Repeat analysis

[mario]

(09-05-2009 06:45 PM)mario Wrote:  It remained 10 hours to uk lotto.
For 1429 draw I searched pairs with a building skip equal with the last skip -1. And I found this:

1 14 9 1
8 31 6 1
8 46 6 3
31 38 6 1
38 46 6 7
6 10 5 5
6 30 5 2
6 41 5 4
6 45 5 6
10 49 5 5
30 49 5 2
41 49 5 4
45 49 5 7
1 36 4 1
5 36 4 7
6 36 4 2
8 36 4 4
12 36 4 15
13 36 4 3
14 36 4 10
15 36 4 15
19 36 4 3
22 36 4 5
24 33 4 6
24 36 4 14
25 36 4 12
29 36 4 7
33 36 4 6
33 43 4 6
33 44 4 1
36 38 4 4
36 42 4 12
36 43 4 11
36 44 4 1
36 49 4 2

Cols 1,2 has pair, col3 has the last skip and col4 has the precedence for the last skip, in fact col3 and col4 has the last two skips for that pair.

All this pair have now a skip in construction, which is not finalize, and this skip is less with 1 in comparation with the last skip.
Ex:
first pair 1-14 has the last skip=9 and the skip in construction, not finalize =8 so, if at least one number for this pair appear in 1430 draw will appear skip=9 equal with the precedence skip, a situation very rare.

So this pair give us a list of elimination numbers. If I can choose from its:

1 14
38 46
6 10
6 45
10 49
45 49
8 36
36 38

because in col4 have a value skip very near to the last skip in col3:
So:
1-14-38-46-6-10-45-49-8-36

it's my bad numbers!

Let's see!

1430 draw was:
12 - 19 - 22 - 26 - 36 - 41 and 39 BB

so in my elimination list was "only" 36; this number gave us for the last skip 4 an another skip 4 and the equality has come.

and this good pairs had 4 good numbers:
12 36
19 36
22 36
it is abnormally fact?

in all pairs exist 26 numbers and 5 good! 36 is very, very strange, is the most repeated numbers in this group.

[mario]

for 1431 draws I found this pairs:



cols A-B is the pair
col C the last skip
col D the precedence skip to the last skip
in fact C-D are the last 2 skips

It's seems that pairs can give us some numbers for elimination.How?
I used the biggest skips which are uncommon.
Col C is very important; this show value skip will be touched.
I've seen 2 facts:
- for a value in col C usually exist one number most repeated: 30 for value=7 and 17 for value=5; if this numbers will appear in the next draw, its generate many pairs with a skip 7 or 5, in fact will generate for the last 2 skips a combination 7-7 or 5-5; too many combinations I say, so put 17-30 in my elimination list;

- for blue cells exist a situation 7-7 and 6-6 for the last 2 skips; if one number from this pairs appear will generate the situation 7-7-7 or 6-6-6, very rare in skip graphics; so put 6-24-38-43-45 in my elimination list;

The elimination list will be for 1431:

30-17- 6-24-38-43-45 and 13-15 (have a big skip 13!)

I atach images with the graphic skip:
30-38, 38-45 and 6-24

[mario]

(09-07-2009 05:26 AM)mario Wrote:  The elimination list will be for 1431:

30-17- 6-24-38-43-45 and 13-15 (have a big skip 13!)

1431: 2-11-23-28-35-39 BB 15

So I don't use BB that my elimination list was correct!

[mario]

I've seen another pairs with the same skip like the last.
In precedence post I found pairs with value= the last -1 but even in that case the elimination list was correct, it's happening probably
I corrected the programm and I've obtained this pair



In A,B is pair, C, D the last 2 skips; in C is the last skip.

For skip=7 I use 30 (most repeated) and 30-38, 38-45 because this pairs have 7-7 the last 2 skips and if one number appear will have 7-7-7 the last 3 skips, very rare case;

For skip 6 I have 6-24, 6-43 with 6-6-6 and 49;
For skip=5 I have 17 the most repeated.

So my elimination list contain:

30-38-45-6-24-43-49,17 and 13-15 (the last skip 13).

I hope that I 've done a good correction.


I post a graphic for 38-45:


I observed that group 7-7 (the last 2 skips) and if one number appear in 1432 it will appear a group 7-7-7, in the skips graphic is a very rare case.

---

2 16 17 23 36 39

This numbers have a chain of 27 consecutive draws which has a property: in each draws exist at least one number from this group; in fact the chain is not longer than 30 consecutive draws;

it's only one comb of 6 in 13 900 000 with this property?????????? indeed??

So I thought that 6 numbers is bad for 1432 draw.

This chain is:

[mario]

(09-11-2009 01:29 PM)mario Wrote:  So my elimination list contain:

30-38-45-6-24-43-49,17 and 13-15 (the last skip 13).






2 16 17 23 36 39

This numbers have a chain of 27 consecutive draws which has a property: in each draws exist at least one number from this group; in fact the chain is not longer than 30 consecutive draws;

it's only one comb of 6 in 13 900 000 with this property?????????? indeed??

So I thought that 6 numbers is bad for 1432 draw.

This chain is:

1432 draw was:
18-20-24-30-43-44 BB 47

so, my elimination list was "bad elimination"!! with 30-24-43;
30 has generated abnormall more pairs with the last 2 skips 7-7!
and 24-43 has generated 6-6!

30 is very very strange it generated even a pair with the last 3 skips 7-7-7!!

but the chain of numbers didn't appeared and gave us 6 bad numbers!!

Frank, where is your elimination list?

[Frank]

I stopped posting mine to avoid confusion, since you are using a different method, and I didn't want to continue posting when they were not always right. The last 3 elimination lists and their results are here. The rows means how many past draws I used to predict the future, based on good skip values and bad skip values, hoping to repeat in th enext draw.
------------------------ RESULT ----------
1430.... 8 rows.... 12 19 22 26 36 41.... 1 7 14 18 22 23 25 27 30 32 36 37 39 40 47 48 49 2 wrong
1431.... 10 rows.... 2 11 23 28 35 39.... 3 7 8 15 17 20 24 27 40 44 45 all correct
1432.... 10 rows.... 18 20 24 30 43 44.... 4 8 12 15 17 18 20 25 30 33 36 45 47 3 wrong !

My next one for Wednesday is orange , possibles , red strongly eliminated.

1 10 14 28 29 37 43 2 8 13 16 24 31 33 38 44 45 47 48

[mario]

(09-14-2009 03:43 AM)Frank Wrote:  I stopped posting mine to avoid confusion, since you are using a different method, and I didn't want to continue posting when they were not always right. The last 3 elimination lists and their results are here. The rows means how many past draws I used to predict the future, based on good skip values and bad skip values, hoping to repeat in th enext draw.
------------------------ RESULT ----------
1430.... 8 rows.... 12 19 22 26 36 41.... 1 7 14 18 22 23 25 27 30 32 36 37 39 40 47 48 49 2 wrong
1431.... 10 rows.... 2 11 23 28 35 39.... 3 7 8 15 17 20 24 27 40 44 45 all correct
1432.... 10 rows.... 18 20 24 30 43 44.... 4 8 12 15 17 18 20 25 30 33 36 45 47 3 wrong !

My next one for Wednesday is orange , possibles , red strongly eliminated.

1 10 14 28 29 37 43 2 8 13 16 24 31 33 38 44 45 47 48

Sorry Frank I have'nt intent to forestall your thread, just I thought that my elimination method add few numbers.
My chain of numbers is like your method, and this is amazing! Both ideeas use a consecutive appearance but mine, generate all combinations and find one with al least one number common with the last 27 draws! This indeed, give us bad numbers I think.

Probably we must find 10-12 different methof for bad numbers to eliminate 25-30 numbers?

[Frank]

well that last set actually trapped 2 good numbers. For 1434 Im eliminating

2 3 11 18 19 21 23 25 28 38 44 46 4 5 6 10 13 20 29 31 32 33 35 37 39 40 41 47 red zone strongly eliminated.

[Len]

(09-19-2009 05:18 AM)Frank Wrote:  well that last set actually trapped 2 good numbers. For 1434 Im eliminating

2 3 11 18 19 21 23 25 28 38 44 46 4 5 6 10 13 20 29 31 32 33 35 37 39 40 41 47 red zone strongly eliminated.

Hi Frank
In your red zone you had 20,31,33 correct

[Frank]

(09-20-2009 09:25 PM)Len Wrote:  

(09-19-2009 05:18 AM)Frank Wrote:  well that last set actually trapped 2 good numbers. For 1434 Im eliminating

2 3 11 18 19 21 23 25 28 38 44 46 4 5 6 10 13 20 29 31 32 33 35 37 39 40 41 47 red zone strongly eliminated.

Hi Frank
In your red zone you had 20,31,33 correct

Yes , I had one in th eblue zone too. The skips I picked to include mostly failed to repeat this time, so making the elimination list into a bankers list