST one pair of the 60 pairs will always be present?

With the same accuracy of 100% of the sweepstakes, but the biggest asset, is to choose a pair
* Without vtrac, and then join with 4 vtrac, will be a combination of normal pair mixed with 4
* Vtrac, this is the idea, the couple may have chosen not to repeat digits this example
* 05xxxx pair 42 has repetition of four digits, the digits initial and final
* Now this example = 05 xxxx45 can not because it has the digit 5 twice, this type
* In normal pair can not be regarded as par pivot,

So, we should find a pair of numbers between 1-49 that we think will be drawn, and then combine with 4 Vtracs which will give us 8 numbers (9 if 24 is one of the Vtracs) to combine with that original pair.

Also, the Vtracs should be chosen so that none of the numbers have a common last digit.

 

[jack]

Hello, perfect attention to differentiate pairs and pairs of vtrac normal 01-49
* Icew in normal pairs are possible in 100% = 1.176, but I managed with the same
* 100% in any future drawings with 276 pairs with the same accuracy of 100%, is still within this 276 pairs 01-49 have pairs filter without repeating digits
* Example of the 276 13.46 this pair can
********************************** 13.36 This pair this pair can not
Icew, you must dismember the draw for the 49/6 in 15 pairs of positions
* 1.2 1.3 .... ate = 5.6 are 15 pairs of positions
* Even by then will serve to put the pair in position
* Example was chosen the pair repeating digits with delay
12.36 its ideal position is 2nd and 4th, then the other will position the 4 vtrac
Yes yes, in 60 pairs vtrac at least a couple vtrac confirm that this is so!

 

[jack]

Hello icew look, the three result you posted
Last 3 draws from Ontario 49:

14.11.2012 B 3,10,21,22,28,42 Vtracs = 9: B = 3,3,10,17,21,22 9

11.17.2012 B = 27 Vtracs 4,21,22,23,26,32: 1,4,7,21,22,23 B = 2

11/21/2012 3,4,18,24,40,46 Vtracs B = 42: 17 B = 3,4,15,18,21,24
* 1st draw par 03, 21 in position 1 and 5th respectively without repeating the initial and final digits
** 2nd draw 04 23 in the 1st and 4th position without repeating digit in the pair
3rd draw 04 18 in the 2nd and 3rd position without repeating digit
* So icew, must dismember a list alongside the sweepstakes are 15 positions in pairs of pairs 1,2 .... up to 5.6 = 15 positions
I have a list of 276 normal peers without the filters of repeated digits, will further reduce, oh let's enjoy the standard of normal peers without repetition
And ainsa puts outro4 vtrac in their upright positions, my god!! icewynd

 

[jack]

Hello, we also this, example the number 01 instead of 25, can be delayed by
Example, when the couple will ride the 1st number in the pair is the reference (fixed) o2 number of the pair is a delay, another 02 in vtrac this example with 26 but could be 02
* With more trasado when it breaks down and a draw from 49/6 TEMSO 15 positions
Just who hit a pair with the following criteria the 1st number of the pair is fixed, the other, the 2nd number is the number fixed by delay, eg if the fixed number 1 or number in the pair is 17
* The 2nd number is for delay we assume that the statistical sweepstakes
* About this delay to 17 is 03, then the pair will be 17.03, icew when catch
* The way we will see pairs with a pair numro and other odd, ok
Goal is to get a couple pairs in normal position and then see
* The four vtrac

 

[jack]

Hello, icew normal pairs 01-49 are 1.176, but I managed
*To 560 pairs without repetition of digits in their first and last 4 digits
*A decrease of 616, or could, take 616 and leave the 560 pairs
99.35 In future raffles, icewm please see if you can use
*Of the 560 pairs without repetition is even 99%! These 560 pairs will be fixed, so now
*It's stats and the other 4 vtrac, fanstastico!!
01 29
01 39
01 38
01 28
01 27
01 26
01 25
01 34
01 24
01 23
01 32
01 35
01 36
01 37
01 42
01 43
01 45
01 46
01 47
01 48
01 49
02 19
02 39
02 38
02 18
02 17
02 16
02 15
02 34
02 14
02 13
02 31
02 35
02 36
02 37
02 41
02 43
02 45
02 46
02 47
02 48
02 49
03 19
03 29
03 28
03 18
03 17
03 27
03 26
03 16
03 15
03 25
03 24
03 14
03 12
03 21
03 41
03 42
03 45
03 46
03 47
03 48
03 49
04 19
04 29
04 39
04 38
04 28
04 18
04 17
04 27
04 26
04 16
04 15
04 25
04 13
04 23
04 32
04 12
04 21
04 31
04 35
04 36
04 37
05 19
05 29
05 39
05 38
05 28
05 18
05 17
05 27
05 26
05 16
05 34
05 24
05 14
05 13
05 23
05 32
05 12
05 21
05 31
05 36
05 37
05 41
05 42
05 43
05 46
05 47
05 48
05 49
06 19
06 29
06 39
06 38
06 28
06 18
06 17
06 27
06 15
06 25
06 34
06 24
06 14
06 13
06 23
06 32
06 12
06 21
06 31
06 35
06 37
06 41
06 42
06 43
06 45
06 47
06 48
06 49
07 19
07 29
07 39
07 38
07 28
07 18
07 26
07 16
07 15
07 25
07 34
07 24
07 14
07 13
07 23
07 32
07 12
07 21
07 31
07 35
07 36
07 41
07 42
07 43
07 45
07 46
07 48
07 49
08 19
08 29
08 39
08 17
08 27
08 26
08 16
08 15
08 25
08 34
08 24
08 14
08 13
08 23
08 32
08 12
08 21
08 31
08 35
08 36
08 37
08 41
08 42
08 43
08 45
08 46
08 47
08 49
10 29
10 39
10 38
10 28
10 27
10 26
10 25
10 34
10 24
10 23
10 32
10 35
10 36
10 37
10 42
10 43
10 45
10 46
10 47
10 48
10 49
09 38
09 28
09 18
09 17
09 27
09 26
09 16
09 15
09 25
09 34
09 24
09 14
09 13
09 23
09 32
09 12
09 21
09 31
09 35
09 36
09 37
09 41
09 42
09 43
09 45
09 46
09 47
09 48
19 20
19 30
19 40
19 38
19 28
19 27
19 26
19 25
19 34
19 24
19 23
19 32
19 35
19 36
19 37
19 42
19 43
19 45
19 46
19 47
19 48
20 39
20 38
18 20
17 20
16 20
15 20
20 34
14 20
13 20
20 31
20 35
20 36
20 37
20 41
20 43
20 45
20 46
20 47
20 48
20 49
29 30
28 30
18 30
17 30
27 30
26 30
16 30
15 30
25 30
24 30
14 30
12 30
21 30
30 41
30 42
30 45
30 46
30 47
30 48
30 49
29 40
29 38
18 29
17 29
16 29
15 29
29 34
14 29
13 29
29 31
29 35
29 36
29 37
29 41
29 43
29 45
29 46
29 47
29 48
39 40
28 39
18 39
17 39
27 39
26 39
16 39
15 39
25 39
24 39
14 39
12 39
21 39
39 41
39 42
39 45
39 46
39 47
39 48
38 40
28 40
18 40
17 40
27 40
26 40
16 40
15 40
25 40
13 40
23 40
32 40
12 40
21 40
31 40
35 40
36 40
37 40
17 38
27 38
26 38
16 38
15 38
25 38
24 38
14 38
12 38
21 38
38 41
38 42
38 45
38 46
38 47
38 49
17 28
16 28
15 28
28 34
14 28
13 28
28 31
28 35
28 36
28 37
28 41
28 43
28 45
28 46
28 47
28 49
18 27
18 26
18 25
18 34
18 24
18 23
18 32
18 35
18 36
18 37
18 42
18 43
18 45
18 46
18 47
18 49
17 26
17 25
17 34
17 24
17 23
17 32
17 35
17 36
17 42
17 43
17 45
17 46
17 48
17 49
16 27
15 27
27 34
14 27
13 27
27 31
27 35
27 36
27 41
27 43
27 45
27 46
27 48
27 49
15 26
26 34
14 26
13 26
26 31
26 35
26 37
26 41
26 43
26 45
26 47
26 48
26 49
16 25
16 34
16 24
16 23
16 32
16 35
16 37
16 42
16 43
16 45
16 47
16 48
16 49
15 34
15 24
15 23
15 32
15 36
15 37
15 42
15 43
15 46
15 47
15 48
15 49
25 34
14 25
13 25
25 31
25 36
25 37
25 41
25 43
25 46
25 47
25 48
25 49
12 34
21 34
13 24
24 31
24 35
24 36
24 37
14 23
14 32
14 35
14 36
14 37
13 42
13 45
13 46
13 47
13 48
13 49
23 41
23 45
23 46
23 47
23 48
23 49
32 41
32 45
32 46
32 47
32 48
32 49
12 35
12 36
12 37
12 43
12 45
12 46
12 47
12 48
12 49
21 35
21 36
21 37
21 43
21 45
21 46
21 47
21 48
21 49
31 42
31 45
31 46
31 47
31 48
31 49
35 41
35 42
35 46
35 47
35 48
35 49
36 41
36 42
36 45
36 47
36 48
36 49
37 41
37 42
37 45
37 46
37 48
37 49
#

 

[jack]

Hello, these ice worth 560 gold, comes along with the study of the position, eg
* The pair is 12.36 in the 2nd and 5th position, the other 4 positions are vtrac 1st 3rd 4th and 6th position
If you use the 560 as fixed in the wheels hit the lottery is 99%, very good
So you can see stats for most couples who come late, of course within the
* 560 pairs, ice was seeing a filter for these pairs of terminations that repeat
Or that are missing in the last draw, to see if there is evidence to predict pairs
* Within the 560, from now on just 4-vtrac, if you ride with games
* All 560 will hit the pivot pair in that wonder 99%!

 

[jack]

Hello, are now ice pairs of vtrac without repetition are 98 groups of vtrac
In 95% of the lottery, or playing all 98 vtrac hits two numbers
Agoara when within the vtrac have two groups with two groups will be
U normal numbers agreed, ice ve the importance of hitting the two numbers
In each of the 24 vtrac if making the accumulation of gain when has two
Number one vtrac, =
Below the 98 possible pairs of vtrac without repeating digits each other
01 23
01 24
02 19
02 18
02 17
02 16
02 15
02 14
02 13
03 19
03 18
03 17
03 16
03 15
03 14
03 12
03 21
03 24
04 19
04 18
04 17
04 16
04 15
04 13
04 12
04 21
04 23
05 19
05 18
05 17
05 16
05 14
05 13
05 12
05 21
05 23
05 24
06 19
06 18
06 17
06 15
06 14
06 13
06 12
06 21
06 23
06 24
07 19
07 18
07 16
07 15
07 14
07 13
07 12
07 21
07 23
07 24
08 19
08 17
08 16
08 15
08 14
08 13
08 12
08 21
08 23
08 24
09 18
09 17
09 16
09 15
09 14
09 13
09 12
09 21
09 23
09 24
10 23
10 24
20 19
20 18
20 17
20 16
20 15
20 14
20 13
19 23
19 24
18 23
18 24
17 23
17 24
16 23
16 24
15 23
15 24
14 23
13 24

 

[jack]

Hello, ice, was mounted below the pairs of vtrac without repeating digits among themselves until
* I could, a few pairs was forced to repeat the digits starting and ending
So instead of riding in 01 15, 02 26, I rode without repetition or 24 vtrac
* Digits, with this pair enjoy the accumulation of correct two numbers in vtrac
So playing with 24 vtrac to see the use of it?
01 49
02 48
12 36
10 25
13 29
14 28
09 42
03 47
04 39
05 46
06 38
07 45
17 20
18 43
35 41
19 24
21 37
16 32
15 31
23 40
08 26
11 22
33 44
27 30 34

 

[jack]

Hello, icew, this last post is an idea to assemble pairs of vtrac, nonrepeating pairs between your 4 digit, to take the hit on a couple of accumulation vtrac
Icew, you can create tools to simulate and test game situations?

 

[Icewynd]

Hello, icew, this last post is an idea to assemble pairs of vtrac, nonrepeating pairs between your 4 digit, to take the hit on a couple of accumulation vtrac
Icew, you can create tools to simulate and test game situations?

Sorry, Jack. That is beyond my capabilities.

 

[jack]

hello, icewynd, what part do not you understand? Easy is almost always a couple of vtrac
Without the 4 digit repetetido not go into a draw, example
The vtrac, 01 23, there is no repetition of digits in the pair vtrac, you have to differentiate
The vtrac, which goes inside the numbers and vtrac and normal peers, understanding this, it's easy
This hybrid mixture seems a good way of normal peers and the rest vtrac because the normal peers may serve as a pivot or reference, icewynd, you can create wheels with the hybrid combination!!